Class 6

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 10 The Other Side of Zero Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 10 The Other Side of Zero Solutions

Ganita Prakash Class 6 Chapter 10 Solutions

Class 6 Maths Ganita Prakash Chapter 10 Solutions The Other Side of Zero

Question 1.
Evaluate these expressions:
(a) (+ 1) + (+ 4) = _______
(b) (+ 4) + (+ 1) = _______
(c) (+ 4) + (- 3) = _______
(d) (- 1) + (+ 2) = _______
(e) (- 1) + (+ 1) = _______
(f) 0 + (+ 2) = _______
(g) 0 + (- 2) = _______
Solutions:
(a) (+ 1) + (+ 4) = + 5
(b) (+ 4) + (+ 1) = + 5
(c) (+ 4) + (- 3) = + 1
(d) (- 1) + (+ 2) = + 1
(e) (- 1) + (+ 1) = 0
(f) 0 + (+ 2) = + 2
(g) 0 + (- 2) = – 2

Question 2.
Try to subtract: – 3 – (+ 5). How many zero pairs will you have to put in? What is the result?
Solution:
– 3 – (+ 5) = – 8
We want to take away 5 positive tokens when we have 3 negative tokens. So, we add 5 zero pairs. After removing the 5 positive tokens, we have 8 negative tokens left.

Question 3.
Suppose you start with 0 rupees in your bank account and then you have debits of ₹ 1, ₹ 2, ₹ 4, ₹ 8, ₹ 16, ₹ 32, ₹ 64 and ₹ 128 and then a single credit of ₹ 256. What is your bank account balance now?
Solution:
Total debits = ₹ 1 + ₹ 2 + ₹ 4 + ₹ 8 + ₹ 16 + ₹ 32 + ₹ 64 + ₹ 128 = ₹ 255
There is a single credit of ₹ 256.
∴ Final balance = Total credit – Total debit = ₹ 256 – ₹ 255 = ₹ 1

Question 4.
Looking at the geographical cross section, fill in the respective heights:

Solution:
A = + 1500 m
B = – 500 m
C = + 300 m
D = – 1200 m
E = + 1200 m
F = – 200 m
G = + 100 m

Question 5.
Leh in Ladakh gets very cold during winter. The following is a table of temperature readings taken during different times of the day and night in Leh on a day in November. Match the temperature with the appropriate time of the day and night.

Solution:
Usually, the coldest time of the day is very early in the morning, before the sun rises, and it starts to get warmer after sunrise, reaching the warmest in the afternoon. At night, the temperature starts to drop again.
So, we can match the temperatures like this:
– 4°C → 2:00 a.m. (This is the coldest time when it is still dark and freezing.)
– 2°C → 11:00 p.m. (It is night time, so it is also very cold, but not as cold as early morning.)
8°C → 11:00 a.m. (It is late morning, the sun is up and it is getting warmer.)
14°C → 2:00 p.m. (This is the afternoon and the warmest time of the day.)
This shows us how the temperature rises as the sun comes up and falls again after sunset.

Question 6.
Complete the grids to make the required border sum

Solution:

Question 7.
There are two dice whose faces have these numbers: – 1, 2, – 3, 4, – 5, 6. The smallest possible sum upon rolling these dice is – 10 = (- 5) + (- 5) and the largest possible sum is 12 = (6) + (6). Some numbers between (- 10) and (+ 12) are not possible to get by adding numbers on these two dice. Find those numbers.
Solution:
Let’s find the sums that are not possible when rolling these two dice.
The faces of the dice are:
– 1, 2, -3, 4, -5, and 6.
First, let’s list all possible sums:
The sum of two negative numbers:
(- 1) + (- 1) = -2; (- 1) + (- 3) = – 4; (- 1) + (- 5) = – 6
(- 3) + (- 3) = – 6; (- 3) + (- 5) = – 8 (- 5) + (- 5) = – 10
The sum of one negative and one positive number:
(-1) + 2 = 1; (-1)+ 4 = 3; (-1) + 6 = 5; (-3) + 2 = – 1; (-3)+ 4=1;
(- 3) + 6 = 3; (- 5) + 2 = – 3; (- 5) + 4 = – 1; (- 5) + 6 = 1
The sum of two positive numbers:
2 + 2 = 4; 2 + 4 = 6; 2 + 6 = 8; 4 + 4 = 8; 4 + 6=10; 6 + 6=12
Now, let’s list all the possible sums in ascending order:
– 10, -8, -6, -4, -3,-2,- 1, 1, 3, 4, 5, 6, 8, 10, 12
The sum of numbers between – 10 and 12 that are not possible to get are: – 9, – 7, – 5, 0, 2, 7, 9, 11 .

Question 8.
This string has a total of 100 tokens arranged in a particular pattern. What is the value of the string?

Solution:
Total tokens =100
Total sets of 5 tokens = \(\frac{100}{5}\) = 20
Value of one set of 5 tokens = 3 + (- 2) = 1
Hence, total value of string = 1 × 20 = 20

InText Questions

Question 1.
Can there be a number less than 0? Can you think of any ways to have less than 0 of something?
Solution:
Yes, there can be numbers less than 0.
They are called negative numbers and written with a minus sign, like – 1, – 2, – 3, etc.
Temperature: In cold places, temperature can go below 0 °C, like – 5 °C or – 10 °C.
Money: If you have ₹ 10 and will have – ₹ 10. you spend ₹ 20, you

Question 2.
Connect the inverses by drawing lines.

Solution:
The inverses of + 5, – 7, – 8 and + 9 are – 5, + 7, + 8 and – 9 respectively.

Question 3.
Should we write – 3 <-4or-4 <-3?
Solution:
We should write – 4 < – 3, because on the number line, – 4 is to the left of – 3, which makes it smaller.
So, – 4 < -3 or -3 > -4.

Question 4.
Evaluate 15 – 5, 100 – 10 and 74 – 34.
Solution:
(i) What should we add to 5 to make 15? i.e.
5 + _______ = 15.
The missing number is 10. So, 15 – 5 = 10.

(ii) What should we add to 10 to make 100? i.e.
10 + ______ = 100.
The missing number is 90. So, 100 – 10 = 90.

(iii) What should we add to 34 to make 74? i.e.
34 + _______ = 74.
The missing number is 40. So, 74 – 34 = 40.

Question 5.
Use unmarked number lines to evaluate these expressions:
(a) _______ – 125 + (- 30) = _______
(b) _______ + 105 – (- 55) = _______
(c) _______ + 80 – (- 150) = _______
(d) _______ – 99 – (- 200) = _______
Solution:
(a) – 125 + (- 30) = – 155
We start at – 125 and move 30 steps left on the number line.

(b) + 105 – (- 55) = + 105 + ( + 55) = + 160
We start at + 105 and move 55 steps right on the number line.

(c) + 80 – (- 150) = + 80 + (+ 150) = + 230
We start at + 80 and move 150 steps right on the number line.

(d) -99-(-200) = -99 + 200 = + 101
We start at – 99 and move 200 steps right on the number line.

The Other Side of Zero Class 6 Extra Questions

The Other Side of Zero Class 6 Very Short Question Answer

Question 1.
In a game, Rishi scored + 10,-5, + 3,-8 and + 4 in five rounds. What is his total score?
Solution:
Given, the scores of Rishi in five rounds are +10, – 5, + 3, – 8 and + 4.
Total score of Rishi
= + 10 + (- 5) + (+ 3) + (- 8) + (+ 4)
= (+ 10 + 3 + 4) + (-5-8)
[Grouping positive and negative integers]
= (+ 17) + (- 13)
= (+ 17)-(+ 13)
[The number that is being added can be replaced by . its additive inverse and then subtracted.]
= + 4
∴ The total score of Rishi is + 4.

Question 2.
Arrange the following integers in ascending order:
9, -7, -4, 0, 3
Solution:
Given integers are: 9, – 7, – 4, 0, 3
For positive integers 3 and 9, we have 3 < 9.
For negative integers – 7 and – 4, we have – 7 < – 4.
We know that on the number line, the numbers right to 0 are greater than 0 and the numbers left to 0 are less than 0. Also, every positive integer is greater than every negative integer.
Hence, the given integers in ascending order: – 7, -4, 0, 3, 9

Question 3.
A boat sailed 40 km to the east of a harbour and then 70 km to the west from there. How far from the harbour is the boat finally?
Solution:
The boat starts from point A at harbour. It sailed 40 km to the east to reach point B. From point B, it sailed 70 km to the west to reach point C, as shown in figure.

Now, the position of boat = (+ 40) + (- 70) = – 30
Hence, the boat is 30 km to the west of the harbour.

Question 4.
Write 5 distinct integers whose sum is 6.
Solution:
We know that the sum of an integer with its additive inverse is zero.
The additive inverse of – 1 is 1 and the additive inverse of – 2 is 2.
∴ 1 + (- 1) = 0 and 2 + (- 2) = 0
Now, [1 + (- 1)] + [2 + (- 2)] + 6 = 0 + 0 + 6 = 6
∴ 5 distinct integers whose sum is 6 are – 1, 1, -2, 2, 6.

Question 5.
Arrange the following integers in descending order:
-8, -3, 5, 1, -1
Solution:
Given integers are – 8, – 3, 5, 1 and – 1.
We know that a number is greater than every number that is to its left on the number line.
For positive integers 1 and 5, we have 5 > 1.
For negative integers -1,-3 and – 8, we have
– 1 > – 3 > – 8.
Also, every positive integer is greater than every negative integer.
Hence, the given integers in descending order:
5, 1, – 1, -3, -8

The Other Side of Zero Class 6 Short Question Answer

Question 1.
Complete the additions using tokens:
(i) (+ 4) + (- 8)
(ii) (- 3) + (+ 4)
(iii) (- 10) + ( + 6)
(iv) (+ 7) + (- 5)
Solution:
(i) (+ 4) + (- 8) = – 4

(ii) (- 3) + (+ 4) = + 1

(iii) (- 10) + (+ 6) = – 4

(iv) (+ 7) + (- 5) = + 2

Question 2.
Complete the subtractions using tokens:
(i) (+ 6) – (+ 3)
(ii) (+ 8) – (+ 7)
(iii) (- 9) – (- 5)
(iv) (- 7) – (- 4)
Solution:
(i) (+ 6) – (+ 3) = + 3

(ii) (+ 8) – (+ 7) = + 1

(iii) (- 9) – (- 5) = – 4

(iv) (- 7) – (- 4) = – 3

Question 3.
Complete the following subtractions using tokens:
(i) (+ 9) – (+ 2)
(ii) (+ 5) – (4- 3)
(iii) (- 8) – (- 3)
(iv) (- 6) – (- 1)
Solution:
(i) (+ 9) – (+ 2) = + 7

(ii) (+ 5) – (+ 3) = + 2

(iii) (- 8) – (- 3) = – 5

(iv) (- 6) – (- 1) = – 5

Question 4.
Complete the following subtractions using tokens:
(i) (+ 3) – (+ 8)
(ii) (+ 7) – (+ 9)
(iii) (+ 4) – (- 4)
(iv) (+ 6) – (- 5)
(v) (- 4) – (+ 3)
(vi) (- 6) – (+ 6)
Solution:
(i) (+ 3) – (+ 8) = – 5

(ii) (+ 7) – (+ 9)

(iii) (+ 4) – (- 4)

(iv) (+ 6) – (- 5)

(v) (- 4) – (+ 3)

(vi) (- 6) – (+ 6)

The Other Side of Zero Class 6 Long Question Answer

Question 1.
There are two dice whose faces have the numbers 1, – 2, 3, – 4, 5 and – 6. If these dice are rolled together, find the smallest possible sum of the numbers so obtained.
Solution:
Given, there are two dice whose faces have the numbers 1, – 2, 3, – 4, 5 and – 6.
To get the smallest possible sum, we need to add the smallest number from each die.
A number is always greater than any number to its left on the number line.
For positive integers 1, 3 and 5, we have 1 < 3 < 5.
For negative integers -2,-4 and – 6, we have -6 < -4 < -2.
We know that every positive integer is greater than every negative integer.
Hence, the given integers in ascending order: -6, -4, -2, 1, 3, 5
The smallest number on each die is – 6.
Hence, smallest possible sum = (- 6) + (- 6) = – 12

Question 2.
Using the number line write the integer which is
(i) 3 more than – 1
(ii) 4 less than – 1
(iii) 4 more than – 9
Solution:
(i) To find the integer which is 3 more than – 1, we start at – 1 on the number line and move 3 steps forward. We reach 2. Hence, the required integer is 2.

(ii) To find the integer which is 4 less than – 1, we start at – 1 on the number line and move 4 steps backward. We reach – 5. Hence, the required integer is -5.

(iii) To find the integer which is 4 more than – 9, we start at – 9 on the number line and move 4 steps forward. We reach – 5. Hence, the required integer is – 5.

Question 3.
There are two dice whose faces have the numbers – 8, – 3, – 1, 2, 4 and 6. If these dice are rolled together, find the largest possible sum of the numbers so obtained.
Solution:
Given, there are two dice whose faces have the numbers – 8, – 3, – 1, 2, 4 and 6.
To get the largest possible sum, we need to add the largest number from each die.
We know that a number is greater than every number that is to its left on the number line.
For positive integers 2, 4 and 6, we have 2 < 4 < 6.
For negative integers -1,-3 and – 8, we have – 8 < – 3 < – 1.
We know that every positive integer is greater than every negative integer.
Hence, the given integers in ascending order: – 8, -3, – 1, 2, 4, 6
The largest number on each die is 6.
So, largest possible sum = (6) + (6) = 12

Question 4.
Complete the subtractions using tokens:
(i) (+ 4) – (+ 7)
(ii) (+ 5) – (+ 8)
(iii) (+ 5) – (- 3)
(iv) (+ 6) – (- 2)
(v) (- 4) – (+ 2)
Solution:
(i) (+ 4) – (+ 7) = – 3

[Adding 3 zero pairs]

(ii) (+ 5) – (+ 8) = – 3

[Adding 3 zero pairs]

(iii) (+ 5) – (- 3) = + 8

[Adding 3 zero pairs]

(iv) (+ 6) – (- 2) = + 8

[Adding 2 zero pairs]

(v) (- 4) – (+ 2) = – 6

[Adding 2 zero pairs]

Question 5.
Complete the grids to make the required border
(i)

-7
		4
	-3

Border sum = + 5

(ii)

	-12
-1
		7

Border sum = – 4

(iii)

3		11
-2		-3

Border sum = 0

(iv)

-15		5
10
		4

Border sum = 3
Solution:
We have to take that number which makes the sum of all the numbers in a row or column equal to the border sum. So, the complete grids are given below.
(i)

-7	3	9
-4		4
16	-3	-8

Border sum = + 5

(ii)

2	-12	6
-1		-17
-5	-6	7

Border sum = – 4

(iii)

3	-14	11
-1		-8
-2	5	-3

Border sum = 0

(iv)

-15	13	5
10		-6
8	-9	4

Border sum = 3

Question 6.
Complete the following additions using tokens:
(i) (+ 7) + (- 2)
(ii) (- 12) + (+ 4)
(iii) (- 9) + (+ 5)
(iv) (-7) + (+ 11)
(v) (- 3) + (+ 8)
(vi) (- 6) + (+ 12)
Solution:
(i) (+ 7) + (-2) = + 5

(ii) (- 12) + (+ 4) = – 8

(iii) (- 9) + (+ 5) = – 4

(iv) (-7) + (+ 11) = + 4

(v) (- 3) + (+ 8) = + 5

(vi) (- 6) + (+ 12) = + 6

Question 7.
Complete the grids to make the required border
(i)

	0
		5
-2
Border sum = – 2

(ii)

		6
-5
	-3
Border sum = + 3

(iii)

	0
3		5
	8
Border sum = 4

(iv)

3
-7		6
		-8
Border sum = 5

Solution:
We have to take that number which makes the sum of all the numbers in a row or column equal to the border sum. So, the complete grids are given below.
(i)

1	0	-3
-1		5
-2	4	-4

(ii)

5	-8	6
-5		-6
3	-3	3

(iii)

3	0	1
3		5
-2	8	-2

(iv)

3	-5	7
-7		6
-2	4	-8

The Other Side of Zero Class 6 Case Based Questions

Question 1.
In Shimla, the temperature in the morning was – 2°C. By afternoon, it rose by 5°C. In another city, Manali, the temperature was – 5°C in the morning and increased by 3°C in the afternoon.
Based on the above information, answer the following questions:
(i) What was the afternoon temperature in Shimla?
(ii) What was the afternoon temperature in Manali?
(iii) What is the difference in afternoon temperatures of both cities?
Solution:
(i) Given, morning temperature in Shimla = – 2 °C
Increase in temperature by afternoon = + 5 °C
To find afternoon temperature, we add the change to the morning temperature.
∴ Afternoon temperature in Shimla = (- 2 °C) + (+ 5°C) = + 3°C

(ii) Given, morning temperature in Manali = – 5°C
Increase in temperature by afternoon = + 3 °C
To find afternoon temperature, we add the change to the morning temperature.
∴ Afternoon temperature in Manali = (- 5 °C) + (+ 3 °C)= – 2 °C

(iii) Afternoon temperature in Shimla = + 3 °C
Afternoon temperature in Manali = – 2 °C
Difference in afternoon temperatures of both cities
= + 3 °C – (- 2 °C)
= + 3 °C + (+ 2 °C) = 5 °C

Question 2.
In a geographical cross-section, some points appear above sea level (marked with positive height values) and others lie below sea level (represented by negative values), using sea level defined as zero as the reference point.
Based on the given information, answer the following questions:

(i) Write the highest and lowest points in the given geographical cross section?
(ii) What is the difference between heights of the highest and lowest points?
(iii) Write the points A, B, …, G in a sequence of increasing order of heights.
OR
(iii) Write the points A, B, …, G in a sequence of decreasing order of heights.
Solution:
(i) E(500 m) is the highest point and C (- 400 m) is the lowest point.

(ii) Difference between heights of E and C
= 500 – (- 400) = 500 + (+ 400) = 900 m

(iii) Points in increasing order are as:
C(- 400 m), B(- 300 m), G(- 200 m), D(0 m), A(200 m), F(300 m), E(500 m)
OR
(iii) Points in decreasing order, are as:
E(500 m), F(300 m), A(200 m), D(0 m), G(- 200 m), B(- 300 m), C(- 400 m)

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 9 Symmetry Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 9 Symmetry Solutions

Ganita Prakash Class 6 Chapter 9 Solutions

Class 6 Maths Ganita Prakash Chapter 9 Solutions Symmetry

Question 1.
For each of the following figures, identify the line(s) of symmetry if it exists.

Solution:

Question 2.
Given the line(s) of symmetry, find the other hole(s).

Solution:

Question 3.
Find the lines of symmetry for the kolam below.

Solution:

Question 4.
Draw the following:
(a) A triangle with exactly one line of symmetry
(b) A triangle with exactly three lines of symmetry
(c) A triangle with no line of symmetry
Is it possible to draw a triangle with exactly two lines of symmetry?
Solution:
(a)

One line of symmetry (Isosceles Triangle)

(b)

Three lines of symmetry
(Equilateral Triangle)

(c)

No line of symmetry (Scalene Triangle)
No, it is not possible to draw a triangle with exactly two lines of symmetry.

Question 5.
Copy the following on a dot grid. For each figure draw two more lines to make a shape that has a line of symmetry.

Solution:

Question 6.
Color the sectors of the circle below so that the figure has i) 3 angles of symmetry, ii) 4 angles of symmetry, iii) what are the possible numbers of angles of symmetry you can obtain by coloring the sectors in different ways?

Solution:
(i) Three angles of symmetry

(ii) Four angles of symmetry

(iii)

For 2 angles of symmetry, the angles are: 180°, 360°
For 3 angles of symmetry, the angles are: 120°, 240°, 360° 1
For 4 angles of symmetry, the angles are: 90°, 180°, 270°, 360°
For 6 angles of symmetry, the angles are: 60°, 120°, 180°, 240°, 300°, 360°
For 12 angles of symmetry, the angles are: 30°, 60°, 90°, 120°, 150°, 180°, 210°, 240°, 270°, 300°, 330°, 360°

Question 7.
In a figure, 60° is an angle of symmetry. The figure has two angles of symmetry less than 60°. What is its smallest angle of symmetry?
Solution:
We know that the angles of symmetry of a figure having rotational symmetry about a point are multiples of the smallest angle of symmetry.

Let the smallest angle of symmetry be x. Then, the other angles of symmetry are 2x, 3x, 4x, 360°.

It is given that the figure has two angles of symmetry less than 60°. This means the third angle of symmetry is 60° itself.
∴ 3x = 60°
⇒ x = 20°
Hence, the smallest angle of symmetry is 20°.

Question 8.
How many lines of symmetry does the shape sequence, the Koch Snowflake sequence, have? Also, find numbers of angles of symmetry.
Solution:
The equilateral triangle has 3 lines of symmetry and 3 angles of symmetry.
In six-pointed star, lines of symmetry are 6 and the angle of symmetry is 6.
The lines of symmetry and angle of symmetry for the rest three figures are the same as six-pointed stars.

InText Questions

Question 1.
Is there any other way to fold the square so that the two halves overlap? How many lines of symmetry does the square shape have?
Solution:
Yes, there are other ways to fold a square so that the two halves overlap exactly. This can be done along its lines of symmetry.

A square has 4 lines of symmetry:
(i) A vertical line (through the midpoints of the top and bottom sides),
(ii) A horizontal line (through the midpoints of the left Square and right sides),
(iii) A diagonal from the top-left to bottom-right corner,
(iv) A diagonal from the top-right to bottom-left corner.
So, we can fold the square along any of these 4 lines, and the two halves will perfectly overlap.

Question 2.
We saw that the diagonal of a square is also a line of symmetry. Let us take a rectangle that is not a square. Is its diagonal a line of symmetry?
Solution:
No, the diagonal of a rectangle is not a line of symmetry. Folding along the diagonal does not produce two matching parts.

So, while a square has both its diagonals as lines of symmetry (because they divide the square into two mirror-image triangles), a non-square rectangle does not have this property.

Question 3.
Consider a figure with radial arms having exactly 7 angles of symmetry. What will be its smallest angle of symmetry? Is the number of degrees a whole number in this case? If not, express it as a mixed fraction.
Solution:
When a figure has rotational symmetry with 7 angles of symmetry, it means it can be rotated around a central point and still look the same 7 times in a full 360° turn.

To find the smallest angle of symmetry, divide 360° by the number of symmetrical positions:
Smallest angle of symmetry = \(\frac{360^{\circ}}{7}\) = \(\left(51 \frac{3}{7}\right)^0\) which is not a whole number.
Mixed fraction form: \(\left(51 \frac{3}{7}\right)^0\)

Symmetry Class 6 Extra Questions

Symmetry Class 6 Very Short Question Answer

Question 1.
In the below figure, line l is the line of symmetry. Complete the figure to be symmetric about line /.

Solution:
The complete figure is as shown below.

Question 2.
For each of the following figure, identify the line(s) of symmetry if it exists.

Solution:
The line(s) of symmetry in the given figures are as follows:

Question 3.
In the following figures if the dotted lines represent the lines of symmetry, find the other hole(s).

Solution:
The other hole(s) in the given figures is(are) as follows:

Question 4.
Observe each of the following sequences of paper folding and cutting. Draw the pattern obtained after unfolding the paper.

Solution:
The pattern obtained after unfolding the paper in each case are as follows:

Question 5.
In the following figure, identify the line of symmetry.

Solution:
We know that a line that cuts a plane figure into two parts that exactly overlap when folded along that line is called a line of symmetry or axis of symmetry of the figure.

In the given figure, we can see the figure is symmetrical about the line l.

Thus, the line of symmetry of the given figure is line l.

Question 6.
In a figure, 60° is the smallest angle of symmetry. What are the other angles of symmetry of this figure?
Solution:
We know that the angles of symmetry of a figure having rotational symmetry about a point are multiples of the smallest angle of symmetry.
So, the other angles of symmetry are
2 × 60° = 120°,
3 × 60° = 180°,
4 × 60° = 240°,
5 × 60° = 300° and
6 × 60° = 360°.

Question 7.
In a given figure, 45° is the smallest angle of symmetry. What are the other angles of symmetry?
Solution:
We know that the angles of symmetry of a figure having rotational symmetry about a point are multiples of the smallest angle of symmetry,

So, the other angles of symmetry are 2 × 45° = 90°, 3 × 45° = 135°, 4 × 45° = 180°, 5 × 45° = 225°, 6 × 45° = 270°, 7 × 45° = 315° and 8 × 45° = 360°.

Symmetry Class 6 Short Question Answer

Question 1.
Copy the following on square paper. Complete them so the dotted line is a line of symmetry.

Solution:
To complete a figure so that it becomes symmetric about dotted line, let us take the mirror image of the figure with respect to the dotted line. Therefore, the completed figures are as follows:

Question 2.
Each of the following figures shows a piece of paper with punched holes. Copy these figures onto a plain sheet of paper and draw the line of symmetry such that, when the paper is folded along this line, the holes on one side match exactly with the holes on the other side.

Solution:
The line(s) of symmetry in the given figures are as follows:

Question 3.
In each letter of the English alphabet given below, find the number of line(s) of symmetry.
(i) U
(ii) E
(iii) I
(iv) V
Solution:
We know that a line of symmetry cuts a plane figure into two equal but flipped figures.

Question 4.
For each of the following figure, identify the line(s) of symmetry if it exists.

Solution:
The line(s) of symmetry in the given figures are as follows:

Question 5.
In the following figures if the dotted lines represent the lines of symmetry, find the other hole(s).

Solution:
We know that a line that cuts a plane figure into two parts that exactly overlap when folded along that line is called a line of symmetry or axis of symmetry of the figure.

Therefore, the other hole(s) in the given figures is(are) as follows:

Question 6.
In a figure, 90° is an angle of symmetry. The figure has two angles of symmetry less than 90°. What is the smallest angle of symmetry?
Solution:
We know that the angles of symmetry of a figure having rotational symmetry about a point are multiples of the smallest angle of symmetry.

Let the smallest angle of symmetry be x. Then, the other angles of symmetry are 2x, 3x, 4x, …, 360°.

It is given that the figure has two angles of symmetry less than 90°. This means the third angle of symmetry is 90° itself.
∴ 3x = 90°
⇒ x = 30°
Hence, the smallest angle of symmetry is 30°.

Question 7.
How many angles of rotational symmetry does the given figure have and what are they?

Solution:
We notice that the figure returns to its original position only after a complete turn or a 360° rotation.

Therefore, the figure does not possess rotational symmetry, as 360° is the only angle at which it maps onto itself.

Hence, it has no angle of rotational symmetry.

Question 8.
Discuss the rotational symmetry of the given figure.

Solution:
We observe that when the given figure is rotated by 180° and 360°, it coincides exactly with its original position.

Thus, the given figure has rotational symmetry of order 2.

Question 9.
Discuss the rotational symmetry of the given figure.

Solution:
We observe that when the given figure is rotated by 120°, 240° and 360°, it coincides exactly with its original position.

Thus, the given figure has rotational symmetry of order 3.

Symmetry Class 6 Long Question Answer

Question 1.
Observe each of the following sequences of paper folding and cutting. Draw the pattern obtained after unfolding the paper.
(i)

(ii)

Solution:
The pattern obtained after unfolding the paper in each case are as follows:

Question 2.
Which of the following figures have rotational symmetry? Also, find the order of rotational symmetry.

Solution:
(i) We observe that the figure fits onto itself when we give it a half turn, i.e. when it is rotated through 180° as shown in given figure.

Thus, the figure has rotational symmetry of order 4.

(ii) We observe that the given figure fits onto itself once only when it is rotated through 360°, i.e. when it takes a full turn.

Thus, the figure has rotational symmetry of order 4.

(iii) We observe that the given figure fits onto itself once only when it is rotated through 360°, i.e. when it takes a full turn.

Thus, it does not have rotational symmetry.

Question 3.
Show that each of the letters H, I and N has a rotational symmetry of order 2. Also, mark the point of rotational symmetry in each case.
Solution:
Each of the letters H, I and N fits into itself when it is rotated through 180° and 360° about the marked point. Thus, each of the letters H, I and N has a
rotational symmetry of order 2.

Regular revision with Class 6 Maths MCQ with Answers and Ganita Prakash Class 6 Maths Chapter 2 Lines and Angles MCQ improves accuracy in objective exams.

MCQ on Lines and Angles Class 6

Lines and Angles MCQ Class 6

Class 6 Maths Lines and Angles MCQ

Question 1.
A line segment has:
(a) No endpoint
(b) One endpoint
(c) Two endpoints
(d) Infinite endpoints
Solution:
(c) Two endpoints
We know that the shortest path from point A to point B (including A and B) is called the line segment AB. It is denoted by either \(\overline{A B}\) or \(\overline{B A}\). The points A and B are called the endpoints of the line segment \(\overline{A B}\).

Question 2.
Which of these represents a ray in real life?
(a) Beam of light from a torch
(b) A thread
(c) A compass needle
(d) A pencil
Solution:
(a) Beam of light from a torch
We know that in geometry, a ray starts at one point and goes on infinitely in one direction.

The light comes out from the head of the torch and travels in a straight line. This is like a ray in real life.
A thread has two ends. In geometry, this is more like a line segment, not a ray.

A compass needle rotates and points in a direction, but it has two ends, and it does not extend infinitely. It’s not fixed in one direction from one point.
A pencil is a physical object with two definite ends. lake a thread, it is a line segment, not a ray.

Question 3.
The shortest path between two points is:
(a) A ray
(b) A line segment
(c) A curve
(d) A line
Solution:
(b) A line segment
We know that the shortest path between two points (including both points) is called a line segment.

Question 4.
Can you draw a complete line on paper?
(a) Yes, only if the line is horizontal
(b) Yes, only if the line is vertical
(c) No
(d) Can’t say
Solution:
(c) No
We know that a line passing through two points A and B is written as \(\overleftrightarrow{A B}\). It extends infinitely in both directions. Therefore, a complete line cannot be drawn on paper.

Question 5.
An angle is formed by two:
(a) Points
(b) Rays with different starting points
(c) Rays with a common starting point
(d) Line segments
Solution:
(c) Rays with a common starting point
We know that an angle is formed by two rays having a common starting point.

Question 6.
Which of the following does not show angle formation by rotation?
(a) Opening a book
(b) Drawing a straight line
(c) Turning the hands of a clock
(d) Opening a pair of scissors
Solution:
(b) Drawing a straight line
Opening a book: Yes, it makes an angle between the two covers.
Drawing a straight line: No turning or rotation happens. It’s just a line.
Turning the hands of a clock: Yes, the hands rotate and form angles.
Opening a pair of scissors: Yes, the blades form an angle when opened.

Question 7.
Superimposition of angles means:
(a) Extending both arms
(b) Placing one angle over the other
(c) Comparing only vertices
(d) Measuring angles with a ruler
Solution:
(b) Placing one angle over the other
The word superimpose means to place one thing exactly on top of another so that you can compare them. So, superimposition of angles means placing one angle over another angle so that their vertices (corners) and one arm (side) coincide.

This helps us see whether the two angles are equal or not, without using a protractor.

Question 8.
In ∠POM, what is the vertex of the angle?
(a) P
(b) M
(c) O
(d) PO
Solution:
(c) O
In ∠POM, 0 is the vertex of the angle as it is written in the middle.

Question 9.
Superimposed angles are equal when:
(a) They look similar
(b) Their vertices match but arms don’t
(c) Both rays and vertex match exactly
(d) They are on the same side
Solution:
(c) Both rays and vertex match exactly
We know that when two angles are superimposed, and the common vertex and the two rays of both angles lie on each other, then the sizes of the angles are equal.

Question 10.
An angle greater than 90° but less than 180° is called:
(a) Right angle
(b) Acute angle
(c) Reflex angle
(d) Obtuse angle
Solution:
(d) Obtuse angle
We know that an angle greater than 90° but less than 180° is called obtuse angle.

Question 11.
An angle that measures exactly 180° is called a:
(a) Right angle
(b) Reflex angle
(c) Obtuse angle
(d) Straight angle
Solution:
(d) Straight angle
An angle that measures exactly 180° is called a straight angle.

Question 12.
Which of the following is a reflex angle?
(a) 85°
(b) 135°
(c) 180°
(d) 220°
Solution:
(d) 220°
An angle that is greater than 180° but less than 360° is known as a reflex angle. Thus, 220° is a reflex angle.

Question 13.
The markings, formed by repeated folding of a semi-circle, divide it into equal parts using the concept of:
(a) Angle scaling
(b) Rotation
(c) Angle bisector
(d) Circular radius
Solution:
(c) Angle bisector
When we fold a semi-circle repeatedly, each fold divides the angle into two equal parts. This is done using the concept of an angle bisector, which divides the angle into two equal halves.

Question 14.
What is the measure of the angle between the hour hand and the minute hand of a clock when it is 4 o’clock?
(a) 180°
(b) 120°
(c) 90°
(d) 60°
Solution:
(b) 120°
We know that the clock has 12 hours and a full circle is 360°.

So, each hour mark is \(\frac{360^{\circ}}{12}\) = 30 apart.
Thus, the angle increases by 30° for each hour.
At 4 o’clock, the hour hand is at 4 and the minute hand is at 12.
From the figure, we can see that there is a gap of 4 hour marks between the hour hand and the minute hand.
Therefore, angle between the hour hand and the minute hand at 4 o’clock = 4 × 30° = 120°

Question 15.
If there are 18 spokes in a motorcycle wheel, then the angle between two adjacent spokes is:
(a) 10°
(b) 12°
(c) 18°
(d) 20°
Solution:
(d) 20°
We know that a complete angle is equal to 360°. Given, there are 18 spokes in a motorcycle wheel.
Therefore, these 18 spokes divide the complete angle into 18 equal parts.
Thus, the angle between two adjacent spokes = \(\frac{360^{\circ}}{18}\) = 20°

Question 16.
Which of the following statements are true?
(i) A line segment has two endpoints.
(ii) A ray extends endlessly in both directions.
(iii) A line can be named using any two points on it.
(iv) A ray has one endpoint and goes on endlessly in one direction.
Choose the correct option from the following:
(a) Only (i) and (iv)
(b) Only (i) and (ii)
(c) (i), (ii) and (iii)
(d) (i), (iii) and (iv)
Solution:
(d) (i), (iii) and (iv)
We know that
A line segment has two endpoints.
A line is named using any two points that lie on the line.
A ray has one endpoint and extends in one direction.
Thus, the statements (i), (iii) and (iv) are correct.

Question 17.
Which of the following are correct ways to name a given angle?

(i) ∠XYZ
(ii) ∠ZXY
(iii) ∠X
(iv) ∠YXZ
Choose the correct option from the following:
(a) Only (i) and (ii)
(b) Only (ii) and (iii)
(c) Only (iii) and (iv)
(d) (ii), (iiii) and (iv)
Solution:
(d) (ii), (iiii) and (iv)
To name an angle, we write the letter of one arm, followed by the letter of the vertex (always in the middle), and then the letter of the other arm.

For example, if O is the vertex and OA and OB are the arms or sides of the angle, then it can be written as ∠AOB or ∠BOA. If there is no other angle formed at the same vertex, we may simply name it ∠O.

Therefore, the given angle can be written as ∠YXZ, ∠ZXY or ∠X.
Thus, (ii), (iii) and (iv) are correct ways of writing the given angle.

Question 18.
While measuring an angle using a protractor, which of the following steps are correct?
(i) Place centre of protractor on the vertex of the angle.
(ii) Align one arm with the 0° mark.
(iii) Always use the outer scale.
(iv) Use subtraction if arms fall on different marks.
Choose the correct option from the following:
(a) Only (i) and (ii)
(b) (i), (ii) and (iv)
(c) Only (iii) and (iv)
(d) (ii), (iii) and (iv)
Solution:
(b) (i), (ii) and (iv)
While measuring an angle using a protractor, place the centre of the protractor on the vertex of the angle and align one arm with the 0° mark.

Now, depending on the orientation of the given angle, either the inner or outer scale of the protractor can he used.

If neither arm lies on the 0° mark, then subtraction can help to determine the actual measurement.
Thus, (i), (ii) and (iv) are correct.

Lines and Angles Class 6 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled .as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): Any two points determine a unique line.
(R): Through two distinct points, only one line can be drawn.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Through two distinct points, only one line can be drawn.

So, any two points determine a unique line.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 2.
(A): An angle is formed when two rays have a common starting point.
(R): The two rays are called the arms of the angle and the common point is called the vertex.
Solution:
(b) Both A and R are true but R is not the correct explanation of A.
We know that an angle is formed by two rays having a common starting point.

Here is an angle formed by rays \(\overrightarrow{O P} \text { and } \overrightarrow{O M}\) where O is the common starting point.
The point 0 is called the vertex of the angle, and the rays \(\overrightarrow{O P} \text { and } \overrightarrow{O M}\) are called the arms of the angle.
Thus, both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

Question 3.
(A): Superimposition helps us to compare the size of two angles accurately.
(R): By placing one angle over another with the same vertex, we can directly compare their
opening.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
The word superimpose means to place one thing exactly on top of another so that you can compare them.
So, superimposition of angles means placing one angle over another angle so that their vertices (corners) and one arm (side) coincide.
This helps us to compare the size of two angles accurately.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 4.
(A): Folding a semi-circle twice gives an angle of 90°.
(R): Each fold halves the angle, allowing us to mark standard angles like 45°, 90°, and 135°.
Solution:
(d) A is false but R is true.

We know that a semi-circle is half of a full circle, which is 180°. If we fold the semi-circle into 2 equal parts, each part is 90°, and if we fold it again, we get 45°. So, each fold halves the angle, allowing us to mark standard angles like 45°, 90°, and 135°.
Thus, Assertion (A) is false, but Reason (R) is true.

Question 5.
(A): The angle between the hands of a clock at 3 o’clock is 90°.
(R): Each hour mark on a clock corresponds to a 30° rotation.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know that the clock has 12 hours and a full circle is 360°.
So, each hour mark is \(\frac{360^{\circ}}{12}\) = 30° apart.

Thus, the angle increases by 30° for each hour.
At 3 o’clock, the hour hand is at 3 and the minute hand is at 12.

From the figure, we can see that there is a gap of 3 hour marks between the hour hand and the minute hand.

Therefore, angle between the hour hand and the minute hand at 3 o’clock = 3 x 30° = 90°

Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Lines and Angles Class 6 Fill in the Blanks

Question 1.
A line has no endpoints and extends infinitely in ___________ directions.
Solution: both
We know that a line is a geometrical figure that is straight, has no width, and extends indefinitely in both directions.
Therefore, a line has no endpoints and extends infinitely in both directions.

Question 2.
If A and B are two points, the line segment between them is written as ____ .
Solution: \(\overline{A B}\) or \(\overline{B A}\)
We know that if A and B are two points, the line segment between them is written as \(\overline{A B}\) or \(\overline{B A}\).

Question 3.
The correct way to name an angle formed by points D, B and E in the given order, is ______ .
Solution: ∠DBE or ∠EBD
An angle is formed by two rays having a common starting point.
Given, the points are D, B and E.
So, the arms of the angle will be \(\overrightarrow{B D} \text { and } \overrightarrow{B E}\).

Thus, The correct way to name an angle formed by points D, B and E in the give order, is ∠DBE or ∠EBD.

Question 4.
The common starting point of the arms of an angle is called the ____ .
Solution: vertex
We know that the common starting point of the arms of an angle is called the vertex.

Question 5.
Any point outside the arms of an angle lies in its ______ .
Solution: exterior
We know that any point outside the arms of an angle lies in its exterior.

Question 6.
While comparing two angles using superimposition, their ____ and _____ must coincide.
Solution: vertices, one arm
While comparing two angles using superimposition, their vertices and one arm must coincide,

Question 7.
There are 360° in a _____ turn.
Solution: full
We know that there are 360° in a full turn.

Question 8.
When we fold a quarter circle in half again, the resulting angle is __________ .
Solution: 45°
We know that a full circle is 360°.
A quarter circle is 90°. If we fold that quarter circle i. e. 90° in half again, we get \(\frac{90^{\circ}}{2}\) = 45°.
So, when we fold a quarter circle in half again, the resulting angle is 45°.

Question 9.
Two times of 45° is a _____ angle.
Solution: right
Two times of 45° = 2 × 45° = 90°, which is a right angle.
Thus, two times of 45° is a right angle.

Question 10.
A line that divides an angle into two equal parts, is called _________ .
Solution: angle bisector
A line that divides an angle into two equal parts, is called angle bisector.

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 8 Playing with Constructions Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 8 Playing with Constructions Solutions

Ganita Prakash Class 6 Chapter 8 Solutions

Class 6 Maths Ganita Prakash Chapter 8 Solutions Playing with Constructions

Question 1.
Draw the rectangle and four squares configuration on a dot paper.

What did you do to recreate this figure so that the four squares are placed symmetrically around the rectangle? Discuss with your classmates.
Solution:
Draw a rectangle using four dots and draw four squares to make sure all . the squares are equal in size and placed around the rectangle in symmetry.

Question 2.
Take a central line of a different length and try to draw the wave on it.
Solution:

Step 1: Draw a central line AB = 10 cm.
Step 2: Since half of AB is 5 cm, mark a point X on AB such that AX = XB = 5 cm.
Step 3: Mark a point C on AX such that AC = CX =2.5 cm. Mark another point D on XB such that-.. . XD = DB = 2.5 cm.
Step 4: With C as centre and radius equal to AC, draw a semicircle above the line AB, Again, with D as centre and radius equal to BD, draw a semicircle below the line AB.
Step 5: The resultant figure is the required wavy wave with central line of length 10 cm.

InText Questions

Question 1.
Is it possible to construct a 4-sided figure in which all the angles are equal to 90° but opposite sides are not equal?
Solution:
Step 1: Draw a line segment AB = 7 cm.
Step 2: At A and B draw perpendiculars with the help of a protractor.
Step 3: fake two points C and D on the two perpendiculars such that AD = 4 cm and BC = 3 cm.
Step 4: Join DC.
Since the opposite sides AD and BC are not equal, it is found that neither ∠D nor ∠C is 90°.
Hence, we conclude that it is not possible to draw a four-sided figure with all angles equal to 90°, when opposite sides are not equal.

Question 2.
Construct a rectangle in which one of the diagonals divides the opposite angles into 45° and 45°. What do you observe about the sides?
Solution:
Step 1: Draw a horizontal line segment AB (say 5 cm). This will be one side of the rectangle.
Step 2: At point A, use a protractor to measure and draw AX at 45° angle.
Step 3: At point B, measure and draw a 90° angle. Draw a line segment BC extending from B at this angle meeting AX at C.
Step 4: At points A and C, draw a 90° angle which meets at the point D.

Thus, ABCD is the required rectangle.
We observe that all sides are equal. Hence, ABCD is a square.

Playing with Constructions Class 6 Extra Questions

Playing with Constructions Class 6 Very Short Question Answer

Question 1.
Construct a circle with the radius 4.5 cm.
Solution:
Steps of construction of a circle with radius 4.5 cm are as follows:
Step 1: Using a ruler, open the compass to a radius of 4.5 cm.
Step 2: Mark a point O on the paper. Place the pointed end of the compass on point O.
Step 3: Rotate the compass around point 0 to draw a circle ensuring that the pencil remains in contact with the paper at all times.

Question 2.
Construct a square with the side 4 cm.
Solution:

Steps of construction of a square with side 4 cm are as follows:
Step 1: Draw a line segment AB = 4 cm using a ruler.
Step 2: At points A and B, construct two perpendiculars to AB using a protractor.
Step 3: From points A and B, mark points D and C on the perpendiculars respectively such that AD = BC = 4 cm.
Step 4: Join CD to complete the square ABCD.

Question 3.
Construct a rectangle with the side lengths 4 cm and 7 cm.
Solution:

Steps of construction of a rectangle with side lengths 4 cm and 7 cm are as follows:
Step 1: Draw a line segment AB = 7 cm using a ruler.
Step 2: At points A and B, draw perpendiculars to AB using a protractor.
Step 3: From points A and B, mark points D and C on the perpendiculars respectively such that AD = BC = 4 cm.
Step 4: Join CD to complete the rectangle ABCD.

Playing with Constructions Class 6 Short Question Answer

Question 1.
Construct a circle with the following radius:
(i) 4 cm
(ii) 5 cm
(iii) 6 cm
Solution:
(i) Step by step construction of a circle with radius 4 cm:

Step 1: Using a ruler, open the compass to a radius of 4 cm.
Step 2: Mark a point 0 on the paper. Place the pointed end of the compass on point 0.
Step 3: Rotate the compass around point 0 to draw a circle ensuring that the pencil remains in contact with the paper at all times.

(ii) Follow the same steps for radius 5 cm as in (i).
(iii) Follow the same steps for radius 6 cm as in (i).

Question 2.
Recreate the following falling squares:

Solution:
In the figure, we have three identical squares of side length 3 cm.

Step 1: Construct a square ABCD with each side measuring 3 cm.
Step 2: Produce AD to a new point G and CD to a new point E such that DC = DE = 3 cm.
Step 3: Construct square DEFG.
Step 4: Produce EE to a new point J and GF to a A new point H such that FJ = FH = 3 cm.
Step 5: Construct square FHIJ.

Question 3.
Recreate the following falling squares:

Solution:
In the figure, we have three squares of side lengths 5 cm, 4 cm and 3 cm respectively.

Step 1: Construct a square ABCD with each side measuring 5 cm.
Step 2: Produce AD to a new point G and CD to a new point E such that DG = DE = 4 cm.
Step 3: Construct square DEFG.
Step 4: Produce EF to a new point J and GF to a new point H such that FJ = FH = 3 cm.
Step 5: Construct square FHIJ.

Question 4.
Recreate the following combinations of square and hole:

Centre of the hole is the same as the centre of the square containing it.
Solution:
Side length of square is not given. You can take any side length of your choice, say 5 cm.

Step 1: Construct a square ABCD of side 5 cm.
Step 2: Draw diagonals AC and BD. Let O be their point of intersection. Point O is the centre of square ABCD.
Step 3: Using a compass, draw a circle with centre O and radius less than half of the side length of the square ABCD.
The figure so obtained is the required square with a hole.

Question 5.
Construct a rectangle which can be divided into 2 identical squares.
Solution:
A rectangle that can be divided into two identical squares must have one side equal to twice of the other.

Let’s take 3 cm and 6 cm as the side lengths of the rectangle.
Steps of construction:
Step 1: Draw a line segment AB = 6 cm using a ruler, (this will be the longer side of the rectangle).
Step 2: Divide AB into two equal parts. Mark the midpoint M, so that TM = MB = 3 cm.
Step 3: At points A, M and B, draw perpendiculars to AB (using a protractor) long enough to mark the height of the rectangle i.e. 3 cm.
Step 4: Mark points D, N and C on the perpendiculars respectively such that
AD = MN = BC = 3 cm.
Step 5: Join CD to complete the rectangle ABCD.
Here, the rectangle ABCD is divided into two squares (AMND and MBCN) each with the side length of 3 cm.

Question 6.
Recreate the shadings as shown in the figure:

Choose measurements of your choice.
Note: The larger 4 sided- figure is a square and so are the smaller ones.
Solution:
Let the side length of the smaller squares be 2 cm.
Then, the side length of the larger square = 4 × 2 = 8 cm.
Steps of construction are as follows:

Step 1: Construct a square PQRS of side length 8 cm.
Step 2: Using a ruler, mark points at distances 2 cm on sides PQ, QR, RS and SP of square PQRS.
Step 3: Draw horizontal and vertical lines through the marked points to get smaller squares with side equal to 2 cm as shown in the figure.
Step 4: Draw diagonals of the smaller squares.
Step 5: In small squares with side equal to 2 cm, draw vertical lines in the portions below the diagonals as shown in the figure.

Question 7.
Construct a rectangle in which one of its sides is 5 cm and the length of a diagonal is 7.5 cm.
Solution:
Steps of constructions are as follows:
Step 1: Draw a line segment AB of length 5 cm.
Step 2: Using protractor, draw perpendicular BX to AB through B.

Step 3: Using compass, draw an arc with centre A and radius 7.5 cm.
Let arc and BX intersect at point C.
Step 4: Using protractor, draw perpendicular AY to AB through A, and perpendicular CZ to BC through C.
Let AY and CZ intersect at point D. ABCD is the required rectangle.

Question 8.
The distance between points A and B is 7 cm. Now, mark the points that are:

(i) 3 cm away from both A and B
(ii) 4.5 cm away from both A and B
Solution:
(i) Step 1: Using a compass, draw a circle with centre A and radius 3 cm.
Step 2: Using a compass, draw a circle with centre B and radius 3 cm.

Since the two circles do not intersect, there is no point that is exattly 3 cm away from both point A and point B.

(ii) Step 1: Using a compass, mark arcs (with centre A and radius 4.5 cm) above and below AB.
Step 2: Using a compass, mark arcs (with centre B and radius 4.5 cm) above and below AB such that they cut the arcs constructed in step 1.
Let above arcs intersect at point P and below arcs intersect at point Q.
P and Q are at distance of 4.5 cm from points A and B.

Question 9.
Construct a square with the following side length:
(i) 3 cm
(ii) 4.5 cm
(iii) 5.5 cm
Solution:
(i) Steps of construction of a square with side 3 cm are as follows:

Step 1: Draw a line segment AB = 3 cm using a ruler.
Step 2: At points A and B, construct two perpendiculars using a protractor.
Step 3: From points A and B, mark points D and C on the perpendiculars respectively such that AD = BC = 3 cm.
Step 4: Join CD to complete the square ABCD.

(ii) Follow the same steps for square with side 4.5 m as in (i).

(iii) Follow the same steps for square with side 5.5 cm as in (i).

Question 10.
Construct a rectangle of the following side lengths:
(i) 3 cm and 5 cm
(ii) 4 cm and 6.5 cm
(iii) 5.5 cm and 7.5 cm
Solution:
(i) Steps of construction of a rectangle with side lengths 3 cm and 5 cm are as follows:

Step 1: Draw a line segment AB = 5 cm using a ruler.
Step 2: At points A and B, draw perpendiculars to AB using a protractor.
Step 3: Mark points D and C on both the perpendiculars respectively such that AD = BC = 3 cm.
Step 4: Join CD to complete the rectangle ABCD.

(ii) Follow the same steps for rectangle with side lengths 4 cm and 6.5 cm as in (i).

(iii) Follow the same steps for rectangle with side lengths 5.5 cm and 7.5 cm as in (i).

Question 11.
Construct a square within a rectangle such that the centre of the square is the same as the centre of the rectangle. The sides of the rectangle are 7 cm and 4 cm.
Solution:
As square lies inside the rectangle, the side of the square is equal to the smaller side of the rectangle.
So, side of the square = 4 cm
Steps of construction:
Step 1: Draw a rectangle ABCD such that AB = 7 cm and BC = 4 cm.

Step 2: With the help of ruler, mark P, Q, R and S as the mid-points of sides AB, BC, CD and DA.
Step 3: Join PR and QS. Their point of intersection 0 is the centre of the rectangle.
Step 4: With P as centre and OP as radius, mark arcs on both sides ofP intersecting AB at E and F respectively.
Step 5: Similarly, with R as centre and OR (= OP) as radius, mark arcs on both sides of R intersecting CD at G and H respectively. Then, join HE and GF.
Thus, EFGH is the required square whose centre is same as the centre of rectangle ABCD.

Question 12.
Recreate the following combinations of square and hole:

Centre of the hole is the same as the centre of the square containing it.
Solution:
Side length of square is not given. You can take any side length of your choice, say 12 cm.

Step 1: Construct a square ABCD of side 12 cm.
Step 2: Using a ruler, mark points P, Q, R and S as the mid-points of AB, BC, CD and DA respectively.
Join PR and QS. Let 0 be their point of intersection.
Step 3: Draw diagonals AO and PS. Let X be their point of intersection. Point X is the centre of square APOS.
Step 4: Using a compass, draw a circle with centre X and radius less than half of the side length of the square APOS.
Step 5: Draw circles in squares PBQO, OQCR and ORDS (same as step 4). Name their centres as Y, Z and W respectively.

Playing with Constructions Class 6 Long Question Answer

Question 1.
Recreate the following combinations of square and curves:

Note: All the 4 arcs bulge uniformly from each side.
Solution:
Side length of square is not given. We can take any side length of our choice, say 6 cm.
Steps of constructions are as follows:
Step 1: Construct a square ABCD of side 6 cm.
Step 2: Using a ruler, mark points P, Q, R and S as the mid-points of AB, BC, CD and DA respectively.
Step 3: Join PR and produce it in both directions. Also, join QS and produce it in both directions.
Step 4: Mark points X,Y, Z and W such that

XP = YQ = ZR = WS ≥ \(\frac{A B}{2}\)
Step 5: Using compass, draw an arc with X as centre and AX as radius such that it passes through A and B.
Similarly, draw arcs with centres Y, Z and W and radius YB, ZC and DW respectively.
The figure so obtained is the required square with curves.

NOTE: If we take XP = YQ = ZR = WS < \(\frac{A B}{2}\), then adjacent arcs will intersect at two points. But that is not the case.

Question 2.
A rectangular orchard of size 20 m × 15 m is to be planned using the square and diagonal planting systems. Each tree requires 5 m spacing. The planner uses only a compass, ruler and protractor to design accurate layouts based on geometric principles.

Based on the above information, answer the following questions:
i) How many trees can be planted using the square system, with 5 in spacing on a 20m × 13 in held?
(ii) Two trees are 6 m apart. How can you find a point that is 4.5 m away from both? Explain using compass method.
Solution:
(i) Given, field dimensions = 20 m × 15 m

Along the 20 m side (length), trees can be planted at the positions 0 m, 5 m, 10 m, 15 m, and 20 m. That is, there are a total of five positions along the 20 m side.
Therefore, the total number of columns of trees = 5.

Along the 15 m side (breadth), trees can be planted at the positions 0 m, 5 m, 10 m, and 15 m. That is, there are a total of four positions along the 15 m side.

Therefore, the total number of rows of trees = 4.

Hence, total trees = 5 × 4 = 20 trees

(ii) Given, two trees are 6 metres apart, we can find a point that is 4.5 metres away from both by following these steps:

Step 1: Mark two points A and B, 6 metres apart and join them using ruler.
Step 2: Set your compass to 4.5 cm.
Step 3: With A as the centre, draw two arcs, one above and one below the line segment AB.
Step 4: With B as centre, draw arcs of the same radius to intersect the previous arcs.
Step 5: Label the intersection points as P and 0.
Hence, points P and Q are both exactly 4.5 m away from A and B.

Question 3.
Recreate the following House:

Note: All the lines forming the border of the house are of length 4 cm.
Solution:

Steps of construction are as follows:
Step 1: Draw a line segment AB = 4 cm.
Step 2: Construct perpendiculars AD and BC to AB such that BC = AD = 4 cm.
Step 3: With D as centre draw an arc of length 4 cm. Also, draw an arc of same length with C as centre meeting previously drawn arc at point P.
Step 4: Join CP and PD, and with P as centre and CP as radius draw an arc joining C and D. The figure so formed represents a house whose all sides are of length 4 cm.
Step 5: Mark points E and F on AB such that
AE = BF = \(\left(\frac{4-1}{2}\right)\) cm = \(\frac{3}{2}\) cm = 1.5 cm.
Step 6: Draw perpendiculars EH and EC on AB at E and F respectively such that EH = FG = 2 cm. Join GH.

Regular revision with Class 6 Maths MCQ with Answers and Ganita Prakash Class 6 Maths Chapter 1 Patterns in Mathematics MCQ improves accuracy in objective exams.

MCQ on Patterns in Mathematics Class 6

Patterns in Mathematics MCQ Class 6

Class 6 Maths Patterns in Mathematics MCQ

Question 1.
Which of the following is a triangular number?
(a) 8
(b) 11
(c) 18
(d) 21
Solution:
(d) 21
We know, the number sequence of triangular numbers is 1, 3, 6, 10, 15, 21, 28 … .
From the given numbers, 21 is present in the number sequence of triangular numbers.
Hence, 21 is a triangular number.

Question 2.
Which of the following is not a cube number?
(a) 8
(b) 27
(c) 125
(d) 169
Solution:
(d) 169
We know, the number sequence of cubes is 1, 8, 27, 64, 125, 216, … .
From the given numbers, 169 is not present in the number sequence of cubes.
Hence, 169 is not a cube.

Question 3.
Which of the following is not a power of 3?
(a) 9
(b) 63
(c) 81
(d) 243
Solution:
(b) 243
We know, the number sequence of powers of 3 is 1, 3, 9, 27, 81, 243, 729 … .
From the given numbers, 63 is not present in the number sequence of powers of 3.
Hence, 63 is not a power of 3.

Question 4.
The pattern in the sequence 1, 4, 9, 16, 25, … is:
(a) n^th term = 2n, where n = 1, 2, 3, 4, …
(b) n^th term = n³, where n = 1, 2, 3, 4, …
(c) n^th term = \(\frac{n}{2}\), where n = 1, 2, 3, 4, …
(d) n^th term = n², where n = 1, 2, 3, 4, …
Solution:
(d) n^th term = n², where n = 1, 2, 3, 4, …
Given sequence is 1, 4, 9, 16, 25, ….
First term = 1 = 1 × 1 = 1²
Second term = 4 = 2 × 2 = 2²
Third term = 9 = 3 × 3 = 3²
Fourth term = 16 = 4 × 4 = 4²
Fifth term = 25 = 5 × 5 = 5²
Therefore, the sequence follows the pattern:
n^th term = n~, where n = 1,2, 3, 4, …

Question 5.
The pattern in the sequence 1, 3, 6, 10, 15, 21, 28, … is:
(a) n^th term = Sum of first n counting numbers + 1, where n = 1, 2, 3, 4, …
(b) n^th term = Sum of first n counting numbers, where n = 1, 2, 3, 4, …
(c) n^th term = Sum of first n counting numbers × 2, where n = 1, 2, 3, 4, …
(d) n^th term = Sum of first n counting numbers ÷ 2, where n = 1, 2, 3, 4, …
Solution:
(b) n^th term = Sum of first n counting numbers, where n = 1, 2, 3, 4, …
Given sequence is 1, 3, 6, 10, 15, 21, 28, … .
First term = 1 (First counting number)
Second term = 3 = 1 + 2 (Sum of first 2 . counting numbers)
Third term = 6 = 1 + 2 + 3 (Sum of first 3 counting numbers)
Fourth term = 10 = 1 + 2 + 3 + 4 (Sum of first 4 counting numbers)
Fifth term = 15 = 1 + 2 + 3 + 4 + 5 (Sum of first 5 counting numbers)
Therefore, the sequence follows the pattern:
n^th term = Sum of first n counting numbers, where n = 1, 2, 3, 4, …

Question 6.
Which of the following sequence is represented by dots forming a square?
(a) 1, 4, 9, 16, 25, …
(b) 1, 3, 5, 7, 9, …
(c) 1, 2, 4, 8, 16, …
(d) 1, 3, 6, 10, 15, …
Solution:
(a) 1, 4, 9, 16, 25, …
We know, the square numbers 1, 4, 9, 16, 25,… are represented by dots forming the squares.

Question 7.
The sequence 1, 8, 27, 64, 125, … is visualised using:
(a) Squares
(b) Circles
(c) Triangles
(d) Cubes
Solution:
(d) Cubes
The given sequence 1, 8, 27, 64, 125, … is asequence of cubes.
We know, the sequence of cubes can be visualised using cubes.

Question 8.
What is the next number in the sequence: 1, 7, 19, 37,…?
(a) 50
(b) 61
(c) 63
(d) 65
Solution:
(b) 61
The given sequence is 1, 7, 19, 37, …, which is the sequence of hexagonal numbers.
The rule followed in the sequence of hexagonal numbers is
1^st term = 1, n^th number (term) = Preceding term + 6 × (n – 1); n = 2, 3, 4, …
∴ Next term, i.e. fifth term = 37 + 6 v (5 – 1) = 37 + 6 × 4 = 61

Question 9.
Triangular numbers can be visualised by arranging dots in the form of:
(a) Rectangles
(b) Squares
(c) Triangles
(d) Hexagons
Solution:
(c) Triangles
We know, the triangular numbers 1, 3, 6, 10, 15, … can be visualised by arranging dots in the form of triangles.

Question 10.
How many dots are there in the 4^th triangular number?
(a) 6
(b) 10
(c) 15
(d) 21
Solution:
(b) 10
We know, the sequence of triangular numbers is 1, 3, 6, 10, 15, 21, 28, 36, 45, … .
So, the 4^th triangular number is 10.
Thus, there are 10 dots in the triangle which represent the 4^th triangular number.

Question 11.
How many dots are there in the 4th hexagonal number?
(a) 19
(b) 37
(c) 61
(d) 91
Solution:
(b) 37
We know, the sequence of hexagonal numbers is 1, 7, 19, 37, 61, 91, ……..
So, the 4^th hexagonal number is 37.

Thus, there are 37 dots in the hexagon which represent the 4^th hexagonal number.

Question 12.
The sum 1 + 7 + 19 + 37 gives which type of number?
(a) Triangular
(b) Virahanka
(c) Square
(d) Hexagonal
Solution:
(c) Square
1 + 7 + 19 + 37 = 64, which is a square number.

Question 13.
A polygon having 6 sides is known as:
(a) Quadrilateral
(b) Pentagon
(c) Hexagon
(d) Heptagon
Solution:
(c) Hexagon
Hexagon has 6 sides.

Question 14.
The geometric pattern 3, 12, 48, 192, 768, …, represents:
(a) The number of sides in the sequence of Koch Snowflakes
(b) The number of line segments in the sequence of Complete Graphs
(c) The number of vertices in the sequence of Regular Polygons
(d) The number of stacked triangles in the sequence of Stacked Triangles
Solution:
(a) The number of sides in the sequence of Koch Snowflakes
The geometric pattern 3, 12, 48, 192, 768, … represents the number of sides in the Koch snowflakes sequence.

Question 15.
The number of line segments in the sequence of complete graphs form a sequence of:
(a) even numbers
(b) virahanka numbers
(c) powers of 2
(d) triangular numbers
Solution:
(d) triangular numbers
The number of line segments in a complete graph follows the sequence: 0, 1, 3, 6, 10, 15…
Therefore, complete graphs sequence is related to triangular number sequence, i.e. i, 3, 6, 10,15 … .

Question 16.
The number of stacked triangles in the stacked triangles sequence form a sequence of:
(a) triangular numbers
(b) square numbers
(c) powers of 2
(d) hexagonal numbers
Solution:
(b) square numbers
We know,

Number of stacked triangles:
1 = 1² 4 = 2² 9 = 3²
16 = 4² 25 = 5²
Thus, the number of stacked triangles in the stacked triangles sequence form a sequence of square numbers.

Question 17.
Which of the following statements is/are true regarding square numbers?
(i) The sum of the first n odd numbers gives a square number.
(ii) Adding counting numbers up from 1 to a number, and then back down again to 1, also gives a square number.
(iii) Adding up consecutive powers of 2, yields square number.
(iv) Multiplying triangular numbers by 6 and adding 1 to each term gives square numbers.
Choose the correct option from the following:
(a) (i) and (iii) only
(b) (i) and (ii) only
(c) (iii) and (iv) only
(d) (i), (iii) and (iv)
Solution:
(b) (i) and (ii) only
We have, the sum of the first two odd numbers = 1 + 3 = 4 = 2²
The sum of the first three odd numbers = 1 + 3 + 5 = 9 = 3²
The sum of the first four odd numbers =1 + 3 + 5 + 7 = 16 = 4² and so on.
Hence, the sum of the first n odd numbers gives a square number.
So, statement (i) is correct.

Also, adding counting numbers up and then down gives a square number.
For example, 1 + 2 + 1 = 4 = 2² and 1 + 2 + 3 + 2 + 1 = 9 = 3² and so on.
So, statement (ii) is also correct.

Now, adding consecutive powers of 2 does not give square numbers. The sequence powers of 2 is given as 1,2, 4, 8, 16, 32, …
Adding two consecutive numbers at a time, we get 3 (1 + 2 ), 6 (2 + 4), etc., which are not square numbers.
So, statement (iii) is incorrect.

We know that if we multiply triangular numbers by 6 and then add 1, we get a new number sequence: 7, 19, 37, 61 and so on. The new sequence is the sequence of hexagonal numbers starting with 7.
So, statement (iv) is also incorrect.

Question 18.
Which of the following statements is/are true?
(i) The regular polygons sequence is related to counting numbers sequence starting with 3.
(ii) The complete graphs sequence is related to triangular numbers sequence, where numbers are given by \(\frac{n(n-1)}{2}\), n = 1, 2, 3, …, i.e. 0, 1, 3, 6, 10, 15 and so on.
(iii) The stacked squares sequence is related to square numbers sequence, i.e. 1, 4, 9, 16, 25 and so on.
(iv) The stacked triangles sequence is related to the square numbers sequence, i.e. 1, 4, 9, 16, 25 and so on.
Choose the correct option from the following:
(a) (i) and (ii) only
(b) (ii) and (iv) only
(c) (i) (ii) and (iii) only
(d) (i), (ii), (iii) and (iv)
Solution:
(d) (i), (ii), (iii) and (iv)
All the statements are correct.

Patterns in Mathematics Class 6 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): All is sequence (1, 1, 1, 1, …) gives counting numbers when added up successively.
(R): Adding 1 repeatedly increases the total by 1 each time.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know, all is sequence is 1, 1, 1, 1, …..
Here, 1 = 1 = 1
1 + 1 = 2 = 1 + 1
1 + 1 + 1 = 3 = 2 + 1
1 + 1 + 1 + 1 = 4 = 3 + 1
So, all is sequence (1, 1, 1, 1, ….) gives the counting numbers when added up successively and adding 1 repeatedly increases the total by 1 each time.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 2.
(A): The pattern 1, 2, 4, 8, 16, … represents the powers of 2.
(R): Each number in this sequence is obtained by adding 2 to the previous term.
Solution:
(c) A is true but R is false.
We know, the number sequence of powers of 2 is 1, 2, 4, 8, 16, 32, 64, … .
First term = 1
Second term = 2 = 1 × 2 (First term × 2)
Third term = 4 = 2 × 2 (Second term × 2)
Fourth term = 8 = 4 × 2 (Third term × 2)
Fifth term = 16 = 8 × 2 (Fourth term × 2)
So, we can say that each term of the number sequence of powers of 2 is obtained by multiplying 2 to the previous term.
Thus, Assertion (A) is true, but Reason (R) is false.

Question 3.
(A): Square numbers can be obtained by adding up the odd numbers.
(R): The sum 1 + 3 + 5 + 7 + 9 = 25, which is a perfect square.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know, the sequence of odd numbers is 1, 3, 5, 7, 9, … .
And, the sequence of squares is 1, 4, 9, 16, 25, ….
Now, 1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25
So, the square numbers can be obtained by adding up the odd numbers.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 4.
(A): The hexagonal number sequence starts with 1 and continues with 7, 19, 37, 61, and so on.
(R): Each number in the sequence adds a growing number of layers of dots in a hexagonal form.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know, the sequence of hexagonal numbers is 1, 7, 19, 37, 61, 91, … .
Pictorial representation of hexagonal numbers is given below:

Here, we observe that each number in the sequence adds a growing number of layers of dots in a hexagonal form.Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Patterns in Mathematics Class 6 Fill in the Blanks

Question 1.
The next term of the sequence 1, 16, 49, 100, 169,… is _______.
Solution: 256
Given sequence is 1, 16, 49, 100, 169, ….
First term = 1 = (3 × 0 + 1)²
Second term = 16 = (3 × 1 + 1)²
Third term = 49 = (3 × 2 + 1)²
Fourth term = 100 = (3 × 3 + 1)²
Fifth term = 169 = (3 × 4 + 1)²
∴ Next term, i.e. sixth term = (3 × 5 + 1)²
= (16)² = 256
Hence, the next term of the sequence 1, 16, 49, 100, 169, … is 256.

Question 2.
A square number sequence can be visualised as dots arranged in shape of a _______ .
Solution: square
We know, a square number sequence can be visualised as dots arranged in the shape of a square.

Question 3.
In the triangular number sequence, the 5^th number is ________ .
Solution: 15
We know, the triangular number sequence is 1,3, 6, 10, 15, 21, 28,… .
So, in the triangular number sequence, the 5^th number is 15.

Question 4.
The difference between consecutive square numbers forms an _______ sequence.
Solution: odd number
The difference between consecutive square numbers forms an odd number sequence: 3, 5, 7, 9, 11, … i.e. 4 – 1 = 3, 9 – 4 = 5, 16 – 9 = 7, and so on.

Question 5.
The pictorial representation of the sequence 1, 8, 27, 64, … is based on _______ .
Solution: cubes
We know, the sequence 1, 8, 27, 64, … is the sequence of cubes.
Thus, the pictorial representation of the sequence 1,8, 27, 64, … is based on cubes.

Question 6.
A polygon having 8 sides is known as __________ .
Solution: Octagon
A polygon having 8 sides is known as Octagon.

Question 7.
The difference between the number of sides of a nonagon and a pentagon is _________ .
Solution: 4
A pentagon has 5 sides whereas a nonagon has 9 sides.
∴ The difference between the number of sides of a nonagon and a pentagon is 9 – 5 = 4.

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 7 Fractions Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 7 Fractions Solutions

Ganita Prakash Class 6 Chapter 7 Solutions

Class 6 Maths Ganita Prakash Chapter 7 Solutions Fractions

Question 1.
Draw a picture and write an addition statement to show:
(a) 5 times \(\frac{1}{4}\) of a roti
(b) 9 times \(\frac{1}{4}\) of a roti
Solution:
(a)

5 times \(\frac{1}{4}\) of a roti = \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\)

(b)

9 times \(\frac{1}{4}\) of a roti = \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\)

Question 2.
Match each fractional unit with the correct picture:

Solution:

Question 3.
On a number line, draw lines of lengths \(\frac{1}{10}\), \(\frac{3}{10}\), and \(\frac{4}{5}\).
Solution:

Question 4.
Write the fraction that gives the lengths of the lines in the respective boxes.

Solution:
\(\frac{6}{5}\), \(\frac{7}{5}\), \(\frac{8}{5}\), \(\frac{9}{5}\)

Question 5.
Figure out the number of whole units in each of the following fractions:
(a) \(\frac{8}{3}\)
(b) \(\frac{11}{5}\)
(c) \(\frac{9}{4}\)
Solution:
(a) \(\frac{8}{3}\) = 2\(\frac{2}{3}\). So, 2 whole units.
(b) \(\frac{11}{5}\) = 2\(\frac{1}{5}\). So, 2 whole units.
(c) \(\frac{9}{4}\) = 2\(\frac{1}{4}\). So, 2 whole units.

Question 6.
Are \(\frac{3}{6}\), \(\frac{4}{8}\) and \(\frac{5}{10}\) equivalent fractions ? Why?
Solution:
Yes, because all of them have the same length i.e., \(\frac{3}{6}\) = \(\frac{4}{8}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\)

Question 7.
\(\frac{4}{6}\) = ___ = ____ = _____ = ____ (write as many as you can)
Solution:
\(\frac{4}{6}\) = \(\frac{2}{3}\) = \(\frac{6}{9}\) = \(\frac{8}{12}\) = \(\frac{10}{15}\)

Question 8.
Rahim mixes \(\frac{2}{3}\) litres of yellow paint with \(\frac{3}{4}\) litres of blue paint to make green paint. What is the volume of green paint he has made?
Solution:
Volume of yellow paint = \(\frac{2}{3}\) litres
Volume of blue paint = \(\frac{3}{4}\)
Volume of green paint = (\(\frac{2}{3}\) + \(\frac{3}{4}\)) litres = (\(\frac{2}{3}\) × \(\frac{4}{4}\) + \(\frac{3}{4}\) × \(\frac{3}{3}\))litres
= (\(\frac{8}{12}\) + \(\frac{9}{12}\))litres = \(\frac{17}{12}\)litres = 1\(\frac{5}{12}\) litres

Question 9.
Geeta bought \(\frac{2}{5}\) metre of lace and Shamim bought \(\frac{3}{4}\) metre of the same lace to put a complete border on a table cloth whose perimeter is 1 metre long. Find the total length of the lace they both have bought. Will the lace be sufficient to cover the whole border?
Solution:
Length of lace bought by Geeta = \(\frac{2}{5}\) metre
Length of lace bought by Shamim = \(\frac{3}{4}\) metre
∴ Total length of lace bought = (\(\frac{2}{5}\) + \(\frac{3}{4}\))metres = (\(\frac{2}{5}\) × \(\frac{4}{4}\) + \(\frac{3}{4}\) × \(\frac{5}{5}\))metres
= (\(\frac{8}{20}\) + \(\frac{15}{20}\))metres = \(\frac{23}{20}\) metres = 1\(\frac{3}{20}\) metres > 1 m
Yes, the lace will be sufficient to cover the whole border as it exceeds the perimeter of table cloth, which is 1 m long.

Question 10.
Solve the following problems:
(a) Jaya’s school is \(\frac{7}{10}\) km from her home. She takes an auto for \(\frac{2}{5}\) km from her home daily, and then walks the remaining distance to reach her school. How much does she walk daily to reach the school?
(b) Jeevika takes \(\frac{10}{3}\) minutes to take a complete round of the park and her friend Namit takes \(\frac{13}{4}\) minutes to do the same. Who takes less time and by bow much?
Solution:
(a) Total distance between school and home = \(\frac{7}{10}\) km
Distance travelled in auto = \(\frac{1}{2}\) km.
∴ Distance she walks daily to reach the school = (\(\frac{7}{10}\) – \(\frac{1}{2}\))km = (\(\frac{7}{10}\) – \(\frac{1}{2}\) × \(\frac{5}{5}\))km
= (\(\frac{7}{10}\) – \(\frac{5}{10}\))km = \(\frac{2}{10}\) km = \(\frac{1}{5}\) km

(b) Time taken by Jeevika = \(\frac{10}{3}\) minutes and time taken by Namit = \(\frac{13}{4}\) minutes
Now, \(\frac{10}{3}\) × \(\frac{4}{4}\) = \(\frac{40}{12}\) and \(\frac{13}{4}\) × \(\frac{3}{3}\) = \(\frac{39}{12}\)
Clearly, \(\frac{10}{3}\) > \(\frac{13}{4}\)
∴ Namit takes less time by (\(\frac{10}{3}\) – \(\frac{13}{4}\)) minutes = (\(\frac{40}{12}\) – \(\frac{39}{12}\)) minutes = \(\frac{1}{12}\) minutes

Fractions Class 6 Extra Questions

Fractions Class 6 Very Short Question Answer

Question 1.
What fraction of a year is 5 months?
Solution:
We know, number of months in a year = 12
So. 5 month is of a \(\frac{5}{12}\) year.

Question 2.
Represent \(\frac{2}{7}\), \(\frac{4}{7}\), \(\frac{6}{7}\) on a number line.
Solution:
In order to represent \(\frac{2}{7}\), \(\frac{4}{7}\), \(\frac{6}{7}\) on a number line. we divide the gap between 0 and 1. i.e. 1 unit, into 7 equal parts and take second, fourth and sixth points from 0, as shown in the figure.

Question 3.
What fraction of a dozen banana is 7 bananas?
Solution:
Number of bananas in 1 dozen = 12
So, 7 bananas is \(\frac{7}{12}\) of a dozen.

Question 4.
Represent \(\frac{1}{8}\), \(\frac{2}{8}\), \(\frac{4}{8}\) on a number line.
Solution:
In order to represent \(\frac{1}{8}\), \(\frac{2}{8}\), \(\frac{4}{8}\) on a number line, we divide the gap between 0 and 1 into 8 equal parts and take first, second and fourth points from 0, as shown in the figure.

Question 5.
Write the fraction that gives the length of the lines in the respective boxes.

Solution:

Question 6.
Write some equivalent fractions which contain all digits from 1 to 9 once only.
Solution:
\(\frac{2}{6}\) = \(\frac{3}{9}\) = \(\frac{58}{174}\),
\(\frac{2}{4}\) = \(\frac{3}{6}\) = \(\frac{79}{158}\)

Question 7.
Write three equivalent fractions of \(\frac{3}{4}\).
Solution:
Equivalent fractions of \(\frac{3}{4}\) are:
\(\frac{3}{4}\) = \(\frac{3 \times 2}{4 \times 2}\) = \(\frac{6}{8}\),
\(\frac{3}{4}\) = \(\frac{3 \times 3}{4 \times 3}\) = \(\frac{9}{12}\),
\(\frac{3}{4}\) = \(\frac{3 \times 4}{4 \times 4}\) = \(\frac{12}{16}\)

Question 8.
Subtract \(\frac{3}{7}\) from \(\frac{6}{7}\).
Solution:
\(\frac{6}{7}\) – \(\frac{3}{7}\) = \(\frac{6-3}{7}\) = \(\frac{3}{7}\)

Question 9.
Subtract 8\(\frac{1}{5}\) from 12\(\frac{2}{5}\).
Solution:
8\(\frac{1}{5}\) from 12\(\frac{2}{5}\) = \(\left(\frac{12 \times 5+2}{5}\right)\) – \(\left(\frac{8 \times 5+1}{5}\right)\)
= \(\left(\frac{60+2}{5}\right)\) = \(\left(\frac{40+1}{5}\right)\)
= \(\frac{62}{5}\) – \(\frac{41}{5}\) = \(\frac{62-41}{5}\) = \(\frac{21}{5}\)

Question 10.
Rohit travelled \(\frac{200}{3}\) km by train and \(\frac{50}{3}\) km by bus. What is the total distance travelled by Rohit?
Solution:
Distance travelled by train = \(\frac{200}{3}\) km;
Distance travelled by bus = \(\frac{50}{3}\) km
∴ Total distance covered
= (\(\frac{200}{3}\) \(\frac{50}{3}\)) km = (\(\frac{200+50}{3}\))km
= \(\frac{250}{3}\) km

Question 11.
Find the difference of \(\frac{19}{24}\) and \(\frac{13}{16}\).
Solution:
We can write 24 = 2 × 2 × 2 × 3 and
16 = 2 × 2 × 2 × 2
So, LCM of 24 and 16 is 2 × 2 × 2 × 2 × 3 = 48.
Now, \(\frac{19}{24}\) = \(\frac{19 \times 2}{24 \times 2}\) = \(\frac{38}{48}\) and
\(\frac{13}{16}\) = \(\frac{13 \times 3}{16 \times 3}\) = \(\frac{39}{48}\)
Clearly, \(\frac{38}{48}\) < \(\frac{39}{48}\) ⇒ \(\frac{19}{24}\) < \(\frac{13}{16}\)
Thus, required difference
= \(\frac{13}{16}\) – \(\frac{19}{24}\) = \(\frac{39}{48}\) – \(\frac{38}{48}\)
= \(\frac{39-38}{48}\) = \(\frac{1}{48}\)

Question 12.
Subtract \(\frac{5}{9}\) from \(\frac{7}{9}\).
Solution:
\(\frac{7}{9}\) – \(\frac{5}{9}\) = \(\frac{7-5}{9}\) = \(\frac{2}{9}\)

Fractions Class 6 Short Question Answer

Question 1.
Write the following fractions as mixed fractions:
(i) \(\frac{10}{3}\)
(ii) \(\frac{12}{5}\)
(iii) \(\frac{16}{7}\)
(iv) \(\frac{11}{3}\)
(v) \(\frac{63}{4}\)
Solution:
(i) \(\frac{10}{3}\) = 3 + \(\frac{1}{3}\) = 3\(\frac{1}{3}\)

(ii) \(\frac{12}{5}\) = 2 + \(\frac{2}{5}\) = 2\(\frac{2}{5}\)

(iii) \(\frac{16}{7}\) = 2 + \(\frac{2}{7}\) = 2\(\frac{2}{7}\)

(iv) \(\frac{11}{3}\) = 3 + \(\frac{2}{3}\) = 3\(\frac{2}{3}\)

(v) \(\frac{63}{4}\) = 15 + \(\frac{3}{4}\) = 15\(\frac{3}{4}\)

Question 2.
Write the following mixed fractions into improper fractions:
(i) 4\(\frac{1}{3}\)
(ii) 2\(\frac{1}{4}\)
(iii) 7\(\frac{3}{10}\)
(iv) 12\(\frac{1}{2}\)
(v) 5\(\frac{3}{7}\)
Solution:
(i) 4\(\frac{1}{3}\) = 4 + \(\frac{1}{3}\) = \(\frac{4 \times 3+1}{3}\)
= \(\frac{12+1}{3}\) = \(\frac{13}{3}\)

(ii) 2\(\frac{1}{4}\) = 2 + \(\frac{1}{4}\) = \(\frac{2 \times 4+1}{4}\)
= \(\frac{8+1}{4}\) = \(\frac{9}{4}\)

(iii) 7\(\frac{3}{10}\) = 7 + \(\frac{3}{10}\) = \(\frac{7 \times 10+3}{10}\)
= \(\frac{70+3}{10}\) = \(\frac{73}{10}\)

(iv) 12\(\frac{1}{2}\) = 12 + \(\frac{1}{2}\) = \(\frac{12 \times 2+1}{2}\)
= \(\frac{24+1}{2}\) = \(\frac{73}{2}\)

(v) 5\(\frac{3}{7}\) = 5 + \(\frac{3}{7}\) = \(\frac{5 \times 7+3}{7}\)
= \(\frac{35+3}{7}\) = \(\frac{35}{7}\)

Question 3.
Replace ☐ in each of the following by the correct number:
(i) \(\frac{2}{5}\) = \(\frac{10}{}\)
(ii) \(\frac{3}{7}\) = \(\frac{27}{}\)
(iii) \(\frac{9}{13}\) = \(\frac{27}{}\)
(iv) \(\frac{6}{7}\) = \(\frac{}{49}\)
(v) \(\frac{5}{7}\) = \(\frac{}{35}\)
Solution:
(i) We know, 10 = 2 × 5

So, we replace ☐ by 25 to get \(\frac{2}{5}\) = \(\frac{10}{25}\)

(ii) We know, 27 = 3 × 7

So, we replace ☐ by 63 to get \(\frac{3}{7}\) = \(\frac{27}{63}\)

(iii) We know, 27 = 9 × 3

So, we replace ☐ by 39 to get \(\frac{9}{13}\) = \(\frac{27}{39}\)

(iv) We know, 49 = 7 × 7

So, we replace ☐ by 42 to get \(\frac{6}{7}\) = \(\frac{42}{49}\)

(v) We know, 27 = 3 × 7

So, we replace ☐ by 63 to get \(\frac{3}{7}\) = \(\frac{27}{63}\)

Question 4.
Find the fraction equivalent to \(\frac{30}{45}\), having:
(i) Numerator 16
(ii) Denominator 30
Solution:
We have, \(\frac{30}{45}\) = \(\frac{2 \times 15}{3 \times 15}\) = \(\frac{2}{3}\)
(i) On dividing 16 by 2, we get 8

So, the fraction with numerator 16 and equivalent to \(\frac{30}{45}\) is \(\frac{16}{24}\).

(ii) On dividing 30 by 3, we get 10

So, the fraction with denominator 30 and equivalent to \(\frac{30}{45}\) is \(\frac{20}{30}\).

Question 5.
Write the following mixed fractions as fractions:
(i) 12\(\frac{3}{10}\)
(ii) 9\(\frac{5}{8}\)
(iii) 7\(\frac{6}{11}\)
(iv) 4\(\frac{2}{9}\)
(v) 6\(\frac{3}{7}\)
Solution:
(i) 12\(\frac{3}{10}\) = 12 + \(\frac{3}{10}\) = \(\frac{12 \times 10+3}{10}\)
= \(\frac{120+3}{10}\) = \(\frac{123}{10}\)

(ii) 9\(\frac{5}{8}\) = 9 + \(\frac{5}{8}\) = \(\frac{9 \times 8+5}{8}\)
= \(\frac{72+5}{8}\) = \(\frac{77}{8}\)

(iii) 7\(\frac{6}{11}\) = 7 + \(\frac{6}{11}\) = \(\frac{7 \times 11+6}{11}\)
= \(\frac{77+6}{11}\) = \(\frac{83}{11}\)

(iv) 4\(\frac{2}{9}\) = 4 + \(\frac{2}{9}\) = \(\frac{4 \times 9+2}{9}\)
= \(\frac{36+2}{9}\) = \(\frac{38}{9}\)

(v) 6\(\frac{3}{7}\) = 6 + \(\frac{3}{7}\) = \(\frac{6 \times 7+3}{7}\)
= \(\frac{42+3}{7}\) = \(\frac{45}{7}\)

Question 6.
Replace ☐ in each of the following by the correct number:
(i) \(\frac{3}{4}\) = \(\frac{15}{}\)
(ii) \(\frac{7}{3}\) = \(\frac{}{15}\)
(iii) \(\frac{10}{7}\) = \(\frac{30}{}\)
(iv) \(\frac{11}{15}\) = \(\frac{44}{}\)
(v) \(\frac{15}{4}\) = \(\frac{}{24}\)
Solution:
(i) On dividing 15 by 3, we get 5.
Now, multiplying the numerator and denominator of \(\frac{3}{4}\) by 5, we get
\(\frac{3}{4}\) = \(\frac{3 \times 5}{4 \times 5}\) = \(\frac{15}{20}\)
So, we replace ☐ by 20.

(ii) On dividing 15 by 3, we get 5.
Now, multiplying the numerator and denominator of \(\frac{7}{3}\) by 5, we get
\(\frac{7}{3}\) = \(\frac{7 \times 5}{3 \times 5}\) = \(\frac{35}{15}\)
So, we replace ☐ by 35.

(iii) On dividing 30 by 10, we get 3.
Now, multiplying the numerator and denominator of \(\frac{10}{7}\) by 3, we get
\(\frac{10}{7}\) = \(\frac{10 \times 3}{7 \times 3}\) = \(\frac{30}{21}\)
So, we replace ☐ by 21.

(iv) On dividing 44 by 11, we get 4.
Now, multiplying the numerator and denominator of \(\frac{11}{15}\) by 4, we get
\(\frac{11}{15}\) = \(\frac{11 \times 4}{15 \times 4}\) = \(\frac{44}{60}\)
So, we replace ☐ by 60.

(v) On dividing 24 by 4, we get 6.
Now, multiplying the numerator and denominator of \(\frac{15}{4}\) by 6, we get
\(\frac{15}{4}\) = \(\frac{15 \times 6}{4 \times 6}\) = \(\frac{90}{24}\)
So, we replace ☐ by 90.

Question 7.
Find the fraction equivalent to \(\frac{18}{45}\), having:
(i) Numerator 50
(ii) Denominator 60
Solution:
We have, \(\frac{15}{45}\) = \(\frac{2 \times 9}{5 \times 9}\) = \(\frac{2}{5}\)
(i) On dividing 50 by 2, we get 25.
Now, multiplying the numerator and denominator of \(\frac{2}{5}\) by 25, we get
\(\frac{2}{5}\) = \(\frac{2 \times 25}{5 \times 25}\) = \(\frac{50}{125}\)
So, the fraction with numerator 50 and equivalent to \(\frac{18}{45}\) is \(\frac{50}{125}\).

(ii) On dividing 60 by 5, we get 12.
Now, multiplying the numerator and ‘ denominator of \(\frac{2}{5}\) by 12, we get
\(\frac{2}{5}\) = \(\frac{2 \times 12}{5 \times 12}\) = \(\frac{24}{60}\)
So, the fraction with denominator 60 and equivalent to\(\frac{18}{45}\) is \(\frac{24}{60}\).

Question 8.
Write the following fractions in the simplest form:
(i) \(\frac{45}{81}\)
(ii) \(\frac{65}{91}\)
(iii) \(\frac{441}{483}\)
Solution:
(i) We have, \(\frac{45}{81}\) = \(\frac{3 \times 3 \times 5}{3 \times 3 \times 3 \times 3}\) = \(\frac{5}{9}\)
(ii) We have, \(\frac{65}{91}\) = \(\frac{5 \times 13}{7 \times 13}\) = \(\frac{5}{7}\)
(iii) We have, \(\frac{441}{483}\) = \(\frac{3 \times 3 \times 7 \times 7}{3 \times 7 \times 23}\) = \(\frac{21}{23}\)

Question 9.
Arrange the following fractions in ascending order:
\(\frac{10}{3}\), \(\frac{21}{6}\), \(\frac{9}{2}\), \(\frac{13}{4}\), \(\frac{25}{8}\)
Solution:
Denominators of the given fractions are 3, 6, 2, 4 and 8.
The smallest common multiple of 3, 6, 2, 4 and 8 is 24.
Now, converting each fraction into equivalent fraction with 24 as its denominator, we get
\(\frac{10}{3}\) = \(\frac{10 \times 8}{3 \times 8}\) = \(\frac{80}{24}\)
\(\frac{21}{6}\) = \(\frac{21 \times 4}{6 \times 4}\) = \(\frac{84}{24}\)
\(\frac{9}{2}\) = \(\frac{9 \times 12}{2 \times 12}\) = \(\frac{108}{24}\)
\(\frac{13}{4}\) = \(\frac{13 \times 6}{4 \times 6}\) = \(\frac{78}{24}\)
\(\frac{25}{8}\) = \(\frac{25 \times 3}{8 \times 3}\) = \(\frac{75}{24}\)
We know, 75 < 78 < 80 < 84 < 108
⇒ \(\frac{75}{24}\) < \(\frac{78}{24}\) < \(\frac{80}{24}\) < \(\frac{84}{24}\) < \(\frac{108}{24}\)
⇒ \(\frac{25}{8}\) < \(\frac{13}{4}\) < \(\frac{10}{3}\) < \(\frac{21}{6}\) < \(\frac{9}{2}\)

Question 10.
There are 30 students in section A & 40 in section B of class VI. Among them, 25 students from section A 8c 34 from section B passed with distinction. Which section performed better?
Solution:
Here, we have to compare \(\frac{25}{30}\) and \(\frac{34}{40}\).
We can write 30 = 3 × 10 and 40 = 4 × 10.
The least common multiple of 30 and 40 is 3 × 4 × 10 = 120.
∴ \(\frac{25}{30}\) = \(\frac{25 \times 4}{30 \times 4}\) = \(\frac{100}{120}\) and \(\frac{34}{40}\) = \(\frac{34 \times 3}{40 \times 3}\) = \(\frac{102}{120}\)
We know, 100 < 102
⇒ \(\frac{100}{120}\) < \(\frac{102}{120}\)
⇒ \(\frac{25}{30}\) < \(\frac{34}{40}\)
So, section B performed better than section A.

Question 11.
The refractive index of stone A and stone B are \(\frac{121}{50}\) and \(\frac{58}{25}\) respectively. Which stone has greater refractive index?
Solution:
Here, we have to compare \(\frac{121}{50}\) and \(\frac{58}{25}\).
The least common multiple of 50 and 25 is 50.
∴ \(\frac{58}{25}\) = \(\frac{58 \times 2}{25 \times 2}\) = \(\frac{116}{50}\)
We know, 121 > 116
⇒ \(\frac{121}{50}\) > \(\frac{116}{50}\)
⇒ \(\frac{151}{50}\) > \(\frac{58}{25}\)
So, the refractive index of stone A is greater than the refractine index of stone B.

Question 12.
Solve the following:
(i) \(\frac{2}{9}\) + \(\frac{3}{9}\)
(ii) \(\frac{1}{7}\) + \(\frac{3}{7}\) + \(\frac{12}{7}\)
(iii) \(\frac{1}{4}\) + \(\frac{3}{4}\) + \(\frac{2}{5}\) + \(\frac{3}{5}\)
(iv) 2\(\frac{3}{5}\) + \(\frac{2}{5}\) + 3\(\frac{1}{5}\)
Solution:
(i) \(\frac{2}{9}\) + \(\frac{3}{9}\)
= \(\frac{2+3}{9}\) = \(\frac{5}{9}\)

(ii) \(\frac{1}{7}\) + \(\frac{3}{7}\) + \(\frac{12}{7}\)
= \(\frac{1+3+12}{7}\) = \(\frac{16}{7}\)

(iii) \(\frac{1}{4}\) + \(\frac{3}{4}\) + \(\frac{2}{5}\) + \(\frac{3}{5}\)
= (\(\frac{1+3}{4}\)) + (\(\frac{2+3}{5}\))
= \(\frac{4}{4}\) + \(\frac{5}{5}\) = 1 + 1 = 2

(iv) 2\(\frac{3}{5}\) + \(\frac{2}{5}\) + 3\(\frac{1}{5}\)
= (\(\frac{2 \times 5+3}{5}\)) + \(\frac{1}{5}\) + (\(\frac{3 \times 5+1}{5}\))
= (\(\frac{10+3}{5}\)) + \(\frac{2}{5}\) + (\(\frac{15+1}{5}\))
= \(\frac{13}{5}\) + \(\frac{2}{5}\) + \(\frac{16}{5}\)
= \(\frac{13+2+16}{5}\) = \(\frac{31}{5}\)

Question 13.
Simplify the following:
(i) 5\(\frac{1}{4}\) + \(\frac{3}{4}\) – 2\(\frac{1}{4}\)
(ii) 7\(\frac{1}{7}\) – \(\frac{13}{7}\) + 2\(\frac{2}{7}\)
Solution:
(i) 5\(\frac{1}{4}\) + \(\frac{3}{4}\) – 2\(\frac{1}{4}\) = (\(\frac{5 \times 4+2}{4}\)) + \(\frac{3}{4}\) – (\(\frac{2 \times 4+1}{4}\))
= (\(\frac{20+2}{4}\)) + \(\frac{3}{4}\) – (\(\frac{8+1}{4}\))
= \(\frac{22}{4}\) + \(\frac{3}{4}\) – \(\frac{9}{4}\)
= \(\frac{22+3-9}{4}\) = \(\frac{16}{4}\) = 4

(ii) 7\(\frac{1}{7}\) – \(\frac{13}{7}\) + 2\(\frac{2}{7}\)
= (\(\frac{7 \times 7+1}{7}\)) – \(\frac{13}{7}\) – (\(\frac{2 \times 7+2}{7}\))
= (\(\frac{49+1}{7}\)) – \(\frac{13}{7}\) – (\(\frac{14+2}{4}\))
= \(\frac{50}{7}\) – \(\frac{13}{7}\) + \(\frac{16}{7}\)
= \(\frac{50-13+16}{7}\) = \(\frac{53}{7}\)

Question 14.
Shikha ate \(\frac{1}{5}\) of the pizza and her friend Sanvi ate \(\frac{3}{5}\) of the pizza. Did they eat whole of the pizza? If not, then what fraction of the pizza is left?
Solution:
Fraction of pizza eaten by Shikha = \(\frac{1}{5}\)
Fraction of pizza eaten by Sanvi = \(\frac{3}{5}\)
Total pizza eaten by both Shikha and Sanvi
= \(\frac{1}{5}\) + \(\frac{3}{5}\) = \(\frac{4}{5}\) < 1 Fraction representing the remaining pizza = 1 – \(\frac{4}{5}\) = \(\frac{5}{5}\) – \(\frac{4}{5}\) = \(\frac{5-4}{5}\) = \(\frac{1}{5}\) So, \(\frac{1}{5}\) of the pizza is left.

Question 15.
Fill in the missing fractions:
(i) \(\frac{3}{8}\) + ☐ = \(\frac{5}{8}\)
(ii) \(\frac{4}{9}\) – ☐ = \(\frac{2}{9}\)
(iii) \(\frac{8}{15}\) – ☐ = \(\frac{1}{5}\)
(iv) ☐ – \(\frac{4}{10}\) = \(\frac{3}{10}\)
Solution:
Let x be the missing fraction.
(i) \(\frac{3}{8}\) + x = \(\frac{5}{8}\)
⇒ x = \(\frac{5}{8}\) – \(\frac{3}{8}\)
⇒ x = \(\frac{5-3}{8}\) = \(\frac{2}{8}\) = \(\frac{1}{4}\)

(ii) \(\frac{4}{9}\) – x = \(\frac{2}{9}\)
⇒ \(\frac{4}{9}\) – \(\frac{2}{9}\) = x
⇒ x = \(\frac{4-2}{9}\)
⇒ x = \(\frac{2}{9}\)

(iii) \(\frac{8}{15}\) – x = \(\frac{1}{5}\)
⇒ \(\frac{8}{15}\) – \(\frac{1}{5}\) = x
⇒ \(\frac{8}{15}\) – \(\frac{3}{5}\) = x
⇒ x = \(\frac{8-3}{15}\)
⇒ x = \(\frac{5}{15}\) ⇒ x = \(\frac{1}{3}\)

(iv) x – \(\frac{4}{10}\) = \(\frac{3}{10}\)
⇒ x = \(\frac{3}{10}\) + \(\frac{4}{10}\)
⇒ x = \(\frac{3+4}{10}\)
⇒ x = \(\frac{7}{10}\)

Question 16.
Arrange the following fractions in descending order:
\(\frac{28}{9}\), \(\frac{55}{18}\), \(\frac{37}{12}\), 3, \(\frac{31}{4}\)
Solution:
Denonimators of the given fractions are 9, 18, 12, 1 and 4.
The smallest common multiple of 9, 18, 12, 1 and 4 is 36.
Now, converting each fraction into equivalent fraction with 36 as its denominator, we get
\(\frac{28}{9}\) = \(\frac{28 \times 4}{9 \times 4}\) = \(\frac{112}{36}\),
\(\frac{55}{18}\) = \(\frac{55 \times 2}{18 \times 2}\) = \(\frac{110}{36}\),
\(\frac{37}{12}\) = \(\frac{37 \times 3}{12 \times 3}\) = \(\frac{11}{36}\),
\(\frac{3}{1}\) = \(\frac{3 \times 36}{1 \times 36}\) = \(\frac{108}{36}\),
\(\frac{31}{4}\) = \(\frac{31 \times 9}{4 \times 9}\) = \(\frac{279}{36}\)
We know, 279 > 112 > 111 > 110 > 108
⇒ \(\frac{279}{36}\) > \(\frac{112}{36}\) > \(\frac{111}{36}\) > \(\frac{110}{36}\) > \(\frac{108}{36}\)
⇒ \(\frac{31}{4}\) > \(\frac{28}{9}\) > \(\frac{37}{12}\) > \(\frac{55}{18}\) > 3

Question 17.
Rahul scored 54 out of 75 marks, while Swati scored 92 out of 125. Who performed better?
Solution:
Here, we have to compare \(\frac{54}{75}\) and \(\frac{92}{125}\).
We can write, 75 = 3 × 25 and 125 = 5 × 25.
Now, the least common multiple of 75 and 125 is 3 × 5 × 25 = 375.
∴ \(\frac{54}{75}\) = \(\frac{54 \times 5}{75 \times 5}\) = \(\frac{270}{375}\) and \(\frac{92}{125}\) = \(\frac{92 \times 3}{125 \times 3}\) = \(\frac{276}{375}\)
We know 270 < 276
⇒ \(\frac{270}{375}\) < \(\frac{276}{375}\)
⇒ \(\frac{54}{75}\) < \(\frac{92}{125}\)
So, Swati performed better than Rahul.

Question 18.
The distance from Delhi to Gurugram is \(\frac{310}{15}\)km, while the distance from Delhi to Noida is \(\frac{415}{20}\) km. Which city, Gurugram or Noida, is
closer to Delhi?
Solution:
Here, we have to compare \(\frac{310}{15}\) and \(\frac{415}{20}\).
We can write, 15 = 3 × 5 and 20 = 4 × 5.
The least common multiple of 15 and 20 is 3 × 4 × 5 = 60.
∴ \(\frac{310}{15}\) = \(\frac{310 \times 4}{15 \times 4}\) = \(\frac{1240}{60}\) and \(\frac{415}{20}\) = \(\frac{415 \times 3}{20 \times 3}\) = \(\frac{1245}{60}\)
We know 1240 < 1245
⇒ \(\frac{1240}{60}\) < \(\frac{1245}{60}\)
⇒ \(\frac{310}{15}\) < \(\frac{415}{20}\)
So, Gurugram is closer to Delhi.

Question 19.
Solve the following:
(i) \(\frac{5}{11}\) + \(\frac{13}{11}\)
(ii) \(\frac{3}{12}\) + \(\frac{4}{12}\) + \(\frac{7}{12}\)
(iii) \(\frac{1}{6}\) + \(\frac{2}{6}\) + \(\frac{3}{6}\)
(iv) 4\(\frac{3}{15}\) + \(\frac{11}{15}\) + 5\(\frac{7}{15}\)
Solution:
(i) \(\frac{5}{11}\) + \(\frac{13}{11}\) = \(\frac{5+13}{11}\) = \(\frac{18}{11}\)

(ii) \(\frac{3}{12}\) + \(\frac{4}{12}\) + \(\frac{7}{12}\)
= \(\frac{3+4+7}{12}\) = \(\frac{14}{12}\)
= \(\frac{7}{6}\)

(iii) \(\frac{1}{6}\) + \(\frac{2}{6}\) + \(\frac{3}{6}\)
= \(\frac{1+2+3}{6}\) = \(\frac{6}{6}\)
= 1

(iv) 4\(\frac{3}{15}\) + \(\frac{11}{15}\) + 5\(\frac{7}{15}\)
= (\(\frac{4 \times 15+3}{15}\)) + \(\frac{11}{15}\) + (\(\frac{5 \times 15+7}{15}\))
= (\(\frac{60+3}{11}\)) + \(\frac{11}{15}\) + (\(\frac{75+7}{15}\))
= \(\frac{63}{15}\) + \(\frac{11}{15}\) + \(\frac{82}{15}\) = \(\frac{63+11+82}{15}\)
= \(\frac{156}{15}\) = \(\frac{3 \times 52}{3 \times 5}\)
= \(\frac{52}{5}\)

Question 20.
Add the following fractions:
(i) \(\frac{3}{4}\) and \(\frac{4}{3}\)
(ii) \(\frac{7}{4}\), \(\frac{2}{3}\) and \(\frac{1}{5}\)
Solution:
(i) LCM of denominators i.e., 4 and 3 is 12.
∴ \(\frac{3}{4}\) = \(\frac{3 \times 3}{4 \times 3}\) = \(\frac{9}{12}\),
\(\frac{4}{3}\) = \(\frac{4 \times 4}{3 \times 4}\) = \(\frac{16}{12}\)
Now, \(\frac{3}{4}\) + \(\frac{4}{3}\) = \(\frac{9}{12}\) + \(\frac{16}{12}\)
= \(\frac{9+16}{12}\) = \(\frac{25}{12}\)

(ii) LCM of denominators i.e., 4, 3 and 5 is 60.
∴ \(\frac{7}{4}\) = \(\frac{7 \times 15}{4 \times 15}\) = \(\frac{105}{60}\),
\(\frac{2}{3}\) = \(\frac{2 \times 20}{3 \times 20}\) = \(\frac{40}{60}\),
\(\frac{1}{5}\) = \(\frac{1 \times 12}{5 \times 12}\) = \(\frac{12}{60}\)
Now, \(\frac{7}{4}\) + \(\frac{2}{3}\) + \(\frac{1}{5}\) = \(\frac{105}{60}\) + \(\frac{40}{60}\) + \(\frac{12}{60}\)
= \(\frac{105+40+12}{60}\) = \(\frac{157}{60}\)

Question 21.
Simplify: 4\(\frac{1}{5}\) – 3\(\frac{2}{3}\)
Solution:
4\(\frac{1}{5}\) – 3\(\frac{2}{3}\) = \(\frac{4 \times 5+1}{5}\) – \(\frac{3 \times 3+2}{3}\)
= \(\frac{20+1}{5}\) – \(\frac{9+2}{3}\) = \(\frac{21}{5}\) – \(\frac{11}{3}\)
L.C.M of denominator i.e. 5 and 3 is 15.
∴ \(\frac{21}{5}\) = \(\frac{21 \times 3}{5 \times 3}\) = \(\frac{63}{15}\),
\(\frac{11}{3}\) = \(\frac{11 \times 5}{3 \times 5}\) = \(\frac{55}{15}\)
Now, 4\(\frac{1}{5}\) – 3\(\frac{2}{3}\) = \(\frac{21}{5}\) – \(\frac{11}{3}\) = \(\frac{63}{15}\) – \(\frac{55}{15}\)
= \(\frac{63-55}{15}\) = \(\frac{8}{15}\)

Question 22.
Rahul, Ashok and Anshika buy a toy. Rahul gives \(\frac{3}{10}\) of the total cost, Anshika gives \(\frac{5}{10}\) of the total cost, and the remaining amount is paid by Ashok. What fraction of the total cost is paid by Ashok?
Solution:
Rahul’s share of total cost = \(\frac{3}{10}\)
Anshika’s share of total cost = \(\frac{5}{10}\)
Total share of Rahul and Anshika
= \(\frac{3}{10}\) + \(\frac{5}{10}\) = \(\frac{3+5}{10}\) = \(\frac{8}{10}\)
∴ Ashok’s share of total cost
= 1 – \(\frac{8}{10}\) = \(\frac{10}{10}\) – \(\frac{8}{10}\) = \(\frac{10-8}{10}\)
= \(\frac{2}{10}\) = \(\frac{2}{2 \times 5}\) = \(\frac{1}{5}\)

Question 23.
Find the difference of \(\frac{17}{12}\) and \(\frac{11}{18}\).
Solution:
We can write 12 = 2 × 2 × 3 and 18 = 2 × 3 × 3.
So, LCM of 12 and 18 is 2 × 2 × 3 × 3 = 36.
∴ \(\frac{17}{12}\) = \(\frac{17 \times 3}{12 \times 3}\) = \(\frac{51}{36}\) and
\(\frac{11}{18}\) = \(\frac{11 \times 2}{18 \times 2}\) = \(\frac{22}{36}\)
We know, 51 > 22
⇒ \(\frac{51}{36}\) > \(\frac{22}{36}\)
⇒ \(\frac{17}{12}\) > \(\frac{11}{18}\)
Thus, the required difference
= \(\frac{17}{12}\) – \(\frac{11}{18}\) = \(\frac{51}{36}\) – \(\frac{22}{36}\)
= \(\frac{51-22}{36}\) = \(\frac{29}{36}\)

Fractions Class 6 Long Question Answer

Question 1.
Write a fraction to represent the shaded part in each of the following diagrams:

Solution:
(i) Number of equal parts or fractional units = 7,
Number of shaded parts = 3
So, the required fraction
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\) = \(\frac{3}{7}\)

(ii) Number of equal parts or fractional units = 9,
Number of shaded parts = 4
So, the required fraction .
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\) = \(\frac{4}{9}\)

(iii) Number of equal parts or fractional units = 8,
Number of shaded parts = 6
So, the required fraction
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\)
= \(\frac{6}{8}\) = \(\frac{6 \div 2}{8 \div 2}\) = \(\frac{3}{4}\)

(iv) Number of equal parts or fractional units = 7,
Number of shaded parts = 4
So, the required fraction
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\) = \(\frac{4}{7}\)

Question 2.
Write a fraction to represent the shaded part in each of the following diagrams:

Solution:
(i) Number of equal parts or fractional units = 12,
Number of shaded parts = 8
So, the required fraction
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\)
= \(\frac{8}{12}\) = \(\frac{2 \times 4}{3 \times 4}\) = \(\frac{2}{3}\)

(ii) Number of equal parts or fractional units = 16,
Number of shaded parts = 8
So, the required fraction
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\)
= \(\frac{8}{16}\) = \(\frac{8 \times 1}{8 \times 2}\) = \(\frac{1}{2}\)

(iii) Number of equal parts or fractional units = 6,
Number of shaded parts = 2
So, the required fraction
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\)
= \(\frac{2}{6}\) = \(\frac{2 \times 1}{2 \times 3}\) = \(\frac{1}{3}\)

(iv) Number of equal parts or fractional units = 8,
Number of shaded parts = 3
So, the required fraction
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\) = \(\frac{3}{8}\)

(v) Number of equal parts or fractional units = 8,
Number of shaded parts = 5
So, the required fraction
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\) = \(\frac{5}{8}\)

Question 3.
Compare the following fractions:
(i) \(\frac{7}{3}\) and \(\frac{11}{5}\)
(ii) \(\frac{4}{7}\) and \(\frac{5}{9}\)
(iii) \(\frac{26}{14}\) and \(\frac{40}{21}\)
(iv) \(\frac{13}{7}\) and \(\frac{23}{11}\)
Solution:
(i) 3 × 5 = 15 is a common multiple of 3 and 5.
∴ \(\frac{7}{3}\) = \(\frac{7 \times 5}{3 \times 5}\) = \(\frac{35}{15}\) and
\(\frac{11}{5}\) = \(\frac{11 \times 3}{5 \times 3}\) = \(\frac{33}{15}\)
We know, 35 > 33
⇒ \(\frac{35}{15}\) > \(\frac{35}{15}\)
⇒ \(\frac{7}{3}\) > \(\frac{11}{5}\)

(ii) 7 × 9 = 63 is a common multiple of 7 and 9.
∴ \(\frac{4}{7}\) = \(\frac{4 \times 9}{7 \times 9}\) = \(\frac{36}{63}\) and
\(\frac{5}{9}\) = \(\frac{5 \times 7}{9 \times 7}\) = \(\frac{35}{63}\)
We know, 35 > 33
⇒ \(\frac{36}{63}\) > \(\frac{35}{63}\)
⇒ \(\frac{4}{7}\) > \(\frac{5}{9}\)

(iii) We can write \(\frac{26}{14}\) = \(\frac{13}{7}\).
21 is a common multiple of 7 and 21.
∴ \(\frac{13}{7}\) = \(\frac{13 \times 3}{7 \times 3}\) = \(\frac{39}{21}\)
We know, 39 < 40
⇒ \(\frac{39}{21}\) > \(\frac{40}{21}\)
⇒ \(\frac{13}{7}\) > \(\frac{40}{21}\)
⇒ \(\frac{26}{14}\) > \(\frac{40}{21}\)

(iv) 7 × 11 = 77 is a common multiple of 7 and 11.
∴ \(\frac{13}{7}\) = \(\frac{13 \times 11}{7 \times 11}\) = \(\frac{143}{77}\) and
\(\frac{23}{11}\) = \(\frac{23 \times 7}{11 \times 7}\) = \(\frac{161}{77}\)
We know, 143 < 161
⇒ \(\frac{143}{77}\) > \(\frac{161}{77}\)
⇒ \(\frac{13}{7}\) > \(\frac{23}{11}\)

Question 4.
Compare the following fractions:
(i) \(\frac{17}{12}\) and \(\frac{11}{6}\)
(ii) \(\frac{13}{5}\) and \(\frac{23}{10}\)
(iii) \(\frac{6}{5}\) and \(\frac{9}{8}\)
(iv) \(\frac{23}{8}\) and \(\frac{17}{6}\)
Solution:
(i) 12 is a common multiple of 12 and 6.
∴ \(\frac{11}{6}\) = \(\frac{11 \times 2}{6 \times 2}\) = \(\frac{22}{12}\)
We know, 17 < 22
⇒ \(\frac{17}{12}\) > \(\frac{22}{12}\)
⇒ \(\frac{17}{12}\) > \(\frac{11}{6}\)

(ii) 10 is a common multiple of 5 and 10.
∴ \(\frac{13}{5}\) = \(\frac{13 \times 2}{5 \times 2}\) = \(\frac{26}{10}\)
We know, 26 < 23
⇒ \(\frac{26}{10}\) > \(\frac{23}{10}\)
⇒ \(\frac{13}{5}\) > \(\frac{23}{10}\)

(iii) 5 × 8 = 40 is a common multiple of 5 and 8.
∴ \(\frac{6}{5}\) = \(\frac{6 \times 8}{5 \times 8}\) = \(\frac{48}{40}\)
\(\frac{9}{8}\) = \(\frac{9 \times 5}{8 \times 5}\) = \(\frac{45}{40}\)
We know, 48 < 4
⇒ \(\frac{48}{40}\) > \(\frac{45}{40}\)
⇒ \(\frac{6}{5}\) > \(\frac{9}{8}\)

(iv) 24 is a common multiple of 8 and 6.
∴ \(\frac{23}{8}\) = \(\frac{23 \times 3}{8 \times 3}\) = \(\frac{69}{24}\) and
\(\frac{17}{6}\) = \(\frac{17 \times 4}{6 \times 4}\) = \(\frac{68}{24}\)
We know, 69 < 68
⇒ \(\frac{69}{24}\) > \(\frac{68}{24}\)
⇒ \(\frac{23}{8}\) > \(\frac{17}{6}\)

Question 5.
Simplify the following:
(i) 3\(\frac{7}{10}\) + \(\frac{3}{10}\) – 2\(\frac{4}{10}\)
(ii) 5\(\frac{1}{6}\) – 3\(\frac{5}{6}\) + 1\(\frac{3}{6}\)
(iii) 4\(\frac{2}{11}\) – 3\(\frac{4}{11}\) + \(\frac{10}{11}\)
(iv) 6\(\frac{3}{13}\) – 2\(\frac{4}{13}\) – 1\(\frac{9}{13}\)
Solution:
(i) 3\(\frac{7}{10}\) + \(\frac{3}{10}\) – 2\(\frac{4}{10}\)
= \(\left(\frac{3 \times 10+7}{10}\right)\) + \(\frac{3}{10}\) – \(\left(\frac{2 \times 10+4}{10}\right)\)
= \(\left(\frac{30+7}{10}\right)\) + \(\frac{3}{10}\) – \(\left(\frac{20+4}{10}\right)\)
= \(\frac{37}{10}\) + \(\frac{3}{10}\) – \(\frac{24}{10}\)
= \(\frac{37+3-24}{10}\) = \(\frac{16}{10}\) = \(\frac{2 \times 8}{2 \times 5}\)
= \(\frac{8}{5}\)

(ii) 5\(\frac{1}{6}\) – 3\(\frac{5}{6}\) + 1\(\frac{3}{6}\)
= \(\left(\frac{5 \times 6+1}{6}\right)\) – \(\left(\frac{3 \times 6+5}{6}\right)\) + \(\left(\frac{1 \times 6+3}{6}\right)\)
= \(\left(\frac{30+1}{6}\right)\) – \(\left(\frac{18+5}{6}\right)\) + \(\left(\frac{6+3}{6}\right)\)
= \(\frac{31}{6}\) – \(\frac{23}{6}\) + \(\frac{9}{6}\)
= \(\frac{31-23+9}{6}\)
= \(\frac{17}{6}\)

(iii) 4\(\frac{2}{11}\) – 3\(\frac{4}{11}\) + \(\frac{10}{11}\)
= \(\left(\frac{4 \times 11+2}{11}\right)\) – \(\left(\frac{3 \times 11+4}{11}\right)\) + \(\frac{10}{11}\)
= \(\left(\frac{44+2}{11}\right)\) – \(\left(\frac{33+4}{11}\right)\) + \(\frac{10}{11}\)
= \(\frac{46}{11}\) – \(\frac{37}{11}\) + \(\frac{10}{11}\)
= \(\frac{46-37+10}{11}\)
= \(\frac{19}{11}\)

(iv) 6\(\frac{3}{13}\) – 2\(\frac{4}{13}\) – 1\(\frac{9}{13}\)
= \(\left(\frac{6 \times 13+3}{13}\right)\) – \(\left(\frac{2 \times 13+4}{13}\right)\) – \(\left(\frac{1 \times 13+9}{13}\right)\)
= \(\left(\frac{78+3}{13}\right)\) – \(\left(\frac{26+4}{13}\right)\) – \(\left(\frac{13+9}{13}\right)\)
= \(\frac{81}{13}\) – \(\frac{30}{13}\) – \(\frac{22}{13}\)
= \(\frac{81-30-22}{13}\)
= \(\frac{29}{13}\)

Question 6.
Simplify the following:
(i) \(\frac{3}{11}\) – \(\frac{7}{9}\) + \(\frac{13}{3}\)
(ii) 5\(\frac{2}{3}\) – 2\(\frac{1}{4}\) + 3\(\frac{1}{6}\)
(iii) 1\(\frac{5}{7}\) – \(\frac{5}{9}\) + 6
(iv) 8\(\frac{1}{4}\) – 2\(\frac{5}{6}\) + 1\(\frac{2}{3}\)
Solution:
(i) \(\left(\frac{3 \times 9}{11 \times 9}\right)\) – \(\left(\frac{7\times 11}{9 \times 11}\right)\) + \(\left(\frac{13 \times 33}{3 \times 33}\right)\)
= \(\frac{29}{13}\) – \(\frac{29}{13}\) + \(\frac{29}{13}\)
[∵ L.C.M of 11, 9 and 3 is 99.]
= \(\frac{27}{99}\) – \(\frac{77}{99}\) + \(\frac{429}{99}\)
= \(\frac{29-77+429}{99}\)
= \(\frac{379}{99}\)

(ii) 5\(\frac{2}{3}\) – 2\(\frac{1}{4}\) + 3\(\frac{1}{6}\)
= \(\left(\frac{5 \times 3+2}{3}\right)\) – \(\left(\frac{2 \times 4+1}{4}\right)\) + \(\left(\frac{3 \times 6+1}{6}\right)\)
= \(\left(\frac{15+2}{3}\right)\) – \(\left(\frac{8+1}{4}\right)\) + \(\left(\frac{18+1}{6}\right)\)
= \(\frac{17}{3}\) – \(\frac{9}{4}\) + \(\frac{19}{6}\)
= \(\left(\frac{17 \times 4}{3 \times 4}\right)\) – \(\left(\frac{9 \times 3}{4 \times 3}\right)\) + \(\left(\frac{19 \times 2}{6 \times 2}\right)\)
[∵ LCM of 3, 4 and 6 is 12.]
= \(\frac{68}{12}\) – \(\frac{27}{12}\) + \(\frac{38}{12}\)
= \(\frac{68-27+38}{12}\) = \(\frac{79}{12}\)

(iii) 1\(\frac{5}{7}\) – \(\frac{5}{9}\) + 6
= \(\left(\frac{1 \times 7+5}{7}\right)\) – \(\frac{5}{9}\) + \(\frac{6}{1}\)
= \(\left(\frac{7+5}{7}\right)\) – \(\frac{5}{9}\) + \(\frac{6}{1}\)
= \(\frac{12}{7}\) – \(\frac{5}{9}\) + \(\frac{6}{1}\)
= \(\left(\frac{12 \times 9}{7 \times 9}\right)\) – \(\left(\frac{5 \times 7}{9 \times 7}\right)\) + \(\left(\frac{6 \times 63}{1 \times 63}\right)\)
[∵ LCM of 7, 9 and 1 is 63.]
= \(\frac{108}{63}\) – \(\frac{35}{63}\) + \(\frac{378}{63}\)
= \(\frac{108-35+378}{63}\) = \(\frac{451}{63}\)

(iv) 8\(\frac{1}{4}\) – 2\(\frac{5}{6}\) + 1\(\frac{2}{3}\)
= \(\left(\frac{8 \times 4+1}{4}\right)\) – \(\left(\frac{2 \times 6+5}{6}\right)\) + \(\left(\frac{1 \times 3+2}{3}\right)\)
= \(\left(\frac{32+1}{4}\right)\) – \(\left(\frac{12+5}{6}\right)\) + \(\left(\frac{3+2}{3}\right)\)
= \(\frac{33}{4}\) – \(\frac{17}{6}\) + \(\frac{5}{3}\)
= \(\frac{33 \times 3}{4 \times 3}\) – \(\frac{17 \times 2}{6 \times 2}\) + \(\frac{5 \times 4}{3 \times 4}\)
[∵ LCM of 4, 6 and 3 is 12.]
= \(\frac{99}{12}\) – \(\frac{34}{12}\) + \(\frac{20}{12}\)
= \(\frac{99-34+20}{12}\) = \(\frac{85}{12}\)

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 6 Perimeter and Area Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 6 Perimeter and Area Solutions

Ganita Prakash Class 6 Chapter 6 Solutions

Class 6 Maths Ganita Prakash Chapter 6 Solutions Perimeter and Area

Question 1.
A rectangle having sidelengths 5 cm and 3 cm is made using a piece of wire. If the wire is straightened and then bent to form a square, what will be the length of a side of the square?
Solutions:
Given, length of rectangle = 5 cm and breadth of rectangle = 3 cm
We know that perimeter of rectangle = 2 × (length + breadth) = 2 × (5 + 3) = 16 cm
Now, if we bend the wire to form a square, the total length of the wire (16 cm) will be divided equally among the four sides of the square.
So, each side of the square = \(\frac{\text { Perimeter }}{4}\) = \(\frac{16}{4}\) = 4 cm

Question 2.
A piece of string is 36 cm long. What will be the length of each side, if it is used to form:
(a) Square
(b) Equilateral Triangle
(c) Regular Hexagon
Solutions:
Given, total length of string = 36 cm
(a) For a square. each side of the square = \(\frac{\text { Total length of string }}{4}\) = \(\frac{36}{4}\) = 12 cm
(b) For a triangle with all si(les of edila! length. each side of the triangle = \(\frac{\text { Total length of string }}{3}\) = \(\frac{36}{3}\) = 12 cm
(c) For a hexagon wit h all sides oI’egual length, each sidle oÍ the hexagon = \(\frac{\text { Total length of string }}{6}\) = \(\frac{36}{6}\) = 6 cm

InText Questions

Question 1.
Deep Dive: In races, usually there is a common finish line for all the runners. Here are two square running tracks with the inner track of 100 m each side and outer track of 150 m each side. The common finishing line for both runners is shown by the flags in the figure which are in the center of one of the sides of the tracks. If the total race is of 350 m, then we have to find out where the starting positions of the two runners should be on these two tracks so that they both have a common finishing fine after they run for 350 m. Mark the starting points of the runner on the inner track as ‘A’ and the runner on the outer track as ‘B’.

Solution:
For the inner square track, each side is 100 m.
For the outer square track, each side is 150 m.
So, inner track perimeter = 4 × side = 4 × 100 m = 400 m
Outer track perimeter = 4 × side = 4 × 150 m = 600 m

Both runners start at different points and run along their respective square tracks, possibly completing full or partial laps. The finish line is fixed at the middle of one side.
The total race distance is 350 m for both runners.
Therefore, each runner must start 350 m before the finish line, in the direction of the run.

Runner A runs on the inner track (400 m) and runner B runs on the outer track (600 m).
Now, we calculate how far before the finish line each runner should start:
Start Point A: 400 m – 350 m = 50 m
Thus, position A is 50 m behind the finish line (in the direction of the run) on the inner track.
Start Point B: 600 m – 350 m = 250 m
Thus, position B is 250 m behind the finish line (in the direction of the run) on the outer track.

Thus, mark point A on the inner track, 50 m behind the flag (finish line), and point B on the outer track, 250 m behind the flag, both measured in the direction of the run.

Question 2.
Look at the figures below and guess which one of them has a larger area.

Solution:
We can estimate the area of any simple closed shape by using a sheet of squared paper or graph paper where every square measures 1 unit X 1 unit or 1 square unit.
To estimate the area, we can trace the shape onto a piece of transparent paper and place the same on a piece of squared or graph paper and then follow the below conventions:

1. Count all the full squares inside the figure. Area of each is 1 square unit.
2. For squares that are more than half filled, count them as 1 square unit.
3. For squares that are exactly half filled, count them as \(\frac{1}{2}\) square unit.
4. Ignore squares that are less than half filled.

Now, adding a square grid, where every square measures 1 unit × 1 unit, we get the following figure.

In figure (a): Number of full squares inside the figure = 31
Number of squares that are more than half filled = 13
Area of figure (a) = 31 × 1 + 13 × 1
= 31 + 13 = 44 sq. units
In figure (b): Number of full squares inside the figure = 16
Number of squares that are more than half filled = 20
Area of figure (b) = 16 × 1 + 20 × 1
= 16 + 20 = 36 sq. units
Thus, figure (a) has a larger area.

Question 3.
Use your understanding from, previous grades to calculate the area of any closed figure using grid paper and

(a) Find the area of triangle BAD. _________
(b) Find the area of triangle ABE. _________
Solution:
From the given figure,

Area of rectangle A BCD = 20 sq. units
Area of rectangle AFED = 12 sq. units
Area of rectangle FBCE = 8 sq. units
(a) Area of triangle BAD
= \(\frac{1}{2}\) × Area of rectangle ABCD
= \(\frac{1}{2}\) × 20 sq. units = 10 sq. units

(b) Area of triangle ABE
= Area of triangle AFE + Area of triangle BEE
= \(\frac{1}{2}\) × Area of rectangle AFED + \(\frac{1}{2}\) × Area of rectangle FBCE
= \(\frac{1}{2}\) × 12 sq. units + \(\frac{1}{2}\) × 8 sq. units
= 6 sq. units + 4 sq. units = 10 sq. units

Question 4.
Using 9 unit squares, solve the following.
(a) What is the smallest perimeter possible?
(b) What is the largest perimeter possible?
(c) Make a figure with a perimeter of 18 units.
(d) Can you make other shaped figures for each of the above three perimeters, or is there only one shape with that perimeter? What is your reasoning?
Solution:
(a) The smallest perimeter possible is 12 units.
This occurs when the 9 unit squares are arranged to form a 3 × 3 square.

Side length = 3 units
Perimeter = 4 × 3 = 12 units

(b) The largest perimeter possible is 20 units.
This occurs when the 9 unit squares are arranged in a single straight line to form 1 × 9 or 9 × 1 rectangle.

Perimeter of rectangle = 2 × (9 + 1) = 20 units

(c) A figure with a perimeter of 18 units is given alongside:

(d) Yes, we can make multiple different shapes for each of the three perimeter values 12, 18 and 20 units as long as the number of external sides (edges not shared with another square) adds up to the correct perimeter.

Reasoning: Each square has 4 edges.

Every time two squares are placed adjacent to each other, they share an edge, reducing the total perimeter by 2 units (1 edge from each square).

By changing how the 9 squares are joined linear, block, zig-zag, L-shape, etc. we can vary how many edges are shared and hence control the perimeter.

The number of shared sides determines the final perimeter, not the specific shape.

Question 5.
Let’s do something tricky now! We have^ a figure below having perimeter 24 units.
Without calculating all over again, observe, think and find out what will be the change in the perimeter if a new square is attached as shown on the right.

Experiment placing this new square at different places and think what the change in perimeter will be. Can you place the square so that the perimeter: (a) increases; (b) decreases; (c) stays the same?
Solution:
When a new square is attached to an existing figure:
If it shares one full side with the existing figure, the perimeter increases by 2 units. (Because one new side is hidden in the shared boundary, and 3 new sides are exposed.
So, net change = + 3 – 1 = + 2).

If it shares two sides (like being inserted into a corner), the perimeter stays the same. (2 new sides added, but 2 sides of the previous figure are now internal and not counted: + 2 – 2 = 0).

If it shares three sides (nestled into a concave corner), the perimeter decreases by 2 units. (Only 1 side added is exposed, 3 sides of the previous figure are hidden: + 1 – 3 = – 2).

(a) Perimeter increases when the new square is added such that only one side touches the original figure.
(b) Perimeter decreases when the square is placed within a corner, sharing three sides with the original figure.
(c) Perimeter stays the same when the square is added in such a way that two sides are shared with the original figure (like being placed into an edge corner).

Perimeter and Area Class 6 Extra Questions

Perimeter and Area Class 6 Very Short Question Answer

Question 1.
Find the perimeter of the given quadrilateral.

Solution:
We know that the perimeter of a polygon is the sum of the lengths of its all sides.
So, the perimeter of given quadrilateral
= AB + BC + CD + DA
= 5 cm + 3 cm + 7.5 cm + 4.5 cm
= 20 cm

Question 2.
Find the perimeter of the following figure:

solution:
We know that the perimeter of a closed figure is the sum of the lengths of its all sides.
So, the perimeter of given figure
= PQ + QR + RS + ST + TU + UP
= 6.4 cm + 2.5 cm + 2.6 cm + 2.6 cm + 2.5 cm + 6.4 cm
= 23 cm

Question 3.
Find the perimeter of a rectangle whose length and breadth are 1.25 m and 75 cm, respectively.

Solution:
Given, length of the rectangle = 1.25 m = 125 cm [∵ 1 m = 100 cm]
Breadth of the rectangle = 75 cm We know,
Perimeter of the rectangle = 2 (Length + Breadth)
= 2 (125 cm + 75 cm) = 2 × 200 cm = 400 cm

Question 4.
Find the perimeter of the following figures:


Solution:
We know that the perimeter of a polygon is the sum of the lengths of its all sides.
(i) Perimeter = 2 cm + 7 cm + 2 cm + 7 cm = 18 cm
(ii) Perimeter = 4 cm + 4 cm + 3 cm = 11 cm
(iii) Perimeter = 3 cm + 3 cm + 3 cm + 3 cm + 3 cm + 3 cm = 18 cm
(iv) Perimeter = 3 cm + 5 cm + 4 cm + 2 cm = 14 cm

Question 5.
Find the perimeter of the following figure:

Solution:
We know that the perimeter of a closed figure is the sum of the lengths of its all sides.
So, the perimeter of given figure = LM + MN + NO + OP + PCI + QR + RS + SL
= 2.9 cm + 2.9 cm + 1.4 cm + 5.5 cm + 2.4 cm + 2.4 cm + 5.5 cm + 1.4 cm
= 24.4 cm

Question 6.
The lengths of two sides of a triangle are 15 cm and 21 cm. The perimeter of the triangle is 46 cm. Find the length of its third side.
Solution:
Given, the perimeter of a triangle is 46 cm.
Length of first side =15 cm

Length of second side = 21 cm
We know, the perimeter of a triangle
= Sum of the lengths of all sides of the triangle
⇒ 46 cm = Length of first side + Length of second side + Length of third side
⇒ 46 cm =15 cm + 21 cm 4 Length of third side
⇒ 46 cm = 36 cm 4 Length of third side
⇒ Length of third side = 46 cm – 36 cm = 10 cm

Question 7.
If the perimeter of a regular heptagon is 63 cm, find the length of its one side.
Solution:
Given, perimeter of a regular heptagon is 63 cm.
We know, number of sides in a regular heptagon = 7
And, perimeter of a regular polygon = Number of sides × Length of one side
So, the perimeter of given regular heptagon
= 7 × Length of one side
⇒ 63 cm = 7 × Length of one side
⇒ Length of one side = \(\frac{63}{7}\) cm = 9 cm

Question 8.
A rectangular carpet has an area of 72 sq. m and a width of 6 m. What is its length?
Solution:
Given, width (breadth) of a rectangular carpet = 6 m
And, area of a rectangular carpet = 72 sq. m
We know, area of the rectangular carpet = Length × breadth
⇒ 72 sq. m = Length × 6 m
⇒ Length = \(\frac{72}{6}\) = 12 m
Thus, the length of the rectangular carpet is 12 m.

Question 9.
If rectangle ABCD has area 42 sq. units, then find the area of triangle ABC, cut along the diagonal of the rectangle ABCD.
Solution:
Given, the area of rectangle ABCD is 42 sq. units.
We know that if a rectangle is cut along one of its diagonals, then the area of each resulting triangle is half the area of the rectangle.
∴ Area of triangle ABC
= \(\frac{1}{2}\) × Area of rectangle ABCD
= \(\frac{1}{2}\) × 42 sq. units = 21 sq. units

Perimeter and Area Class 6 Short Question Answer

Question 1.
The perimeter of a rectangle is 72 cm and its breadth is 8 cm. Find its length.

Solution:
Given, perimeter of the rectangle = 72 cm
Breadth of the rectangle = 8 cm
We know,
Perimeter of the rectangle = 2 (Length + Breadth )
⇒ 72 cm = 2(Length + 8 cm)
⇒ 2(Length + 8 cm) = 72
⇒ Length + 8 cm = \(\frac{72}{2}\) = 36 cm
⇒ Length = 36 cm – 8 cm = 28 cm

Question 2.
In the adjoining figure, the length of each side is 2.75 cm. Find the perimeter of the figure.

Solution:
Given figure is a polygon and the length of its each side is 2.75 cm.
Also, the number of sides is 10.
So, the perimeter of given polygon
= 10 × Length of one side
= 10 × 2.75 cm
= 27.5 cm

Question 3.
If the perimeter of a regular pentagon is 55 cm, find the length of its one side.

Solution:
Given, perimeter of a regular pentagon is 55 cm.
We know, the number of sides in a regular pentagon is 5.
And, perimeter of a regular polygon
= Number of sides × Length of each side
So, the perimeter of regular pentagon = 5 × Length of each side
⇒ 55 cm = 5 × Length of each side
⇒ Length of each side = \(\frac{55}{5}\) cm = 11 cm

Question 4.
The lid of a rectangular box of length 45 cm and breadth 30 cm is sealed all around with tape. Find the required length of tape.

Solution:
Given, length of the rectangular box is 45 cm.
Breadth of the rectangular box is 30 cm.
And, the lid of the rectangular box is sealed all around with tape.
Therefore, length of the tape
= Perimeter of the rectangle
= 2(Length + Breadth)
= 2(45 cm + 30 cm)
= 2 × 75 cm = 150 cm

Question 5.
A wire is in the shape of an equilateral triangle of side 26 cm. It is rebent into the shape of rectangle whose length is 23 cm, find its breadth.
Solution:
Given, length of the side of an equilateral triangle is 26 cm.
Therefore, length of the wire = Perimeter of the equilateral triangle of side 26 cm
= 3 × 26 cm [∵ Number of sides in an equilateral triangle is 3.]
= 78 cm

Given that the wire is rebent into the shape of rectangle whose length is 23 cm.
∴ Perimeter of the rectangle = Length of wire
⇒ 2(Length + Breadth) = 78 cm
⇒ 2(23 cm + Breadth) = 78 cm
⇒ 23 cm + Breadth = \(\frac{78}{2}\) = 39 cm
⇒ Breadth = 39 cm – 23 cm = 16 cm

Question 6.
Priya went to a rectangular field 160 m long and 110 m wide. She took 4 complete rounds on its boundary. Find the distance covered by her.

Solution:
Given, length of the rectangular field = 160 m
Breadth of the rectangular field = 1 10 m
∴ Distance covered by Priya in one round
= Perimeter of the rectangular field
= 2(Length + Breadth)
= 2(160 m + 110 m)
= 2 × 270 m = 540 m
Since she took 4 complete rounds on its boundary,
Distance covered by Priya in four rounds
= 4 × Distance covered by Priya in one round
= 4 × 540 m = 2160 m
Thus, Priya covers 2160 m distance in 4 rounds.

Question 7.
In below figure, the length of each side is 3.25 cm. Find the perimeter of the figure.

Solution:
In the given figure, number of sides is 14.
Also, the length of each side is 3.25 cm.
So, the perimeter of given polygon = 14 × Length of one side = 14 × 3.25 cm = 45.5 cm

Question 8.
The perimeter of a rectangle is 56 cm and its length is 20 cm. Find its breadth.
Solution:
Given, perimeter of the rectangle = 56 cm

Length of the rectangle = 20 cm
We know, perimeter of the rectangle = 2 (Length 4- Breadth)
⇒ 56 cm = 2(20 cm + Breadth)
⇒ 28 cm = 20 cm + Breadth
⇒ Breadth = 28 cm – 20 cm = 8 cm

Question 9.
A rectangular piece of land measures 0.95 km by 0.65 km. Each side of the land is to be fenced with 5 rows of wires. Find the required length of the wire.
Solution:
Given, length of the rectangular piece of land = 0.95 km

Breadth of the rectangular piece of land = 0.65 km
∴ Perimeter of the rectangular piece of land
= 2(Length + Breadth)
= 2(0.95 km + 0.65 km)
= 2 × 1.60 km = 3.20 km
Since each side of the land is to he fenced with 5 rows of wires, length of the wire is five times the perimeter of the land.
∴ Required length of wire = 5 × 3.20 km = 16 km

Question 10.
Find the perimeter of a rectangular field whose length is 345 cm and which has an area equal to 56925 sq. cm.
Solution:
Given, length of rectangular field = 345 cm and area of the field = 56925 sq. cm
We know, area of rectangular field
= Length × Breadth
⇒ 56925 sq. cm = 345 cm × Breadth
⇒ Breadth = \(\frac{56925}{345}\) = 165 cm
So, breadth of the field is 165 cm.
Hence,
perimeter of the field = 2 (Length + Breadth)
= 2 (345 cm + 165 cm) = 2 × 510 cm = 1020 cm
Thus, the perimeter of the rectangular field is 1020 cm.

Question 11.
In the figure, find the area of the path which is 2.5 m wide all around.

Solution:
Given, length of outer rectangle = 70 m
Breadth of outer rectangle = 44 m
As the path is 2.5 m wide,
Length of inner rectangle = 70 m – 2.5 m – 2.5 m
= 65 m
Breadth of inner rectangle = 44 m – 2.5 m – 2.5 m
= 39 m
Now, the area of path = Area of outer rectangle – Area of inner rectangle
= 70 m × 44 m – 65 m × 39 m
= 3080 sq. m – 2535 sq. m = 545 sq. m
Thus, the area of the path is 545 sq. m

Question 12.
Four square tiles of side 2.5 m each are placed together without any gaps. What is the total area covered?
Solution:
Given, the side of a square tile is 2.5 m.
We know, area of a square = Side × Side
∴ Area of a square tile = 2.5 m × 2.5 m
= 6.25 sq. m
As four square tiles are placed together without any gaps,
Total covered area = 4 × Area of one square tile
= 4 × 6.25 sq. m = 25 sq. m
Thus, total area covered by the four square tiles of side 2.5 m is 25 sq. m.

Question 13.
A floor measuring 15 m by 9 m is covered with a carpet of size 12 m × 8 m. Find the area that remains uncovered.
Solution:
Given, length of the floor = 15 m,
Breadth of the floor = 9 m,
Length of the carpet = 12 m
and breadth of the carpet = 8 m
Therefore, area of the floor
= Length of the floor × Breadth of the floor
= 15 m × 9 m = 135 sq. m
And, area of the carpet
= Length of the carpet × Breadth of the carpet
= 12 m × 8 m = 96 sq. m
Thus, area that remains uncovered = Area of the floor – Area of the carpet
= 135 sq. m – 96 sq. m = 39 sq. m

Question 14.
Find the area of the given shaded region (figure alongside) drawn on a square paper, taking the area of each square as 1 cm².

Solution:
The figure drawn on square paper contains 35 complete squares and 9 half squares.
∴ Area of given closed figure = Area of 35 complete squares + Area of 9 half squares
= 35 × 1 cm² + 9 × \(\frac{1}{2}\) cm²
= 35 cm² + 4.5 cm² = 39.5 cm²

Perimeter and Area Class 6 Long Question Answer

Question 1.
A wire is 39 cm long. What will be the length of each side if the wire is used to form
(i) an equilateral triangle?
(ii) a regular hexagon?
Solution:
Given, a wire is 39 cm long.
(i) The wire is used to form an equilateral triangle.
We know that there are 3 equal sides in an equilateral triangle.
∴ Perimeter of equilateral triangle = Length of wire
⇒ 3 × Length of side = 39 cm
⇒ Length of side = \(\frac{39}{3}\) cm = 13 cm

(ii) The wire is used to form a regular hexagon.
We know that there are 6 equal sides in a regular hexagon.
∴ Perimeter of regular hexagon = Length of wire
⇒ 6 × Length of side = 39 cm
⇒ Length of side = \(\frac{39}{6}\) cm = 6.5 cm

Question 2.
A farmer has a rectangular field of length 350 m and breadth 175 m. He wants to fence it with 4 rounds of rope as shown in the figure. If cost of rope is ₹ 4.5 per metre, then find the cost of fencing the field.

Solution:
Given, length of the rectangular field = 350 m
Breadth of the rectangular field = 175 m
∴ Perimeter of the rectangular field = 2(Length + Breadth)
= 2(350 m + 175 m)
= 2 × 525 m = 1050 m
Since the farmer wants to fence the field with 4 rounds of rope, length of the rope is four times the perimeter of the field.
∴ Required length of rope = 4 × 1050 m = 4200 m
Given, the cost of 1 m rope is ₹ 4.5.
∴ Total cost of 4200 m rope = ₹ 4.5 × 4200
= ₹ 18,900
Thus, the cost of fencing the field is ₹ 18,900.

Question 3.
Priya runs around a square field of side 90 m. Rishi runs around a rectangular field of length 120 m and breadth 75 m. Who covers more distance and by how much in one round?

Solution:
Given, Priya runs around a square field of side 90 m.
∴ Distance covered by Priya in one round
= Perimeter of the square field
= 4 × length of a side of the square field
= 4 × 90 m = 360 m
And, Rishi runs around a rectangular field of length 120 m and breadth 75 m.
∴ Distance covered by Rishi in one round
= Perimeter of the rectangular field
= 2(Length + Breadth)
= 2(120 m + 75 m) = 2 × 195 m = 390 m
So, difference in the distance covered in one round = 390 m – 360 m = 30 m
Thus, Rishi covers more distance by 30 m in one round.

Question 4.
A rectangle, having side lengths 19 cm and 13 cm, is made using a piece of wire. If the wire is straightened and then bent to form a square, what will be the length of a side of the square?
Solution:
Given, length of the rectangle is 19 cm.
Breadth of the rectangle is 13 cm.

Therefore, length of the wire
= Perimeter of the rectangle
= 2(Length + Breadth)
= 2(19 cm + 13 cm)
= 2 × 32 cm = 64 cm
Given that the wire is rebent into the shape of a square.
∴ Perimeter of the square = Length of wire
⇒ 4 × Length of side = 64 cm
⇒ Length of side = \(\frac{64}{4}\) cm = 16 cm

Question 5.
A wire is 39 cm long. What will be the length of each side if the wire is used to form
(i) a square?
(ii) a regular pentagon?
Solution:
Given, a wire is 39 cm long.
(i) The wire is used to form a square.
We know that there are 4 equal sides in a square.
∴ Length of wire = 4 × Length of side
⇒ 39 cm = 4 × Length of side
⇒ Length of side = \(\frac{39}{4}\) cm = 9.75 cm

(ii) The wire is used to form a regular pentagon.
We know that there are 5 equal sides in a regular pentagon.
∴ Length of wire = 5 × Length of side
⇒ 39 cm = 5 × Length of side
⇒ Length of side = \(\frac{39}{5}\) cm = 7.8 cm

Question 6.
The length of a rectangular field is four times its breadth. A man runs around it 4 times and covered a distance of 5 km. What is the length of the field?
Solution:
Given, the length of a rectangular field is four times its breadth.
∴ Length = 4 × Breadth ……. (i)
And, a man runs around the field 4 times and covered a distance of 5 km.
∴ Distance covered in 4 rounds
= 4 × Perimeter of the field
⇒ 5 km = 4 × 2 (Length + Breadth)
⇒ 5000 m = 8 × (4 × Breadth + Breadth) [From (i)] [∵ 1 km = 1000 m]
⇒ 5000 m = 8 × 5 × Breadth
⇒ 5000 m = 40 × Breadth
⇒ Breadth = \(\frac{5000 \mathrm{~m}}{40}\) = 125 m
Substituting I he value of breadth in (I). we gel
Length = 4 × 125 m = 500 m
Thus, the length of the field is 500 m.

Question 7.
Seema runs 6 times around a rectangular park with length 70 m long and breadth 45 m while Ramesh runs 5 times around a square park of side 65 m. Who covers more distance and by how much?

Solution:
Given, Seema runs around a rectangular park of length 70 m and breadth 45 m.
∴ Distance covered by Seema in one round
= Perimeter of the rectangular park
= 2(Length + Breadth)
= 2(70 m + 45 m) = 2 × 115 m = 230 m
So, distance covered by Seema in 6 rounds
= 6 × Distance covered by Seema in one round
= 6 × 230 m = 1380 m
And Ramesh runs around a square park of side 65 m.
∴ Distance covered by Ramesh in one round
= Perimeter of the square park
= 4 × length of a side of the square park
= 4 × 65 m = 260 m
So, distance covered by Ramesh in 5 rounds
= 5 × Distance covered by Ramesh in one round
= 5 × 260 m = 1300 m
So, difference in the distance covered
= 1380 m- 1300 m = 80 m
Thus, Seema covers 80 m more distance than Ramesh.

Question 8.
Match the closed figure given in Column I with their perimeter given in Column II.


Solution:
We know that the perimeter of a polygon/closed figure is the sum of the lengths of its all sides.
(A) Perimeter = 40 cm + 40 cm + 40 cm + 40 cm = 160 cm
(B) Perimeter = 30 cm + 60 cm + 30 cm + 60 cm = 180 cm
(C) Perimeter = 35 cm + 35 cm + 35 cm = 105 cm
(D) Perimeter = 30 cm + 20 cm + 25 cm + 24 cm = 99 cm
(E) Perimeter = 31 cm + 40 cm + 20 cm + 22 cm + 40 cm = 153 cm
(F) Perimeter = 15 cm + 28 cm + 15 cm + 2 cm + 25 cm + 10 cm + 5 cm + 40 cm = 140 cm
Thus, (A) – (t), (B) – (s), (C) – (p), (D) – (q), (E) – (u), (T) – (r)

Question 9.
Rishi wants to cover the floor of a room 4 m wide & 8 m long by square tiles. If each square tile is of side 0.4 m, then find the number of tiles required.
Solution:
Given, length of the room = 8 m and breadth of the room = 4 m
As the room is in rectangular shape,
Area of the room = Length × Breadth
= 8 m × 4 m = 32 sq. m
And, side length of each square tile = 0.4 m
∴ Area of each tile = Side × Side
= 0.4 m × 0.4 m = 0.16 sq. m
Now, number of tiles = \(\frac{\text { Area of floor of the room }}{\text { Area of one tile }}\)
= \(\frac{32}{0.16}\) = 200
Thus, the number of required tiles is 200.

Question 10.
A square and a rectangle have equal area. If the side of the square is 36 cm and the length of the rectangle is 54 cm, then find:
(i) the breadth of the rectangle)
(ii) the perimeter of die rectangle.
Solution:
(i) Given, side of the square
= 36 cm and length of the rectangle
= 54 cm
∴ Area of the square = Side × Side
= 36 cm × 36 cm = 1296 sq. cm
As area of the rectangle = area of the square
⇒ Length × Breadth = 1296 sq. cm
⇒ 54 cm × Breadth = 1296 sq. cm
⇒ Breadth = \(\frac{1296}{54}\) = 24 cm
Thus, the breadth of the rectangle is 24 cm.

(ii) We know, perimeter of the rectangle
= 2 (Length + Breadth)
= 2 (54 cm + 24 cm)
= 2 × 78 cm = 156 cm

Question 11.
Find the area of the given figure.

Solution:
To find the required area, we divide the enclosed region of given figure into rectangles and squares as shown in the figure.

Part I is a rectangle of length 200 m and breadth 100 m.
Part II is a square of side length 100 m.
Part III is a rectangle of length 400 m and breadth 200 m.
Now, area of part I = 200 m × 100 m = 20000sq. m
Area of part II = 100 m × 100 m = 10000 sq.m
And area of part III = 400 m × 200 in = 80000 sq. m
Total area of given closed figure
= 20000 sq. m + 10000 sq. m + 80000 sq. m
= 110000 sq.m

Question 12.
In the following figure, find the missing value of the area of a region.

Solution:
We name the given figure as shown.
Here,
breadth of the rectangle II = 7 m
And area of the rectangle II = 63 sq. m
∴ Length of rectangle II × Breadth of rectangle
II = 63 sq. m
⇒ Length of rectangle II × 7 m = 63 sq. m

⇒ Length of rectangle II = \(\frac{63}{7}\) = 9 m
= IG = HD
So, JC = IC – IJ = 9m – 5m = 4 m
GD = GH + HD = 2m + 9m = 11m
Now, area of rectangle III = 33 sq. m
∴ Length of rectangle III × Breadth of rectangle III = 33 sq. m
⇒ GD × Breadth of rectangle III = 33 sq. m
⇒ 11 m × Breadth of rectangle III = 33 sq. m
⇒ Breadth of rectangle III = \(\frac{33}{11}\) = 3 m
= FG = DE
Given, BE = 14 m
⇒ BC + CD + DE = 14 m
⇒ BC + 7 m + 3 m = 14 m
⇒ BC = 14 m – 7 m – 3 m = 4 m = AJ
As AJ = JC = 4 m, ABCJ is a square.
∴ Area of square I = AJ × JC = 4 × 4
= 16 sq. m.

Question 13.
A square and a rectangle have equal area. If the side of the square is 27 cm and length of rectangle is 81 cm, then find:
(i) the breadth of the rectangle.
(ii) the perimeter of the rectangle.
Solution:
(i) Given, side of the square = 27 cm
and length of the rectangle = 81 cm
∴ Area of the square = Side × Side
= 27 cm × 27 cm
= 729 sq. cm
As area of the rectangle = Area of the square
⇒ Length × Breadth = 729 sq. cm
⇒ 81 cm × Breadth = 729 sq. cm
⇒ Breadth = \(\frac{729}{81}\) = 9 cm
Thus, the breadth of the rectangle is 9 cm.

(ii) We know, perimeter of the rectangle
= 2 (Length + Breadth)
= 2 (81 cm + 9 cm)
= 2 × 90 cm = 180 cm

Question 14.
If the perimeter of the square is thrice the perimeter of a triangle whose sides are 3 cm, 4 cm and 5 cm, then find the area of the square.
Solution:
Given, sides of a triangle are 3 cm, 4 cm and 5 cm.
∴ Perimeter of the triangle
= 3 cm + 4 cm + 5 cm = 12 cm
As the perimeter of the square is thrice the perimeter of a triangle,
Perimeter of the square = 3 × Perimeter of the triangle
⇒ 4 × Side = 3 × 12 cm
⇒ Side = \(\frac{3 \times 12}{4}\) = 9 cm 4
Now, area of the square = Side × Side
= 9 cm × 9 cm
= 81 sq. cm
Thus, the area of the square is 81 sq. cm.

Question 15.
Find the area of the given figure (a) alongside.

Solution:
To find the required area, we divide the enclosed region of given figure into rectangles as shown in the figure.
Part I is a rectangle of length 10 m and breadth 2 m.
Part II is a rectangle of length 6 m and breadth 2 m.
Part III is a rectangle of length 6 m and breadth 2 m.
Part IV is a rectangle of length 10 m and breadth 2 m.
Now, area of part I = 10 m × 2 m = 20 sq. m
Area of part II = 6 m × 2 m = 12 sq. m
Area of part III = 6 m × 2 m = 12 sq. m
And area of part IV = 10m × 2m = 20 sq. m
Total area of given dosed figure
= 20 sq. m + 12 sq. m + 12 sq. m + 20 sq. m
= 64 sq. m

Question 16.
In the following figure, find the missing value of the area of a region.

Solution:
To find the required area, we divide the enclosed region of given figure into squares and rectangles as shown in the figure.

Now, breadth of the rectangle I = 3 cm
And area of the rectangle I = 21 sq. cm
∴ Length of rectangle I × Breadth of rectangle I
= 21 sq. cm
⇒ Length of rectangle I × 3 cm = 21 sq. cm
⇒ Length of rectangle I = \(\frac{21}{3}\) = 7 cm
So, Breadth of rectangle II = 7 cm – 4 cm
= 3 cm
Now, area of rectangle II = 27 sq. cm
∴ Length of rectangle II × Breadth of rectangle
II = 27 sq. cm
⇒ Length of rectangle II × 3 = 27 sq. cm
⇒ Length of rectangle II = \(\frac{27}{3}\) = 9 cm
So, breadth of region III = 9 cm – 4 cm
= 5 cm
Now, area of square III
= Side length × Side length
= 5 cm × 5 cm = 25 sq. cm

Question 17.
In the following figure, find the missing value of the area of a region.

Solution:
We name the given figure as shown.

Area of the rectangle I is 17 sq. cm, which can be written as 1 cm × 17 cm or 17 cm × 1 cm.
And area of the rectangle II is 51 sq. cm, which can be written as 3 cm × 17 cm or 17 cm × 3 cm.
Front figure, we can see that the breadth of the rectangles I and II are equal.
Therefore,
Length of rectangle I = 1 cm
Breadth of rectangle I = 17 cm
Length of rectangle II = 3 cm
Breadth of rectangle II = 17 cm
For rectangle III, length = 1 cm
Now, area of rectangle III = 30 sq. cm [Given]
⇒ Length of rectangle III × Breadth of rectangle III = 30 sq. cm
⇒ 1 cm × Breadth of rectangle III = 30 sq. cm
⇒ Breadth of rectangle III = 30 cm
So, length of rectangle IV = 3 cm and breadth of rectangle IV = 30 cm
∴ Area of rectangle IV = 3 cm × 30 cm
= 90 sq. cm

Question 18.
Find the area of the given figure.

Solution:
To find the required area, we divide the enclosed region of given figure into rectangles and squares as shown in the figure.
Part I is a square of side 6 cm.
Part II is a rectangle of length 6 cm and breadth 2 cm.
Part III is a square of side 6 cm.
Part IV is a rectangle of length 7 cm and breadth 2 cm.
Now, area of part I = 6 cm × 6 cm = 36 sq. cm

Area of part II = 6 cm × 2 cm = 12 sq. cm
Area of part III = 6 cm × 6 cm = 36 sq. cm
And area of part IV = 7 cm × 2 cm = 14 sq. cm
Total area of given closed figure
= 36 sq. cm + 12 sq. cm + 36 sq. cm + 14 sq. cm
= 98 sq. cm

Perimeter and Area Class 6 Case Based Questions

Question 1.
Priya and Rishi start running along the rectangular tracks as shown in the figure. Rishi runs along the outer track. Priya runs along the inner track. Now, they are wondering who ran more.

Based on the above information, answer the following questions:
(i) Find the distance covered by Priya in one round.
(ii) Find the distance covered by Rishi in one round.
(iii) Find out who ran the longer distance when Rishi runs 6 rounds and Priya runs 7 rounds.
Solution:
(i) Given, Priya runs along the rectangular track of length 130 m and breadth 80 m.
∴ Distance covered by Priya in one round
= Perimeter of the inner rectangular track
= 2(Length + Breadth)
= 2(130 m + 80 m) = 2 × 210 m = 420 m

(ii) Given, Rishi runs along the rectangular track of length 150 m and breadth 100 m.
∴ Distance covered by Rishi in one round
= Perimeter of the outer rectangular track
= 2(Length + Breadth)
= 2(150 m + 100 m) = 2 × 250 m = 500 m

(iii) Distance covered by Rishi in 6 rounds
= 6 × Distance covered by Rishi in one round
= 6 × 500 m = 3000 m
And, distance covered by Priya in 7 rounds
= 7 × Distance covered by Priya in one round
= 7 × 420 m = 2940 m

So, difference in the distance covered = 3000 m – 2940 m = 60 m
Thus, Rishi covers 60 m more distance than Priya.

Question 2.
Look at the plan of a house build on a rectangular plot as shown in the below figure.

Based on the above information, answer the following questions:
(i) Find the missing measurements.
(ii) Compare areas of the small bedroom & kitchen. Which one has a bigger area? By how much?
(iii) Compare areas of small bedroom & drawing room. Which one has a bigger area? By how much?
Solution:
(i) For store room, length = 8 ft and breadth = 7 ft
∴ Area of store room = 8 ft × 7 ft = 56 sq. ft

For toilet:
Length = Length of store room = 8 ft
And breadth = 6 ft
∴ Area of toilet = 8 ft × 6 ft = 48 sq. ft

For master bedroom:
Length = 15 ft
Breadth = 26 ft – Breadth of store room – Breadth of toilet
= 26 ft – 7 ft – 6 ft = 13 ft
∴ Area of master bedroom = 15 ft × 13 ft
= 195 sq. ft

For entrance:
Breadth = 8 ft
And, Area = 80 sq. ft
⇒ Length × Breadth = 80 sq. ft
⇒ Length × 8 ft = 80 sq. ft
⇒ Length = \(\frac{80}{8}\) = 10 ft

For drawing room:
Breadth = Breadth of master bedroom = 13 ft
Length = 38 ft – Length of master bedroom – Length of entrance
= 38 ft – 15 ft – 10 ft = 13 ft
∴ Area of drawing room = 13 ft × 13 ft = 169 sq. ft

For hall:
Breadth = Length of entrance =10 ft
Length = 26 ft – Breadth of entrance = 26 ft – 8 ft = 18 ft
∴ Area of hall = 18 ft × 10 ft = 180 sq. ft

For kitchen:
Breadth = 8 ft
Length = 26 ft – Breadth of drawing room
= 26 ft – 13 ft = 13 ft
∴ Area of kitchen = 13 ft × 8 ft = 104 sq. ft

For small bedroom:
Length = Length of kitchen = 13 ft
Breadth = 38 ft – Length of toilet – Breadth of kitchen – Breadth of hall ,
= 38 ft – 8 ft – 8 ft – 10ft = 12ft
∴ Area of small bedroom = 13 ft × 12 ft = 156 sq. ft

(ii) Area of small bedroom =156 sq. ft
Area of kitchen =104 sq. ft
Difference of areas = 156 sq. ft – 104 sq. ft = 52 sq. ft
Thus, the small bedroom has 52 sq. ft more area than the kitchen.

(iii) Area of small bedroom = 156 sq. ft
Area of drawing room = 169 sq. ft
Difference of areas = 169 sq. ft – 156 sq. ft = 13 sq. ft
Thus, the drawing room has 13 sq. ft. area more than small bedroom.

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 5 Prime Time Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 5 Prime Time Solutions

Ganita Prakash Class 6 Chapter 5 Solutions

Class 6 Maths Ganita Prakash Chapter 5 Solutions Prime Time

Question 1.
Who am I?
(a) I am a number less than 40. One of my factors is 7. The sum of my digits is 8.
(b) I am a number less than 100. Two of my factors are 3 and 5. One of my digits is 1 more than the other.
Solution:
(a) The numbers less than 40 whose one of the factor is 7 are 7, 14, 21, 28, 35. Out of these numbers the sum of digits of 35 is 8.
Hence, the number less than 40 whose one of the factors is 7 and sum of digits equals to 8 is 35.

(b) The number less than 100 whose two factors 3 and 5 are 15, 30, 45, 60, 75, 90. Out of these numbers, the number 45 has one of its digits 1 more than the other.
Hence, the number less than 100 whose two factors are 3 and 5 and one of its digits is 1 more than the other is 45.

Question 2.
In the diagram below, Guna has erased all the numbers except the common multiples. Find out what those numbers could be and fill in the missing numbers in the empty regions.

Solution:

Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.
Common factors of 24, 48 and 72 are 1, 2, 3, 4, 6, 8, 12, 24.
So, in multiples of 6 and 8, common multiples are 24, 48 and 72.
Other possibilities of the numbers are 3 and 8, 3 and 24, 4 and 6, 4 and 24, 6 and 24, etc.

Question 3.
Find the smallest number that is a multiple of all the numbers from 1 to 10.
Solution:
To find the smallest number that is a multiple of all numbers from 1 to 10, we need to determine least common multiple (LCM) of the numbers from 1 to 10.
Prime factorisation of numbers from 1 to 10.
1 = 1; 2 = 2; 3 = 3; 4 = 2 × 2; 5 = 5; 6 = 2 × 3; 7 = 7; 8 = 2 × 2 × 2; 9 = 3 × 3; 10 = 2 × 5
LCM (1 to 10) = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520 Thus, the required smallest number is 2520.

Question 4.
Write three pairs of prime numbers less than 20 whose sum is a multiple of 5.
Solution:
Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17 and 19.
Now, 2 + 3 = 5 (multiple of 5)
2 + 13 = 15 (multiple of 5)
7 + 13 = 20 (multiple of 5)
Thus, three pairs are (2, 3), (2, 13) and (7, 13).

Question 5.
The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3 . Find such pairs of prime numbers up to 100.
Solution:
Pairs of prime numbers having same digits upto 100 are:
17 and 71 (both have digits 1 and 7);
37 and 73 (both have digits 3 and 7)
79 and 97 (both have digits 7 and 9)

Question 6.
Twin primes are pairs of primes having a difference of 2. For example, 3 and 5 are twin primes. So are 17 and 19. Find the other twin primes between 1 and 100.
Solution:
Twin primes between 1 and 100 are
(3, 5) → 5 – 3 = 2;
(5, 7) → 7 – 5 = 2;
(17, 19) → 19- 17 = 2;
(29, 31) → 31 – 29 = 2;
(59, 61) → 61 – 59 = 2;
(71, 73) → 73 – 71 = 2

Question 7.
Observe that 3 is a prime number and 2 × 3 + 1 = 7 is also a prime. Are there other primes for which doubling and adding 1 give another prime? Find atleast five such examples.
Solution:
Five such examples are:
2 × 2+ 1 = 5 (prime number)
2 × 5 + 1 = 11 (prime number)
2 × 11 + 1 = 23 (prime number)
2 × 23 + 1 = 47 (prime number)
and 2 × 29 + 1 = 59 (prime number)

Question 8.
What is the smallest number whose prime factorisation has
(a) three different prime numbers?
(b) four different prime numbers?
Solution:
(a) The smallest three prime numbers are 2, 3 and 5.
Thus, the smallest number with exactly three different prime factors = 2 × 3 × 5 = 30

(b) The smallest four prime numbers are 2, 3, 5 and 7.
Thus, the smallest number with exactly four different prime factors = 2 × 3 × 5 × 7 = 210

Question 9.
The first number has prime factorisation 2 × 3 × 7 and the second number has prime factorisation 3 × 7 × 11. Are they co-prime? Does one of them divide the other?
Solution:
Prime factorisation of two numbers are 2 × 3 × 7 and 3 × 7 × 11.
Both the numbers have common factors 3 and 7. Therefore, they are not co-prime.
Since the prime factors of one number are not the prime factors of another number, therefore one of them does not divide the other.

Question 10.
Find the largest and smallest 4-digit numbers that are divisible by 4 and are also palindromes.
Solution:
The smallest 4-digit palindrome is 1001 .
Checking divisibility by 4:
1001 (not divisible by 4);
1111 (not divisible by 4);
1221 (not divisible by 4)
1331 (not divisible by 4);
1441 (not divisible by 4);
1551 (not divisible by 4);
1661 (not divisible by 4);
1771 (not divisible by 4);
1881 (not divisible by 4);
1991 (not divisible by 4);
2002 (not divisible by 4);
2112 (divisible by 4)
Therefore, die smallest 4-digit palindromic number divisible by 4 is 2112 .
The largest 4-digit palindrome is 9999.
Checking divisibility by 4:
9999 (not divisible by 4);
9669 (not divisible by 4);
9339 (not divisible by 4);
9009 (not divisible by 4):
9889 (not divisible by 4);
9559 (not divisible by 4);
9229 (not divisible by 4);
8998 (not divisible by 4);
9779 (not divisible by 4);
9449 (not divisible by 4);
9119 (not divisible by 4);
8888 (divisible by 4)
Therefore, the largest 4 -digit palindromic number divisible by 4 is 8888.

Question 11.
The teacher asked if 14560 is divisible by all of 2, 4, 5, 8 and 10. Guna checked for divisibility of 14560 by only two of these numbers and then declared that it was also divisible by all of them. What could those two numbers be?
Solution:
If a number is divisible by 8 and 5, then it will also be divisible by 2, 4 and 10.
Divisibility by 8 ensures it is always divisible by 2 and 4.
Divisibility by 5 ensures it ends with 0 or 5. But since the number also needs to be divisible by 8 then it must end with 0.
A number ending with 0 is also divisible by 10.
Therefore, 8 and 5 are the required numbers.

InText Questions

Question 1.
Look at the table below. What do you notice?

(i) Is there anything common among the shaded numbers?
(ii) Is there anything common among the circled numbers?
(iii) Which numbers are both shaded and circled? What are these numbers called?
Solution:
(i) All shaded numbers are multiples of 3.
(ii) All circled numbers are multiples of 4.
(iii) Numbers (both shaded and circled) are 36, 48 and 60.
They are called ‘common multiples of 3 and 4’.

Question 2.
Grumpy and Jumpy are playing treasure finding game. Treasures are kept on two numbers. Jumpy gets the treasures only if he is able to reach both the numbers with the same jump size. Also, a jump size of 1 is not allowed.
Where should Grumpy place the treasures so that Jumpy cannot reach both the treasures? Check if these pairs are safe:
(a) 15 and 39
(b) 4 and 15
(c) 18 and 29
(d) 20 and 55
Solution:
Grumpy should place the treasures at co-prime numbers.
(a) Pair (15 and 39) is not safe because 15 and 39 are not co-prime as 3 is their common factor.
(b) Pair (4 and 15) is safe because 4 and 15 are co-prime.
(c) Pair (18 and 29) is safe because 18 and 29 are co-prime.
(d) Pair (20 and 55) is not safe because 20 and 55 are not co-prime as 5 is their common factor.

Question 3.
Observe the following thread art. The first diagram has 12 pegs and the thread is tied to every fourth peg (we say that the thread-gap is 4). The second diagram has 13 pegs and the thread- gap is 3. What about the other diagrams? Observe these pictures, share and discuss your findings in class. In some diagrams, the thread is tied to every peg. In some, it is not. Is it related to the two numbers (the number of pegs and the thread- gap) being co-prime?

Make such pictures for the following:
(a) 15 pegs, thread-gap of 10
(b) 10 pegs. thread-gap of’ 7
(c) 14 pegs, thread-gap of 6
(d) 8 pegs, thread -gap of 3
Solution:
Yes, when two numbers are co-prime, the thread is tied to every peg.

Question 4.
Find numbers between 330 and 340 that are divisible by 4. Also, find numbers between 1730 and 1740, and 2030 and 2040, that are divisible by 4. What do you observe?
Solution:
Numbers between 330 and 340 that are divisible by 4 are 332 and 336
Numbers between 1730 and 1740 that are divisible by 4 are 1732 and 1736
Numbers between 2030 and 2040 that are divisible by 4 are 2032 and 2036
We observe that if number formed by last two digits of a number is divisible by 4, then the number is divisible by 4.

Question 5.
Find numbers between 120 and 140 that are divisible by 8. Also find numbers between 1120 and 1140, and 3120 and 3140, that are divisible by 8. What do you observe?
Solution:
Numbers between 120 and 140 that are divisible by 8 are 128 and 136.
Numbers between 1120 and 1140 that are divisible by 8 are 1128 and 1136.
Numbers between 3120 and 3140 that are divisible by 8 are 3128 and 3136.
We observe that if last three digits of a number is divisible by 8, then the number is divisible by 8.

Question 6.
Fill the grid with prime numbers only so that the product of each row is the number to the right of the row and the product of each column is the number below the column.

Solution:
(i)

7	5	3	105
2	5	2	20
2	5	3	30
28	125	18

(ii)

2	2	2	8
3	5	7	105
5	7	2	70
30	70	28

Prime Time Class 6 Extra Questions

Prime Time Class 6 Very Short Question Answer

Question 1.
Write all factors of each of the following numbers:
(i) 24
(ii) 32
Solution:
(i) We can write 24 = 1 × 24 = 2 × 12 = 3 × 8 = 4 × 6
∴ 1,2,3, 4, 6, 8, 12 and 24 are the factors of 24.

(ii) We can write 32 = 1 × 32 = 2 × 16 = 4×8
1, 2, 4, 8, 16 and 32 are the factors of 32.

Question 2.
Write first five multiples of each of the following numbers.
(i) 9
(ii) 13
Solution:
(i) To obtain first five multiples of 9, we multiply it by 1,2, 3, 4 and 5 respectively.
9 × 1 = 9;
9 × 2 = 18;
9 × 3 = 27;
9 × 4 = 36;
9 × 5= 45
Hence, the first five multiples of 9 are 9, 18, 27, 36 and 45.

(ii) To obtain first five multiples of 13, we multiply it by 1, 2, 3, 4 and 5 respectively.
13 × 1 = 13;
13 × 2 = 26;
13 × 3 = 39;
13 × 4 = 52;
13 × 5 = 65
Hence, the first five multiples of 13 are 13, 26, 39, 52 and 65.

Question 3.
Show that 17 is a factor of 170017 without actual division.
Solution:
We can write
170017 = 170000 + 17 = 17 × 10000 + 17 × 1 = 17 × (10000 + 1) = 17 × 10001
Hence, 17 is a factor of 170017.

Question 4.
Write any three numbers that are multiples of 15 but not of 30.
Solution:
Multiples of 15 are 15, 30, 45, 60, 75, 90, … Multiples of 30 are 30, 60, 90, …
Thus, three numbers that are multiples of 15 but not of 30 are 15, 45 and 75.

Question 5.
Find the common factors of 16 and 28.
Solution:
We can write 16 = 1 × 16 = 2 × 8 = 4 × 4
∴ The factors of 16 are 1, 2, 4, 8 and 16.
And, 28 = 1 × 28 = 2 × 14 = 4 × 7
∴ The factors of 28 are 1, 2, 4, 7, 14 and 28.
Hence, the common factors of 16 and 28 are 1, 2 and 4.

Question 6.
Write five prime triplets such that the difference between the greatest and the smallest prime number is 6.
Solution:
{7, 11, 13}, (11, 13, 17}, {13, 17, 19}, {17, 19, 23} and {37, 41, 43}.

Question 7.
Write all prime numbers less than 100.
Solution:
The prime numbers less than 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.

Question 8.
Write five pairs of co-prime numbers.
Solution:
Two numbers are said to be co-prime if they do not have a common factor other than 1. Thus, five pairs of co-primes are (2, 3), (4, 5), (5, 6), (9, 10), (10, 21).

Question 9.
Write two numbers whose product is 10000. The two numbers should not have 0 as the units digit.
Solution:
We can write 10000 = 100 × 100
= (2 × 2 × 5 × 5) × (2 × 2 × 5 × 5)
= (2 × 2 × 2 × 2) × (5 × 5 × 5 × 5)
= 16 × 625

Question 10.
Check the divisibility 100100 by 10:
Solution:
Since 100100 ends with the digit 0, it is divisible by 10.

Question 11.
Check the divisibility 412030 by 10:
Solution:
Since 412030 ends with the digit 0, it is divisible by 10.

Question 12.
Check the divisibility 718005 by 10:
Solution:
Since 718005 does not end with the digit 0, it is not divisible by 10.

Question 13.
Check the divisibility 123456 by 4:
Solution:
The number formed by the last two digits of 123456 is 56.
Since 56 = 4 × 14, it is divisible by 4.
∴ 123456 is divisible by 4.

Question 14.
Check the divisibility 725976 by 4:
Solution:
The number formed by the last two digits of 725976 is 76.
Since 76 = 4 × 19, it is divisible by 4.
∴ 725976 is divisible by 4.

Question 15.
Check the divisibility 985442 by 4:
Solution:
The number formed by the last two digits of 985442 is 42.
Clearly, 42 is not divisible by 4.
∴ 985442 is not divisible by 4.

Question 16.
Give an example of a number which is divisible by 2 but not by 4
Solution:
6

Question 17.
Give an example of a number which is divisible by 4 but not by 8.
Solution:
20

Question 18.
Give an example of a number which is divisible by both 4 and 8 but not by 32.
Solution:
48

Question 19.
Express each of the following as a sum of three unique prime numbers.
(i) 10
(ii) 25
(iii) 35
(iv) 49
Solution:
(i) 10 = 2 + 3 + 5
(ii) 25 = 5 + 7 + 13
(iii) 35 = 5 + 7 + 23
(iv) 49 = 13 + 17 + 19

Question 20.
Express each of the following as a sum of two unique prime numbers:
(i) 15
(ii) 30
(iii) 54
(iv) 80
Solution:
(i) 15 = 2 + 13
(ii) 30 = 7 + 23
(iii) 54 = 13 + 41
(iv) 80 = 19 + 61

Question 21.
Express each of the following as a sum of twin primes:
(i) 12
(ii) 36
(iii) 60
(iv) 84
Solution:
(i) 12 = 5 + 7
(ii) 36 = 17 + 18
(iii) 60 = 29 + 31
(iv) 84 = 41 + 43

Question 22.
Check the divisibility of the following numbers by 5:
(i) 235965
(ii) 783120
(iii) 415812
Solution:
(i) Since 235965 ends with the digit 5, it is divisible by 5.
(ii) Since 783120 ends with the digit 0, it is divisible by 5.
(iii) Since 415812 does not end with the digits 0 or 5, it is not divisible by 5.

Prime Time Class 6 Short Question Answer

Question 1.
Write all factors of each of the following numbers:
(i) 60
(ii) 420
Solution:
(i) We can write
60 = 1 × 60 = 2 × 30 = 3 × 20 = 4 × 15
= 5 × 12 = 6 × 10
∴ 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60 are the factors of 60.

(ii) We can write
420 = 1 × 420 = 2 × 210 = 3 × 140 = 4 × 105 = 5 × 84 = 6 × 70 = 7 × 60 = 10 × 42
= 12 × 35 = 14 × 30 = 15 × 28 = 20 × 21
∴ 1, 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, 210 and 420 are the factors of 420.

Question 2.
The product of two numbers is 42. Their sum is 17. What are the numbers?
Solution:
We can write 42 = 1 × 42 = 2 × 21 = 3 × 14 = 6 × 7.
∴ 1, 2, 3, 6, 7, 14, 21 and 42 are the factors of 42.
Out of all the above factors, only 3 and 14 add up to a total of 17.
Hence, the required numbers are 3 and 14.

Question 3.
Find the smallest number that is a multiple of all the numbers from 1 to 10 except 7.
Solution:
We know, 8 is a multiple of 1, 2, 4 and 8.
9 is a multiple of 1, 3 and 9.
8 × 9 is a multiple of 6.
5 × 8 × 9 is a multiple of 5 and 10.
[As 5 × 8 = 40 is a multiple of 5 and 10.]
∴ Required smallest number is 5 × 8 × 9 = 360.

Question 4.
Find the common factors of 12, 18 and 24.
Solution:
We can write 12 = 1 × 12 = 2 × 6 = 3 × 4
∴ The factors of 12 are 1, 2, 3, 4, 6 and 12.
18 = 1 × 18 = 2 × 9 = 3 × 6
∴ The factors of 18 are 1, 2, 3, 6, 9 and 18.
24 = 1 × 24 = 2 × 12 = 3 × 8 = 4 × 6
∴ The factors of 24 are 1,2, 3, 4, 6, 8, 12 and 24.
Hence, the common factors of 12, 18 and 24 are 1, 2, 3 and 6.

Question 5.
Find the common multiples of 6, 9 and 12.
Solution:
The multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, …
The multiples of 9 are 9, 18, 27, 36, 45, 54, 63, 72, ……
The multiples of 12 are 12, 24, 36, 48, 60, 72, 84, ………
Hence, the common multiples of 6, 9 and 12 are 36, 72, 108, … .

Question 6.
A number is divisible by both 6 and 14. By which other number will that number be always divisible?
Solution:
Since the number is divisible by 6 and 14, the number is a common multiple of 6 and 14. So, the number will always be divisible by the smallest common multiple of 6 and 14.
Now, the multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, … .
And, the multiples of 14 are 14, 28. 42, 56, 70, 84,
So, the common multiples of 6 and 14 are 42, 84, 126,….
∴ The smallest common multiple of 6 and 14 is 42.
Hence, 42 is the required number.

Question 7.
How many numbers between 1 and 100 have exactly three factors?
Solution:
As we know, prime numbers have exactly two factors, 1 and itself. So, squares of prime numbers will have exactly three factors.
2² = 4 has factors 1, 2 and 4.
3² = 9 has factors 1,3 and 9. .
5² = 25 has factors 1,5 and 25.
7² = 49 has factors 1, 7 and 49.
Thus, four numbers i.e. 4, 9, 25 and 49 between 1 and 100 have exactly three factors.

Question 8.
The product of two numbers is 36. Their difference is 9. What are the numbers?
Solution:
We can write 36 = 1 × 36 = 2 × 18 = 3 × 12 = 4 × 9 = 6 × 6.
∴ 1, 2, 3, 4, 6, 9, 12, 18 and 36 are the factors of 36.
The difference between factors 12 and 3 is 9.
Hence, the required numbers are 3 and 12.
Note: The difference between factors 18 and 9 is also 9, but their product is not equal to 36.

Question 9.
Without actual division, show that 13 is a factor of each of the following numbers:
(i) 1313
(ii) 13013
(iii) 131313
Solution:
(i) We can write
1313 = 1300 + 13 = 13 × 100 + 13× 1 = 13(100 + 1) = 13 × 101
Hence, 13 is a factor of 1313.

(ii) We can write
13013 = 13000 + 13 = 13 × 1000 + 13 × 1 = 13(1000 + 1) = 13 × 1001
Hence, 13 is a factor of 13013.

(iii) We can write
131313 = 130000 + 1300 + 13 = 13 × 10000 + 13 × 100 + 13 × 1
= 13(10000 + 100 + 1) = 13 × 10101
Hence, 13 is a factor of’ 131313.

Question 10.
Find the common factors of:
(i) 12 and 18
(ii) 35 and 63
(iii) 60 and 210
Solution:
(i) We can write 12 = 1 × 12 = 2 × 6 = 3 × 4
∴ The factors of 12 are 1, 2, 3, 4, 6 and 12.
And, 18 = 1 × 18 = 2 × 9 = 3 × 6
∴ The factors of 18 are 1,2, 3, 6, 9 and 18.
Hence, the common factors of 12 and 18 are 1, 2, 3 and 6.

(ii) We can write 35 = 1 × 35 = 5 × 7
∴ The factors of 35 are 1, 5, 7 and 35.
And, 63 = 1 × 63 = 3 × 21 = 7 × 9
∴ The factors of 63 are 1. 3, 7, 9, 21 and 63.
Hence, the common factors of 35 and 63 are 1 and 7.

(iii) We can write 60 = 1 × 60 = 2 × 30 = 3 × 20 = 4 × 15 = 5 × 12 = 6 × 10
∴ The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
210 = 1 × 210 = 2 × 105 = 3 × 70 = 5 × 42 = 6 × 35 = 7 × 30 = 10 × 21 = 14 × 15
∴ The factors of 210 are 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105 and 210.
Hence, the common factors of 60 and 210 are 1, 2, 3, 5, 6, 10, 15 and 30.

Question 11.
Find first three common multiples of:
(i) 10 and 20
(ii) 25 and 50
(iii) 40 and 60
Solution:
(i) The multiples of 10 are 10, 20, 30, 40, 50, 60,
The multiples of 20 are 20, 40, 60, 80, …
Hence, the first three common multiples of 10 and 20 are 20, 40 and 60.

(ii) The multiples of 25 are 25, 50, 75, 100, 125, 150, …
The multiples of 50 are 50, 100, 150, 200, …
Hence, the first three common multiples of 25 and 50 are 50, 100 and 150.

(iii) The multiples of 40 are 40; 80, 120, 160, 200, 240, 280, 320, 360 …
The multiples of 60 are 60, 120, 180, 240, 300, 360 …
Hence, the first three common multiples of 40 and 60 are 120, 240 and 360.

Question 12.
A number is divisible by both 8 and 12. By which other numbers will that number be always divisible?
Solution:
Since the number is divisible by 8 and 12, the number is a common multiple of 8 and 12. So, the number will always he divisible by the smallest common multiple of 8 and 12.
Now, the multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72 …
The multiples of 12 are 12, 24, 36, 48, 60, 72, …
So, the common multiples of 8 and 12 are 24, 48, 72, …
∴ The smallest common multiple of 8 and 12 is 24.
Hence, the number is divisible by 24 and hence by all factors of 24, i.e., 1, 2, 3, 4, 6, 8, 12 and 24.

Question 13.
Write all pairs of twin primes between 1 and 100.
Solution:
The prime numbers less than 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83,89 and 97.
The required pairs of twin primes are (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61) and (71, 73).

Question 14.
Write all pairs of prime numbers less than 30 whose sum is a multiple of 5.
Solution:
The prime numbers less than 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.
Now, 2 + 3 = 5, 2 + 13 = 15, 2 + 23 = 25, 3 + 7=10,
3 + 17 = 20, 7 + 13 = 20, 7 + 23 = 30, 11 + 19 = 30,
11 + 29 = 40, 13 + 17 = 30, 17 + 23 = 40
∴ The required pairs are (2, 3), (2, 13), (2, 23), (3, 7), (3, 17), (7, 13), (7, 23), (1 1, 19), (11, 29), (13, 17) and (17, 23).

Question 15.
Determine the prime factorisation of the following numbers:
(i) 3094
(ii) 5082
(iii) 10010
Solution:
(i) The prime factorisation of 3094 is 2 × 7 × 13 × 17.
(ii) The prime factorisation of 5082 is 2 × 3 × 7 × 11 × 11.
(iii) The prime factorisation of 10010 is 2 × 5 × 7 × 11 × 13.

Question 16.
Find prime factorisation of the following numbers without multiplying first:
(i) 48 × 60
(ii) 56 × 75
(iii) 81 × 25
Solution:
(i) Prime factorisation of 48 = 2 × 2 × 2 × 2 × 3
Prime factorisation of 60 = 2 × 2 × 3 × 5
Thus, prime factorisation of 48 × 60 is 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5.

(ii) Prime factorisation of 56 = 2 × 2 × 2 × 7
Prime factorisation of 75 = 3 × 5 × 5
Thus, prime factorisation of 56 × 75 is 2 × 2 × 2 × 3 × 5 × 5 × 7.

(iii) Prime factorisation of 81 = 3 × 3 × 3 × 3
Prime factorisation of 25 = 5 × 5
Thus, prime factorisation of 81 × 25 is 3 × 3 × 3 × 3 × 5 × 5.

Question 17.
The first number has prime factorisation 2 × 5 × 7 and the second number has prime factorisation 5 × 7 × 11. Are they co-prime? Does one of them divide the other?
Solution:
Since 5 and 7 are the common prime factors of two numbers, they are not co-prime.
Clearly, 2 is a prime factor of first number but not a prime factor of second number.
Hence, second number is not divisible by first number.
And, 1 1 is a prime factor of second number but not a prime factor of first number.
Hence, first number is not divisible by second number. So, none of them divide the other number.

Question 18.
Check the divisibility of the following numbers by 8:
(i) 6245826
(ii) 2727272
(iii) 2491664
Solution:
(i) The number formed by the last three digits of 6245826 is 826.
Clearly, 826 is not divisible by 8.
∴ 6245826 is not divisible by 8.

(ii) The number formed by the last three digits of 2727272 is 272.
Since 272 = 8 x 34, it is divisible by 8.
∴ 2727272 is divisible by 8.

(iii) The number formed by the last three digits of 2491664 is 664.
Since 664 = 8 X 83, it is divisible by 8.
∴ 2491664 is divisible by 8.

Prime Time Class 6 Long Question Answer

Question 1.
Who am I?
(i) I am a number less than 50. One of my factors is 11. The sum of my digits is 8.
(ii) I am an even number less than 100. One of my factors is 7. One of my digits is 1 more than the other.
Solution:
(i) Numbers less than 50 and having 11 as a factor are
11 × 1 = 11, 11 × 2 = 22, 11 × 3 = 33, 11 × 4 = 44
Out of these numbers, only 44 is the number whose sum of digits is 8.
Hence, the required number is 44.

(ii) Even numbers less than 100 and having 7 as a factor are
7 × 2 = 14, 7 × 4 = 28, 7 × 6 = 42, 7 × 8 = 56, 7 × 10 = 70, 7 × 12 = 84, 7 × 14 = 98
Out of these numbers, only 56 and 98 are the numbers whose one digit is 1 more than the other.
Hence, the required number is either 56 or 98.

Question 2.
Using prime factorisation, check whether the following pairs of numbers are co-prime or not:
(i) 154 and 195
(ii) 132 and 225
(iii) 357 and 286
Solution:
(i) Prime factorisations of 154 and 195 are as follows:
154 = 2 × 7 × 11 and 195 = 3 × 5 × 13
Since there are no common prime factors, 154 and 195 are co-prime.

(ii) Prime factorisations of 132 and 225 are as follows:
132 = 2 × 2 × 3 × 11 and 225 = 3 × 3 × 5 × 5
Since 3 is a common prime factor, 132 and 225 are not co-prime.

(iii) Prime factorisations of 357 and 286 are as follows:
357 = 3 × 7 × 17 and 286 = 2 × 11 × 13
Since there are no common prime factors, 357 and 286 are co-prime.

Question 3.
Using prime factorisation method, determine whether the first number is divisible by the second number or not:
(i) 420 and 105
(ii) 693 and 78
(iii) 990 and 90
Solution:
(i) Prime factorisations of 420 and 105 are as follows:
420 = 2 × 2 × 3 × 5 × 7 and 105 = 3 × 5 × 7
Clearly, all prime factors of 105 are prime factors of 420 and prime factorisation of 105 is included in the prime factorisation of 420.
We can write, 420 = (3 × 5 × 7) × 2 × 2 = 105 × 4
Hence, 420 is divisible by 105.

(ii) Prime factorisations of 693 and 78 are as follows:
693 = 3 × 3 × 7 × 11 and 78 = 2 × 3 × 13 Clearly, 13 is a prime factor of 78 but not a prime factor of 693.
Hence, 693 is not divisible by 78.

(iii) Prime factorisations of 990 and 90 are as follows:
990 = 2 × 3 × 3 × 5 × 11 and 90 = 2 × 3 × 3 × 5
Clearly, all prime factors of 90 are prime factors of 990 and prime factorisation of 90 is included in the prime factorisation of 990.
We can write, 990 = (2 × 3 × 3 × 5) × 11 = 90 × 11
Hence, 990 is divisible by 90.

Question 4.
Using prime factorisation, check whether the following pairs of numbers are co-prime or not:
(i) 308 and 585
(ii) 396 and 450
(iii) 1071 and 1430
Solution:
(i) Prime factorisations of 308 and 585 are as follows:
308 = 2 × 2 × 7 × 11 and 585 = 3 × 3 × 5 × 13
Since there are no common factors in these prime factorisations, 308 and 585 are co-prime.

(ii) Prime factorisations of 396 and 450 are as follows:
396 = 2 × 2 × 3 × 3 × 11 and 450 = 2 × 3 × 3 × 5 × 5
Since 2 and 3 are common factors in these prime factorisations, 396 and 450 are not co-prime.

(iii) Prime factorisations of 1071 and 1430 are as follows:
1071 = 3 × 3 × 7 × 17 and 1430 = 2 × 5 × 11 × 13
Since there are no common factors in these prime factorisations, 1071 and 1430 are co-prime.

Question 5.
Using prime factorisation method, determine whether the first number is divisible by the second number or not:
(i) 1260 and 315
(ii) 7623 and 198
(iii) 3780 and 126
Solution:
(i) Prime factorisations of 1260 and 315 are as follows:
1260 = 2 × 2 × 3 × 3 × 5 × 7 and 315 = 3 × 3 × 5 × 7
Clearly, all prime factors of 315 are prime factors of 1260 and prime factorisation of 315 is included in the prime factorisation of 1260.
We can write, 1260 = (3 × 3 × 5 × 7) × 2 × 2 = 315 × 4
Hence, 1260 is divisible by 315.

(ii) Prime factorisations of 7623 and 198 are as follows:
7623 = 3 × 3 × 7 × 11 × 11 and 198 = 2 × 3 × 3 × 11
Clearly, 2 is a prime factor of 198 but not a prime factor of 7623.
Hence, 7623 is not divisible by 198.

(iii) Prime factorisations of 3780 and 126 are as follows:
3780 = 2 × 2 ×3 × 3 × 3 × 5 × 7 and 126 = 2 × 3 × 3 × 7
Clearly, all prime factors of 126 are prime factors of 3780 and prime factorisation of 126 is included in the prime factorisation of 3780.
We can write,
3780 = (2 × 3 × 3 × 7) × 2 × 3 × 5 = 126 × 30
Hence, 3780 is divisible by 126.

Question 6.
State whether the following statements are true or false. Give reasons.
(i) If a number is divisible by 4, then it must be divisible by 8.
(ii) If a number is divisible by 8, then it must be divisible by 4.
(iii) If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.
(iv) The sum of two consecutive odd numbers is always divisible by 4.
(v) Sum of two even numbers gives a multiple of 4.
(vi) Sum of two odd numbers gives a multiple of 4.
Solution:
(i) False.
For example, 28 is divisible by 4 but not by 8.

(ii) True.
If a number is divisible by 8, then it must be divisible by 4 because 4 is a factor of 8.

(iii) False.
For example, let us take 14 and 6. Their sum, 14 + 6 = 20, is divisible by 4. But neither 14 nor 6 are divisible by 4.

(iv) True.
Let 2n + 1 and 2n + 3 be two consecutive odd numbers, where n = 0, 1,2,3, … .
Then, their sum,
(2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1), is divisible by 4.
For example, 3 and 5 are two consecutive odd numbers, whose sum is 8 and 8 is divisible by 4.

(v) False.
For example, 12 and 6 are even numbers but their sum 18 is not a multiple of 4.

(vi) False.
For example, 13 and 5 are odd numbers but their sum 18 is not a multiple of 4.

Question 7.
Solve the prime puzzle given below:

Solution:
385 = 5 × 7 × 11;
190 = 2 × 5 × 19;
42 = 2 × 3 × 7;
285 = 3 × 5 × 19;
154 = 2 × 7 × 11;
70 = 2 × 5 × 7
∴ Solution of given prime puzzle is as follows:

Question 8.
Solve the prime puzzle given below:

Solution:
Prime factorisations of the given numbers are as follows:
231 = 3 × 7 × 11;
130 = 2 × 13 × 5;
170 = 2 × 5 × 17;
102 = 3 × 2 × 17;
182 = 2 ×7 × 13;
275 = 5 × 5 × 11
∴ Solution of given prime puzzle is as follows:

Question 9.
Solve the prime puzzle given below:

Solution:
Prime factorisations of the given numbers are as follows:
286 = 2 × 11 × 13;
30 = 2 × 3 × 5;
70 = 2 × 5 × 7;
42 = 2 × 3 × 7;
110 = 2 × 5 × 11;
130 = 2 × 5 × 13
∴ Solution of given prime puzzle is as follows:

Prime Time Class 6 Case Based Questions

Question 1.
A Greek mathematician Eratosthenes, who lived around 2200 years ago, gave a method to list the prime numbers. The method is called the Sieve of Eratosthenes. In the given figure, encircled numbers arc prime numbers.

Based on the above information, answer the following questions:
(i) How many prime numbers are there between 1 and 100?
(ii) Which digits can never appear in the units place of a prime number?
(iii) Which digit most frequently appears in the units place of prime numbers less than 100?
(iv) Write live consecutive composite numbers less than 100 so that there is no prime number between them.
Solution:
(i) There are 25 prime numbers between 1 and 100.
(ii) 0, 4, 6 and 8 never appear in the unit place of a prime number.
(iii) Digit 3 appears 7 times in the units place of prime numbers less than 100.
(iv) 24, 25, 26, 27, 28 or 32, 33, 34, 35, 36 or 48, 49, 50, 51, 52 or 54, 55,56,57, 58 or 62, 63, 64, 65, 66 or 74, 75, 76, 77, 78 or 84, 85, 86, 87, 88
or 90, 91,92, 93, 94 or 91,92, 93, 94, 95 or 92, 93, 94, 95,96

Question 2.
The school gardening club has 210 marigold plants and 150 sunflower plants. The club members want to create equal rows for each type of flower such that each row contains the same number of one kind of plant and no plants are left unused.
Based on the above information, answer the following questions:
(i) Can they arrange the marigold plants in rows of 2, 5 or 7?
(ii) What is the largest number of plants per row they can use for each type so that all plants are used and plants in each row are equal?
(iii) They decide to pack the flowers into minimum flower boxes, and each box must contain the same number of plants. Which number less than 10 can they choose so that both 210 and 150 divide evenly by it?
(iv) Perform prime factorisation of 150 and 210 and find the common prime factors.
Solution:
(i) It is given that the school gardening club has 210 marigold plants. For 210 marigold plants to be arranged in rows of 2, 5 or 7, 210 must be divisible by 2, 5 or 7.
As 210 ends with 0, it is divisible by 2 and 5 both.
Now, 210 ÷ 7 = 30. Thus, 210 is also divisible by 7.
Thus, the marigold plants can be arranged in rows of 2, 5 or 7.

(ii) We need the largest number of plants of the same type per row such that

• Each row has the same number of plants.
• There are no leftover plants.
• Thus, we are looking for the largest number that can:
• Evenly divide 210 (so marigold plants are used up completely)
• Evenly divide 150 (so sunflower plants are used up completely)

This number will be the highest common factor of 210 and 150.
Now, the factors of 210 are 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105 and 210.
And, the factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75 and 150.
The highest common factor is 30. Thus, if we use 30 plants in each row we get 210 ÷ 30 = 7 rows of marigold plant and 150 ÷ 30 = 5 rows of sunflower plant and no plants are left.

(iii) It is given that each flower box must contain equal number of plants and the number must be less than 10. Thus, we need to find the common factors of 150 and 210 which are less than 10.
Now, the factors of 210 are 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105 and 210.
And, the factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75 and 150.
Thus, the common factors less than 10 are 1,2, 3, 5 and 6. ‘
For minimum boxes, the club members must place maximum plants in each box.
Thus, club members can keep 6 plants in each box, so that the flower boxes are minimum and there are equal plants in each box.

(iv) Prime factorisation of 210 = 2 × 3 × 5 × 7.
Prime factorisation of 150 = 2 × 3 × 5 × 5.
Thus, the common prime factors of 210 and 150 are 2, 3 and 5.

MCQ Questions for Class 6 Maths with Answers

Class 6 Maths MCQ Chapter Wise

Ganita Prakash Class 6 MCQ Questions

• Chapter 1 Patterns in Mathematics Class 6 MCQ
• Chapter 2 Lines and Angles Class 6 MCQ
• Chapter 3 Number Play Class 6 MCQ
• Chapter 4 Data Handling and Presentation Class 6 MCQ
• Chapter 5 Prime Time Class 6 MCQ
• Chapter 6 Perimeter and Area Class 6 MCQ
• Chapter 7 Fractions Class 6 MCQ
• Chapter 8 Playing with Constructions Class 6 MCQ
• Chapter 9 Symmetry Class 6 MCQ
• Chapter 10 The Other Side of Zero Class 6 MCQ

Also Read Ganita Prakash Class 6 Solutions

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 4 Data Handling and Presentation Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 4 Data Handling and Presentation Solutions

Ganita Prakash Class 6 Chapter 4 Solutions

Class 6 Maths Ganita Prakash Chapter 4 Solutions Data Handling and Presentation

Question 1.
Shri Nilesh is a teacher. He decided to bring sweets to the class to celebrate the new year. The sweets shop nearby has jalebi, gulab jamun, gujiya, barfi, and rasgulla. He wanted to know the choices of the children. He wrote the names of the sweets on the board and asked each child to tell him their preference. He put a tally mark ‘ | ’ for each student and when the count reached 5, he put a line through the previous four and marked it as

Complete the table to help Shri Nilesh to purchase the correct numbers of sweets.
(a) How many students chose jalebi? _______
(b) Barfi was chosen by _______ students?
(c) How many students chose gujiya? _______
(d) Rasgulla was chosen by _______ students?
(e) How many students chose gulab jamun? _______
Solution:

(a) It is clear from the table, Jalebi was chosen by 6 students.
(b) It is clear from the table, Barfi was chosen by 3 students.
(c) Gujiya was chosen by 13 students.
(d) Rasgulla was chosen by 7 students.
(e) Gulab jamun was chosen by 9 students.

Question 2.
Magan Bhai sells kites at Jamnagar. Six shopkeepers from nearby villages come to purchase kites from him. The number of kites he sold to these six shopkeepers are given below.

Shopkeeper	Number of Kites sold
Chaman	250
Rani	300
Rukhsana	100
Jasmeet	450
Jetha Lai	250
Poonam Ben	700

Prepare a pictograph using the symbol to represent 100 kites.
Answer the following questions:
(a) How many symbols represent the kites that Rani purchased?
(b) Who purchased the maximum number of kites?
(c) Who purchased more kites, Jasmeet or Chaman?
(d) Rukhsana says Poonam Ben purchased more than double the number of kites that Rani purchased. Is she correct? Why?
Solution:
Required pictograph is given below.

(a) Since Rani purchased 300 kites, it is clear from the pictograph, 3 symbols represent the kites that Rani purchased.

(b) The shopkeeper who purchased the maximum number of kites is the one with the most symbols in pictograph. It is clear from the pictograph that Poonam Ben purchased the maximum number of kites (7 × 100 = 700 kites).

(c) Number of kites purchased by Jasmeet = 450 Number of kites purchased by Chaman = 250 Jasmeet purchased more kites.

(d) Number of kites purchased by Poonam Ben = 700 Number of kites purchased by Rani = 300
Hence, Rukhsana is correct as 700 kites is more than double of 300 kites (300 × 2 = 600).

Question 3.
Samantha visited a tea garden and collected data of the insects and critters she saw there. Here is the data she collected —

Help her prepare a bar graph representing this data.
Solution:
To prepare a bar graph representing the given data, follow the point given below.
(i) Draw a horizontal line labeled, ‘Insects and Critters’ with each type of insect or critter evenly spaced.

(ii) Draw a vertical line labeled ‘Number of insects and critters seen’ with numbers starting from 0 up to the maximum number seen (in this case, 10) evenly spaced.

(iii) For each insect or critter, draw a bar that rises to the number seen. For example, the bar of caterpillars should reach up to 10.
Required bar graph is given alongside.

Question 4.
Chinu listed the various means of transport that passed across the road in front of his house from 9 a.m. to 10 a.m.

(a) Prepare a frequency distribution table for the data.
(b) Which means of transport was used the most?
(c) If you were there to collect this data, how could you do it? Write the steps or process.
Solution:
(a) Frequency distribution table for the given data is given below:

Means of transport	Bike	Car	Bus	Auto rickshaw	Bicycle	Bullock Cart	Scooter
Frequency	13	6	4	8	8	2	9

(b) The bike has the highest frequency in the table, indicating it was the most common means of transport observed.

(c) To collect this data, you could follow the steps given below. ;
Observation Timeframe: I will choose a specific timeframe, such as 9 a.m. to 10 a.m., to observe the road traffic.
Recording Data: 1 will use a tally chart or counting app to record the type of transport passing by during that hour.
Categorisation: Then I will organise the data into categories (e.g., bike, car, scooter, bus, etc.).
Final Count: After the observation period, I will get the total the number of occurrences for each category.
Analysis: Finally, I prepare a frequency distribution table based on the recorded data for analysis.

Question 5.
The number of girl students in each class of a school is depicted by the pictograph.

Observe this pictograph and answer the following questions.
(a) Which class has the least number of girl students?
(b) What is the difference between the number of girls in Class 5 and 6?
(c) If two more girls were admitted in Class 2, how would the graph change?
(d) How many girls are there in Class 7?
Solution:
(a) The pictograph shows the least number of symbols for Class 8.
Hence, Class 8 has the least number of girl students.

(b) Girl students in Class 5 = 2.5 × 4 = 10 Girl students in Class 6 = 4 × 4 = 16
Hence, there are 6 more girls in Class 6 than in Class 5.

(c) Here, adding 2 more girls would increase the count by 2 and requires a half additional symbol.
The pictograph would show half additional symbol for class 2, assuming each symbol represent 4 girls.

(d) Girl students in Class 7 = 3 × 4 = 12.

Data Handling and Presentation Class 6 Extra Questions

Data Handling and Presentation Class 6 Very Short Question Answer

Question 1.
Define the following terms:
(i) Observations
(ii) Data
Solution:
(i) Observations are the individual pieces of information we collect.
For example, if we ask five friends their ages, each friend’s age is one observation.

(ii) Any collection of facts, numbers, measures, observations or other descriptions of things that convey information about those things is called data.

Question 2.
Define the following terms:
(i) Tabulation of data
(ii) Tally marks
Solution:
(i) Tabulation of data is the process of systematically arranging data into rows and columns within a table to make it easier to understand, compare and analyse.

(ii) Tally marks are short vertical lines used to count items. Each time we count one item, we draw one line (|). After four lines (| | | |), we draw a oblique line (\) across them to show five.

Question 3.
Define the following terms:
(i) Raw data
(ii) Array
Solution:
(i) Raw data is the information as we first collect it (original form), before we sort or arrange it.
For example, writing ages on a sheet of paper as friends tell you, without putting them in any order.

(ii) Organising the data by arranging it in ascending or descending order is called an array.
For example, Meenal arranged the shoe sizes of the students in ascending order as follows:
3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7

Question 4.
Define the following terms:
(i) Frequency of observation
(ii) Frequency distribution
Solution:
(i) Frequency of observation tells us how many times a particular value appears in the data. For example, if the age ‘12’ appears three times in the list, then the frequency of age ‘12’ is 3.
(ii) Frequency distribution is a table that shows each value (or group of values) and how often it appears (frequency). For example:

Age	Frequency
10	2
11	1
12	3

Question 5.
A die was thrown 20 times and the following outcomes were noted:
1, 2, 5, 1, 6, 2, 3, 4, 2, 4, 2, 5, 6, 6, 6, 2, 3, 1, 1, 5
Represent the above data in the form of frequency distribution.
Solution:

Outcome	Frequency
1	4
2	5
3	2
4	2
5	3
6	4

Question 6.
Enlist 1 advantage and 1 disadvantage of presentation of data through pictograph.
Answer:
1 advantage and 1 disadvantage of pictograph are as follows:

Advantage	Disadvantage
1. Pictorial representation makes it easier to understand the data.	1. Drawing a pictograph is time consuming.

Data Handling and Presentation Class 6 Short Question Answer

Question 1.
Given below is the data showing the number of children in 15 families of a colony.
2, 3, 1, 1, 2, 3, 2, 3, 3, 4, 1, 2, 2, 1, 3
Arrange the above data in ascending order and then form a tally table.
Solution:
Given data (number of children in 15 families) is as follows: 2, 3, 1, 1, 2, 3, 2, 3, 3, 4, 1, 2, 2, 1, 3
(i) Data in ascending order: 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4.
(ii) Given data in tabular form is shown below:

Question 2.
The number of cricket bat and ball pairs, sold by a shopkeeper during a week are given below:

Day	Mon.	Tue.	Wed.	Thu.	Fri.	Sat.
Number of bat and ball pair sold	12	9	15	15	3	27

Decide the scaling and draw a pictograph to represent the given data.
Solution:

Question 3.
Total number of cows in five villages are as follows:

Village	A	B	C	D	E
Number of cows	40	60	20	100	120

Decide the scale and draw a pictograph to represent the given data.
Solution:

Question 4.
Following pictograph shows the data of the number of students who like a particular sport.

Answer the following questions:
(i) How many students like cricket?
(ii) How many more students like basketball than volleyball?
Solution:
(i) Number of symbols for cricket = 4
Students who like cricket = 4 × 3= 12

(ii) Number of symbols for basketball = 3
∴ Students who like basketball = 3 × 3 = 9
Number of symbols for volleyball = 2
∴ Students who like volleyball = 2 × 3 = 6
Difference = 9 – 6 = 3
Hence, 3 more students like basketball than volleyball.

Question 5.
The bar graph below shows the production of Kharif and Rabi crops (in tons) in 5 different states in India. Draw a table that represents the data in the bar graph.

Solution:
Table for the given bar graph can be created as follows:

States	Rabi crops (in tons)	Kharif crops (in tons)
West Bengal	30	60
Karnataka	45	40
Madhya Pradesh	20	54
Tamil Nadu	36	22
Uttar Pradesh	50	42

Data Handling and Presentation Class 6 Long Question Answer

Question 1.
The weights of newborn babies (in Kg) in a hospital on a particular day are as follows:
2.1, 2.5, 3.5, 2.7, 2.9, 3.1, 2.6, 2.5, 2.8, 2.3, 2.9, 3.3, 3.4, 2.7, 2.8
Answer the following questions:
(i) How many babies are born on that day?
(ii) Arrange the above data in descending order.
(iii) Determine the highest weight.
(iv) Determine the lowest weight.
(v) Determine the range of weight.
(vi) How many babies weigh below 2.5 kg?
Solution:
(i) There are 15 babies born on that day.

(ii) Writing the weights from highest to lowest:
3.5, 3.4, 3.3, 3.1, 2.9, 2.9, 2.8, 2.8, 2.7, 2.7, 2.6, 2.5, 2.5, 2.3, 2.1

(iii) The highest weight is the first weight in the descending order of weights.
Highest weight = 3.5 kg

(iv) The lowest weight is the last weight in the descending order of weights.
Lowest weight = 2.1 kg

(v) We know, range = highest value – lowest value
∴ Range of weight = highest weight – lowest weight
= 3.5 kg – 2.1 kg = 1.4 kg

(vi) The weights below 2.5 kg are 2.1 kg and 2.3 kg. So, the required number of babies is 2.

Question 2.
Study the following pictograph:

Answer the following questions:
(i) In which city was the rainfall maximum and how much?
(ii) In which city was the rainfall minimum and how much?
(iii) Which cities had rainfall of more than 50 cm?
Solution:
(i) Chennai has three full icons, which is maximum.
∴ Rainfall in Chennai = 3 × 50 cm = 150 cm

(ii) Delhi has one full icon, which is minimum.
∴ Rainfall in Delhi = 1 × 50 cm = 50 cm.

(iii) Mumbai has 2 full icons.
∴ Rainfall in Mumbai = 2 × 50 cm = 100 cm (which is greater than 50 cm)
Hyderabad has 1 full and 1 half icons.
∴ Rainfall in Hyderabad = 50 cm + 25 cm = 75 cm (which is greater than 50 cm) Chennai has 3 full icons.
∴ Rainfall in Chennai = 50 × 3 = 150 cm (which is greater than 50 cm)
Kolkata has 2 full and 1 half icons.
∴ Rainfall in Kolkata =100 cm + 25 cm =125 cm (which is greater than 50 cm)
Hence, Mumbai, Hyderabad, Chennai, and Kolkata had rainfall of more than 50 cm.

Question 3.
In a shop, there are 5 different types of toys. The number of each toy is given in the table below. Construct a bar graph representing this data. Use appropriate scaling.

Toys Name	Blocks	Cars	Dolls	Balls	Bikes
Number of Toys	40	48	60	35	30

Solution:
The required bar graph can be easily created using the following steps:
Step 1: On a graph sheet draw two mutually perpendicular lines, a horizontal and a vertical line.

Step 2: Label horizontal line as ‘Toys name’ & write names of toys from the table with equal gaps between them: Blocks, Cars, Dolls, Balls, Bikes. Label vertical line as ‘Number of Toys’.

Step 3: Scaling: Choose a suitable scale to show the number of toys.
Here, we take: 1 unit (1 big division) = 10 toys. Each big division is further divided into 10 sub-divisions, representing 1 toy.

Step 4: Calculate the height of each bar:
10 toys = 1 large divisions
∴ 1 toy = \(\frac{1}{10}\) large divisions
= 1 sub-division
Now, the height of the bar for Blocks
= 40 × (\(\frac{1}{10}\)) = 4 large divisions
The height of the bar for Cars
48 × (\(\frac{1}{10}\)) = 4 large divisions and 8 sub-divisions
The height of the bar for Dolls = 60 × (\(\frac{1}{10}\)) = 6 large divisions
The height of the bar for Balls = 35 × (\(\frac{1}{10}\)) = 3 large divisions and 5 sub-divisions
The height of the bar for Bikes = 30 × (\(\frac{1}{10}\))

Step 5: Now draw vertical bars for each toy using the heights calculated above. Make sure the bars are of equal width and there is equal spacing between them.

Question 4.
The zookeeper in Delhi is preparing a presentation for higher officers on the number of animals in the zoo. He intends to create a bar graph for better visualisations. The number of each animal is given below in the table. Construct a bar graph for the table.

Animals	Number of Animals
Lion	8
Elephant	12
Gorilla	9
Zebra	15
Giraffe	7
Cheetah	6

Solution:
Step 1: Take a graph sheet and draw two mutually perpendicular lines, a horizontal line and a vertical line.

Step 2: Label the horizontal line as ‘Animals’ and write the names of the animals from the table with equal gaps between them: Lion, Elephant, Gorilla, Zebra, Giraffe, Cheetah Label the vertical lines ‘ Number of Animals’.

Step 3 (Scaling) : Choose a suitable scale to show the number of animals.
Here, we take: 1 unit (1 large division)
= 2 animals
∴ 1 animal = \(\frac{1}{2}\) unit = 1 sub-division
This scale helps us draw the graph clearly.

Step 4: Calculate the height of each bar using the scale:
Height of the bar for the Lion = 8 ÷ 2 = 4 large divisions
Height of the bar for the Elephant = 12 ÷ 2 = 6 large divisions
Height of the bar for the Gorilla = 9 ÷ 2 = 4 large divisions and 1 subdivision
Height of the bar for the Zebra = 15 ÷ 2 = 7 large divisions and 1 sub-division
Height of the bar for the Giraffe = 7 + 2 = 3 large divisions and 1 subdivision
Height of the bar for the Cheetah = 6 + 2 = 3 large divisions

Step 5: Now draw vertical bars for each animal using the heights calculated above. Make sure the bars are of equal width and there is equal spacing between them.

Question 5.
The following bar graph shows the revenue of your country from exports of various items in 1 year. (1 unit = 10 crore rupees)

(a) Write the information given by the bar graph in a table.
(b) What is the difference between the maximum revenue and the minimum revenue?
(c) What is the total revenue from exports?
Solution:
(a) The required table is shown below:

Items	Revenue (Rupees in crores)
Electronics	30
Foodgrains	70
Livestock	25
Metals	50
Softwares	80

(b) From the table, it can be observed that the maximum revenue was generated by the export of software, i.e. 80 crore rupees.
Also, the minimum revenue was generated by the exports of livestock, i.e. 25 crore rupees.
∴ The difference between the revenue from software and livestock = 80 – 25 = 55 crore

(c) Total revenue from exports can be calculated by summing up the revenues from all the commodities.
∴ Total revenue from exports = 30 + 70 + 25 + 50 + 80 = 255 crore rupees

Question 6.
The table shows how much money Imran’s family spends each month on different items:

Items	Expenditure (in ₹)
House rent	3,000
Food	3,400
Education	800
Electricity	400
Transport	600
Miscellaneous	1,200

Answer the following questions:
(a) Represent the given data in the form of a bar graph
(b) On which item does the Imran’s family spend the most and the second most?
(c) Is the cost of electricity about one half of the cost of education?
Solution:
(a) To represent this data in the form of a bar graph, here are the steps:

Step 1: Draw two perpendicular lines, one horizontal and one vertical.
Step 2: Along the horizontal line, mark the ‘items’ with equal spacing between them and mark the corresponding expenditures along the vertical line.
Step 3: Take bars of the same width, keeping a uniform gap between them.
Step 4: Choose a suitable scale along the vertical line. Let, 1 unit length = ₹ 200, and then mark and write the corresponding values (₹ 200, ₹400, etc.) representing each unit length and calculate the heights of the bars for various items as shown below:

House rent	3000 ÷ 200	15 units
Food	3400 ÷ 200	17 units
Education	800 ÷ 200	4 units
Electricity	400 ÷ 200	2 units
Transport	600 ÷ 200	3 units
Miscellaneous	1200 ÷ 200	6 units

Here is the bar graph that we obtained based on the above steps.

(b) From the given data, we can say that on food Imran’s family spend the most and on house rent they spend the second most.

(c) Yes. As the cost of electricity is ₹ 400 and the cost of eduction is ₹ 800, we can say that the cost of electricity is about one-half the cost of education.

Question 7.
Your school is collecting data on various transportation options that students used to opt. The table below shows the data collected by you. Prepare a bar graph to be given to your school principal, also use appropriate scaling.

Transportation	Number of Students
Bus	200
Cycle	50
Walk	80
Auto	120
Car	40

Solution:
The required bar graph can be easily created using the following steps:

Step 1: Take a graph sheet and draw two mutually perpendicular lines, a horizontal line and a vertical line.

Step 2: Label the horizontal line as‘Transportation’ and write the names from the table with equal gaps between them: Bus, Cycle, Walk, Auto, Car.
Label the vertical line as ‘Number of Students’.

Step 3: (Scaling): Choose a suitable scale to show the number of students.
Here, we take:
1 unit (1 big division) = 20 students (This helps us fit large numbers like 200 easily on the graph.)

Step 4: Calculate the height of each bar:
= 200 × (\(\frac{1}{20}\)) = 10 large divisions
Height of the bar for the cycle
= 50 × (\(\frac{1}{20}\))= 2 large divisions and 5 sub-divisions
Height of the bar for the walk
= 80 × (\(\frac{1}{20}\)) = 4 large divisions
Height of the bar for the auto
= 120 × (\(\frac{1}{20}\)) = 6 large divisions
Height of the bar for the car
= 40 × (\(\frac{1}{20}\)) = 2 large divisions

Step 5: Now draw vertical bars for each mode of transportation using the heights calculated above. Make sure the bars are of equal width and there is equal spacing between them.

Question 8.
The table below displays the number of bicycles produced at a factory from 1998 to 2002.

Year	1998	1999	2000	2001	2002
No. of Bicycles	800	600	900	1100	1200

Answer the following questions:
(a) Create a bar graph to represent this information. Select your preferred scale.
(b) Which year had the greatest number of bicycles produced?
(c) Which year had the least number of bicycles produced?
Solution:
(a) To represent this data in the form of a bar

Step 1: Draw two perpendicular lines, one horizontal and one vertical.
Step 2: Along the horizontal line, mark the ‘years’ with equal spacing between them and mark the corresponding ‘No. of bicycles’ along the vertical line.
Step 3: Take bars of the same width, keeping a uniform gap between them.
Step 4: Choose a suitable scale along the vertical line. Let, 1 unit length =100 bicycles, and then mark and write the corresponding values (100 bicycles, 200 bicycles, etc.) representing each unit length and calculate the heights of the bars for various years as shown below:

1998	800 ÷ 100	8 units
1999	600 ÷ 100	6 units
2000	900 ÷ 100	9 units
2001	1100 ÷ 100	11 units
2002	1200 ÷ 100	12 units

(b) The greatest number of bicycles produced in 2002.
(c) The least number of bicycles produced in 1999.

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 3 Number Play Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 3 Number Play Solutions

Ganita Prakash Class 6 Chapter 3 Solutions

Class 6 Maths Ganita Prakash Chapter 3 Solutions Number Play

Question 1.
Fill the table below with only 4-digit numbers such that the supercells are exactly the coloured cells.

Solution:

5346

8643

1166

1258

1056

2012

8000

9635

9905

Question 2.
Fill the table below such that we get as many supercells as possible. Use numbers between 100 and 1000 without repetitions.

Solution:

999

102

909

110

918

210

928

350

958

Question 3.
Can you fill a supercell table without repeating numbers such that there are no supercells? Why or why not?
Solution:
No, we cannot fill a supercell table without repeating numbers such that there are no supercells because at least one number will always be larger than its adjacent cell unless repetition is allowed.
For example,

3

6

9

12

15

18

21

20 or 22

Here, if the last number is greater than 21, it is a supercell and if less than 21, then 21 itself is a supercell number.

Question 4.
Will the cell having the largest number in a table always be a supercell? Can the cell having the smallest number in a table be a supercell? Why or why not?
Solution:
Yes, the cell having the largest number in a table will always be a supercell, because largest number will always be greater than its neighbouring numbers.
But, the cell having smallest number in a table can never be a supercell, because it will always be smaller than its neighbouring numbers.

Question 5.
Fill a table such that the cell having the second largest number is not a supercell but the second smallest number is a supercell. Is it possible?
Solution:
Yes, it is possible.

245

147

368

313

696

758

532

590

485

In this table, the second largest number is 696, it is not a supercell number and second smallest number is 245, which is a supercell number.

Question 6.
What is the sum of the smallest and largest 5-digit palindrome? What is their difference?
Solution:
Smallest 5-digit palindrome = 10001
Largest 5-digit palindrome = 99999
Sum = 10001 + 99999 = 110000
Difference = 99999 – 10001 = 89998

Question 7.
Write an example for each of the below scenarios whenever possible.

Could you find examples for all the cases? If not, think and discuss what could be the reason. Make other such questions and challenge your classmates.
Solution:
(i) 45000 + 50000 = 95000 > 90250
(ii) 99999 + 900 = 100899
(iii) 4-digit 4- 4-digit to give a 6-digit sum. Since, the maximum sum for two 4-digit numbers (9999 + 9999) is 19998. Thus, it is not possible to get a 6 -digit number by adding two 4-digit numbers.
(iv) 50000 + 61000 = 111000
(v) The minimum sum of two 5-digit numbers (10000 + 10000) is 20000, which is greater than 18500. Therefore, it is not possible to get a sum of 18500 with two 5-digit numbers.
(vi) 70000 – 15000 = 55000 < 56503
(vii) 10000 – 999 = 9001
(viii) 12000 – 8000 = 4000
(ix) 20000 – 19500 = 500
(x) The difference between the largest 5-digit number and smallest 5-digit number (99999 – 10000) is 89999 which is less than 91500.
Therefore, it not possible to get the difference of 91500 between two 5-digit numbers.

Question 8.
There is only one supercell (number greater than all its neighbours) in this grid. If you exchange two digits of one of the numbers, there will be 4 supercells.

Figure out which digits to swap.
Solution:
Swap the digits 6 and 1 in the number 62,871. The number becomes 12,876 which will be the smallest among all the numbers.

16,200	39,344	29,765
23,609	12,876	45,306
19,381	50,319	38,408

Now, there are four supercells.

Question 9.
We are the group of 5-digit numbers between 35000 and 75000 such that all of our digits are odd. Who is the largest number in our group? Who is the smallest number in our group? Who among us is the closest to 50,000?
Solution:
The possible odd digits are: 1, 3, 5, 7 and 9.
Largest odd number is 73,999. (between 35000 and 75000)
Smallest odd number is 35,111. (between 35000 and 75000)
Closest odd number to 50,000 is 51,111.

Question 10.
Recall the sequence of powers of 2 from chapter 1. Why is the Collatz conjecture correct for all the starting numbers in this sequence?
Solution:
Powers of 2 → 2, 4, 8, 16,
For 2 (even) → 2 divide by 2 = 1
For 4 (even) → 4 divide by 2 = 2 (even) → 2 divide by 2 = 1
For 8 (even) → 8 divide by 2 = 4 (even) → 4 divide by 2 = 2 (even) → 2 divide by 2 = 1
Thus, for starting numbers that are powers of 2, the Collatz conjecture holds true because the sequence of operations simply involves a series of divisions by 2, which eventually leads to 1.

InText Questions

[Instruction for Q1 to Q7]
Some Children in a park are standing in a line. Each one says a number.

• A child says ‘1’ if there is only one taller child standing next to them.
• A child says ‘2’ if both the children standing next to them are taller.
• A child says ‘O’, if neither of the children standing next to them are taller.

That is each person says the number of taller neighbours they have.

Question 1.
Can the children rearrange themselves so that the children standing at the ends say ‘2’?
Solution:
No, the children cannot line up in a way that the ones at the ends say ‘2′ because a child only says ‘2’ when both of their neighbours are taller and children at the ends have only one neighbour, not two.

Question 2.
Can we arrange the children in a line so that all would say only 0s?
Solution:
No, it’s not possible to arrange the children in a line so that they all say only ‘0’, because a child says ‘0’ only when neither of their neighbours is taller. This would only happen if all the children are the same height.

Question 3.
Can two children standing next to each other say the same number?
Solution:
Yes, two children standing next to each other can say the same number i.e. 1 because there can be one taller and one smaller child standing next to them.

Question 4.
There are 5 children in a group, all of different heights. Can they stand such that four of them say 1 and the last one says ‘0’? Why or why not?
Solution:
Yes, because a child can say ‘0’ only if neither of the children standing next to them are taller. Arranging them in increasing or decreasing order of heights will eventually put tallest child at one of the end. So it is obvious that last child will say ‘0’.

Question 5.
For this group of 5 children, is the sequence 1, 1, 1, 1, 1 possible?
Solution:
No, the sequence 1, 1, 1, 1, 1 is not possible because there are 5 children and first four will say 1 only if they are arranged in increasing order of height. So, the last child will be tallest and will say 0 (not 1).

Question 6.
Is the sequence 0, 1,2, 1,0 possible? Why or why not?
Solution:
Yes, the sequence 0, 1, 2, 1, 0 is possible. It represents a situation where the middle child is the smallest (has two taller neighbours). The two children at the end are taller than the second last child from both the ends.

Question 7.
How would you rearrange the five children so that the maximum number of children say ‘2’?
Solution:
We can rearrange the five children so that the maximum number of children say ‘2’ as 0, 2, 0, 2, 0.
Thus, maximum number of children who can say ‘2’ is 2 . ,

Question 8.
Complete Table 2 with 5-digit numbers whose digits are ‘1’, ‘0’, ‘6’, ‘3’, and ‘9’ in some order. Only a coloured cell should have a number greater than all its neighbours.

Once you have filled the table above, put commas appropriately after the thousands digit.
The biggest number in the table is _______.
The smallest even number in the table is ______.
The smallest number greater than 50,000 in the table is _______.
Solution:

96,310	96,301	36,109	36,190′
93,610.	13,609	60,319	19,306
93,106	10,639	60,193	30,196
10,369	10,963	10,396	31,906

The biggest number in the table is 96,310.
The smallest even number in the table is 10,396.
The smallest number greater than 50,000 in the table is 60,193.

Question 9.
Among the numbers 1-100, how many times will the digit 7’ occur? Among the numbers 1-1000, how many times will the digit ‘7’ occur?
Solution:
Among the numbers 1-100, we can divide the range into two parts: When ‘7’ appears in the ones place and when ‘7’ appears in the tens place.

In the ones place: The numbers that have ‘7’ in the ones place are: 7, 17, 27, 37, 47, 57, 67, 77, 87, and 97. So, there are 10 numbers where ‘7’ appears in the ones place.

In the tens place: The numbers 70 to 79 contain the digit ‘7’ in the tens place. So, there are 10 numbers where ‘7’ appears in the tens place.

Total occurrences of ‘7’ between 1 and 100:
Ones place: 10 times and Tens place: 10 times
Total = 10 + 10 = 20
So, digit ‘7’ appears 20 times among the numbers 1-100.

Among the numbers 1-1000, we can divide the range into three parts:
hundreds place, tens place and ones place
In the hundreds place: The numbers 700 to 799 contain the digit ‘7’ in the hundreds place. There are 100 numbers front 700 to 799.

In the tens place: We already know that in each set of 100 numbers (i.e. 0-99, 100-199, …, 900-999),
‘7’ will appear in the tens place 10 times (just like in the 1-100 range). Since we have 10 sets of 100 numbers, ‘7’ will appear in the tens place, 10 × 10 = 100 times.

In the ones place: Similarly, for each set of 100 numbers (i.e. 0-99, 100-199, …. 900-999), 7’ will appear in the ones place 10 times (like in the 1-100 range). So, across 10 sets of 100 numbers, 7’ will appear in the ones place 10 × 10 = 100 times.

Total occurrences of 7’ between 1 and 1000:
Hundreds place: 100 times, Tens place: 100 times, Ones place: 100 times
Total = 100 + 100 + 100 = 300
So, digit ‘7’ appears 300 times among the numbers 1-1000.

Question 10.
Write all possible 3-digit palindromes using 1, 2, 3.
Solution:
All the possible 3-digit palindromes using the digits ‘1’, ‘2’, and ‘3’ are: 111, 121, 131, 212, 222, 232, 313, 323 and 333.
There are 9 possible palindromes using the digits ‘1’, ‘2’, and ‘3’.

Question 11.
Puzzle time:

I am a 5-digit palindrome.
I am an odd number.
My ‘t’ digit is double of my ‘M’ digit.
My ‘h’ digit is double of my ‘t’ digit.
Who am I? _______
Solution:
Since palindrome is an odd number, u digit would be 1, 3, 5, 7 or 9 and tth digit would be same as u digit.
Since double of 5, 7 and 9 is not a digit, u digit would be either 1 or 3.
Double of 1 is 2 and double of 3 is 6. So, t digit would be either 2 or 6.
Since double of 6 is not a digit, u digit is 1 and t digit is 2.
Then, h digit would be 4.
So, the required 5-digit odd palindrome number is 12421.
In words: Twelve thousand four hundred twenty one

Number Play Class 6 Extra Questions

Number Play Class 6 Very Short Question Answer

Question 1.
What is the largest 4-digit number with non-zero digits such that its digits add up to 14?
Solution:
The largest 4-digit number must have largest digit ‘9’ at thousands place and the remaining digits in descending order such that the sum of all digits is 14.
Hence, the required number is 9311.

Question 2.
Write all 2-digit palindromic numbers.
Solution:
The 2-digit palindromic numbers are: I 1, 22, 33, 44, 55, 66, 77, 88 and 99.

Question 3.
Write all possible 3-digit palindromes using the digits 7, 8, 9.
Solution:
Using digits 7, 8 and 9, we get the following palindromes:
777, 787, 797, 878, 888, 898, 979, 989, 999

Question 4.
Write two 4-digit numbers such that their difference is a 2-digit number.
Solution:
Let us take 5500 and 5460 as two 4-digit numbers.
Clearly, 5500 – 5460 = 40 which is a 2-digit number.

Question 5.
Write two 5-digit numbers whose sum is 32500.
Solution:
Let us take 12500 and 20000 as two 5-digit numbers.
Clearly, 12500 + 20000 = 32500

Question 6.
Write two 4-digit numbers whose difference is 4680.
Solution:
Let us take 7500 and 2820 as two 4-digit numbers.
Clearly, 7500 – 2820 = 4680

Question 7.
Write the smallest and largest 4-digit palindromes. Find their sum and difference.
Solution:
Smallest 4-digit palindrome = 1001 and largest 4-digit palindrome = 9999
Required sum = 1001 + 9999 = 11000
Required difference = 9999 – 1001 = 8998

Question 8.
Write the second smallest and second largest 5-digit palindromes. Find their sum.
Solution:
Second smallest 5-digit palindrome = 10101
Second largest 5-digit palindrome = 99899
Required sum = 10101 + 99899 = 110000

Question 9.
Write a 6-digit number and a 4-digit number such that their difference is a 5-digit number.
Solution:
Let us take 103200 as a 6-digit number and 8200 as a 4-digit number.
Clearly, 103200 – 8200 = 95000 which is a 5-digit number.

Question 10.
Write a 6-digit number and a 5-digit number such that their difference is a 5-digit number.
Solution:
Let us take 120500 as a 6-digit number and 40500 as a 5-digit number.
Clearly, 120500 – 40500 = 80000, which is a 5-digit number.

Question 11.
Rearrange the digits of 48900125 to get the largest 8-digit number?
Solution:
In order to get the largest 8-digit number using the digits of 48900125, digits should be in descending order on moving from left to right.
Therefore, the required largest number is 98542100.

Question 12.
What is the smallest number whose digit sum is 16?
Solution:
There is no single digit number to give digit sum equal to 16.
The 2-digit numbers whose digits add up to 16 are: 79, 88 and 97.
We see that 79 is the smallest number among these numbers.
Therefore, 79 is the smallest number whose digit sum is 16.

Question 13.
Rearrange the digits of 4500731 to get the smallest 7-digit number?
Solution:
In order to get the smallest 7-digit number using the digits of 4500731, the first digit from left to right should be the smallest digit except 0. So, in this case, it is 1. Then, on moving from left to right further, digits should be in ascending order.
Therefore, the required smallest number is 1003457.

Question 14.
What is the largest 4-digit number whose digits add up to 15?
Solution:
The largest 4-digit number should have the largest digit at thousands place. Since 9 is the largest digit, it takes thousands place. Then, 6 takes the hundreds place because 9 + 6 = 15. As the sum is exhausted by thousands and hundreds place digits, rest places will take digit ‘0’.
Therefore, the required largest 4-digit number is 9600.

Number Play Class 6 Short Question Answer

Question 1.
Fill the given table such that the cell having the 2nd largest number is not a supercell but the cell having 2nd smallest number is a supercell. Use numbers between 10 and 100.

Solution:
To make sure the cell with the second largest number is not a supercell, it has to be next to the cell with the largest number. This is because the cell with the largest number is always a supercell, and two supercells cannot be next to each other.

To make the cell with the second smallest number a supercell, it should be one of the extreme end cells and the smallest number should be in the adjacent cell.

53

88

72

64

22

32

Here, 72 is the second largest number and 32 is the second smallest number
Activity: Think of another numbers by yourself.

Question 2.
Mark the supercells in the table below.

6828	670	9435	2180
3780	3708	7308	9225
8000	5583	52	5001

Solution:
6828 is a supercell because it is larger than its neighbours 670 and 3780.
9435 is a supercell because it is larger than its neighbours 670, 2180 and 7308.
9225 is a supercell because ii is larger than its neighbours 2180, 7308 and 5001.
8000 is a supercell because it is larger than its neighbours 3780 and 5583.

6828	670	9435	2180
3780	3708	7308	9225
8000	5583	52	5001

Question 3.
Fill in the table using only 3-digit numbers, making sure that the supercells line up exactly with the coloured cells.

Solution
Since 2^nd cell is a supercell, it must contain a number greater than 576. Let it he 832.

Since 3^rd and 5^th cells are not supercells and 4^th cell is a supercell, the numbers in 3^rd and 5^th cells should be smaller than the number in 4^th cell i.e., 188. Let them he 112 and 176 respectively.

576

832

112

188

176

912

The last cell is a supercell, and its neighbouring cell contains 912. So, the number in the last cell should be greater than 912. Let it be 950.

Since 7^th cell is not a supercell, it must contain number smaller than 912. Let it be 492.

Also, 6^th cell is not a supercell, it must contain a number smaller than the number present in 7^th cell, i.e. 492. Let it be 350.

Now, the complete table is as follows:

576

832

112

188

176

350

492

912

950

Activity: Think of another numbers by yourself.

Question 4.
Fill in the table using only 3-digit numbers, making sure that the supercells line up exactly with the coloured cells.

Solution:
Since 2^nd cell is a supercell, it must contain a number greater than 576. Let it be 832. Then, number in 1^st cell should be less than 432. Let it be 100.

100

432

120

312

731

512

Since 4^th cell is not supercell, it may Contain any 3-digit number less than 312. Let it be 230.

Also. 7^th cell is not supercell and second last cell is a supercell. it must contain number smaller than 731 and 512. I et it be 219.

The second last cell is a supercell, and it contains 512. So, the number in the last cell should be smaller than 512. Let it he 194.

Now, the complete table is as follows:

100

432

120

230

312

731

219

512

494

Question 5.
Calculate the digit sums of 2-digit numbers whose digits are consecutive. Do you observe a pattern?
Solution:
2-digit numbers having consecutive digits are: 12, 23, 34, 45, 56, 67, 78 and 89 .
The sums of digits of these 8 numbers arc:
1 + 2 = 3
2 + 3 = 5
3 + 4 = 7
4 + 5 = 9
5 + 6 = 11
6 + 7 = 13
7 + 8 = 15
8 + 9 = 17
We find that each sum is 2 more than the preceding sum.

Question 6.
What is the largest 5-digit number with non-zero digits whose digits add up to 17?
Solution:
Non-zero digits are 1, 2, 3, 4, 5, 6, 7, 8, 9.
The largest 5-digit number must have largest digit ‘9’ at ten thousands place (tth) and the remaining digits in descending order such that the sum of all digits is 17.

tth

th

h

t

u

Question 7.
Calculate the digit sums of 4-digit numbers whose digits are consecutive. Do you observe any pattern?
Solution:
The 4-digit numbers having consecutive digits are 1234,2345, 3456, 4567, 5678 and 6789.
The sum of digits of these 6 numbers are 10, 14, 18, 22, 26 and 30 respectively.
We find that each sum is 4 more than the preceding sum.

Question 8.
What is the smallest 5-digit number whose digits add up to 18?
Solution:
In order to write smallest 5-digit number, ten thousands place must be occupied by digit ‘1’. Then, sum of the rest of the digits should be 18 – 1 = 17.

The smallest 5-digit number should have the largest digit at units place. Since 9 is the largest digit, it takes units place. Then, 8 takes the tens place because 9 + 8 = 17. As the sum is exhausted by ten thousands, tens and units place digits, rest places will take digit ‘0’.

Therefore, the required smallest 5-digit number is 10089.

Question 9.
Convert the following times from 12-hour format to 24-hour format.
(i) 02:00 PM
(ii) 10:00 AM
(iii) 07:30 PM
(iv) 12:20 AM
(v) 12:15 PM
Solution:

	12-hour Format	24-hour Format
(i)	02:00 PM	14:00 hours
(ii)	10:00 AM	10:00 hours
(iii)	07:30 PM	19:30 hours
(iv)	12:20 AM	00:20 hours
(v)	12:15 PM	12:15 hours

Question 10.
What can be the previous number in the Collatz sequence to 40?
Solution:
Let x be the required number.
If x is an even number, then
\(\frac{x}{2}\) = 40 ⇒ x = 2 × 40 ⇒ x = 80
If x is an odd number, then
3x + 1 = 40 ⇒ 3x = 40 – 1
⇒ 3x = 39 ⇒ x = \(\frac{39}{3}\) = 13
Thus, either 80 or 13 can be the number preceding 40 in a Collatz sequence.

Question 11.
Observe the following numbers written in a pattern. Find their sum.

Solution:
We find that there are 3 rows each consisting of 5 boxes written with 30 and 2 rows each consisting of 6 boxes written with 70.
∴ Total sum = 3(5 × 30) + 2(6 × 70)
= 3 × 150 + 2 × 420
= 450 + 840 = 1290

Question 12.
Apply Kaprekar’s routine on the number 3562 to get Kaprekar constant.
Solution:
The number is 3562.
Round 1: A = Largest number using digits of the number 3562 = 6532
B = Smallest number using digits of the number 3562 = 2356
C = A – B = 6532 – 2356 = 4176

Round 2: A = Largest number using digits of the number 4176 = 7641
B = Smallest number using digits of the number 4176 = 1467
C = A – B = 7641 – 1467 = 6174

Question 13.
Apply Kaprekar’s routine on the number 8825 to get Kaprekar constant.
Solution:
The number is 8825.
Round 1:
A = Largest number using digits of the number 8825 = 8852
B = Smallest number using digits of the number 8825 = 2588
C = A – B = 8852 – 2588 = 6264

Round 2:
A = Largest number using digits of the number 6264 = 6642
R = Smallest number using digits of the number 6264 = 2466
C = A – B = 6642 – 2466 = 4176

Round 3:
A = Largest number using digits of the number 4176 = 7641
B = Smallest number using digits of the number 4176 = 1467
C = A – B = 7641 – 1467 = 6174

Question 14.
Convert the following times from 24-hour format to 12-hour format.
(i) 18:00 hours
(ii) 06:00 hours
(iii) 12:30 hours
(iv) 00:45 hours
(v) 23:15 hours
Solution:

	24-hour Format	12-hour Format
(i)	18:00 hours	06:00 PM
(ii)	06:00 hours	06:00 AM
(iii)	12:30 hours	12:30 PM
(iv)	00:45 hours	12:45 AM
(v)	23:15 hours	11:15 PM

Question 15.
Count all the suns in the following pattern.

Solution:
There are 26 boxes with 1 sun each and 26 boxes with 4 suns each.
∴ Total number of suns = (26 × 1) + (26 × 4)
= 26 + 104 = 130

Question 16.
Find the sum of the numbers in the number pattern shown in the figure.

Solution:
We note that the number 75 occurs 20 times around the outer edge, the number 40 appears 12 times along the second layer, the number 50 is present 4 times on the edge of the inner layer, and the number 1500 is located at the centre.
∴ Required sum
= (20 × 75) + (12 × 40) + (4 × 50) + 1500
= 1500 + 480 + 200 + 1500 = 3680

Question 17.
Consider the numbers in the following boxes:

Using operations addition/subtraction on these numbers, we can get
13100 = 32000 – 20000 + 700 + 200 + 200
Similarly, obtain the following numbers by using the numbers in the boxes.
(i) 25800
(ii) 28000
(iii) 7500
(iv) 14400
(v) 53000
Solution:
(i) 25800 = 20000 + 6000 – 200
(ii) 28000 = 32000 – 3500 – 700 + 200
(iii) 7500 = 3500 + 3500 + 700 – 200
(iv) 14400 = 20000 – 6000 + 200 + 200
(v) 53000 = 32000 + 20000 – 6000 + 3500 + 3500

Question 18.
Consider the numbers in the following boxes:

Using operations addition/subtraction on these numbers, we can get
23300 = 16000 + 16000 – 9000 + 300
Similarly, obtain the following numbers by using the numbers in the boxes.
(i) 60200
(ii) 2500
(iii) 7700
(iv) 64600
(v) 17300
Solution:
(i) 60200 = 45000 4 16000-800
(ii) 2500 = 9000 + 9000 – 16000 + 800 – 300
(iii) 7700 = 9000 – 800 – 800 + 300
(iv) 64600 = 45000 + 9000 + 9000 + 800 + 800
(v) 17300 = 16000 + 800 + 800 – 300

Number Play Class 6 Long Question Answer

Question 1.
Colour or mark the supercells in the given table:

6235

970

8145

4780

4708

7084

9000

160

Solution:
Clearly, 6235 is greater than the number 970 in the neighbouring cell.
Hence, the cell containing 6235 is the supercell.
970 is smaller than 6235 and 8145, so the cell containing 970 is not a supercell.
As 8145 is greater than the numbers 970 and 4780 in the neighbouring cells, the cell containing 8145 is a supercell.
As 4780 is smaller than 8145, the cell containing 4780 is not a supercell.
As 4708 is smaller than 4780 and 7084, the cell containing 4708 is not a supercell.
As 7084 is smaller than 9000, the cell containing 7084 is not a supercell.
As 9000 is greater than the numbers 7084 and 160 in the neighbouring cells, the cell containing 9000 is a supercell.
As 160 is smaller than 9000, the cell containing 160 is not a supercell.

6235

970

8145

4780

4708

7084

9000

160

Question 2.
Mark supercells in the following grid:

3462	2198	6757	5678
1001	5982	4723	2345
2723	5600	7210	2316
3298	1465	3467	4621

Solution:
The cell containing 3462 is a supercell as the numbers 2198 and 1001 in neighbouring cells are smaller than 3462.
The cell containing 6757 is a supercell as the numbers 2198, 4723 and 5678 in neighbouring cells are smaller than 6757.
The cell containing 5982 is a supercell as the numbers 2198, 4723, 5600 and 1001 in neighbouring cells are smaller than 5982.
The cell containing 7210 is a supercell as the numbers 4723, 2316, 3467 and 5600 in neighbouring cells are smaller than 7210.
The cell containing 3298 is a supercell as the numbers 2723 and 1465 in neighbouring cells are smaller than 3298.
The cell containing 4621 is a supercell as the numbers 2316 and 3467 in neighbouring cells are smaller than 4621.

3462	2198	6757	5678
1001	5982	4723	2345
2723	5600	7210	2316
3298	1465	3467	4621

Question 3.
Identify the numbers marked on the number lines below, and label the remaining positions.

Put a circle around the smallest number and a box around the largest number in each of the sequences above.
Solution:
(i) There are 3 sub-divisions between 729 and 732.
Therefore, each sub-division represents \(\frac{732-729}{3}\) = \(\frac{3}{3}\) = 1 number.

(ii) There are 7 sub-divisions between 6236 and 6250.
Therefore, each sub-division represents \(\frac{6250-6250}{7}\) = \(\frac{14}{7}\) = 2 numbers.

(iii) There are 6 sub-divisions between 12100 and 12160.
Therefore, each sub-division represents \(\frac{12100-12160}{6}\) = \(\frac{60}{6}\) = 10 numbers.

(iv) There is 1 sub-division between 53312 and 54312.
Therefore, each sub-division represents \(\frac{53312-54312}{1}\) = 1000 numbers

Question 4.
Among the numbers 100 – 1000, how many times will the digit ‘3’ occur?
Solution:
There are 901 numbers from 100 to 1000.

When digit ‘3’ is at hundreds place:
Numbers with ‘3’ in the hundreds place look like: 300 – 399
So, total count of digit ‘3’ at hundreds place = 100

When digit ‘3’ is at tens place:
To count how often ‘3’ appears in the tens place, fix the hundreds and units digit, and loop through all possibilities:
There are:

• 9 choices for hundreds digit (1 to 9)
• 1 choice for tens digit (3)
• 10 choices for units digit (0 to 9)

So, total count of digit ‘3’ at tens place = 9 × 1 × 10 = 90

When digit ‘3’ is at units place:
To count how often ‘3’ appears in the units place, fix the hundreds and tens digit, and loop through all possibilities:
There are:

• 9 choices for hundreds digit (1 to 9)
• 10 choices for tens digit (0 to 9)
• 1 choice for units digit (3)

So, total count of digit ‘3’ at units place = 9 × 1 × 10 = 90
Total count of digit ‘3’ = 100 + 90 + 90 = 280
Thus, digit ‘3’ occurs 280 times.

Question 5.
How big a number can you form having the digit sum 16? Can you make an even bigger number?
Solution:
2-digit numbers having the digit sums 16 are: 79, 88 and 97.
3-digit numbers having the digit sums 16 are: 169, 178, 187, 196, 259, 268, 277, 286, 295, 349, 358, 367, 376, 385, 394, 439, 448, 457, 466, 475, 484, 493, 529, 538, 547, 556, 565, 574, 583, 592, 619, 628, 637, 646, 655, 664, 673, 682, 691, 709, 718, 727, 736, 745, 754, 763, 772, 781, 790, 808, 817, 826, 835, 844, 853, 862, 871, 880, 907, 916, 925, 934, 943, 952, 961 and 970.
To form a 4-digit number having the digit sum 16, put one 0 in any of the above 3-digit numbers anywhere after first digit from left or two 0 in any of the above 2-digit numbers anywhere after first digit from left.

Similarly, we can form w-digit numbers by putting an appropriate number of 0 in any of the above numbers.

Thus, there will be infinitely many numbers with digit sum 16.

Yes, we can make even much bigger numbers.

Question 6.
Solve the puzzle:
My father’s salary is a 6-digit palindrome.
It is an even number.
Its tens digit is double of units digit and hundreds digit is double of tens digit.
What is my father’s salary?
Solution:
Units digit of an even number is 0 or 2 or 4 or 6 or 8. Since it is a palindrome, 0 cannot be the units digit. It is given that tens digit is double of units digit. So, 6 and 8 cannot be the units digit as 6 × 2 = 12 and 8 × 2 = 16 but 12 and 16 are not digits. If 4 is the units digit, then 8 would be the tens digit. But it is given that hundreds digit is double of tens digit, then hundreds digit would be 16, which is not possible.

So, units digit must be 2 only. Then, tens digit would be 4 and hundreds digit would be 8.

2

4

8

8

4

2

Thus, the palindromic number is 248842.
Hence, father’s salary is 248842.

Question 7.
List all 3-digit palindromic numbers and highlight those whose digit sum is greater than 15.
Solution:
3-digit palindromic numbers are given in the below table:

101	111	121	131	141	151	161	171	181	191
202	212	222	232	242	252	262	272	282	292
303	313	323	333	343	353	363	373	383	393
404	414	424	434	444	454	464	474	484	494
505	515	525	535	545	555	565	575	585	595
606	616	626	636	646	656	666	676	686	696
707	717	727	737	747	757	767	777	787	797
808	818	828	838	848	858	868	878	888	898
909	919	929	939	949	959	969	979	989	999

Highlighted (bold) palindromes have their digit sum greater than 15.

Question 8.
Write a 5-digit number, a 4-digit number and a 3-digit number such that their sum is 24680.
Solution:
Let us take 20000 as a 5-digit number. Then,
20000 + Sum of a 4-digit number and a 3-digit number = 24680
⇒ Sum of a 4-digit number and a 3-digit number
= 24680 – 20000 = 4680
Let us take 4500 as a 4-digit number. Then,
4500 + a 3-digit number = 4680
⇒ 3-digit number = 4680 – 4500 = 180
Hence, 20000 is a 5-digit number, 4500 is a 4-digit number and 180 is a 3-digit number such that their sum is 24680.
There can also be other numbers like
19000 + 5100 + 580 = 24680,
17500 + 7000 4 180 = 24680 etc.
Think of another numbers by yourself.

Question 9.
State whether the following statements are True/ False.
(i) The sum of two 5-digit numbers is always a 5-digit number.
(ii) The sum of a 5-digit number and a 3-digit number is always a 5-digit number.
(iii) 4’be sum of a 5-digit number and a 3-digit number is always a 6-digit number.
(iv) The difference of two 4-digit numbers is always a 4-digit number.
(ii) The difference between a 5-digit number and a 3-digit number is always a 4-digit number.
Solution:
(i) False; 26000 + 42000 = 68000 → a 5-digit number
76000 + 84000 = 160000 + not → a 5-digit number
Thus, the sum of two 5-digit numbers may or may not be a 5-digit number.

(ii) False; 10000 + 100 = 10100 → a 5-digit number
99900 + 900 = 100800 → not a 5-digit number
Thus, the sum of a 5-digit number and a 3-digit number may or may not be a 5-digit number.

(iii) False; 10000 + 100 = 10100 → not a 6-digit number
99900 + 900 = 100800 → a 6-digit number
Thus, the sum of a 5-digit number and a 3-digit number may or may not be a 6-digit number.

(iv) False; 8040 – 6080 = 1960 → a 4-digit number
1980 – 1200 = 780 → not a 4-digit number
Thus, the difference of two 4-digit numbers may or may not be a 4-digit number.

(v) False; 10250 – 750 = 9500 → a 4-digit number
15500 – 900 = 14600 → not a 4-digit number
Thus, the difference between a 5-digit number and a 3-digit number may or may not be a 4-digit number.

Question 10.
Apply Kaprekar’s routine on a 3-digit number. What number will start repeating?
Solution:
Let the number be 729.
Round 1:
A = Largest number using digits of the number 729 = 972
B = Smallest number using digits of the number 729 = 279
C = A – B = 972 – 279 = 693

Round 2:
A = Largest number using digits of the number 693 = 963
B = Smallest number using digits of the number 693 = 369
C = A – B = 963 – 369 = 594

Round 3:
A = Largest number using digits of the number 594 = 954
B = Smallest number using digits of the number 594 = 459
C = A – B = 954 – 459 = 495

Round 4:
A = Largest number using digits of the number 495 = 954
B = Smallest number using digits of the number 495 = 459
C = A – B = 954 – 459 = 495
Clearly, 495 repeats after 3 iterations.

Question 11.
How many rounds does the number 7443 take to reach the Kaprekar constant.
Solution:
The number is 7443.
Round 1:
A = Largest number using digits of the number 7443 = 7443
B = Smallest number using digits of the number 7443 = 3447
C = A – B = 7443 – 3447 = 3996

Round 2:
A = Largest number using digits of the number 3996 = 9963
B = Smallest number using digits of the number 3996 = 3699
C = A – B = 9963 – 3699 = 6264

Round 3:
A = Largest number using digits of the number 6264 = 6642
B = Smallest number using digits of the number 6264 = 2466
C = A – B = 6642 – 2466 = 4176

Round 4:
A = Largest number using digits of the number 4176 = 7641
B = Smallest number using digits of the number 4176 = 1467
C = A – B = 7641 – 1467 = 6174, which is the Kaprekar constant
Thus, we reach at the Kaprekar constant in four rounds.

Question 12.
Make the Collatz sequence by starting with number 24.
Solution:
The rule followed by a Collatz sequence is: Start with any number; if the number is even, take half of it; if the number is odd, multiply it by 3 and add 1; repeat till 1 is obtained.
First term: 24
Second term: \(\frac{24}{2}\) = 12 [As 24 is an even number.]
Third term: \(\frac{12}{2}\) = 6 [As 12 is an even number.] 0
Fourth term: \(\frac{6}{2}\) = 3 [As 6 is an even number.]
Fifth term: 3 × 3 + 1 = 10 [As 3 is an odd number.]
Sixth term: \(\frac{10}{2}\) = 5 [As 10 is an even number.]
Seventh term: 3 × 5 + 1 = 16 [As 5 is an odd number.]
Eighth term: \(\frac{16}{2}\) = 8 [As 16 is an even number.]
Ninth term: \(\frac{8}{2}\) = 4 [As 8 is an even number.]
Tenth term: \(\frac{4}{2}\) = 2 [As 4 is an even number.]
Eleventh term: \(\frac{2}{2}\) = 1 [As 2 is an even number.]
Hence, the Collatz sequence starting with 24 is:
24, 12, 6, 3, 10, 5, 16, 8, 4, 2, 1

Question 13.
Make the Collatz sequence by starting with number 17.
Solution:
The rule followed by a Collatz sequence is: Start with any number; if the number is even, take half of it; if the number is odd, multiply it by 3 and add 1; repeat till I is obtained.
1^st term : 17
2^nd term : 3 × 17 + 1 = 52 [As 17 is an odd number]
3^rd term : \(\frac{52}{2}\) = 26 [As 52 is an even number]
4^th term : \(\frac{26}{2}\) = 13 [As 26 is an even number]
5^th term : 3 × 13 + 1 = 40 [As 13 is an odd number]
6^th term : \(\frac{40}{2}\) = 20 [As 40 is an even number]
7^th term : \(\frac{20}{2}\) = 10 [As 20 is an even number]
8^th term : \(\frac{10}{2}\) = 5 [As 10 is an even number]
9^th term : 3 × 5 + 1 = 16 [As 5 is an odd number]
10^th term : \(\frac{16}{2}\) = 8 [As 16 is an even number]
11^th term : \(\frac{8}{2}\) = 4 [As 8 is an even number]
12^th term : \(\frac{4}{2}\) = 2 [As 4 is an even number]
13^th term : \(\frac{2}{2}\) = 1 [As 2 is an even number]
Hence, the Collatz sequency starting with 17 is: 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.

Question 14.
Find the Collatz sequence by starting with number 11.
Solution:
The rule followed by a Collatz sequence is: Start with any number; if the number is even, take half of it; if the number is odd, multiply it by 3 and add 1; repeat till 1 is obtained.
1^st term : 11
2^nd term : 3 × 11 + 1 = 34 [As 11 is an odd number]
3^rd term : \(\frac{34}{2}\) = 17 [As 34 is an even number]
4^th term : 3 × 17 + 1 = 52 [As 1 7 is an odd number]
5^th term : \(\frac{52}{2}\) = 26 [As 52 is an even number]
6^th term : \(\frac{26}{2}\) = 13 [As 26 is an even number]
7^th term : 3 × 13 + 1 = 40 [As 13 is an odd number]
8^th term : \(\frac{40}{2}\) = 20 [As 40 is an even number]
9^th term : \(\frac{20}{2}\) = 10 [As 20 is an even number]
10^th term : \(\frac{10}{2}\) = 5 [As 10 is an even number]
11^th term : 3 × 5 + 1 = 16 [As 5 is an odd number]
12^th term : \(\frac{16}{2}\) = 8 [As 16 is an even number]
13^th term : \(\frac{8}{2}\) = 4 [As 8 is an even number]
14^th term : \(\frac{4}{2}\) = 2 [As 4 is an even number]
15^th term : \(\frac{2}{2}\) = 1 [As 2 is an even number]

Question 15.
Make a quick estimate. Take about 30 seconds. Then compare your answer with your friends.
(i) Number of tiles on your classroom floor:
(a) More than 200
(b) Less than 200
(Hint: Count how many tiles there are*in one row and one column.)

(ii) Time taken to write your full name:
(a) More than 10 seconds
(b) Less than 10 seconds

(iii) Estimate the number of books in your school library.
(a) More than 2000
(b) Less than 2000

(iv) Weight of a fully packed school bag:
(a) More than 4 kg
(b) Less than 4 kg

(v) Estimate the number of leaves on a big tree near your school.
(a) Around 1000
(b) Around 10,000
(c) More than 50,000
Solution:
(i) Estimate the number of tiles in one row and one column. Multiply the two to get the total. If it is 20 rows and 15 columns, then 20 × 15 = 300 tiles.
(ii) Say your name aloud and imagine writing each letter. Most students take 5-12 seconds depending on length.
(iii) Think of how many shelves there are and how many books per shelf. For example, 50 shelves × 50 books = 2500 books.
(iv) Lift your bag and guess based on how heavy it feels. Most packed school bags weigh around 3-5 kg.
(v) A large tree has thousands of branches and leaves. It’s very hard to count, but an estimate can be made.

Number Play Class 6 Case Based Questions

Question 1.
During a school trip, Class 6 students noticed a unique design featuring a 9-box row on a wall. Inspired, they decided to fill the boxes with numbers between 300 and 700, using each number only once.
They were excited to experiment with different number arrangements and see what interesting patterns might emerge.

Based on the above information, answer the following:
(ii) Fill the table such that we get maximum number of supercells.
(ii) How many supercells are there in the table.
Solution:
(i) If there are n cells in a row, then maximum number of supercells = \(\left\{\begin{array}{c}
\frac{n}{2}, \text { if } n \text { is even } \\
\frac{n+1}{2}, \text { if } n \text { is odd }
\end{array}\right.\)
Since total number of cells is 9, maximum 9 +1
number of supercell = \(\frac{9+1}{2}\) = 5
We know that two adjacent cells can never be supercells. So, the maximum number of supercells would be obtained when every alternate cell is a supercell.

In order to get the maximum number of supercells, always consider the 1st cell to be the supercell.
One such table with numbers between 300 and 700 is as follows:

699

301

698

302

697

303

696

304

695

(ii) Clearly, there are 5 supercells.

Question 2.
One day, Raghu bought a packet of pencils and found a lottery scratch coupon tucked inside. Curious, he scratched it off and revealed six two-digit numbers. The first number was 34, and to his surprise, the rest of the numbers followed the mysterious pattern of the Collatz sequence.

Based on the above information, answer the following questions:
(i) What number is written in second box?
(ii) What number is written in last box?
(iii) How many boxes are supercells?
Solution:
(i) As the numbers on the lottery ticket follow the Collatz sequence, and the first number is 34, which is even, the number in second box = \(\frac{34}{2}\) = 17

(ii) In the given Collatz sequence
First term = 34
Second term = \(\frac{34}{2}\) = 17 [As 34 is an even number]
Third term = 3 × 17 + 1 = 52 [As 17 is an odd number]
Fourth term = \(\frac{52}{2}\) = 26 [As 52 is an even number]
Fifth term = \(\frac{26}{2}\) = 13 [As 26 is an even number]
Sixth term = 3 × 13 + 1 = 40 [As 13 is an odd number]
Thus, the number in the last i.e. sixth box is 40.
The list of numbers obtained on the lottery ticket is given below.

(iii) The list of numbers obtained on the lottery ticket is given below.

34

17

52

26

13

40

Clearly, 3 boxes are supercells.