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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 use Class 8 Math Solution Odia Medium and Class 8 Maths Chapter 2 Question Answer Odia Medium ଘାତର ଖେଳ to check their answers after solving exercises.

8th Class Maths Chapter 2 Question Answer Odia Medium

Class 8 Maths Chapter 2 Odia Medium

2.1 ଘାତଖେଳର ଅଭିଜ୍ଞତା (Page No. 19)

Question 1.
ଖଣ୍ଡିଏ କାଗଜ ଫର୍ଦ ନିଅ। ଏହାକୁ ଗୋଟିଏ ଥର ଭାଙ୍ଗ। ପୁଣି ଏହାକୁ ବାରମ୍ବାର ଭାଙ୍ଗ। ଏହାକୁ ବାରମ୍ବାର କେତେଥର ଭାଙ୍ଗି ହେବ?
Solution:
କାଗଜର ଆରମ୍ଭ ବିନ୍ଦୁ (୦ ଭାଙ୍ଗ) : କାଗଜର ପ୍ରକୃତ ମୋଟେଇ 0.001 ସେ.ମି.।
ପ୍ରଥମ ଭାଙ୍ଗ : କାଗଜକୁ ପ୍ରଥମ ଭାଙ୍ଗରେ ଏହାର ମୋଟେଇ ଅଧାର ଦୁଇଗୁଣ।
ତେଣୁ ଏହା 0.001 ସେ.ମି. × 2 = 0.002 ସେ.ମି. ।
ଦ୍ଵିତୀୟ ଭାଙ୍ଗ : ଏହାକୁ ପୁଣି ଭାଙ୍ଗିଲେ ଏହା ହେବ = 0.001 ସେ.ମି. × 2 × 2 = (0.001 × 2²) = 0.004 ସେ.ମି.।
ଏହି କ୍ରମରେ ‘n’ ଥର ଭାଙ୍ଗିଲେ = 0,001 × 2ⁿ ହେବ।

Question 2.
ଗୋଟିଏ ଫର୍ଦ୍ଦ କାଗଜକୁ ତୁମ ଇଚ୍ଛାନୁସାରେ ଯେତେଥର ଚାହୁଁଛ ସେତେଥର ଭାଙ୍ଗ କରିପାରିବ। 30 ଭାଙ୍ଗ କରିବା ପରେ ଏହାର ମୋଟେଇ କେତେ ହେବ? ଅନୁମାନ କର।
Solution:
ଆମେ ଜାଣୁ କାଗଜର ମୋଟେଇ = 0.001 ସେ.ମି.।
ଏହାକୁ 30 ଥର ଭାଙ୍ଗିଲେ ଏହାର ମୋଟେଇ = 0.001 × 2³⁰ ସେ.ମି.।
= 0.001 × 1,073,741,824 ସେ.ମି.
= 1,073,741.824 ସେ.ମି.
= 10,737,41824 ମିଟର (1 ମିଟର = 100 ସେ.ମି.)
= 10.74 କି.ମି. (1 କି.ମି. = 1000 ମିଟର)
∴ 30 ଭାଙ୍ଗ କରିବା ପରେ କାଗଜର ମୋଟେଇ 10.74 କି.ମି. ହେବ।

Question 3.
ଗୋଟିଏ ଫର୍ଦ କାଗଜକୁ 46 ଭାଙ୍ଗ କରିବା ପରେ ଏହା କେତେ ମୋଟା ହେବ? ଧରନିଅ କାଗଜର ମୋଟେଇ 0.001 ସେ.ମି. ।
Solution:
ଆମେ ଜାଣୁ କାଗଜର ମୋଟେଇ = 0.001 ସେ.ମି.।
ଏହାକୁ 46 ଥର ଭାଙ୍ଗିଲେ ଏହାର ମୋଟେଇ = 0.001 × 2⁴⁶ ସେ.ମି.। = 703687,442 କି.ମି. ହେବ।

Page No. 20

Question 1.
ନିମ୍ନ ସାରଣୀରେ ପ୍ରତ୍ୟେକ ଭାଙ୍ଗପରେ କାଗଜ ମୋଟେଇର ତାଲିକା ଦିଆଯାଇଛି। ପ୍ରତ୍ୟେକ ଭାଙ୍ଗପରେ ଏହାର ମୋଟେଇ ଦୁଇଗୁଣ ହେଉଛି।

(≈ ଚିହ୍ନଟିକୁ ଆମେ ‘ପ୍ରାୟତଃ ସମାନ’ ବୋଲି ବ୍ୟବହାର କରିଥାଉ ।)
Solution:

2.2 ଘାତାଙ୍କୀୟ ସଂକେତ ଏବଂ ପ୍ରକ୍ରିୟା

Page No. 22

Question 1.
ଏକ କାଗଜ ଫର୍ଦ 10 ଥର ଭାଙ୍ଗିବା ପରେ ତାହାର ମୋଟେଇ ପରିପ୍ରକାଶ କିଭଳି ହେବ? ପ୍ରାରମ୍ଭିକ ମୋଟେଇ v ଅକ୍ଷର ସଂଖ୍ୟାଦ୍ଵାରା ସୂଚିତ କରାଯାଇଛି।
(i) 10v
(ii) 10 + v
(iii) 2 × 10 × v
(iv) 2¹⁶
(v) 2¹⁰ v
(vi) 10² v
Solution:
ପ୍ରତିଥର ଭାଙ୍ଗିବା ପରେ ଏହାର ମୋଟେଇ ଦୁଇଗୁଣ ବୃଦ୍ଧି ହେଲା।
∴ 1 ଥର ଭାଙ୍ଗ, ମୋଟେଇ = 2 × v
2 ଥର ଭାଙ୍ଗ, ମୋଟେଇ = 2² × v
ସେହିପରି 10 ଥର ଭାଙ୍ଗ, ମୋଟେଇ = 2¹⁰ × v
ତେଣୁ ଠିକ୍ ଉତ୍ତରଟି ହେଉଛି (v) 2¹⁰ × v

Question 2.
32400 ସଂଖ୍ୟାକୁ ଏହାର ମୌଳିକ ଗୁଣନୀୟକରେ ପ୍ରକାଶ କର ଏବଂ ମୌଳିକ ଗୁଣନୀୟକଗୁଡ଼ିକୁ ସେମାନଙ୍କ ଘାତାଙ୍କ ରୂପରେ ଚିହ୍ନଟ କର।
Solution:

32400 = 2 × 2 × 2 × 2 × 5 × 5 × 3 × 3 × 3 × 3
ଏହାର ଘାତାଙ୍କୀୟ ପରିପ୍ରକାଶଟି ହେବ = 32400 = 2⁴ × 5² × 3⁴

Question 3.
(-1)⁵ ଟି କ’ଣ? ଏହା ଧନାତ୍ମକ ନା ଋଣାତ୍ମକ? (-1)⁵⁶ ଟି କ’ଣ?
Solution:
(−1)⁵ = -1 (ଏହା ଋଣାତ୍ମକ) [(-1)^{ସଂଖ୍ୟ} ‍= -1 (ଋଣାତ୍ମକ)]
(-1)⁵⁶ = +1 (ଏହୀ ଧନାତ୍ମକ) [(-1)^{ଯୁଗ ସଂଖ୍ୟ} = 1 (ଧନାତ୍ମକ)]

Question 4.
(-2)⁴ = 16 କି? ଯାଞ୍ଚକର।
Solution:
(-2)⁴ = (-2) × (-2) × (-2) × (-2) = 4 × 4 = 16
∴ (-2)⁴ = 16, ଏହା ଠିକ୍।

ନିଜେ କରି ଦେଖ (Page No. 22)

Question 1.
ନିମ୍ନରେ ଦିଆଯାଇଥବା ପରିପ୍ରକାଶକୁ ଘାତାଙ୍କୀୟ ରୂପରେ ଲେଖି :
(i) 6 × 6 × 6 × 6
(ii) y × y
(iii) b × b × b × b
(iv) 5 × 5 × 7 × 7 × 7
(v) 2 × 2 × a × a
(vi) а × а × а × c × c × c × c × d
Solution:
(i) 6 × 6 × 6 × 6 = 6⁴
(ii) y × y = y²
(iii) b × b × b × b = b⁴
(iv) 5 × 5 × 7 × 7 × 7 = 5² × 7³
(v) 2 × 2 × a × a = 2² × a²
(vi) a × a × a × c × c × c × c × d = a³ × c⁴ × d¹

Question 2.
ନିମ୍ନଲିଖତ ପ୍ରତ୍ୟେକ ସଂଖ୍ୟାକୁ ସେମାନଙ୍କର ମୌଳିକ ଗୁଣନୀୟକଗୁଡ଼ିକର ଘାତର ଗୁଣଫଳ ରୂପରେ ପ୍ରକାଶ କର।
(i) 648
(ii) 405
(iii) 540
(iv) 3600
Solution:
(i) 648 ର ମୌଳିକ ଗୁଣନୀୟକ = 2 × 2 × 2 × 3 × 3 × 3 × 3
ଗୁଣନୀୟକଗୁଡ଼ିକର ଘାତର ଗୁଣଫଳ = 2³ × 3⁴
(ii) 405 ର ମୌଳିକ ଗୁଣନୀୟକ = 3 × 3 × 3 × 3 × 5
ଗୁଣନୀୟକଗୁଡ଼ିକର ଘାତର ଗୁଣଫଳ = 3⁴ × 5
(iii) 540 ର ମୌଳିକ ଗୁଣନୀୟକ = 2 × 2 × 3 × 3 × 3 × 5
ଗୁଣନୀୟକଗୁଡ଼ିକର ଘାତର ଗୁଣଫଳ = 2² × 3³ × 5
(iv) 3600 ର ମୌଳିକ ଗୁଣନୀୟକ = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
ଗୁଣନୀୟକଗୁଡ଼ିକର ଘାତର ଗୁଣଫଳ = 2⁴ × 3² × 5²

Question 3.
ନିମ୍ନଲିଖତ ପ୍ରତ୍ୟେକର ସାଂଖ୍ୟକ ମାନ ଲେଖ।
(i) 2 × 10³
(ii) 7² × 2³
(iii) 3 × 4⁴
(iv) (-3)² × (-5)²
(v) 3² × 10⁴
(vi) (-2)⁵ × (-10)⁶
Solution:
(i) 2 × 10³ = 2 × 10 × 10 × 10
= 2 × 1000
= 2000

(ii) 7² × 2³ = 7 × 7 × 2 × 2 × 2
= 49 × 8
= 392

(iii) 3 × 4⁴ = 3 × 4 × 4 × 4 × 4
= 3 × 256
= 768

(iv) (-3)² × (-5)² = -3 × -3 × -5 × -5
= 9 × 25
= 225

(v) 3² × 10⁴ = 3 × 3 × 10 × 10 × 10 × 10
= 9 × 10,000
= 90,000

(vi) (-2)⁵ × (-10)⁶ = −2 × −2 × −2 × -2 × −2 × 10 × 10 × 10 × 10 × 10 × 10
= (-32) × 10,00,000
= -3,20,00,000

Page No. 23

Question 1.
ମୋଟ କେତୋଟି କୋଠରି ଥିଲା?

Solution:
ଦତ୍ତ ଚିତ୍ରରୁ ଆମେ ଦେଖୁଲୁ ଯେ କୋଠରି ସଂଖ୍ୟା ହେଉଛି 34 ।
3 × 3 × 3 × 3 = 243
27 × 3 = 81
81 × 3 = 243

Question 2.
ମୋଟ କେତୋଟି ହୀରା ଥୁଲା? ଉପରୋକ୍ତ ଗୁଣଫଳ ବ୍ୟବହାର କରି କେବଳ ଗୋଟିଏ ଗୁଣନଦ୍ଵାରା ଆମେ ଏହା ଜାଣିପାରିବା କି?
Solution:
ମୋଟ ହୀରାର ସଂଖ୍ୟା = 3 × 3 × 3 × 3 × 3 × 3 × 3 = 3⁷
3⁷ = (3 × 3 × 3 × 3) × (3 × 3 × 3)
3 × 3 × 3 × 3 = 3⁴ = 243
3 × 3 × 3 = 3³
∴ 3⁴ × 3³ = 81 × 27 = 2187

Page No. 24

Question 1.
3⁷ କୁ 3² × 3⁵ ଭାବରେ ମଧ୍ୟ ଲେଖାଯାଇପାରିବ। ତୁମେ ଏହାର କାରଣ ଦର୍ଶାଇ ପାରିବ କି?
Solution:
ଏହାକୁ ସମାନ ଅକ୍ଷର-ସଂଖ୍ୟା ଥ‌ିବା ଘାତଗୁଡ଼ିକର ଗୁଣନ ଭାବରେ ସହଜରେ ବିସ୍ତାରିତ କରି ଏହାକୁ ଲେଖୁ ପାରିବା।

Question 2.
p⁴ × p⁶ ର ଗୁଣଫଳକୁ ଘାତାଙ୍କୀୟ ରୂପରେ ଲେଖ।
Solution:
p⁴ × p⁶ = (p × p × p × p) × (p × p × p × p × p × p) = p¹⁰
ଏହାକୁ ଆମେ ସାଧାରଣ ରୂପରେ ନିମ୍ନମତେ ପ୍ରକାଶ କରିପାରିବା।
n^a × n^b = n^a+b, ଯେଉଁଠାରେ a ଓ b ଗଣନସଂଖ୍ୟା ଅଟନ୍ତି।

Question 3.
ଏହାକୁ ପର୍ଯ୍ୟବେକ୍ଷଣ କରି ନିମ୍ନଲିଖତ ପରିପ୍ରକାଶଗୁଡ଼ିକୁ ଗଣନା କର।
(i) 2⁹
(ii) 5⁷
(iii) 4⁶
Solution:
(i) 2⁹ = 2³ × 2³ × 2³
= 8 × 8 × 8
= 512

(ii) 5⁷ = 5² × 5² × 5² × 5
= 25 × 25 × 25 × 5
= 625 × 125
= 78125

(iii) 4⁶ = 4² × 4² × 4²
= 16 × 16 × 16
= 256 × 16
= 4096

Question 4.
2¹⁰ ମଧ୍ୟ (2⁵)² ସହିତ ସମାନ କି? ଏହାକୁ ଏକ ଗୁଣଫଳ ଭାବରେ ଲେଖ।
Solution:
2¹⁰ = (2 × 2 × 2 × 2 × 2) × (2 × 2 × 2 × 2 × 2)
= (2⁵) × (2⁵)
= (2⁵)², ଏହା ସମାନ।

Question 5.
ନିମ୍ନଲିଖୁତ ପରିପ୍ରକାଶଗୁଡ଼ିକୁ ଅତିକମ୍‌ରେ ଦୁଇଟି ଭିନ୍ନଭିନ୍ନ ଉପାୟରେ ଏକ ଘାତର ଘାତ ଭାବରେ ଲେଖ।
(i) 8⁶
(ii) 7¹⁵
(iii) 9¹⁴
(iv) 5⁸
Solution:

Page No. 25

Question 1.
ଏକ କୁହୁକ ପୋଖରୀ ମଝିରେ ଗୋଟିଏ ଗୋଲାପୀ ପଦ୍ମଫୁଲ ଅଛି। ଏହି ପୋଖରୀରେ ପ୍ରତିଦିନ ପଦ୍ମଫୁଲର ସଂଖ୍ୟା ଦ୍ଵିଗୁଣିତ ହୁଏ। 30 ଦିନ ପରେ ପୋଖରାଟି ପଦ୍ମଫୁଲରେ ସମ୍ପୂର୍ଣ୍ଣଭାବେ ଭର୍ତ୍ତି ହୋଇଯାଏ। କେଉଁଦିନ ପୋଖରୀଟି ଅଧାପୂର୍ଣ୍ଣ ଥିଲା?
Solution:
ଏହି ପୋଖରୀରେ ପ୍ରତିଦିନ ପଦ୍ମଫୁଲର ସଂଖ୍ୟା ଦ୍ଵିଗୁଣିତ ହୁଏ।
30 ଦିନ ପରେ ପୋଖରୀଟି ପଦ୍ମଫୁଲରେ ସମ୍ପୂର୍ଣ୍ଣଭାବେ ଭର୍ତ୍ତି ହୋଇଯାଏ।
30 ଦିନ ପରେ ପୋଖରୀଟି ଅଧାପୂର୍ଣ୍ଣ ଥିଲା।
ପ୍ରଥମ ଦିନ → 1 = 2⁰
ଦ୍ଵିତୀୟ ଦିନ → 2 = 2¹
ତୃତୀୟ ଦିନ → 2²
ଚତୁର୍ଥ ଦିନ → 2³
………………………….
………………………….
୨୯ ଦିନ → 2²⁸
୩୦ ଦିନ → 2²⁹

Question 2.
ଯଦି ପୋଖରୀଟି 30 ତମ ଦିନରେ ପଦ୍ମଫୁଲରେ ସମ୍ପୂର୍ଣ୍ଣଭାବେ ଭର୍ତ୍ତି ହୋଇଯାଏ, ତେବେ 29 ତମ ଦିନରେ ଏହାର କେତେ ଅଂଶ ପଦ୍ମଫୁଲରେ ଭର୍ତ୍ତି ହୋଇଥିଲା?
Solution:
ଯଦି ପୋଖରୀଟି 30 ତମ ଦିନରେ ପଦ୍ମଫୁଲରେ ସମ୍ପୂର୍ଣ୍ଣଭାବେ ଭର୍ତ୍ତି ହୋଇଯାଏ, ତେବେ 29 ତମ ଦିନରେ ଏହାର ଅଧା ଅଂଶ ପଦ୍ମଫୁଲରେ ଭର୍ତ୍ତି ହୋଇଥିଲା।

Question 3.
ପଦ୍ମଫୁଲର ସଂଖ୍ୟା (ଘାତାଙ୍କୀୟ ରୂପରେ) ଲେଖ, ଯେତେବେଳେ ପୋଖରୀଟି–
(i) ସମ୍ପୂର୍ଣ୍ଣ ଭର୍ତ୍ତି ହୋଇଥିଲା।
(ii) ଅଧା ଭର୍ତ୍ତି ହୋଇଥିଲା।
Solution:
(i) 30 ତମ ଦିନରେ ପୋଖରୀଟି ସମ୍ପୂର୍ଣ୍ଣ ଭର୍ତ୍ତି ହୋଇଥିଲା = 2²⁹
(ii) 29 ତମ ଦିନରେ ପୋଖରୀଟି ଅଧା ଭର୍ତ୍ତି ହୋଇଥିଲା = 2²⁸
ଆଉ ଏକ ପୋଖରୀ ଅଛି ଯେଉଁଥରେ ପଦ୍ମଫୁଲର ସଂଖ୍ୟା ପ୍ରତିଦିନ ତିନିଗୁଣ ହୁଏ । ଯେତେବେଳେ ଉଭୟ ପୋଖରୀରେ କୌଣସି ଫୁଲ ନଥୁଲା, ଶିବାନୀ ଦୁଇଗୁଣ ହେଉଥିବା ପୋଖରୀରେ ଗୋଟିଏ ପଦ୍ମଫୁଲ ରଖୁଲେ । 4 ଦିନପରେ ସେ ସେଠାରୁ ସମସ୍ତ ପଦ୍ମଫୁଲ ନେଇ ତିନିଗୁଣ ହେଉଥ‌ିବା ପୋଖରୀରେ ରଖିଲେ ।

Question 4.
4 ଦିନପରେ ତିନିଗୁଣ ହେଉଥ‌ିବା ପୋଖରୀରେ କେତୋଟି ପଦ୍ମଫୁଲ ଥୁବ?
Solution:
ପ୍ରଥମ 4 ଦିନପରେ, ପଦ୍ମଫୁଲର ସଂଖ୍ୟା ହେଉଛି – 1 × 2 × 2 × 2 × 2 = 2⁴
ପରବର୍ତୀ 4 ଦିନପରେ, ପଦ୍ମଫୁଲର ସଂଖ୍ୟା ହେଉଛି – 2⁴ × 3 × 3 × 3 × 3 = 2⁴ × 3⁴

Question 5.
ଯଦି ଶିବାନୀ ଫୁଲଗୁଡ଼ିକୁ ପୋଖରୀରେ ରଖୁବାର କ୍ରମ ବଦଳାଇଥାନ୍ତେ, ତେବେ କେତୋଟି ପଦ୍ମଫୁଲ ରହିଥାନ୍ତା?
Solution:
1 × 3⁴ × 2⁴ = (3 × 3 × 3 × 3) × (2 × 2 × 2 × 2)

Question 6.
ଏହି ଗୁଣଫଳକୁ ଘାତାଙ୍କୀୟ ସଙ୍କେତ mn ଭାବରେ ପ୍ରକାଶ କରାଯାଇପାରିବ କି ? ଯେଉଁଠାରେ m ଏବଂ n ଗଣନ ସଂଖ୍ୟା ଅଟନ୍ତି ।
Solution:
ସଂଖ୍ୟାଗୁଡ଼ିକୁ ପୁନଃଗୋଷ୍ଠୀଭୁକ୍ତ କରାଗଲେ = (3 × 2) × (3 × 2) × (3 × 2) × (3 × 2)
= (3 × 2)⁴
= 6⁴

Question 7.
2⁵ × 5⁵ ର ମୂଲ୍ୟ ନିର୍ଣ୍ଣୟ କର।
Solution:
2⁵ × 5⁵ = (2 × 5)⁵
= 10⁵
= 1,00,000

Question 8.
\(\frac{10^4}{5^4}\) କୁ ସରଳକର ଏବଂ ଏହାକୁ ଘାତାଙ୍କୀୟ ରୂପରେ ଲେଖ।
Solution:
\(\frac{10^4}{5^4}=\left(\frac{10}{5}\right)^4=(2)^4=2^4\) [∵ \(\frac{\mathrm{a}^{\mathrm{m}}}{\mathrm{~b}^{\mathrm{m}}}=\left(\frac{\mathrm{a}}{\mathrm{~b}}\right)^{\mathrm{m}}\)]

Page No. 26

Question 1.
ଉଦାହରଣ : ଇତୁ ପାଖରେ 4ଟି ପୋଷାକ ଓ 3ଟି ଟୋପି ଅଛି। ଇଡୁ କେତେ ପ୍ରକାରରେ ପୋଷାକ ଓ ଟୋପିକୁ ମିଶାଇ ପାରିବେ?

Solution:
ପ୍ରତ୍ୟେକ ଟୋପିପାଇଁ, ସେ 4ଟି ପୋଷାକ ମଧ୍ୟରୁ ଯେକୌଣସି ଗୋଟିଏ ବାଛିପାରିବେ। ତେଣୁ 3ଟି ଟୋପି ପାଇଁ 4 + 4 + 4 = 4 × 3 = 12 ସମାବେଶ ସମ୍ଭବ ଅଟେ। ଆମେ ଏହାକୁ ଅନ୍ୟ ଏକ ଉପାୟରେ ମଧ୍ୟ ଲକ୍ଷ୍ୟ କରିପାରିବା, ପ୍ରତ୍ୟେକ ପୋଷାକ ପାଇଁ ଇତୁ 3ଟି ଟୋପି ମଧ୍ୟରୁ ଯେକୌଣସି ଗୋଟିଏ ବାଛିପାରିବେ, ତେଣୁ 4ଟି ପୋଷାକ ପାଇଁ 3 + 3 + 3 + 3 = 3 × 4 = 12 ସମାବେଶ ସମ୍ଭବ।

Question 2.
ରକି ପାଖରେ 7ଟି ପୋଷାକ, 2ଟି ଟୋପି ଏବଂ 3 ଯୋଡ଼ା ଜୋତା ଅଛି । ରକି କେତେ ପ୍ରକାରରେ ଏଗୁଡ଼ିକୁ ପିନ୍ଧିପାରିବ?
Solution:
7 × 2 × 3 = 42 ପ୍ରକାର।
ରକି 42 ପ୍ରକାରରେ ଏଗୁଡ଼ିକୁ ପିନ୍ଧିପାରିବ।

Question 3.
ଇତୁ ଏବଂ ରକିଙ୍କୁ ଏକ ପୁରୁଣା ଷ୍ଟାମ୍ପ ଏବଂ ମୁଦ୍ରା ଥିବା ବାକ୍ସ ମିଳିଲା, ଯାହାକୁ କି ତାଙ୍କ ଜେଜେ ସଂଗ୍ରହ କରିଥିଲେ । ଏହାର ଚାବି 5 ଅଙ୍କ ବିଶିଷ୍ଟ ପାସୱାର୍ଡ଼ ଦ୍ଵାରା ସୁରକ୍ଷିତ ଥିଲା । ଏହାର ପାସ୍ୱାର୍ଡ଼ କାହାରିକୁ ଜଣା ନଥୁବାରୁ ସେମାନଙ୍କୁ ପ୍ରତ୍ୟେକ ପାସ୍ୱାର୍ଡ଼ଦ୍ୱାରା ଚେଷ୍ଟା କରିବା ବ୍ୟତୀତ ବାକ୍ସକୁ ଖୋଲିବା ପାଇଁ ଅନ୍ୟ କୌଣସି ବିକଳ୍ପ ଉପାୟ ନଥିଲା । ଦୁର୍ଭାଗ୍ୟ ଏହି ଯେ, ସବୁ ସମ୍ଭାବ୍ୟ ସମାବେଶଥର ଚେଷ୍ଟା କରିବା ପରେ ହିଁ ଶେଷ ପାସ୍ୱାର୍ଡ଼ରେ ଲିକ୍ ଖୋଲିଲା ; ସେମାନେ କେତୋଟି ପାସ୍ୱାର୍ଡ଼ ନେଇ ଚେଷ୍ଟା କରିଥିଲେ ?
Solution:
ପ୍ରତ୍ୟେକ ସଂଖ୍ୟା 0 ରୁ ୨ ପର୍ଯ୍ୟନ୍ତ ଯେକୌଣସି ସଂଖ୍ୟା ହୋଇପାରେ,
ପ୍ରତି ଅଙ୍କରେ 10ଟି ବିକଳ୍ପ ସମ୍ଭବ ହେବ ।
ଦୁଇ ଅଙ୍କ ବିଶିଷ୍ଟ ପାସ୍ୱାର୍ଡ଼ ପାଇଁ = 10² = 100 ଟି ବିକଳ୍ପ ଅଛି ।
ପାଞ୍ଚ ଅଙ୍କ ବିଶିଷ୍ଟ ପାସ୍ୱାର୍ଡ଼ ପାଇଁ ମୋଟ ମିଶ୍ରଣ = 10⁵ = 1,00,000 ବିକଳ୍ପ ଅଛି ।

Question 4.
ମନେକର ଆମ ପାଖରେ 2 ଅଙ୍କବିଶିଷ୍ଟ ତାଲା ଅଛି ଏବଂ ଏଥ‌ିପାଇଁ କେତୋଟି ପାସ୍ୱାର୍ଡ଼ ସମ୍ଭବ ତାହା ଜାଣିବାକୁ ଆମେ ଚେଷ୍ଟା କରିବା ।
Solution:
ପ୍ରଥମ ଅଙ୍କ ପାଇଁ 10 ଟି ବିକଳ୍ପ ଅଛି (ଯଦି ) (ପ୍ରଥମ ଅଙ୍କ, ତେବେ 00, 01, 02, 03. 09) ସମ୍ଭବ । ଯଦି ।
ପ୍ରଥମ ଅଙ୍କ, ତେବେ 10, 11, 12, 13, …, 19 ସମ୍ଭବ ।
ତେଣୁ ଏକ 2 ଅଙ୍କ ବିଶିଷ୍ଟ ଲକ୍ ପାଇଁ ସମୁଦାୟ ସମାବେଶ ସଂଖ୍ୟା ହେଉଛି 10 × 10 = 100

Question 5.
ମନେକର ଆମ ପାଖରେ ୩ ଅଙ୍କ ବିଶିଷ୍ଟ ତାଲା ଅଛି ।
Solution:
ପୂର୍ବରୁ ଥିବା 100 (2 ଅଙ୍କ ବିଶିଷ୍ଟ) ପାସ୍ୱାର୍ଡ଼ ମଧ୍ୟରୁ ପ୍ରତ୍ୟେକ ଥର ତୃତୀୟ ଅଙ୍କ ପାଇଁ
100 × 10 = 1000ଟି ସମାବେଶ ଅଛି ।
ସେଗୁଡ଼ିକୁ ତାଲିକାଭୁକ୍ତ କଲେ = 000, 001, 002, ……, 997, 998, 999

Question 6.
କେତୋଟି 5 ଅଙ୍କ ବିଶିଷ୍ଟ ପାସ୍ୱାର୍ଡ଼ ସମ୍ଭବ?
Solution:
ପ୍ରତ୍ୟେକ ଅଙ୍କର 10 ଟି ପସନ୍ଦ ଅଛି । ତେଣୁ ଏକ 5 ଅଙ୍କ ବିଶିଷ୍ଟ ତାଲାରେ ଥିବ = 10 × 10 × 10 × 10 × 10 = 10⁵ = 1,00,000 ସଂଖ୍ୟକ ପାସ୍ୱାର୍ଡ଼ ।
ତେଣୁ 99,999 ପର୍ଯ୍ୟନ୍ତ 5 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ଲେଖାଯାଇପାରିବ ।
00000, 00001, 00002, ……… 00010, 00011, ………. 00100, 00101, ………. 30456, ………. 99998, 99999

2.3 ଘାତର ଅପରପାର୍ଶ୍ବ

Page No. 27

Question 1.
2 ର ଘାତ ଆକାରରେ 2¹⁰⁰ ÷ 2²⁵ର ମାନ କେତେ ହେବ ?
Solution:
2¹⁰⁰ ÷ 2²⁵ = 2^100-25 = 2⁷⁵

Page No. 28

Question 1.
n କାହିଁକି 0 ହୋଇପାରିବ ନାହିଁ ?
Solution:
n ର ମୂଲ୍ୟ 0 ହେଲେ, ଏହାର ମୂଲ୍ୟ। ସଙ୍ଗେ ସମାନ।

Question 2.
2⁰ ମୂଲ୍ୟ କେତେ?
Solution:
ଆମେ 2⁰ କୁ ଏପରି ଭାବରେ ବ୍ୟାଖ୍ୟା କରିବା ଯେ ଉପରୋକ୍ତ ସାଧାରଣ ରୂପ ସତ୍ୟ ହେବ।
\(2^0=2^{4-4}=2^4 \div 2^4=\frac{2 \times 2 \times 2 \times 2}{2 \times 2 \times 2 \times 2}\) = 1

Page No. 29

Question 1.
ଆମେ \(10^3=\frac{1}{10^{-3}}\) ଲେଖୁପାରିବା କି?
Solution:
ହଁ, ଆମେ ଲେଖୁପାରିବା = \(\frac{1}{10^{-3}}=\frac{1}{1 / 10^3}=1 \div \frac{1}{10^3}=1 \times 10^3=10^3\)

Question 2.
ନିମ୍ନଲିଖୁତ ପଦଗୁଡ଼ିକର ସମତୁଲ୍ୟ ରୂପ ଲେଖ :
(i) 2^-4
(ii) 10^-5
(iii) (-7)^-2
(iv) (-5)^-3
(v) 10^-100
Solution:
(i) \(2^{-4}=\frac{1}{2^4}=\frac{1}{16}\)
(ii) \(10^{-5}=\frac{1}{10^5}=\frac{1}{1,00,000}\)
(iii) \((-7)^{-2}=\frac{1}{(-7)^2}=\frac{1}{49}\)
(iv) \((-5)^{-3}=\frac{1}{(-5)^3}=\frac{1}{-125}\)
(v) \(10^{-100}=\frac{1}{10^{100}}\)

Question 3.
ସରଳକର ଏବଂ ଉତ୍ତରଗୁଡ଼ିକୁ ଘାତଙ୍କୀୟ ରୂପରେ ଲେଖ :
(i) 2^-4 × 2⁷
(ii) 3² × 3^-5 × 3⁶
(iii) p³ × p^-10
(iv) 2⁴ × (-4)^-2
(v) 8^p × 8^q
Solution:
(i) \(2^{-4} \times 2^7=\frac{1}{2^4} \times 2^7=2^{7-4}=2^3\)
(∵ a^m ÷ aⁿ = a^m-n)

(ii) \(3^2 \times 3^{-5} \times 3^6=3^2 \times \frac{1}{3^5} \times 3^6=3^2 \times 3^6 \times \frac{1}{3^5}=3^{2+6-5}=3^{8-5}=3^3\)
(∵ a^m × aⁿ = a^m+n & a^m ÷ aⁿ = a^m-n)

(iii) \(p^3 \times p^{-10}=p^3 \times \frac{1}{p^{10}}=\frac{p^3}{p^{10}}=p^{3-10}=p^{-7}\)
(∵ a^m ÷ aⁿ = a^m-n)

(iv) \(2^4 \times(-4)^{-2}=2^4 \times \frac{1}{(-4)^2}=2^4 \times \frac{1}{16}=2^4 \times \frac{1}{2^4}=2^{4-4}=2^0=1\)
(∵ a^m ÷ aⁿ = a^m-n)

(v) 8^p × 8^q = 8^p+q (∵ a^m × aⁿ = a^m+n)

Page No. 30

Question 1.
16384 (= 4⁷) ସଂଖ୍ୟାଟି 1024 (= 4⁵) ଠାରୁ 16 (= 4²) ଗୁଣ ବଡ଼ କି?
Solution:
ହଁ, ଯେହେତୁ 4⁷ ÷ 4⁵ = 4²
ତେଣୁ 16384(= 4⁷) ସଂଖ୍ୟାଟି 1024 (= 4⁵) ଠାରୁ 16 (= 4²) ଗୁଣ ବଡ଼।

Question 2.
4^-2 ଠାରୁ 4² କେତେଗୁଣ ବଡ଼?
Solution:
4² = 16
⇒ 4^-2 = \(\frac{1}{4^2}\)
⇒ 4^-2 = \(\frac{1}{16}\)
⇒ \(\frac{4^2}{4^{-2}}\) = 4² × 4² = 4²⁺² = 4⁴ = 256
4^-2 ଠାରୁ 4² ହେଉଛି 256 (4⁴) ଗୁଣ ବଡ଼।

Question 3.
7ର ଘାତରେଖା ବ୍ୟବହାର କରି ନିମ୍ନଲିଖତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଲେଖ।

Solution:

2.4 10 ର ଘାତ

Page No. 30

Question 1.
ଉପରୋକ୍ତ ଉପାୟରେ ଏହି ସଂଖ୍ୟାଗୁଡ଼ିକୁ ଲେଖ :
(i) 172
(ii) 5642
(iii) 6474
Solution:
(i) 172 = 1 × 100 + 7 × 10 + 2 × 1 = (1 × 10²) + (7 × 10¹) + 2 × 10⁰)
(ii) 5642 = 5 × 1000 + 6 × 100 + 4 × 10 + 2 × 1 = (5 × 10³) + (6 × 10²) + (4 × 10¹) + (2 × 10⁰)
(iii) 6474 = 6 × 1000 + 4 × 100 + 7 × 10 + 4 × 1 = (6 × 10³) + (4 × 10²) + (7 × 10¹) + (4 × 10⁰)

Question 2.
561,903 କୁ ଆମେ କିପରି ଲେଖିପାରିବା?
Solution:
561.903 = (5 × 100) + (6 × 10) + 1 + (9 × \(\frac {1}{10}\)) + (0 × \(\frac {1}{100}\)) + (3 × \(\frac {1}{1000}\))
10 ର ଘାତ ବ୍ୟବହାର କରି ଆମେ ଏହାକୁ ଲେଖୁବା :
561.903 = (5 × 10²) + (6 × 10¹) + (1 × 10⁰) + (9 × 10^-1) + (0 × 10^-2) + (3 × 10^-3)

Page No. 32

Question 1.
ଏହି ତିନୋଟି ଦୂରତା ମଧ୍ଯରୁ କେଉଁ ଦୂରତାଟି ସବୁଠାରୁ ସାନ?

Solution:
(i) 1.4335 × 10¹² ମିଟର = 14.335 × 10¹¹ ମିଟର।
(ii) 1.439 × 10¹² ମିଟର = 14.39 × 10¹¹ ମିଟର।
(iii) 1.496 × 10¹¹ ମିଟର।
ତୁଳନା କଲେ 14.39 × 10¹¹ ମିଟର > 14.335 × 10¹¹ ମିଟର > 1.496 × 10¹¹ ମିଟର।
∴ (iii) 1.496 × 10¹¹ ମିଟର ସବୁଠାରୁ ସାନ।
ସୂର୍ଯ୍ୟ ଓ ପୃଥ‌ିବୀ ମଧ୍ୟରେ ଦୂରତା = 1.496 × 10¹¹ ମିଟର।

Question 2.
ନିମ୍ନଲିଖୁତ ସଂଖ୍ୟାଗୁଡ଼ିକୁ ମାନକ ରୂପରେ ପ୍ରକାଶ କର।
(i) 59,583
(ii) 65,950
(iii) 34,30,000
(iv) 70,04,00,00,000
Solution:
(i) 59,583 = \(\frac {59853}{10000}\) × 10000 = 5.9853 × 10⁴
(ii) 65,950 = \(\frac {65950}{10000}\) × 10000 = 6.595 × 10⁴
(iii) 34,30,000 = \(\frac {3430000}{1000000}\) × 1000000 = 3.43 × 10⁶
(iv) 70,04,00,00,000 = \(\frac {70040000000}{10000000000}\) × 10000000000 = 7.004 × 10¹⁰

2.5 ତୁମେ କେବେ ଭାବିଛ କି?

Page No. 33

Question 1.
ଦାନ କରାଯାଉଥ‌ିବା ଗୁଡ଼ର ମୂଲ୍ୟ (ଟଙ୍କାରେ) କେତେ ହେବ ? ଦାନ କରାଯାଇଥିବା ଗହମର ମୂଲ୍ୟ (ଟଙ୍କାରେ) କେତେ ହେବ?
Solution:
ଗୁଡ଼ର ମୂଲ୍ୟ (ଟଙ୍କାରେ) = ରକିର ଓଜନ (କିଲୋଗ୍ରାମ୍‌ରେ) × 1 କି.ଗ୍ରା. ଗୁଡ଼ର ମୂଲ୍ୟ।
ଗହମର ମୂଲ୍ୟ (ଟଙ୍କାରେ) = ଇତୁର ଓଜନ (କିଲୋଗ୍ରାମିଂରେ) × 1 କି.ଗ୍ରା. ଗହମର ମୂଲ୍ୟ।

Question 2.
ଅଜ୍ଞାତ ରାଶି ପାଇଁ ଆବଶ୍ୟକ ଏବଂ ଉପଯୁକ୍ତ ଆନୁମାନିକ ଆକଳନ କର ଏବଂ ଉତ୍ତର ନିର୍ଣ୍ଣୟ କର । ମନେରଖ ରକିକୁ 13 ବର୍ଷ ଏବଂ ଇତୁକୁ 11 ବର୍ଷ ।
Solution:
ମନେକର ରକିର ବୟସ = 13 ବର୍ଷ ଓ ଓଜନ = 45 କି.ଗ୍ରା.।
1 କି.ଗ୍ରା. ଗୁଡ଼ର ମୂଲ୍ୟ = ଟ.70.00
ଦାନ କରାଯାଇଥିବା ଗୁଡ଼ର ମୂଲ୍ୟ = 45 × 70 ଟଙ୍କା = ଟ.3150.00 ଅଟେ।
ମନେକର ଇତୁର ବୟସ = 11 ବର୍ଷ ଓ ଓଜନ = 50 କି.ଗ୍ରା.।
1 କି.ଗ୍ରା. ଗହମର ମୂଲ୍ୟ = ଟ.50.00
ଦାନ କରାଯାଇଥ‌ିବା ଗହମର ମୂଲ୍ୟ = 50 × 50 ଟଙ୍କା = ଟ.2500.00 ଅଟେ।

Question 3.
ଅନୁମାନ : କୌଣସି ଗଣନା | ହିସାବ ନକରି, ଉତ୍ତର କ’ଣ ହୋଇପାରେ ତାହା ବିଷୟରେ ସ୍ଵତଃସ୍ଫୁର୍ଭ (ଶୀଘ୍ର ଅନୁମାନ କର)
Solution:
ରକିର ବୟସ = 13 ବର୍ଷ ।
ତେଣୁ ଆମେ ଅନୁମାନ କରିପାରିବା ଯେ ତା’ର ଓଜନ ପ୍ରାୟ 40ରୁ 50 କି.ଗ୍ରା. ମଧ୍ଯରେ ହୋଇପାରେ ।
ଯଦି ଆମେ ଧରିନେବା । କି.ଗ୍ରା. ଗୁଡ଼ର ମୂଲ୍ୟ = 60 ଟଙ୍କା ।
ତେବେ 1 ଟଙ୍କିଆ ମୁଦ୍ରାର ସଂଖ୍ୟା = 45 × 60 = 2700

Page No. 34

Question 1.
ଆକଳନ ଏବଂ ଆନୁମାନିକ ଗଣନା :
(i) ଉତ୍ତର ପାଇବା ପାଇଁ ଆବଶ୍ୟକ ପରିମାଣଗୁଡ଼ିକ ମଧ୍ୟରେ ଥିବା ସମ୍ପର୍କ ବର୍ଣ୍ଣନା କର ।
(ii) ଯଦି ଆବଶ୍ୟକ ତଥ୍ୟ ଉପଲବ୍‌ଧ ନଥାଏ, ତେବେ ଉପଯୁକ୍ତ ଅନୁମାନ ଓ ଆକଳନ କର ।
(iii) ହିସାବ କର ଏବଂ ଉତ୍ତର ଖୋଜ (ଏବଂ ତୁମର ଅନୁମାନ, ଠିକ୍ ଉତ୍ତରରେ କେତେ ନିକଟତର ଥୁଲା ଦେଖ ।)
Solution:
(i) ଉତ୍ତର ପାଇବା ପାଇଁ ଆବଶ୍ୟକ ପରିମାଣଗୁଡ଼ିକ ମଧ୍ୟରେ ଥିବା ସମ୍ପର୍କ ବର୍ଣ୍ଣନା :
ରକିର ଓଜନ କେତେ । ଟଙ୍କା ମୁଦ୍ରା ମେଳ ଖାଉଛି ତାହା ଜାଣିବା ପାଇଁ ଆମକୁ ଚିହ୍ନଟ କରିବାକୁ ପଡ଼ିବ : ରକିର ଓଜନ (କିଲୋଗ୍ରାମରେ), 1 କି.ଗ୍ରା. ଗୁଡ଼ର ମୂଲ୍ୟ ।
ଦାନର ମୂଲ୍ୟ = ଓଜନ କିଲୋଗ୍ରାମରେ × 1 କି.ଗ୍ରା.ଗୁଡ଼ର ମୂଲ୍ୟ।
ଆବଶ୍ୟକ 1 ଟଙ୍କିଆ ମୁଦ୍ରା ସଂଖ୍ୟା = ଦାନର ମୂଲ୍ୟ ÷ 1

(ii) ଯଦି ଆବଶ୍ୟକ ତଥ୍ୟ ଉପଲବ୍‌ଧ ନଥାଏ, ତେବେ ଉପଯୁକ୍ତ ଅନୁମାନ ଓ ଆକଳନ :
ଆସ ଯୁକ୍ତିଯୁକ୍ତ ଆକଳନ ବ୍ୟବହାର କରିବା :
ରକିର ଓଜନ ପ୍ରାୟ 45 କି.ଗ୍ରା. (ସାଧାରଣ 13 ବର୍ଷର ପିଲା)
1 କିଲୋଗ୍ରାମ୍ ଗୁଡ଼ର ଓଜନ ପ୍ରାୟ ଟ.60.00 (ହାରାହାରି ବଜାର ଦର)
45 କି.ଗ୍ରା ଗୁଡ଼ର ମୂଲ୍ୟ = 45 × 60 = ଟ.2700.00

(iii) ହିସାବ କର ଏବଂ ଉତ୍ତର ଖୋଜ : (ତୁମର ଅନୁମାନ, ଠିକ୍ ଉତ୍ତରର କେତେ ନିକଟତର ଥୁଲା ଦେଖ।)
1 ଟଙ୍କିଆ ମୁଦ୍ରା ଆବଶ୍ୟକ = 6.2700 ÷ ଟ.1 = 2700

Question 2.
ମୁଦ୍ରା ସଂଖ୍ୟା ଶହ ଶହ, ହଜାର ହଜାର, ଲକ୍ଷ ଲକ୍ଷ, କୋଟି କୋଟି କିମ୍ବା ତା’ଠାରୁ ଅଧିକ ହେବ କି ? ଅନୁମାନ କରି କୁହ।
Solution:
ଆମେ ଶୀଘ୍ର ବୁଦ୍ଧିମାନ କରି ଅନୁମାନ କରିବା।
ରକିର ଓଜନ ପ୍ରାୟ 45 କି.ଗ୍ରା. ଏବଂ 1 କି.ଗ୍ରା. ଗୁଡ଼ର ମୂଲ୍ୟ 60 ଟଙ୍କା
∴ ମୋଟ ମୂଲ୍ୟ = 45 × 60 ଟଙ୍କା = 2700 ଟଙ୍କା ।
ଯଦି ଆମେ 1 ଟଙ୍କିଆ ମୁଦ୍ରା ବ୍ୟବହାର କରୁ, ତେଣୁ 2700 ମୁଦ୍ରା ଆବଶ୍ୟକ,
ଯାହା ହଜାର ପରିସର ମଧ୍ୟରେ ପଡ଼ିଥାଏ ।
ତେଣୁ ହଜାର ମୁଦ୍ରା ଆବଶ୍ୟକ । ଲକ୍ଷ କିମ୍ବା କୋଟି ନୁହେଁ, କିନ୍ତୁ ଶହେରୁ ଅଧିକ।

Question 3.
ଅଜ୍ଞାତ ରାଶି ପାଇଁ ଆବଶ୍ୟକୀୟ ଅନୁମାନ ଏବଂ ଆକଳନ କରି ଉତ୍ତର ଖୋଜ ।
Solution:
ପିଲାମାନେ ଶିକ୍ଷକଙ୍କ ସହାୟତାରେ ନିଜେ କରିବେ ।

Question 4.
ଇଡ଼ୁ ପଚାରିଲା, ଯଦି ଆମେ 5 ଟଙ୍କିଆ ମୁଦ୍ରା କିମ୍ବା 10 ଟଙ୍କିଆ ନୋଟ୍ ବ୍ୟବହାର କରୁ, ଏହା କେତେ ଟଙ୍କା ହୋଇପାରେ?
Solution:
ମୋଟ ଦାନର ମୂଲ୍ୟ 1140 ଟଙ୍କା ।
5 ଟଙ୍କିଆ ମୁଦ୍ରାର ସଂଖ୍ୟା = 1140 ÷ 5 = 228
10 ଟଙ୍କିଆ ମୁଦ୍ରାର ସଂଖ୍ୟା = 1140 ÷ 10 = 114

Question 5.
ଇଡୁ କହିଲା, ‘‘ଯେତେବେଳେ ମୁଁ ବଡ଼ ହେବି, ପ୍ରତିବର୍ଷ ମୁଁ ମୋ ଓଜନର ନୋଟବୁକ୍ ଦାନ କରିବି ।’’ ରକି କହିଲା, “ଯେତେବେଳେ ମୁଁ ବଡ଼ ହେବି ପ୍ରତିବର୍ଷ ମୁଁ ମୋ ଓଜନର ଅନ୍ନଦାନ କରିବି ।’’ ଏହି ଦାନରୁ ବର୍ଷକୁ କେତେ ଲୋକ ଉପକୃତ ହେବେ ? ପୁନଶ୍ଚ ଖୋଜିବା ପୂର୍ବରୁ ପ୍ରଥମେ ଅନୁମାନ କର ।
Solution:
ଇତୁର ନୋଟବୁକ ଦାନ ପ୍ରତିବର୍ଷ 100ରୁ 200 ଛାତ୍ର ଉପକୃତ ହୋଇପାରନ୍ତି ।
ରକିର ଖାଦ୍ୟ ପ୍ରଦାନ : ପ୍ରାୟ ବାର୍ଷିକ 150 ରୁ 300
ଲୋକଙ୍କୁ ଖାଦ୍ୟ ଦେବାପାଇଁ ଯଥେଷ୍ଟ ଶସ୍ୟ କିମ୍ବା ଆନୁମାନିକ ଆକଳନ :
ଇତୁର ଓଜନ ପ୍ରାୟ 38 କି.ଗ୍ରା.।
1 ଟି ନୋଟବୁକ୍‌ର ମୂଲ୍ୟ ପ୍ରାୟ 20 ଟଙ୍କା ।
ରକିର ଓଜନ ପ୍ରାୟ 45 କି.ଗ୍ରା.।
ଦାନ କରାଯାଇଥିବା 1 କି.ଗ୍ରା. ଶସ୍ୟ କିମ୍ବା ଖାଦ୍ୟର ମୂଲ୍ୟ ପ୍ରାୟ 30 ଟଙ୍କା ।
ପ୍ରତିବ୍ୟକ୍ତି ପାଇଁ ଖାଦ୍ୟ ଖର୍ଚ୍ଚ ପ୍ରାୟ 15 ଟଙ୍କା ।
ଇତ୍ରୁର ନୋଟବୁକ୍ ଦାନ : ତାଙ୍କ ଓଜନ ମୂଲ୍ୟର ନୋଟ୍‌ବୁକ୍ ଦାନର ମୂଲ୍ୟ = 38 × 20 ଟଙ୍କା = 760 ଟଙ୍କା।
ଯଦି ପ୍ରତ୍ୟେକ ଛାତ୍ରଙ୍କୁ ଦୁଇଟି ନୋଟବୁକ୍‌ ଆବଶ୍ୟକ ହୁଏ,
ସାହାଯ୍ୟ ପାଇପାରୁଥିବା ଛାତ୍ରସଂଖ୍ୟା = 760 ÷ 40 = 19 ଜଣ।
∴ ପ୍ରତିବର୍ଷ 19 ଜଣ ଛାତ୍ର ନୋଟ୍‌ବୁକ୍‌ ପାଇପାରିବେ ।
ରକିର ଖାଦ୍ୟଦାନ : ତାଙ୍କ ଓଜନ ମୂଲ୍ୟର ଶସ୍ୟ କିମ୍ବା ଖାଦ୍ୟ ଦାନର ମୂଲ୍ୟ = 45 × 30 = 1350 ଟଙ୍କା।
ପ୍ରତ୍ୟେକ ଖାଦ୍ୟର ମୂଲ୍ୟ = 15 ଟଙ୍କା ।
ଖାଦ୍ୟ ଯୋଗାଇ ପାରୁଥୁବା ଲୋକଙ୍କ ସଂଖ୍ୟା = 1350 ÷ 15 = 90 ଜଣ।
∴ 90 ଜଣ ଲୋକ ଗୋଟିଏ ଥର ଖାଦ୍ୟ ପାଇବେ କିମ୍ବା ବର୍ଷକରେ 90 ରୁ କମ୍ ଏକାଧିକ ଥର ଖାଇପାରିବେ ।

Question 6.
ରକି ଓ ଇଡୁ ଆଉ କେହି କହୁଥିବାର ଶୁଣିଲେ – ଆମେ ଏହି ସ୍ଥାନରେ ପହଞ୍ଚିବା ପାଇଁ 400 କି.ମି. ପଦଯାତ୍ରା କଲୁ । ଆମେ ଆଜି ସକାଳୁ ପହଞ୍ଚିଲୁ । ସେମାନେ କେତେ ସମୟ ପୂର୍ବରୁ ଯାତ୍ରା ଆରମ୍ଭ କରିଥିବେ ?
Solution:
ଯଦି କେହିଜଣେ ପାଦରେ 400 କି.ମି. ଚାଲିଥାନ୍ତି, କେତେ ସମୟ ଲାଗିପାରେ ?
ଆସ ଅନୁମାନ କରିବା ଯେ, ସେମାନେ ପ୍ରତିଦିନ ପ୍ରାୟ 25 ରୁ 30 କି.ମି. ଚାଲେ ।
ଆନୁମାନିକ ଦିନ ସଂଖ୍ୟା = 400 ÷ 25 = 16 ଦିନ ।
କିମ୍ବା ତା’ର ଗତି ଓ ବିଶ୍ରାମ ଉପରେ ନିର୍ଭର କରି 13 ରୁ 20 ଦିନ ହୋଇପାରେ ।
ଆକଳନ ଓ ଗଣନା :
ଦୈନିକ ଅତିକ୍ରାନ୍ତ ଦୂରତା = 30 କି.ମି. (ଧରାଯାଉ)
ଅତିକ୍ରାନ୍ତ ଦୂରତା = 400 କି.ମି. ।
ଦିନସଂଖ୍ୟା = 400 ÷ 30 = 13.3 ଦିନ।
∴ ପ୍ରାୟତଃ 13 ରୁ 14 ଦିନ ।

Page No. 35

Question 1.
ଯଦି ଜଣେ ବ୍ୟକ୍ତି ନିରନ୍ତର ଚାଲିଥାଏ, ତେବେ ସେ ନିଜ ଜୀବନକାଳରେ କେତେଥର ପୃଥିବୀକୁ ପରିକ୍ରମା (ବିଶ୍ଵ ପରିକ୍ରମଣ) କରିପାରିବେ ? ପୃଥ‌ିବୀର ପରିଧ୍ଵକୁ 40,000 କି.ମି. ବୋଲି ଧରି ନିଆଯାଉ ।
Solution:
ଆନୁମାନିକ : ପୃଥ‌ିବୀର ପରିଧ୍ଵ = 40,000 କି.ମି. ।
ହାରାହାରି ଚାଲିବାର ଗତି ଘଣ୍ଟାପ୍ରତି 5 କି.ମି. ।
ପ୍ରତିଦିନ ସର୍ବାଧ‌ିକ ଚାଲିବା ପ୍ରାୟ = 8 ଘଣ୍ଟା ।
ଆସ ଅନୁମାନ କରିବା, 60 ବର୍ଷର (ବୟସ 15ରୁ 75 ମଧ୍ୟରେ)
ଗୋଟିଏ ଦିନରେ ଅତିକ୍ରାନ୍ତ ଦୂରତା = 8 × 5 = 40 କି.ମି. ।
ମୋଟ ଚାଲିଯାଇଥବା ଦିନ = 60 × 365 = 21,900 ଦିନ ।
ତା’ର ଜୀବନକାଳରେ ମୋଟ ଚାଲିଥିବା ଦୂରତା = 21,900 × 40 = 8,76,000 କି.ମି.।
∴ ମୋଟ ପୃଥ‌ିବୀ ପରିକ୍ରମା = 8,76,000 ÷ 40,000 = 21.9 ଥର ।

Page No. 36

Question 1.
3,84,400 କି.ମି.ରେ କେତେ 20 ସେ.ମି. ଅଛି ?

Solution:
ଯଦି ଆମେ ହିସାବ କରୁ, ତେବେ ଫଳାଫଳ 1,92,20,00,000 ପାହାଚ ମିଳିବ । ଏହା 192 କୋଟି 20 ଲକ୍ଷ ପାହାଚ କିମ୍ବା 1 ବିଲିୟନ 992 ମିଲିୟନ୍ ପାହାଚ ସହ ସମାନ । ପ୍ରତ୍ୟେକ ପଦକ୍ଷେପ ପରେ ପୃଥ‌ିବୀଠାରୁ ଦୂରତାରେ ନିର୍ଦ୍ଦିଷ୍ଟ ବୃଦ୍ଧି (ପ୍ରତ୍ୟେକ ପଦକ୍ଷେପ 20) ସେ.ମି. ବୃଦ୍ଧି)କୁ କୈକ ବୃଦ୍ଧି କୁହାଯାଏ ।

Question 2.
ତୁମେ ରେଖକ ବୃଦ୍ଧି ଏବଂ ଘାତାଙ୍କୀୟ ବୃଦ୍ଧିର କିଛି ଉଦାହରଣ ଦେଇପାରିବ କି ?
Solution:
ନିଜେ ଅଭ୍ୟାସ କର ।

Page No. 38-39

Question 1.
ବିଶ୍ଵ ଜନସଂଖ୍ୟା ପ୍ରାୟ 8 × 10⁹ ଓ ଆଫ୍ରିକୀୟ ହାତୀ ସଂଖ୍ୟା ହେଉଛି 4 × 10⁵ ଆମେ କହିପାରିବା କି, ପାଇଁ ପ୍ରାୟ 20,000 ଲୋକ ଅଛନ୍ତି ।
Solution:
ବିଶ୍ଵ ଜନସଂଖ୍ୟା ପ୍ରାୟ = 8 × 10⁹ = 8,00,00,00,000
ଆଫ୍ରିକୀୟ ହାତୀ ସଂଖ୍ୟା = 4 × 10⁵ = 4,00,000
∴ ପ୍ରତ୍ୟେକ ହାତୀ ପାଇଁ ଲୋକ = \(\frac {8,00,00,00,000}{4,00,000}\) = 20,000

Question 2.
ବୈଜ୍ଞାନିକ ସଂକେତ ବ୍ୟବହାର କରି ଗଣନା କର ଏବଂ ଉତ୍ତର ଲେଖ।
(i) ବିଶ୍ଵରେ ମନୁଷ୍ୟ ଓ ପିମ୍ପୁଡ଼ିମାନଙ୍କ ସଂଖ୍ୟାର ଅନୁପାତ କେତେ?
(ii) ଯଦି ଷ୍ଟଲିଂ ପକ୍ଷୀମାନଙ୍କର ଏକ ଦଳରେ 10,000 ପକ୍ଷୀ ଥାଆନ୍ତି, ତେବେ ବିଶ୍ଵରେ କେତେ ଦଳ ପକ୍ଷୀ ଥାଇପାରନ୍ତି?
(iii) ଯଦି ଗୋଟିଏ ଗଛରେ 10⁴ ସଂଖ୍ୟକ ପତ୍ର ଥାଏ, ତେବେ ପୃଥ‌ିବୀର ସମସ୍ତ ଗଛରେ ଥିବା ପତ୍ର ସଂଖ୍ୟା କେତେ ନିର୍ଣ୍ଣୟ କର।
(iv) ଯଦି ତୁମେ କାଗଜ–ଫର୍ଦଗୁଡ଼ିକୁ ଉପରକୁ ଉପର ଥାକ କରି ରଖୁବ, ତେବେ ଚନ୍ଦ୍ରରେ ପହଞ୍ଚିବା ପାଇଁ ତୁମକୁ କେତେ ଫର୍ଦ୍ଦ କାଗଜ’ ଆବଶ୍ୟକ ହେବ?
Solution:
(i) ବିଶ୍ଵରେ ପିମ୍ପୁଡ଼ିମାନଙ୍କ ସଂଖ୍ୟା = 20 କ୍ୱାଡ୍ରିଲିୟନ୍ = 2 × 10¹⁶ ପ୍ରାୟ ।
ବିଶ୍ୱରେ ମନୁଷ୍ୟମାନଙ୍କ ସଂଖ୍ୟା = 8 ବିଲିୟନ୍ = 8 × 10⁹ ପ୍ରାୟ ।
ବିଶ୍ଵରେ ମନୁଷ୍ୟ ଓ ପିମ୍ପୁଡ଼ିମାନଙ୍କ ସଂଖ୍ୟାର ଅନୁପାତ = \(\frac{2 \times 10^{16}}{8 \times 10^9}\)
= 0.25 × 107
= 2.5 × 106

(ii) ଷ୍ଟଲିଂ ପକ୍ଷୀସଂଖ୍ୟା = 3.1 × 10⁸ ପ୍ରାୟ ।
ଷ୍ଟଲିଂ ପକ୍ଷୀମାନଙ୍କର ଗୋଟିଏ ଦଳରେ ପକ୍ଷୀସଂଖ୍ୟା = 1 × 10⁴
ବିଶ୍ଵରେ ଥ‌ିବା ଷ୍ଟଲିଂ ପକ୍ଷୀମାନଙ୍କର ଦଳସଂଖ୍ୟା = \(\frac{3.1 \times 10^8}{1 \times 10^4}\) = 3.1 × 10⁴

(iii) ବିଶ୍ଵର ଗଛସଂଖ୍ୟା ପ୍ରାୟ = 3 × 10¹²
ଗୋଟିଏ ଗଛର ପତ୍ର ସଂଖ୍ୟା ପ୍ରାୟ = 1 × 10⁴
ବିଶ୍ଵର ସମୁଦାୟ ଗଛର ପତ୍ର ସଂଖ୍ୟା = 3 × 10¹² × 1 × 10⁴ = 3 × 10¹⁶

(iv) ଚନ୍ଦ୍ରର ଦୂରତା = 3.84 × 10⁸ ମିଟର।
ଗୋଟିଏ କାଗଜର ମୋଟେଇ = 1 × 10^-4 ମିଟର।
ଚନ୍ଦ୍ରରେ ପହଞ୍ଚିବା ପାଇଁ କାଗଜ ଦରକାର = \(\frac{3.84 \times 10^8}{1 \times 10^{-4}}\) = 3.84 × 10¹²

Page No. 39

Question 1.
‘‘ତୁମର ବୟସ କେତେ ?” ଇଡୁ ପଚାରିଲା :
‘‘ମୁଁ କିଛି ସପ୍ତାହ ପୂର୍ବରୁ 13 ବର୍ଷ ପୂରଣ କରିଛି’’ ରକି ଉତ୍ତର ଦେଲା ।
‘‘ତୁମର ବୟସ କେତେ ?” ଇଡୁ ପୁଣି ପଚାରିଲା ।
‘‘ଆଜି ମୋତେ 4840 ଦିନ’’ ରକି କହିଲା ।
‘‘ତୁମକୁ କେତେ ବୟସ’’ ଇଡୁ ଆଉ ଥରେ ପଚାରିଲା,
‘‘ମୋର ବୟସ ————– ଘଣ୍ଟା !’’ ରକି କହିଲା ।
Solution:
1 ବର୍ଷ = 365 ଦିନ ।
4840 ଦିନ = \(\frac {4840}{365}\) ବର୍ଷ = 13.26 ବର୍ଷ |
1 ଦିନ = 24 ଘଣ୍ଟା ।
4840 ଦିନ = 24 × 4840 = 116160 ଘଣ୍ଟା
∴ ରକିର ବୟସ 13,26 ବର୍ଷ ଓ 116160 ଘଣ୍ଟା ।

Question 2.
ଇତୁ ‘‘ଆଜି ମୋର ବୟସ 4070 ଦିନ । ତୁମେ ମୋର ଜନ୍ମତାରିଖ ଖୋଜି ପାରିବ କି ?’’
Solution:
ନିଜେ କର ।

Question 3.
ଯଦି ତୁମେ ଏକ ନିୟୁତ ସେକେଣ୍ଡ ବଞ୍ଚୁଛ, ତେବେ ତୁମକୁ କେତେ ବୟସ ହେବ?
Solution:
ମିନିଟ୍ = 60 ସେକେଣ୍ଡ ।
1 ଘଣ୍ଟା = 60 ମିନିଟ୍ = 60 × 60 ସେକେଣ୍ଡ = 3600 ସେକେଣ୍ଡ ।
1 ଦିନ = 24 ଘଣ୍ଟା = 24 × 3600 = 86,400 ସେକେଣ୍ଡ ।
ତୁମେ ବଞ୍ଚ = 1 ନିୟୁତ ସେକେଣ୍ଡ = 10,00,000 ସେକେଣ୍ଡ ।
ଦିନ ସଂଖ୍ୟା = \(\frac {10,00,000}{86,400}\) = 11.57 ଦିନ।

Page No. 40

Question 1.
10⁵ ସେକେଣ୍ଡ = 1.16 ଦିନ ଏବଂ 10⁶ ସେକେଣ୍ଡ = 11.57 ଦିନ । କିଛି ଘଟଣା ବିଷୟରେ ଚିନ୍ତାକର ଯାହାର ସମୟ (i) 10⁵ ସେକେଣ୍ଡ ଏବଂ (ii) 10⁶ ସେକେଣ୍ଡର କ୍ରମରେ ଅଛନ୍ତି, ସେଗୁଡ଼ିକୁ ବୈଜ୍ଞାନିକ ସଂକେତରେ ଲେଖ ।
Solution:
(i) 10⁵ ସେକେଣ୍ଡର ଘଟଣାଗୁଡ଼ିକ (ପ୍ରାୟ 1.16 ଦିନ)
(କ) ବହୁସ୍ତରୀୟ ସାଇକେଲ ଚାଳନା ଦୌଡ଼ ପର୍ଯ୍ୟାୟ (ଟୁର ଡି ଫ୍ରାନ୍ସର ଗୋଟିଏ ପର୍ଯ୍ୟାୟ ପରି)
ସମୟ ବ୍ୟବଧାନ : 1 × 10⁵ ସେକେଣ୍ଡ ।
(ଖ) ଏକ କ୍ଷୁଦ୍ର ଚଳଚ୍ଚିତ୍ର ବା ଡକୁମେଣ୍ଟାରୀ ଫିଲ୍ମ କରିବା ।
ସୁଟିଂ କାର୍ଯ୍ୟସୂଚୀ : 1.2 × 10⁵ ସେକେଣ୍ଡ ।

(ii) 10⁶ ସେକେଣ୍ଡର ଘଟଣାଗୁଡ଼ିକ (ପ୍ରାୟ 11.57 ଦିନ)
(କ) ବଡ଼ ଉତ୍ସବର ଅବଧୂ (ଯଥା- କୁମ୍ଭମେଳା କିମ୍ବା ଅଲମ୍ପିକ୍ସ ଉଦ୍‌ଘାଟନୀ ଇଭେଣ୍ଟ)
ସମ୍ପୂର୍ଣ୍ଣ ଉତ୍ସବ ଅବଧୂ ପ୍ରାୟ 1 × 10⁶ ସେକେଣ୍ଡ ।
(ଖ) କୁକୁଡ଼ା ଅଣ୍ଡାର ଉଷୁମାଇବା ସମୟ (ଛୁଆ ହେବା ପର୍ଯ୍ୟନ୍ତ) ପ୍ରାୟ 1.2 × 10⁶ ସେକେଣ୍ଡ।

Page No. 42

Question 1.
ବୈଜ୍ଞାନିକ ସଂକେତ ବ୍ୟବହାର କରି ଗଣନ କର ଏବଂ ଉତ୍ତର ଲେଖ।
(i) ଯଦି ପ୍ରତି ସେକେଣ୍ଡରେ ଗୋଟିଏ ତାରା ଗଣାଯାଏ, ତେବେ ବ୍ରହ୍ମାଣ୍ଡର ସମସ୍ତ ତାରା ଗଣନା କରିବା ପାଇଁ କେତେ ସମୟ ଲାଗିବ? ବୈଜ୍ଞାନିକ ‘ସଂକେତ ବ୍ୟବହାର କରି ସେକେଣ୍ଡ ଏକକରେ ଉତ୍ତର ଦିଅ।
(ii) ଯଦି ଜଣେ ବ୍ୟକ୍ତି 10 ସେକେଣ୍ଡରେ ଏକ ଗ୍ଲାସ ପାଣି (200 ମି.ଲି.) ପିଇପାରନ୍ତି, ତେବେ ପୃଥ‌ିବୀରେ ଥିବା ସବୁ ପାଣି ପିଇ ଶେଷ କରିବାକୁ ତାଙ୍କୁ କେତେ ସମୟ ଲାଗିବ?
Solution:
(i) ପର୍ଯ୍ୟବେକ୍ଷଣ ଯୋଗ୍ୟ ସ୍ଥାନରେ ଥିବା ତାରାଗୁଡ଼ିକର ଆନୁମାନିକ ସଂଖ୍ୟା :
ବ୍ରହ୍ମାଣ୍ଡରେ ତାରା ସଂଖ୍ୟା : 1 × 10²⁴ (1 ସେକ୍ସଟିଲିୟନ୍)
ଯଦି ଆମେ ପ୍ରତି ସେକେଣ୍ଡରେ ଗୋଟିଏ ତାରା ଗଣନା କରନ୍ତି,
ତେବେ ଆବଶ୍ୟକ ସମୟ = 1 × 10²⁴ ସେକେଣ୍ଡ ।
(ii) ପୃଥିବୀରେ ଥିବା ମୋଟ ଜଳର ଆୟତନ ହେଉଛି 1.38 × 10⁹ କି.ମି.³
ଘନ କି.ମି.କୁ ମିଲିଲିଟରରେ ପରିଣତ କଲେ = 1.386 × 10⁹ × 10¹⁵ ମି.ଲି.। = 1.386 × 10²⁴ ମି.ଲି.।
ପ୍ରତ୍ୟେକ ଗ୍ଲାସ୍ଟର କ୍ଷମତା = 200 ମି.ଲି.।
ଏକ ଗ୍ଲାସ୍ ପାଣି ପିଇବାକୁ ସମୟ ଲାଗେ = 10 ସେକେଣ୍ଡ ।
ସମୁଦାୟ ଗ୍ଲାସ୍ ସଂଖ୍ୟା = 6.93 × 10²¹
∴ ସବୁ ପାଣି ପିଇ ଶେଷ କରିବାକୁ ସମୟ ଲାଗିବ = 6.93 × 10²¹ × 10 = 6.93 × 10²² ସେକେଣ୍ଡ।

2.5 ଇତିହାସ ପୃଷ୍ଠାରୁ

ନିଜେ କରି ଦେଖ : (Page No. 44-45)

Question 1.
2²²⁴ ÷ 4³² ମାନର ଏକକ ସ୍ଥାନରେ ଥ‌ିବା ଅଙ୍କଟି କେତେ? (ସୂଚନା 4 = 2²)
Solution:
2²²⁴ ÷ 4³² = 2²²⁴ ÷ (2²)³² = 2²²⁴ ÷ 2⁶⁴ = 2^(224-64) = 2¹⁶⁰
[∵ (a^m)ⁿ = a^mn]
2 ଘାତର ଏକକ ଅଙ୍କ ପ୍ରତି ଚାରି ଘାତରେ ଏକ ପୁନରାବୃତ୍ତି ଢାଞ୍ଚା ଅନୁସରଣ କରେ ।
2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16 (ଏକକ ଅଙ୍କ 6),
2⁵ = 32 (ଏକକ ଅଙ୍କ 2), 2⁶ = 64 (ଏକକ ଅଙ୍କ 4)
2ର ଘାତ ହେଉଛି 160। ତେଣୁ = \(\frac {160}{4}\) = 40
ଯେତେବେଳେ ଭାଗଶେଷ 0, ଚକ୍ରରେ ଏକକ ସ୍ଥାନରେ ଥିବା ଶେଷ ଅଙ୍କଟି 6 ହେବ।
∴ 2²²⁴ ÷ 4³² ମାନର ଏକକ ସ୍ଥାନରେ ଥିବା ଅଙ୍କଟି 6।

Question 2.
ଗୋଟିଏ ପାତ୍ରରେ 5ଟି ବୋତଲ ଅଛି । ପ୍ରତିଦିନ ଗୋଟିଏ ନୂଆପାତ୍ର ଅଣାଯାଉଥାଏ । 40 ଦିନପରେ ସେଥିରେ କେତେ ନୂଆ ବୋତଲ ଥିବ?
Solution:
ଗୋଟିଏ ପାତ୍ରରେ 5ଟି ବୋତଲ ଅଛି ।
40 ଦିନ ପର୍ଯ୍ୟନ୍ତ ପ୍ରତିଦିନ ଗୋଟିଏ ନୂଆପାତ୍ର ଅଣାଯାଉଥାଏ ।
40 ଦିନପରେ ବୋତଲ ସଂଖ୍ୟା = 40ଟି ପାତ୍ର × 5ଟି ବୋତଲ = 200 ବୋତଲ ।

Question 3.
ନିମ୍ନରେ ଦିଆଯାଇଥବା ସଂଖ୍ୟାକୁ ଦୁଇ କିମ୍ବା ଅଧିକ ଘାତାଙ୍କର ଗୁଣଫଳ ଭାବରେ ତିନୋଟି ଭିନ୍ନଭିନ୍ନ ଉପାୟରେ ଲେଖ । ଘାତାଙ୍କଗୁଡ଼ିକ ଯେକୌଣସି ପୂର୍ବସଂଖ୍ୟା ହୋଇପାରିବ ।
(i) 64³
(ii) 192⁸
(iii) 32^-5
Solution:

ବିକଳ୍ପ ଉତ୍ତର :
ଆମେ ଜାଣିଛେ (a^m)ⁿ = a^mn ଓ a^m+n = a^m × aⁿ
(i) 64³ = (8²)³ = 8^2×3 = 8² × 8³
64³ = (4³)³ = 4^3×3 = 4³ × 4³

(ii) 192⁸ = (64 × 3)⁸ = (8² × 3¹)⁸ = 8^2×8 × 3⁸ = 8¹⁶ × 3⁸
192⁸ = (64 × 3)⁸ = (4³ × 3¹)⁸ = 4²⁴ × 3⁸
192⁸ = (64 × 3)⁸ = (2⁶ × 3¹)⁸ = 2⁴⁸ × 3⁸

(iii) 32^-5 = (2⁵)^-5 = 2^5×(-5)
(32)^-5 = (2⁵)^-5 = 2^-25 = 2^(-5-10-10) = 2^-5 × 2^-10 × 2^-10
(32)^-5 = (2⁵)^-5 = 2^-25 = 2^(10-35) = 2¹⁰ × 2^-35

Question 4.
ନିମ୍ନରେ ଦିଆଯାଇଥିବା ପ୍ରତ୍ୟେକ ଉକ୍ତିକୁ ପରୀକ୍ଷା କର ଏବଂ କେଉଁଗୁଡ଼ିକ ‘ସର୍ବଦା ସତ୍ୟ’, ‘ବେଳେବେଳେ ସତ୍ୟ’ କିମ୍ବା ‘ଆଦୌ’ ସତ୍ୟ ନୁହେଁ? ନିର୍ଣ୍ଣୟ କର । ତୁମର ଯୁକ୍ତିକୁ ବର୍ଣ୍ଣନା କର ।
(i) ଘନ ସଂଖ୍ୟାଗୁଡ଼ିକ ମଧ୍ୟ ବର୍ଗ ସଂଖ୍ୟା ଅଟନ୍ତି ।
(ii) ଚତୁର୍ଥ ଘାତାଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାଗୁଡ଼ିକ ମଧ୍ୟ ବର୍ଗ ସଂଖ୍ୟା ଅଟନ୍ତି ।
(iii) ଗୋଟିଏ ସଂଖ୍ୟାର ପଞ୍ଚମ ଘାତାଙ୍କ ସେହି ସଂଖ୍ୟାର ଘନଦ୍ଵାରା ବିଭାଜ୍ୟ ।
(iv) ଦୁଇଟି ଘନସଂଖ୍ୟାର ଗୁଣଫଳ ଏକ ଘନସଂଖ୍ୟା ଅଟେ ।
(v) q⁴⁶ ସଂଖ୍ୟାଟି ଉଭୟ ଚତୁର୍ଥ ଘାତାଙ୍କ ଓ ଷଷ୍ଠ ଘାତାଙ୍କ ଅଟେ । (q ଏକ ମୌଳିକ ସଂଖ୍ୟା ଅଟେ)।
Solution:
(i) ଘନ ସଂଖ୍ୟାଗୁଡ଼ିକ ମଧ୍ୟ ବର୍ଗ ସଂଖ୍ୟା ଅଟନ୍ତି ।
ଉଦାହରଣ : 64 = 4³ = 8² ହେଉଛି ଗୋଟିଏ ଘନ ଏବଂ ବର୍ଗ ସଂଖ୍ୟା ।
ଯେପରିକି 8 = 2³ ହେଉଛି ଗୋଟିଏ ଘନସଂଖ୍ୟା କିନ୍ତୁ ବର୍ଗସଂଖ୍ୟା ନୁହେଁ ଏବଂ
9 = 3² ହେଉଛି ଗୋଟିଏ ବର୍ଗସଂଖା କିନ୍ତୁ ଘନସଂଖ୍ୟା ନୁହେଁ ।
ତେଣୁ ଏହି ଉକ୍ତିଟି ବେଳେବେଳେ ସତ୍ୟ ।

(ii) ଚତୁର୍ଥ ଘାତାଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାଗୁଡ଼ିକ ମଧ୍ୟ ବର୍ଗ ସଂଖ୍ୟା ଅଟନ୍ତି ।
ଏହି ଉକ୍ତିଟି ସର୍ବଦା ସତ୍ୟ ।
ଉଦାହରଣ : ଏକ ଚତୁର୍ଥ ଘାତକୁ x⁴ ଲେଖାଯାଇପାରିବ ।
ଏହାକୁ (x²)² ଭାବରେ ପୁନଃ ଲେଖାଯାଇପାରେ, ଯାହା x²ର ବର୍ଗ ।

(iii) ଗୋଟିଏ ସଂଖ୍ୟାର ପଞ୍ଚମ ଘାତାଙ୍କ ସେହି ସଂଖ୍ୟାର ଘନଦ୍ଵାରା ବିଭାଜ୍ୟ।
ମନେକର ସଂଖ୍ୟାଟି = n
ଏହାର ପଞ୍ଚମ ଘାତ = n⁵ ଏବଂ ଘନ = n³
ଯେହେତୁ n⁵ = n³ × n²
∴ n⁵ ଯେକୌଣସି ପୂର୍ବସଂଖ୍ୟା ପାଇଁ n³ ଦ୍ଵାରା ବିଭାଜ୍ୟ।
ତେଣୁ ଏହି ଉକ୍ତିଟି ସର୍ବଦା ସତ୍ୟ।

(iv) ଦୁଇଟି ଘନସଂଖ୍ୟାର ଗୁଣଫଳ ଏକ ଘନସଂଖ୍ୟା ଅଟେ ।
ମନେକର ଘନସଂଖ୍ୟାଦ୍ଵୟ a³ ଓ b³
ସେମାନଙ୍କର ଗୁଣଫଳ = a³ × b³ = (a × b)³
ଯେହେତୁ ଗୁଣଫଳ ଅନ୍ୟ ସଂଖ୍ୟା (a × b)ର ଘନ ଭାବରେ ପ୍ରକାଶ କରାଯାଇପାରେ,
ଏହା ସର୍ବଦା ଗୋଟିଏ ଘନସଂଖ୍ୟା ହେବ।
ତେଣୁ ଉକ୍ତିଟି ସର୍ବଦା ସତ୍ୟ ଅଟେ।

(v) q⁴⁶ ସଂଖ୍ୟାଟି ଉଭୟ ଚତୁର୍ଥ ଘାତାଙ୍କ ଓ ଷଷ୍ଠ ଘାତାଙ୍କ ଅଟେ । (q ଏକ ମୌଳିକ ସଂଖ୍ୟା ଅଟେ) ।
q⁴⁶ କୁ ଚତୁର୍ଥ ଘାତରେ ପରିଣତ କରିବା ପାଇଁ 46 ସଂଖ୍ୟାଟି 4 ଦ୍ଵାରା ବିଭାଜ୍ୟ ହେବ, ଯାହା ସମ୍ଭବ ନୁହେଁ ।
q⁴⁶ କୁ ଷଷ୍ଠ ଘାତରେ ପରିଣତ କରିବା ପାଇଁ 46 ସଂଖ୍ୟାଟି 6 ଦ୍ଵାରା ବିଭାଜ୍ୟ ହେବ, ଯାହା ସମ୍ଭବ ନୁହେଁ ।
∴ q⁴⁶ ସଂଖ୍ୟାଟି ଉଭୟ ଚତୁର୍ଥ ଘାତାଙ୍କ ଓ ଷଷ୍ଠ ଘାତାଙ୍କ ହୋଇପାରିବ ନାହିଁ । ତେଣୁ ଉକ୍ତିଟି ସତ୍ୟ ନୁହେଁ ।

Question 5.
ନିମ୍ନ ରାଶିଗୁଡ଼ିକୁ ସରଳ କର ଏବଂ ଘାତାଙ୍କୀୟ ରୂପରେ ଲେଖ :
(i) 10^-2 × 10^-5
(ii) 5⁷ ÷ 5⁴
(iii) 9^-7 ÷ 9⁴
(iv) (13^-2)^-3
(v) m⁵n¹²(mn)⁹
Solution:
(i) 10^-2 × 10^-5 = 10^-2-5 = 10^-7 [∵ a^m × aⁿ = a^m+n]
(ii) 5⁷ ÷ 5⁴ = 5^7-4 = 5³ [∵ a^m ÷ aⁿ = a^m-n]
(iii) 9^-7 ÷ 9⁴ = 9^-7-4 = 9^-11 [∵ a^m ÷ aⁿ = a^m-n]
(iv) (13^-2)^-3 = (13)^-2×(-3) = (13)⁶ [∵ (a^m)ⁿ = a^mn]
(v) m⁵n¹²(mn)⁹ = m⁵n¹²m⁹n⁹ = m⁵⁺⁹ . n¹²⁺⁹ = m¹⁴n²¹ [∵ (a^m)ⁿ = a^mn, a^m × aⁿ = a^m+n]

Question 6.
ଯଦି 12² = 144, ତେବେ ନିମ୍ନଲିଖୂତ ପରିପ୍ରକାଶଗୁଡ଼ିକର ମାନ କେତେ ହେବ?
(i) (1.2)²
(ii) (0.12)²
(iii) (0.012)²
(iv) 120²
Solution:
(i) (1.2)² = \(\left(\frac{12}{10}\right)^2=\frac{144}{100}\) = 1.44
(ii) (0.12)² = \(\left(\frac{12}{100}\right)^2=\frac{144}{10000}\) = 0.0144
(iii) (0.012)² = \(\left(\frac{12}{1000}\right)^2=\frac{144}{10,00,000}\) = 0.000144
(iv) 120² = (12 × 10)² = 144 × 100 = 14400

Question 7.
ସମାନ ମୂଲ୍ୟବିଶିଷ୍ଟ ସଂଖ୍ୟାଗୁଡ଼ିକ ଗୋଲ ବୁଲାଅ-
2⁴ × 3⁶, 6⁴ × 3², 6¹⁰, 18² × 6², 6²⁴
Solution:

Question 8.
ନିମ୍ନଲିଖୁତ ପ୍ରତ୍ୟେକ ସ୍ଥଳରେ ବୃହତ୍ତର ସଂଖ୍ୟାଟିକୁ ଚିହ୍ନଟ କର ।
(i) 4³ କିମ୍ବା 3⁴
(ii) 2⁸ କିମ୍ବା 8²
(iii) 100² କିମ୍ବା 2¹⁰⁰
Solution:
(i) 4³ = 6⁴, 3⁴ = 81
⇒ 81 > 64
∴ 3⁴ > 4³

(ii) 2⁸ = 256, 8² = 64
⇒ 256 > 64
∴ 2⁸ > 8²

(iii) 100² = 10000, 2¹⁰⁰ = (2¹⁰)¹⁰ = (1024)¹⁰
1024 ସଂଖ୍ୟାଟି 100 ଠାରୁ ବଡ଼,
ଏହାର ଘାତ 10 ବୃଦ୍ଧି ହେଲେ, 10000 ଠାରୁ ବଡ଼ ହେବ ।
∴ 2¹⁰⁰ > 100²

Question 9.
ଏକ ଡାଏରୀ ଫାର୍ମ ବର୍ଷକୁ 8.5 ବିଲିୟନ ପ୍ୟାକେଟ୍ କ୍ଷୀର ଉତ୍ପାଦନ କରିବାକୁ ଯୋଜନା କରୁଛି । ସେମାନେ ପ୍ରତ୍ୟେକ ପ୍ୟାକେଟ୍‌ ପାଇଁ ଏକ ଅନନ୍ୟ ID (ପରିଚୟ ପତ୍ର) କୋର୍ଡ଼ ଚାହୁଁଛନ୍ତି । ଯଦି ସେମାନେ 0 – 9 ଅଙ୍କଗୁଡ଼ିକୁ ବ୍ୟବହାର କରିବାକୁ ସ୍ଥିର କରନ୍ତି, ତେବେ କୋଡ଼ରେ କେତୋଟି ଅଙ୍କ ରହିବ?
Solution:
ଗୋଟିଏ ବର୍ଷରେ ଉତ୍ପାଦିତ ମୋଟ ପ୍ୟାକେଟ୍ ସଂଖ୍ୟା = 8.5 ବିଲିୟନ୍ ।
ଆବଶ୍ୟକ କୋଡ଼୍ ସଂଖ୍ୟା = 8.5 × 10⁹ କୋଡ଼୍ ।
0 ରୁ 9 ପର୍ଯ୍ୟନ୍ତ ଅଙ୍କଗୁଡ଼ିକ ବ୍ୟବହାର କରି ID କୋଡ଼ପାଇଁ ପ୍ରତ୍ୟେକ ସଂଖ୍ୟାର 10ଟି ବିକଳ୍ପ ଅଛି ।
n ସଂଖ୍ୟା ବିଶିଷ୍ଟ କୋଡ଼୍ ତିଆରି କରିବାକୁ ସମ୍ଭାବ୍ୟ ମୋଟ କୋଡ୍ ସଂଖ୍ୟା = 10ⁿ
∴ 10ⁿ ≥ 8.5 × 10⁹
10⁹ = 1,00,00,00,000
10ⁿ ≥ 8.5 × 10⁹ର ସର୍ବନିମ୍ନ ମୂଲ୍ୟ ସନ୍ତୁଷ୍ଟ କରିବାକୁ nର ମୂଲ୍ୟ = 10
∴ କୋଡ଼ର ଅଙ୍କସଂଖ୍ୟା 10¹⁰

Question 10.
64 ଏକ ବର୍ଗ ସଂଖ୍ୟା (8²) ଏବଂ ଘନସଂଖ୍ୟା (4³) ସହ ସମାନ । ଏହିପରି ଅନ୍ୟ କୌଣସି ସଂଖ୍ୟା ଅଛି କି ଯାହା ଉଭୟ ବର୍ଗ ଓ ଘନ ଅଟେ ? ସାଧାରଣ ଭାବେ ଏହିପରି ସଂଖ୍ଯାଗୁଡ଼ିକୁ ପ୍ରକାଶ କରିବାର କୌଣସି ଉପାୟ ଅଛି କି?
Solution:

ହଁ ଏହିପରି ଅନେକ ସଂଖ୍ୟା ଅଛି ଯାହା ଉଭୟ ବର୍ଗ ଓ ଘନ ଅଟେ।
ସାଧାରଣ ନିୟମ : ଏକ ସଂଖ୍ୟା ଉଭୟ ପୂର୍ବବର୍ଗ ଏବଂ ପୂର୍ବଘନ ସଂଖ୍ୟା ହେବ, ଯଦି ଏହା ଏକ ପୂର୍ଣ୍ଣ ଷଷ୍ଠ ଘାତ ହୁଏ ।
ଅର୍ଥାତ୍ କୌଣସି ପୂର୍ବସଂଖ୍ୟା x ପାଇଁ ଏହାକୁ x⁶ ଭାବରେ ଲେଖାଯାଇପାରିବ।
ତେବେ ଏକ ସଂଖ୍ୟା ଉଭୟ ପୂର୍ବବର୍ଗ ଓ ପୂର୍ଣ୍ଣ ଘନ ହେବ।

Question 11.
ଏକ ଡିଜିଟାଲ ଲକର୍‌ରେ 5 ଅକ୍ଷର ବିଶିଷ୍ଟ ଆଲଫାନ୍ୟୁମେରିକ୍ (ଏଥିରେ ଉଭୟ ଅଙ୍କ ଏବଂ ଅକ୍ଷର ରହିପାରିବ) ପାସ୍କୋର୍ଡ଼ ଅଛି, କୋର୍ଡ଼ଗୁଡ଼ିକର କିଛି ଉଦାହରଣ ହେଉଛି G89PO, 38098, BRJKW ଏବଂ 003AZ । ଏହିପରି କେତୋଟି କୋର୍ଡ଼ ସମ୍ଭବ?
Solution:
ଅକ୍ଷର ସଂଖ୍ୟା (A – Z) = 26
ଅଙ୍କ ସଂଖ୍ୟା (0 – 9) = 10
ତେଣୁ ପ୍ରତ୍ୟେକ ଅକ୍ଷର 36ଟି ଆଲ୍‌ଫାନ୍ୟୁମେରିକ୍ ଅକ୍ଷର ମଧ୍ୟରୁ ଯେକୌଣସି ହୋଇପାରେ ।
ଏହା ସହିତ ପ୍ରତ୍ୟେକ 5ଟି ସ୍ଥାନ ପାଇଁ 36ଟି ବିକଳ୍ପ ଅଛି ।
ମୋଟ୍ କୋଡ଼୍ = 36⁵ = 6,04,66,176
ତେଣୁ, 5 ଅକ୍ଷର ବିଶିଷ୍ଟ 6,04,66,176ଟି ଆଲଫାନ୍ୟୁମେରିକ୍ ପାସ୍‌ର୍ଡ଼ ସମ୍ଭବ ।

Question 12.
ସମଗ୍ର ବିଶ୍ଵରେ ମେଣ୍ଢାମାନଙ୍କ ସଂଖ୍ୟା 2024 ମସିହା ପ୍ରାୟ 10⁹ ଏବଂ ଛେଳିମାନଙ୍କର ସଂଖ୍ୟା ପ୍ରାୟ ସମାନ । ମେଣ୍ଢା ଓ ଛେଳିମାନଙ୍କର ମୋଟ ସଂଖ୍ୟା ନିମ୍ନୋକ୍ତ ବିକଳ୍ପଗୁଡ଼ିକ ମଧ୍ୟରୁ କେଉଁଟି ?
(i) 20⁹
(ii) 10¹¹
(iii) 10¹⁰
(iv) 10¹⁸
(v) 2 × 10⁹
(vi) 10⁹ + 10⁹
Solution:
ମେଣ୍ଢାମାନଙ୍କ ସଂଖ୍ୟା = 10⁹
ଛେଳିମାନଙ୍କ ସଂଖ୍ୟା = 10⁹
ମେଣ୍ଢା ଓ ଛେଳିମାନଙ୍କର ମୋଟ ସଂଖ୍ୟା = 10⁹ + 10⁹ = 2 × 10⁹
∴ (v) ବିକଳ୍ପଟି ଠିକ୍ ।

Question 13.
ହିସାବ କର ଏବଂ ବୈଜ୍ଞାନିକ ସଂକେତରେ ଉତ୍ତର ଲେଖ ।
(i) ଯଦି ବିଶ୍ଵର ପ୍ରତ୍ୟେକ ବ୍ୟକ୍ତିଙ୍କ ପାଖରେ 30 ଖଣ୍ଡ ପୋଷାକ ଥାଏ, ତେବେ ମୋଟ ପୋଷାକ ସଂଖ୍ୟା ନିର୍ଣ୍ଣୟ କର।
(ii) ବିଶ୍ଵରେ ପ୍ରାୟ 100 ନିୟୁତ ମହୁଫେଣା ଅଛି; ପ୍ରତ୍ୟେକ ଫେଣାରେ ପ୍ରାୟ 50,000 ମହୁମାଛି ଥାଆନ୍ତି, ତେବେ ମହୁମାଛି ସଂଖ୍ୟା ନିର୍ଣ୍ଣୟ କର।
(iii) ମାନବ ଶରୀରରେ ପ୍ରାୟ 38 ଟ୍ରିଲିୟନ୍ ବ୍ୟାକ୍ଟେରିଆ କୋଷ ଅଛି, ପୃଥୁବୀରେ ଥିବା ସମସ୍ତ ମଣିଷଙ୍କ ଶରୀରରେ ଥିବା ବ୍ୟାକ୍ଟେରିଆ ସଂଖ୍ୟା ନିର୍ଣ୍ଣୟ କର।
(iv) ଜୀବନକାଳରେ ଖାଇବାରେ ବିତାଇଥବା ମୋଟ ସମୟକୁ ସେକେଣ୍ଡ ଏକକରେ ନିର୍ଣ୍ଣୟ କର।
Solution:
(i) ଆମେ ଜାଣୁ ଯେ, ବିଶ୍ଵର ଜନସଂଖ୍ୟା = 8.2 ବିଲିୟନ୍ = 8.2 × 10⁹
ପ୍ରତ୍ୟେକ ବ୍ୟକ୍ତିଙ୍କ ପାଖରେ ଥିବା ପୋଷାକ ସଂଖ୍ୟା = 30 ଖଣ୍ଡ ।
∴ ବିଶ୍ଵର ମୋଟ ପୋଷାକ ସଂଖ୍ୟା = 8.2 × 10⁹ × 30
= 246 × 10⁹
= 2.46 × 10¹¹ ଟି ପୋଷାକ।

(ii) ବିଶ୍ଵର ମହୁଫେଣା ସଂଖ୍ୟା = 100 ନିୟୁତ = 1 × 10⁸
ପ୍ରତ୍ୟେକ ଫେଣାରେ ମୁହମାଛି ଥାଆନ୍ତି = 5,000 = 5 × 10⁴
ମୋଟ ମହୁମାଛି ସଂଖ୍ୟା = 1 × 10⁸ × 5 × 10⁴ = 5 × 10¹² ମହୁମାଛି ।

(iii) ମାନବ ଶରୀରରେ ବ୍ୟାକ୍ଟେରିଆ କୋଷ ସଂଖ୍ୟା = 38 ଟ୍ରିଲିୟନ୍ = 3.8 × 10¹³
ପୃଥ‌ିବୀର ଜନସଂଖ୍ୟା = 8.2 ବିଲିୟନ୍ = 8.2 × 10⁹
ପୃଥ‌ିବୀରେ ଥ‌ିବା ସମସ୍ତ ମଣିଷଙ୍କ ଶରୀରରେ ଥିବା ବ୍ୟାକ୍ଟେରିଆ ସଂଖ୍ୟା = 3.8 × 10¹³ × 8.2 × 10⁹
= 31.16 × 10¹³⁺⁹
= 31.16 × 10²²

(iv) ମନେକର ଦୈନିକ ଖାଇବାରେ ବିତାଉଥବା ହାରାହାରି ସମୟକୁ 1.5 ଘଣ୍ଟା ଧରାଯାଉ ।
ସେକେଣ୍ଡରେ ପ୍ରକାଶ କଲେ = 1.5 × 60 × 60 = 5400 ସେକେଣ୍ଡ = 5.4 × 10³ ସେକେଣ୍ଡ ।
ଜଣେ ବ୍ୟକ୍ତିର ଜୀବନକାଳ ସମୟ ଆନୁମାନିକ = 70 ବର୍ଷ ।
ଜୀବନକାଳ ସମୟକୁ ସେକେଣ୍ଡରେ ପ୍ରକାଶ କଲେ = 70 × 365 × 24 × 60 × 60
= 161,148,960,000
= 1.61 × 10¹¹
ଖାଇବାରେ ବିତାଇଥବା ମୋଟ ସମୟ = 5.4 × 10³ × 1.61 × 10¹¹
= 8.694 × 10³⁺¹¹
= 8.7 × 10¹⁴ ସେକେଣ୍ଡ।

Question 14.
1 ଅରବ / 1 ବିଲିୟନ ସେକେଣ୍ଡ ପୂର୍ବରୁ ତାରିଖ କେତେ ଥିଲା?
Solution:
1 ଅରବ = 10⁹
ମିନିଟ୍ = \(\frac{1,000,000,000}{60}\)
ଘଣ୍ଟା = \(\frac{1,000,000,000}{60 \times 60}\)
ଦିନ = \(\frac{1,000,000,000}{60 \times 60 \times 24}\)
ବର୍ଷ = \(\frac{1,000,000,000}{365 \times 60 \times 60 \times 24}\)
ଆମେ କ୍ୟାଲେଣ୍ଡର ଦେଖୁଲେ ଏହା ପ୍ରାୟ 31 ବର୍ଷ, 8 ମାସ ଓ 15 ଦିନ ହେବ ।
ମନେକର ଆଜିର ଦିନ 1 ଜାନୁୟାରୀ 2026
ଆଜିର ଦିନର 31 ବର୍ଷ, 8 ମାସ ଓ 15 ଦିନ ପୂର୍ବେ ଥୁଲା = 24 ଏପ୍ରିଲ୍ 1994

Class 8 Maths Chapter 2 MCQ Odia Medium

ସମ୍ଭାବ୍ୟ ଚାରୋଟି ଉତ୍ତର ମଧ୍ୟରୁ ଠିକ୍ ଉତ୍ତରଟି ବାଛି ଲେଖ ।

Question 1.
2⁴ ର ମୂଲ୍ୟ କେତେ?
(a) 8
(b) 4
(c) 16
(d) 32
Answer:
(c) 16

Question 2.
10^-10 ର ମୂଲ୍ୟ କେତେ?
(a) \(\frac{1}{(10)^{10}}\)
(b) \(\frac{1}{(100)^{10}}\)
(c) (100)¹⁰
(d) 1000
Answer:
(a) \(\frac{1}{(10)^{10}}\)

Question 3.
5² × 5^-2 ର ମୂଲ୍ୟ କେତେ ?
(a) 1
(b) 5
(c) 5⁴
(d) 5^-4
Answer:
(a) 1

Question 4.
(-9)³ × (-9)⁸ ର ମୂଲ୍ୟ କେତେ?
(a) (-9)⁵
(b) (-9)^-5
(c) (-9)¹¹
(d) (-9)^-11
Answer:
(c) (-9)¹¹

Question 5.
625 କୁ ଘାତ ରାଶିରେ ପ୍ରକାଶ କଲେ କେତେ ହେବ ?
(a) 5³
(b) 5²
(c) 5⁴
(d) 5⁵
Answer:
(c) 5⁴

Question 6.
2.08 × 10^-8 ର ସାଧାରଣ ରୂପ କେତେ?
(a) 0.0000208
(b) 0.000028
(c) 0.000208
(d) 0.0002080
Answer:
(a) 0.0000208

Question 7.
ଯଦି 3 × 10² ଟି ଗଛ ଅଛି ଓ ପ୍ରତ୍ୟେକ ଗଛରେ 10⁴ ଟି ପତ୍ର ଅଛି, ତେବେ ମୋଟ ପତ୍ର ସଂଖ୍ୟା କେତେ ?
(a) 3 × 10¹⁶
(b) 3 × 10⁸
(c) 30 × 10¹²
(d) 10¹⁸
Answer:
(a) 3 × 10¹⁶

Question 8.
5⁰ ର ମୂଲ୍ୟ କେତେ?
(a) 1
(b) 0
(c) -1
(d) 2
Answer:
(a) 1

Question 9.
1 କୋଟିକୁ 10 ର ଘାତରେ ପ୍ରକାଶ କଲେ କେତେ ହେବ?
(a) 10⁵
(b) 10⁶
(c) 10⁷
(d) 10⁸
Answer:
(c) 10⁷

Question 10.
(2^-1 × 3^-1)^-1 ର ମୂଲ୍ୟ କେତେ?
(a) 4
(b) 5
(c) 6
(d) 1
Answer:
(c) 6

Question 11.
3⁴ × 3³ ÷ 3⁵ = _________
(a) -9
(b) \(\frac {1}{9}\)
(c) 3⁴
(d) 9
Answer:
(d) 9

Question 12.
3⁹ × 3⁵ ÷ 9⁷ = ?
(a) 2
(b) 3
(c) 5
(d) 1
Answer:
(d) 1

Question 13.
1000 ଯେଉଁ ଆଧାରର ତୃତୀୟ ଘାତ, ସେହି ଆଧାରର nତମ ଘାତ କେତେ ?
(a) n
(b) 10ⁿ
(c) 10n
(d) n¹⁰
Answer:
(b) 10ⁿ

Question 14.
(-2)ⁿ = -512 ଏଠାରେ n ଏକ-
(a) ଗଣନ ସଂଖ୍ୟା
(b) ପୂର୍ଣ୍ଣସଂଖ୍ୟା
(c) ପରିମେୟ ସଂଖ୍ୟା
(d) ବାସ୍ତବ ସଂଖ୍ୟା
Answer:
(a) ଗଣନ ସଂଖ୍ୟା

Question 15.
8³ × 4² ÷ 16² = ___________
(a) 2
(b) 4
(c) 8
(d) 16
Answer:
(b) 4

Question 16.
256 ଯେଉଁ ଆଧାରର ଚତୁର୍ଥ ଘାତ ତା’ର ଘନ କେତେ ?
(a) 8
(b) 16
(c) 32
(d) 64
Answer:
(d) 64

Question 17.
625କୁ ଘାତ ରାଶି ରେ ପରି ଣତ କଲେ _________ ହେବ ।
(a) 25²
(b) 5³
(c) 5⁴
(d) କେଉଁଟି ନୁହେଁ
Answer:
(c) 5⁴

Question 18.
(-8)³ ର ମାନ _________
(a) 512
(b) -512
(c) 64
(d) -64
Answer:
(b) -512

Question 19.
10 ର ଚତୁର୍ଥ ଘାତ = _________
(a) 10
(b) 1000
(c) 10000
(d) କେଉଁଟି ନୁହେଁ
Answer:
(c) 10000

Question 20.
5ର _________ ଘାତ 625
(a) 2
(b) 4
(c) 3
(d) 5
Answer:
(b) 4

Question 21.
5 ଆଧାରର ଚତୁର୍ଥ ଘାତ _________ ଆଧାରର ଦ୍ବିତୀୟ ଘାତ ସହ ସମାନ ।
(a) 5
(b) 15
(c) 25
(d) 20
Answer:
(c) 25

Question 22.
256 ଯେଉଁ ଆଧାରର ଚତୁର୍ଥ ଘାତ, ତାହାର 2ୟ ଘାତ _________
(a) 4
(b) 8
(c) 32
(d) 16
Answer:
(d) 16

Question 23.
(4² × 4³) ÷ 4⁵ କୁ ସରଳ କଲେ _________ ହେବ ।
(a) 4⁰
(b) 1
(c) 4¹
(d) କେଉଁଟି ନୁହେଁ
Answer:
(b) 1

Question 24.
(64)³ କୁ ମୌଳିକ ଆଧାର ବିଶିଷ୍ଟ ଘାତରାଶିରେ ପ୍ରକାଶ କଲେ _________ ହେବ ।
(a) 4³
(b) 4⁹
(c) 2⁶
(d) 2¹⁸
Answer:
(d) 2¹⁸

Question 25.
(125)^m-1 = _________
(a) 5^3m-3
(b) 5^3m-1
(c) 5^m-3
(d) 5^m-1
Answer:
(a) 5^3m-3

Question 26.
କେଉଁ କ୍ଷେତ୍ରରେ n ଏକ ଗଣନ ସଂଖ୍ୟା ହେବ?
(a) 5ⁿ = 100
(b) 4ⁿ = 1024
(c) 5ⁿ = 1250
(d) \(\left(\frac{2}{3}\right)^n=\frac{32}{15}\)
Answer:
(b) 4ⁿ = 1024

ସଂକ୍ଷେପରେ ଉତ୍ତର ଲେଖ।

Question 1.
ନିମ୍ନ ଘାତରାଶିମାନଙ୍କର ଆଧାର ଓ ଘାତାଙ୍କ ଦର୍ଶାଇ ମାନ ନିର୍ଣ୍ଣୟ କର ।
(i) (1)¹⁵
(ii) (-1)¹¹
(iii) (-1)¹⁸
(iv) (9)⁵
(v) (-2)⁵
(vi) \(\left(\frac{1}{2}\right)^6\)
(vii) \(\left(\frac{2}{3}\right)^5\)
(viii) (5 × 2)⁴
(ix) (10)⁷
(x) (-10)⁵
Answer:
(i) (1)¹⁵ ରେ ଆଧାର 1 ଓ ଘାତାଙ୍କ 15 ଅଟେ ।
(1)¹⁵ ର ମାନ = 1 × 1 × 1 × 1 × ….× 1 (15 ଥର) = 1

(ii) (-1)¹¹ ରେ ଆଧାର -1 ଓ ଘାତାଙ୍କ 11
(-1)¹¹ ର ମାନ = -1 [∵ (−1)^m = -1, ଯେଉଁଠାରେ m ଏକ ଅଯୁଗ୍ମ ସଂଖ୍ୟା]

(iii) (-1)¹⁸ ରେ ଆଧାର -1 ଓ ଘାତାଙ୍କ 18
(-1)¹⁸ ର ମାନ = 1 [∵ (-1)^m = 1, ଯେଉଁଠାରେ m ଏକ ଯୁଗ୍ମ ସଂଖ୍ୟା)

(iv) (9)⁵ ରେ ଆଧାର 9 ଓ ଘାତାଙ୍କ 5
(9)⁵ = 9 × 9 × 9 × 9 × 9 = 59049
∴ (9)⁵ ର ମାନ = 59049

(v) (-2)⁵ ରେ ଆଧାର -2 ଓ ଘାତାଙ୍କ 5।
(-2)⁵ = (-2) × (-2) × (-2) × (-2) × (-2) = -32
∴ (-2)⁵ ର ମାନ = -32

(vi) \(\left(\frac{1}{2}\right)^6\) ରେ ଆଧାର \(\frac {1}{2}\) ଓ ଘାତାଙ୍କ 6।
\(\left(\frac{1}{2}\right)^6=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{64}\)
∴ \(\left(\frac{1}{2}\right)^6\) ର ମାନ = \(\frac {1}{64}\)

(vii) \(\left(\frac{2}{3}\right)^5\) ରେ ଆଧାର \(\frac {2}{3}\) ଓ ଘାତାଙ୍କ 5।
\(\left(\frac{2}{3}\right)^5=\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)=\frac{32}{243}\)
∴ \(\left(\frac{2}{3}\right)^5\) ର ମାନ = \(\frac {32}{243}\)

(viii) (5 × 2)⁴ = (10)⁴
= 10 × 10 × 10 × 10
= 10000
(5 × 2)⁴ ରେ ଆଧାର 10 ଓ ଘାତାଙ୍କ 4
∴ ମାନ = 10000

(ix) (10)⁷ ରେ ଆଧାର 10 ଓ ଘାତାଙ୍କ 7।
(10)⁷ = 10 × 10 × 10 × 10 × 10 × 10 × 10 = 10000000
∴ (10)⁷ ର ମାନ = 10000000

(x) (-10)⁵ ରେ ଆଧାର -10 ଓ ଘାତାଙ୍କ 5
(-10)⁵ = (-10)(-10)(-10)(-10)(-10) = -100000
∴ (-10)⁵ ର ମାନ = -100000

Question 2.
\(\frac{2^3 \times 3^4}{3 \times 2^5}\) କୁ ସରଳ କର।
Answer:
\(\frac{2^3 \times 3^4}{3 \times 2^5}=\left(\frac{2^3}{2^5}\right) \times\left(\frac{3^4}{3}\right)=\frac{1}{2^{5-3}} \times 3^{4-1}\) = \(\frac{1}{2^2} \times 3^3=\frac{27}{4}\)

Question 3.
ନିମ୍ନଲିଖିତ ରାଶିଗୁଡ଼ିକୁ ଏକ ଆଧାର ବିଶିଷ୍ଟ ଘାତ ରାଶି ରୂପେ ପ୍ରକାଶ କର।
(i) (9)³ × (27)⁴
(ii) (8)³ × (-4)⁴
(iii) (7)⁸ × (-7)⁵
Answer:
(i) (9)³ × (27)⁴ = (3²)³ × (3³)⁴
= 3^2×3 × 3^3×4 [∵ (a^m)ⁿ = a^m×n]
= 3⁶ × 3¹²
= 3⁶⁺¹²
= 3¹⁸

(ii) (8)³ × (-4)⁴ = 8³ × (-1 × 4)⁴
= 8³ × (-1)⁴ × 4⁴ [∵ (ab)^m = a^m b^m]
= (-1)⁴ × 8³ × 4⁴
= 1 × (2³)³ × (2²)⁴
= 2⁹ × 2⁸ [∵ (-1)⁴ = 1]
= 2⁹⁺⁸
= 2¹⁷

(iii) (7)⁸ × (-7)⁵
= 7⁸ × (-1 × 7)⁵ [∵ (ab)^m = a^m . b^m]
= 7⁸ × (-1)⁵ × 7⁵
= (-1)⁵ × 7⁸ × 7⁵
= -1 × 7⁸⁺⁵
= -1 × 7¹³
= (-1)¹³ × 7¹³
= (-1 × 7)¹³
= (-7)¹³

Question 4.
ମୌଳିକ ଆଧାର ବିଶିଷ୍ଟ ଘାତରାଶିରେ ପ୍ରକାଶ କର।
(i) (9)⁷
(ii) (125)^m-1
Answer:
(i) (9)⁷ = (3²)⁷ = 3^2×7 = 3¹⁴
[∵ 9 ର ଏକ ମୌଳିକ ଗୁଣନୀୟକ 3]

(ii) (125)^m-1 = (5³)^m-1 = 5^3(m-1) = 5^3m-3
[∵ 125 ର ଏକ ମୌଳିକ ଗୁଣନୀୟକ 5]

Question 5.
ମାନ ନିର୍ଣ୍ଣୟ କର ।
(i) (3¹¹ × 4⁵) ÷ (4⁴ × 3⁶)
(ii) (4³ × 4² × 4) ÷ (2⁴ × 2³ × 2²)
Answer:
(i) (3¹¹ × 4⁵) ÷ (4⁴ × 3⁶)
= \(\frac{3^{11} \times 4^5}{4^4 \times 3^6}\)
= \(\frac{3^{11}}{3^6} \times \frac{4^5}{4^4}\)
= 3^11-6 × 4^5-4
= 3⁵ × 4
= 243 × 4
= 972

(ii) (4³ × 4⁵ × 4) ÷ (2⁴ x 2³ x 2²)
= 4³⁺²⁺¹ ÷ 2⁴⁺³⁺²
= 4⁶ ÷ 2⁹
= (2²)⁶ ÷ 2⁹
= 2¹² ÷ 2⁹
= 2^12-9
= 2³
= 8

Question 6.
ନିମ୍ନ ଘାତରାଶି ବିଶିଷ୍ଟ ପରିପ୍ରକାଶଗୁଡ଼ିକୁ ସରଳ କର।
(i) \(\left(\frac{2}{9}\right)^5 \div\left(-\frac{2}{9}\right)^4\)
(ii) \(\left(\frac{1}{25}\right)^4 \div 5^4\)
(iii) \(\frac{3^8 \times a^5}{27 \times a^2}\) (a ≠ 0)
(iv) (4² × 4³) ÷ 4⁵
(v) \(\left(\frac{-2}{3}\right)^9 \div\left(\frac{2}{3}\right)^7\)
(vi) \(\left\{\left(\frac{1}{2}\right)^3\right\}^2 \div\left(\frac{1}{4}\right)^3\)
Answer:

Question 7.
ନିମ୍ନଲିଖୂତ ରାଶିଗୁଡ଼ିକୁ ଏକ ଆଧାର ବିଶିଷ୍ଟ ଘାତ ରାଶି ରୂପେ ପ୍ରକାଶ କର।
(i) \(\frac{7^4}{3^4}\)
(ii) 3⁹ ÷ 4⁹
(iii) \(\left(\frac{\mathrm{a}}{\mathrm{~b}}\right)^7+\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^3\)
(iv) \(\left(\frac{a}{b}\right)^4 \div\left(-\frac{b}{a}\right)^3\)
Answer:

Question 8.
ସରଳ କର
(i) (2² × 2)³
(ii) (ab)⁵ × a³ × b²
(iii) \(\left(\frac{a}{b}\right)^7 \times a^6 \times b^5 \times\left(\frac{b}{a}\right)^6\)
(iv) 3⁹ × 3⁵ ÷ 9⁷
(v) \(\left(\frac{2}{3}\right)^5 \div\left(\frac{2}{3}\right)^8 \times\left(\frac{2}{3}\right)^3\)
Answer:

Go through BSE Odisha Class 8 Science Solutions Chapter 11 Keeping Time with the Skies Question Answer to understand textbook questions more clearly.

Class 8 Science Curiosity Chapter 11 Question Answer

Class 8 Science Ch 11 Keeping Time with the Skies Question Answer

Class 8 Science Chapter 11 Keeping Time with the Skies Question Answer

Probe and Ponder Questions

Question 1.
Have you ever seen the Moon during the day? Why do you think it is sometimes visible when the sun is up?
Answer:
Yes, the Moon can often be seen during the day, especially in its waxing and waning phases. This happens because the Moon reflects sunlight and during certain phases it is high enough in the sky while the Sun is also above the horizon, so the bright part of the Moon is visible to us.

Question 2.
Imagine you lived on the Moon instead of Earth. What would you mean by a day, a month or a year?
Answer:

• A day on the Moon (from one sunrise to the next) would be about 29.5 Earth days.
• A month might be defined by one complete orbit around Earth (which is also about 27.3 Earth days).
• A year is the time the Earth takes to complete one orbit around the Sun, which you could still observed from the Moon in about 365 days.

Question 3.
What would happen if Earth had two moons instead of one? How would that change the night sky?
Answer:
If Earth had two moons:

• The night sky would be brighter and more dynantic.
• There could be more frequent eclipses.
• The gravitational pull on Earth would be different, possibly affecting tides.
• The two moons might cross paths, creating fascinating views or even risks of collision over long periods.

Question 4.
If we didn’t have clocks or calendars, how else could we measure time?
Answer:
We could measure time by:

• Observing the position of the Sun (sunrise, noon, sunset).
• Using the phases of the Moon to count months.
• Tracking stars and constellations that change with the seasons.
• Using natural events, like plant flowering or animal behaviour to mark the passage of time.

Question 5.
Share your questions …………
Answer:
Here are some fun questions from the chapter.

• Why does the Moon change shapes?
• How do festivals link to Moon phases?
• Can we see statellites at night?
• Why do some calendars add extra days?

InText Questions

Question 1.
Why does the illuminated portion of the Moon seen from the Earth decrease when it appears closer to the Sun?(Page 174)
Answer:
The Moon does not have its own light; it shines because it reflects sunlight that falls on it. When the Moon appears closer to the Sun in the sky, the side of the Moon that is illuminated by the Sun faces away from the Earth. As a result, from Earth we can see only a smaller part of the bright portion and most of the Moon’s surface visible to us is dark. That is why the illuminated portion of the Moon seen from Earth decreases when it comes closer to the Sun – it appears as a cresent Moon and finally becomes invisible on the new Moon day.

Question 2.
So, changing phases of the Moon is a natural periodic event, with a cycle of almost a month, which can also be used for time keeping. Yes, along with the natural periodic events of day and night and the changing seasons about which we learnt earlier. But how are these periodic events used for keeping time? (Page 178)
Answer:
The natural periodic events – like day and night, phases of the Moon and changing seasons – helps measure time because they repeat regularly after fixed intervals.

• Day and Night: The rotation of the Earth on its axis causes day and night. One complete rotation takes 24 hours which is used to define one day.
• Phases of the Moon: The Moon revolves around the Earth and completes one full cycle of phases (from one full Moon to the next) in about 29.5 days, which forms the basis of a month.
• Changing Seasons: The Earth revolves around the Sun once in about 365 days. This regular change in the position of the Earth causes seasons and gives us a year.
• Thus, these periodic motions of celestial bodies – the Earth and the Moon acts as natural clocks and calendars that humans have used for keeping time since ancient times.

Question 3.
Why do most Indian festivals fall on different dates every year? (Page 183)
Answer:
Many Indian festivals are linked to the phases of the Moon and hence are based on either lunar or luni-solar calendars, not the regular Gregorian solar calendar that we use daily. A lunar month is about 29.5 days and a lunar year (12 lunar months) has about 354 days, which is 11 days shorter than the solar year of 365 days.

Because of this difference, festivals like Diwali, Holi, Eid-ul-Fitr and Buddha Purnima which follow the Moon’s phases appear on different Gregorian calendar dates each year. To adjust this difference, the luni-solar calendars sometimes add an extra month called Adhika Maasa (extra month) every few years which helps keep the festivals in the same season.

Question 4.
When I look at the night sky in early evening, I see some moving stars. What are they? Is their motion also periodic? (Page 185)
Answer:
The moving ‘stars’ seen in the night sky are actually artificial satellites, not real stars. These are man-made objects launched into space that revolve around the Earth. They reflect sunlight which is why they appear as small moving points of light in the sky. Yes, their motion is periodic. Each satellite orbits the Earth in a fixed path and takes about 100 minutes to complete one round. Artificial satellites are used for communication, weather forecasting, navigation, disaster management and scientific research.

Keeping Time with the Skies Class 8 Questions and Answers

Keep the Curiosity Alive (Pages 187-188)

Question 1.
State whether the following statements are True or False.
(i) We can only see that part of the Moon which reflects sunlight towards us.
(ii) The shadow of Earth blocks sunlight from reaching the Moon, causing phases.
(iii) Calendars are based on various astronomical cycles which repeat predictably.
(iv) The Moon can only be seen at night.
Answer:
(i) True: We can only see the part of the Moon that reflects sunlight towards Earth.
(ii) False: The Earth’s shadow causes lunar eclipses, not the regular phases of the Moon.
(iii) True: Calendars are based on repeating astronomical events like day-night, Moon phases, and seasons.
(iv) False: The Moon can also be seen during the daytime, depending on its phase and position in the sky.

Question 2.
Amol was born on the 6th of May on a full Moon day. Does his birthday fall on the full Moon day every year? Explain your answer.
Answer:
No, Amol’s birthday does not fall on a full Moon Day every year. This is because the Moon’s phases follow a lunar cycle of about 29.5 days, while the calendar year follows the solar cycle of about 365 days. So, the date of the full Moon changes each year in the Gregorian calendar.

Question 3.
Name two things that are incorrect in the figure.

Answer:
Two incorrect things are :

• Stars are shown near the Moon during the daytime, which is incorrect because stars are not visible in the daytime sky.
• The Moon’s dark part is shaded incorrectly to show a phase. The shadow in the figure suggests it’s caused by Earth’s shadow, which is not true for regular Moon phases; they are caused by the Moon’s position relative to the Earth and Sun, not a shadow.

Question 4.
Look at the pictures of the Moon in the figure, and answer the following questions.

(i) Write the correct panel number corresponding to the phases of the Moon shown in the pictures above.

(ii) List the picture labels of the phases of the Moon that are never seen from Earth. Hint: You can use your observations from Activity 11.1 or Figure as reference.
Answer:
(i)

Picture label	Phase of Moon
C	Three days after New Moon
E	Full Moon
F	Three days after the Full Moon
A	A week after the Full Moon
B	Day of New Moon

(ii) Picture B (New Moon phase) is never seen from Earth because the illuminated side of the Moon is facing away from us.

Question 5.
Malini saw the Moon overhead in the sky at sunset.
(i) Draw the phase of the Moon that Malini saw.
(ii) Is the Moon in the waxing or the waning phase?
Answer:
(i) At sunset, the Moon is overhead only during the first quarter (a week after New Moon), when the right half is illuminated. So, we need to draw a half Moon (right half bright, left half dark).
(ii) Waxing phase (because it occurs after New Moon and the bright part is increasing).

Question 6.
Ravi said, “I saw a crescent Moon, and it was rising in the East when the Sun was setting.” Kaushalya said, “Once I saw the gibbous Moon during the afternoon in the East.” Who out of the two is telling the truth?
Answer:
Kaushalya is telling the truth because gibbous Moons can be seen in the East during the afternoon. Ravi’s statement is incorrect because Crescent Moons do not rise in the East at sunset. A crescent Moon appears just after the New Moon (waxing crescent) or just before the New Moon (waning crescent). Waxing crescent is visible after sunset in the western sky, not rising in the east, and waning crescent rises just before sunrise, not at sunset.

Question 7.
Scientific studies show that the Moon is getting farther away from the Earth and slower in its revolution. Will luni-solar calendars need an intercalary month more often or less often?
Answer:
Luni-solar calendars will need an intercalary month more often as the Moon moves farther and slower, and it takes longer to complete a cycle. So, a lunar year becomes even shorter compared to the solar year.

Question 8.
A total of 37 full Moons happen during 3 years in a solar calendar. Show that at least two of the 37 full moons must happen during the same month of the solar calendar.
Answer:
Yes, at least two full Moons must happen in the same solar month.

• A solar calendar has 12 months × 3 years =36 months.
• 37 full Moons in 36 months, at least one month must have 2 full Moons.

Question 9.
On a particular night, Vaishali saw the Moon in the sky from sunset to sunrise. What phase of the Moon would she have noticed?
Answer:
As the Moon is visible all night long only on a Full Moon, it is a Full Moon.

Question 10.
If we stopped having leap years, in approximately how many years would the Indian Independence Day happen in winter?
Answer:
One leap year adds ∼1 day every 4 years.
Without leap years, the calendar shifts by 1 day every 4 years.
There are roughly 183 days between 15 August (monsoon) and winter (mid-February).
183 days × 4=732 years
In approximately 730-732 years, 15 August would occur in winter.

Question 11.
What is the purpose of launching artificial satellites?
Answer:
Artificial satellites are launched for:

• Communication
• Navigation
• Weather monitoring -Scientific research
• Disaster management
• Earth observation

Question 12.
On which periodic phenomenon are the following measures of time based :
(i) day
(ii) month
(iii) year
Answer:
(i) Day → Earth’s rotation
(ii) Month → Moon’s revolution (phases of the Moon)
(iii) Year → Earth’s revolution around the Sun.

Class 8 Science Chapter 11 Question Answer

Activity 1

Activity 1.

Let us explore
Aim: To observe the shape of the moon from full moon day to new moon day.

Procedure:
1. Observe the Moon at sunrise in the western direction starting from the first day after the full Moon.
2. Construct a table same as Table -1 in your notebook. Complete the following details:

• Date
• When you saw the Moon (at sunrise or sunset)?
• Shade the corresponding Circle with pencil to show the bright portion of the Moon as shown in figure.

Table 1: Documenting changes in the Moon’s appearance

Answer:

Observations:

1. From the second day onwards also document the following:

• Is the size of bright portion of the Moon increasing or decreasing from the previous day.
• Yes, there is a change in shape of the moon everyday. Bright portion goes on decreasing.
• Whether the Moon appears closer to or farther from the Sun in the sky than the day before.

2. After about 15 days, you may not be able to see the Moon at sunrise or sunset. For the next 15 days, carry out this activity at sunset.

Inferences:

• Moon appear different each day.
• No, after 15 days, you may not be able to see the moon at sunrise or sunset.
• After the full moon, the bright face of the moon goes on decreasing every night. By another fifteen days again new moon is formed.
• The crescent moon goes on increasing everyday, till on the fifteenth day (from the new moon), the full face of the moon is visible.
• Each morning at sunrise, moon appears closer to the sun’s position in the sky.

Activity 2.

Let us explore
Aim: To show the Moon’s phases with a simple model.
Materials Required: A soft ball, a stick, a torch or lamp as the sun, your own head as the Earth.

Procedure:

• Take a small soft ball and insert a stick into it [(Figure (a)]. This represents the Moon.
• Move to a dark open place (at night), and ask your teacher or guardian to shine a torchlight towards you from about 3 m which will represent that light is coming from the Sun or stand near an electric lamp. Your head represents the Earth.
• Hold the ball in your hand, slightly above your head.
• Shine the torch toward the ball to represent sunlight.
• As you turn in a circle, the ball (“Moon”) shows a changing illuminated portion to your eyes.

Observations:

What does the model show

• Full Moon: When the ball is held opposite the lamp (behind you compared to the Sun), the side facing you is fully lit-just like a full Moon.
• New Moon: When the ball is held between your head and the lamp (towards the Sun), you see only the dark side-like a new Moon.
• Crescent and Gibbous Phases: Turning the ball slowly, the visible portion transitions. Sometimes you see a crescent (less than half lit), other times gibbous (more than half lit).
• The line between the bright and dark parts is always curved-this matches what we see in the real Moon.

Inference:
The Science behind the phasses

• Half illuminated, Half Dark: At every moment, half the Moon is lit by sunlight and half is in darkness.
• Moon’s Revolution: As the Moon revolves around the Earth, the angle between Earth, Moon and Sun changes. The part of the Moon we see as bright changes accordingly.

Activity 3.

Let us measure a day!

Aim: To find the duration of a day by observing the length of the shadow.
Materials Required: A 1m stick.

Procedure:

• Choose a small flat and open area in a ground which receives sunlight during the day. Fix a 1 m stick vertically in it as shown in figure.
• Let us start observing the shadow at 11:00 a.m. Every minute, mark a dot on the ground at the tip of the stick’s shadow. Keep marking dots until around 1:10 p.m.
• Identify when the shadow was shortest and find out its time by counting the number of dots. Record this time in Table. Repeat this experiment for the next few days.
• Now calculate the duration of the solar day. This can be done by finding a difference in time on two consecutive days as shown in Table.

Observations:
Table1: Finding the duration of a solar day

Answer:

Date	Time of shortest shadow (hh:mm)	Duration of day (hh:mm)
22 March 2025 23 March 2025 24 March 2025 25 March 2025 26 March 2025	12:20 12:20 12:19 12:19 12:18	…………… 24:00 23:59 23:59 24:00

1. The shortest shadow during the day marks when the sun is at highest point in the sky (noon).
2. Measuring from one day’s noon to the next day gives the length of a day.
3. The average solar day i.e. average duration of the day is about 24 hours.

Inference:
The Month and the Moon

• The phases of the Moon create another natural cycleone complete phase cycle (from full Moon to next full Moon) takes about 29.5 days (approximately one month).
• This lunar cycle is the basis for measuring a month. (see figure)

The Year and the Seasons

• One year is the duration for Earth to make a full revolution around the Sun, which is about 365 \(\frac{1}{4}\) days.
• The repetition of seasons (spring, summer, autumn, winter) marks the annual cycle.
• The Earth undergoes one cycle of seasons during this time, which can be used to define a solar year (see figure).

Activity 4.

Let us identify
Aim: To observe artificial satellites.
Materials Required: A telescope.

Procedure:

• Go to a location that has a clear view of sky; along with an adult. There should not be any obstruction of trees or tall buildings.
• To identify satellites in the sky, look for a small, bright, continuously moving dot in the sky, typically before sunrise or after sunset.

Observations:

• Can be seen without a telescope or with binoculars.
• Satellite-tracking mobile apps or websites help identify visible satellites in your location and when they will be passing above you in the sky.

Keeping Time with the Skies Class 8 Extra Questions and Answers

Short Answer Type Questions

Question 1.
How does the Moon’s position in the sky change each day?
Answer:
The Moon’s position shifts slightly eastward each day, so it is not in the same place at the same time. This is why its rise and set times also change daily.

Question 2.
Why does the Moon sometimes appear in the daylight?
Answer:
The Moon can rise before sunset, sometimes in the afternoon. In such cases, it is visible in the sky while the Sun is still up.

Question 3.
What does the simple ball-and-lamp model of the Moon help us understand?
Answer:
The model shows how sunlight falls on the Moon and creates different phases. By turning with the ball, we can see changes in the illuminated portion, similar to what happens in reality.

Question 4.
Why is the line between the bright and dark portions of the Moon always curved ?
Answer:
The Moon is spherical, so the dividing line between sunlight and shadow is curved. This curve is visible from Earth during all phases.

Question 5.
Why are Moon phases not caused by Earth’s shadow ?
Answer:
Moon phases happen because we see varying parts of its sunlit side as it orbits. Earth’s shadow only causes a lunar eclipse, which is rare.

Question 6.
Why don’t we have eclipses every full Moon or new Moon?
Answer:
The Moon’s orbit is tilted compared to Earth’s orbit. This tilt means the Sun, Earth and Moon usually don’t line up perfectly.

Long Answer Type Questions

Question 1.
Explain the waxing and waning periods of the Moon and how they form a monthly cycle.
Answer:

• The waxing period is when the bright portion of the Moon increases, starting from the new Moon and becoming full in about two weeks. The waning period is when the bright portion decreases, starting from the full Moon and becoming new Moon in about two weeks.
• Together, waxing and waning make a repeating cycle every month. This cycle takes about 29.5 days from one full Moon to the next.
• These changes occur because of the Moon’s revolution around Earth and the changing angle between the Sun, Earth and Moon.

Question 2.
Explain how the motion of the Sun in the sky helps in measuring a day.
Answer:

• The Sun appears to rise in the east, move across the sky and set in the west because Earth rotates on its axis. The highest position of the Sun in the sky, when shadows are shortest, marks noon.
• The time from one noon to the next is called a mean solar day. This length is about 24 hours, which is the basic unit for measuring time in days. This daily cycle has been used since ancient times to keep track of time.

Case-Study Based Questions

1. Read the following passage carefully and answer the questions that follow :

Anita kept a Moon observation diary for a week. On Day 1, she saw a thin crescent on the right side. By Day 4, half of the Moon was visible. By Day 7, almost the full Moon was visible.

(a) Which phase did Anita observe on Day 1?
Answer:
Waxing Crescent

(b) What is the name of the phase on Day 4?
Answer:
First Quarter

(c) Was the Moon waxing or waning during these days?
Answer:
Waxing

Picture Based Questions

I. Look at the pictures and answer the following questions:

(a) Identify the pictures (i) and (ii).
Answer:
(i) Warli painting
(ii) Dhokra Brass Sculpture

(b) Write the name of painting given below and the state from given where it is associated ?
Answer:

Madhubani painting. It is a famous painting associated to Bihar.

Keeping Time with the Skies Class 8 MCQ

Multiple Choice Questions (Mcqs)

Question 1.
Phases of the moon occur because:
(a) We can see only that part of the Moon which reflects light towards us.
(b) Our distance from the Moon keeps changing.
(c) The shadow of the earth covers only a part of the Moon’s surface.
(d) The thickness of the Moon’s atmosphere is not constant.
Answer:
(a) We can see only that part of the Moon which reflects light towards us.

Question 2.
What do they call the Moon when it’s more than half lit but not full ?
(a) Crescent
(b) Gibbous
(c) Quarter
(d) Eclipse Moon
Answer:
(b) Gibbous

Question 3.
What is the sequence of phases starting from New Moon?
(a) New Moon → Full Moon → Waxing → Crescent → First quarter
(b) New Moon → Waxing Crescent → First Quarter → Full Moon
(c) Full Moon → Waxing Gibbous → New Moon
(d) First quarter → Full Moon → New Moon
Answer:
(b) New Moon → Waxing Crescent → First Quarter → Full Moon

Question 4.
What causes the phases of the Moon ?
(a) The Earth’s rotation
(b) The Moon’s rotation
(c) The Moon’s orbit around the Earth
(d) The Sun’s movement
Answer:
(c) The Moon’s orbit around the Earth

Question 5.
How long does one complete cycle of Moon phases take?
(a) 7 days
(b) 15 days
(c) 29.5 days
(d) 365 days
Answer:
(c) 29.5 days

Assertion and Reasoning

These questions consist of two statements, each printed as Assertion (A) and Reason (R). While answering these questions, you are required to choose any one of the following four responses.

(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.
(b) Assertion (A) and Reason (R) both are correct and reason is not correct explanation for assertion.
(c) Assertion (A) is correct but the Reason (R) is wrong.
(d) Assertion (A) is wrong but the Reason (R) is correct.

1. Assertion (A): The Moon appears to change shape throughout the month.
Reason (R): The Earth casts different shadows on the Moon each night.
Answer:
(c) Assertion (A) is correct but the Reason (R) is wrong.

2. Assertion (A): The full Moon rises at sunset and sets at sunrise.
Reason (R): The full Moon is directly opposite the sun in the sky.
Answer:
(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.

Fill in the blanks

1. The Moon does not have its own light, it………… the sunlight.
Answer:
reflects

2. The changing shapes of the Moon that we see from Earth are called ………….
Answer:
Phases

3. A full cycle of the Moon’s phases takes about ……….. days.
Answer:
29.5

4. Waxing Moon is best seen at ………….
Answer:
Sunset

5. Waning Moon is best seen at ………….
Answer:
sunrise.

True or False

1. A lunar eclipse occurs when the Moon comes between the Earth and the Sun.
Answer:
False

2. The Full Moon appears once every lunar cycle.
Answer:
True

3. A Moon’s surface is smooth and shiny.
Answer:
False

4. We can sometimes see the Moon during the daytime.
Answer:
True

5. The Moon always shows the same side to the Earth.
Answer:
True

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.

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Part 2 Chapter 7 Finding the Unknown MCQ improves accuracy in objective exams.

MCQ on Finding the Unknown Class 7

Finding the Unknown MCQ Class 7

Class 7 Maths Finding the Unknown MCQ

Question 1.
A weighing scale is balanced when two identical boxes and a 4-unit weight are placed on one pan, and one identical box with a 12-unit weight is placed on the other pan. What is the weight of one box?
(a) 6
(b) 4
(c) 8
(d) 12
Solution:
(c) 8
Let the weight of one box be x units. Then,
Total weight on one pan = (2x + 4) units
Total weight on other pan = (x + 12) units
As weighing scale is balanced, weights on boththe pans are equal.
∴ 2x + 4 = x + 12
⇒ 2x – x = 12 – 4 ⇒ x = 8
Hence, the weight of one box is 8 units.

Question 2.
A sequence of arrangements uses 4 sticks in the first step and increases by 3 sticks in every next step. Which expression gives the number of sticks in step n?
(a) 3n
(b) 3n + 1
(c) 3n + 2
(d) 3n – 1
Solution:
(b) 3n + 1
Number of sticks in:
1^st step = 4
2^nd step = 4 + 3 × 1
3^rd step = 4 + 3 × 2
4^th step = 4 + 3 × 3
.
.
.
.
.
n^th step = 4 + 3 × (n – 1) = 4 + 3n – 3
= 3n + 1
Hence, the number of sticks in step n is 3n + 1.

Question 3.
The value of m so that 0.6 m – (-0.8 m + 0.4) = 0.2 – 0.5 m, is:
(a) \(\frac{7}{19}\)
(b) \(\frac{6}{19}\)
(c) \(\frac{9}{19}\)
(d) \(\frac{8}{19}\)
Solution:
(b) \(\frac{6}{19}\)
Given, 0.6m – (- 0.8 m + 0.4) = 0.2 – 0.5m
⇒ 0.6m + 0.8m – 0.4 = 0.2 – 0.5m
⇒ 1.4m – 0.4 = 0.2 – 0.5m
⇒ 1.4m + 0.5m = 0.2 + 0.4
⇒ 1.9m = 0.6
⇒ m = \(\frac{0.6}{1.9}\) ⇒ m = \(\frac{6}{19}\)

Question 4.
Which of the following is correct about x = x?
(a) The equation has no solution.
(b) The equation has exactly one solution.
(c) The equation has infinitely many solutions.
(d) The equation is incorrect.
Solution:
(c) The equation has infinitely many solutions.
We have, x = x
It means it is true (LHS = RHS) for all values of the variable. Hence, it has infinitely many solutions.

Question 5.
It is given that \(\frac{45}{56} \times \frac{14}{9} \times \frac{7}{3} \times 5=\frac{22050}{1512}\), then the value of \(\frac{45}{56} \times \frac{14}{9} \times 5\) is:
(a) \(\frac{3150}{504}\)
(b) \(\frac{2850}{564}\)
(c) \(\frac{3150}{564}\)
(d) \(\frac{2850}{504}\)
Solution:
(a) \(\frac{3150}{504}\)
Given, \(\frac{45}{56} \times \frac{14}{9} \times \frac{7}{3} \times 5=\frac{22050}{1512}\)
⇒ \(\frac{45}{56} \times \frac{14}{9} \times 5=\frac{22050}{1512} \times \frac{3}{7}=\frac{3150}{504}\)

Question 6.
If 3x – \(\frac{4}{5}=\frac{5 x}{3}+\frac{16}{5}\), then the value of x which satisfies the given equation is:
(a) Natural number greater than 7
(b) Natural number between 5 and 7
(c) Natural number less than 2
(d) Natural number between 2 and 4
Solution:
(d) Natural number between 2 and 4
Given, \(3 x-\frac{4}{5}=\frac{5 x}{3}+\frac{16}{5}\)
⇒ \(3 x-\frac{5 x}{3}=\frac{16}{5}+\frac{4}{5}\) ⇒ \(\frac{9 x-5 x}{3}=\frac{16+4}{5}\)
⇒ \(\frac{4 x}{3}=\frac{20}{5}\) ⇒ \(\frac{4 x}{3}=4\)
⇒ x = \(\frac{4 \times 3}{4}\) ⇒ x = 3

Question 7.
The sum of two consecutive even numbers is 70. The greater one is:
(a) 32
(b) 34
(c) 36
(d) 38
Solution:
(c) 36
Let x and x + 2 be two consecutive even numbers.
According to question, we have
x + (x + 2) = 70
⇒ 2x + 2 = 70 ⇒ 2x = 70 – 2
⇒ 2x = 68 ⇒ x = \(\frac{68}{2}\) = 34
So, smaller even number = 34 and greater even number = 34 + 2 = 36

Question 8.
Rajini has three balls, A, B and C. A is twice as heavy as B, B weighs one third of C. The heaviest ball is:
(a) A
(b) B
(c) C
(d) A and C
Solution:
(c) C
Let the weight of ball C be x units. Then,
Weights of ball B and ball A are \(\frac{x}{3}\) units and \(\frac{2x}{3}\) units respectively. Clearly, x > \(\frac{2x}{3}\) > \(\frac{x}{3}\)
[As weight cannot be negative, x > 0]
Thus, the heaviest ball is C.

Question 9.
Which of the following equations represents the statement “twice of a number is 25 more than one-third of the number”?
(a) \(2 x=\frac{x}{3}+25\)
(b) \(2 x+25=\frac{x}{3}\)
(c) \(2 x=\frac{25+x}{3}\)
(d) \(3 x=25+\frac{x}{2}\)
Solution:
(a) \(2 x=\frac{x}{3}+25\)
Let the number be x. Then,
Twice of the number = 2x and one-third of the number = \(\frac{x}{3}\)
According to question, we have 2x = \(\frac{x}{3}\) + 25

Question 10.
Adding 7 to the thrice of a whole number gives 34. The whole number is:
(a) 13
(b) 9
(c) 14
(d) 11
Solution:
(b) 9
Let the whole number be x. Then, thrice of the whole number = 3x
According to question, we have
3x + 7 = 34 ⇒ 3x = 34 – 7
⇒ 3x = 27 ⇒ x = \(\frac{27}{3}\) = 9

Question 11.
Which of the following equations have t = – 3 as a solution?
(i) 2t + 7 = 1
(ii) 4(t + 5) = 8
(iii) 2t + \(2\left(t+\frac{1}{4}\right)=\frac{-5}{2}\) + t
(iv) -2 = \(\frac{3}{5}\) – 7t
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Solution:
(a) (i) and (ii)
(i) LHS = 2t + 7 = 2 × (-3) + 7 = -6 + 7
= 1 = RHS
∴ t = – 3 is the solution of equation 2t + 7 = 1.

(ii) LHS = 4(t + 5) = 4(-3 + 5) = 4 × 2 = 8 = RHS
∴ t = – 3 is the solution of equation
4 (t + 5) = 8.

(iii) LHS = 2t + 2\(\left(t+\frac{1}{4}\right)\)
= 2 x (-3) + 2 \(\left(-3+\frac{1}{4}\right)\)
= -6 + \(2\left(\frac{-11}{4}\right)\) = -6 – \(\frac{11}{2}\)
= \(-\frac{23}{2}\)
RHS = \(\frac{3}{5}-7 t=\frac{3}{5}-7 \times(-3)\)
= \(\frac{3}{5}+21=\frac{3+105}{5}=\frac{108}{5}\) ≠ -2 = LHS

Question 12.
“48 is divided into 2 parts such that three times the greater part is 10 less than four times the smaller part”.
Identify the correct statements about the given situation.
(i) The smaller part is 16.
(ii) The smaller part is 22.
(iii) The greater part is 26.
(iv) The greater part is 32.
Choose the correct option from the following:
(a) (i) and (iii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
Solution:
(c) (ii) and (iii)
Let the greater part be x. Then, the smaller part = 48 – x
According to given situation, we have
3x = 4(48 – x) – 10 ⇒ 3x = 4 × 48 – 4x – 10
⇒ 3x = 192 – 4x – 10 ⇒ 3x + 4x = 192 – 10
⇒ 7x = 182 ⇒ x = \(\frac{182}{7}\) = 26
So, the greater part is 26 and the smaller part is 48 – 26 = 22.
Thus, statements (ii) and (iii) are correct.

Question 13.
The sum of three consecutive odd numbers is 63.
Identify the correct statements about the given situation.
(i) Smallest odd number is 17.
(ii) Largest odd number is 23.
(iii) Smallest odd number is 19.
(iv) Largest odd number is 21.
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Solution:
(b) (ii) and (iii)
We know, difference between two consecutive odd numbers is 2.
Let x, x + 2 and x + 4 be the three consecutive odd numbers.
∴ x + (x + 2) + (x + 4) = 63
[Given, sum of three consecutive odd numbers is 63.]
⇒ 3x + 6 = 63 ⇒ 3x = 63 – 6
⇒ 3x = 57 ⇒ x = \(\frac{57}{3}\) = 19
So, the three consecutive odd numbers are 19, 21 and 23.
Thus, the smallest odd number is 19 and the largest odd number is 23.

Finding the Unknown Class 7 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): Only addition can be done on both sides of an equation, not subtraction.
(R): An equation remains balanced when we perform the same operation on both sides.
Solution:
(d) A is false but R is true.
An equation remains balanced when we perform the same operation (addition, subtraction, multiplication, division) on both sides.
∴ Assertion (A) is false, but Reason (R) is true.

Question 2.
(A): Two complementary angles differ by 8° can be represented by equation x – (90° – x) = 8°.
(R): Sum of complementary angles is 180°.
Solution:
(c) A is true but R is false.
We know, sum of complementary angles is 90°.
Given, two complementary angles differ by 8°.
Let x be one of the two complementary angles. Then, the other angle = 90° – x
∴ x – (90° – x) = 8° is the required equation.
Thus, Assertion (A) is true, but Reason (R) is false.

Question 3.
(A): For the given triangle, y + 2 (y + 15°) = 180°.
(R): In an isosceles triangle, angles opposite to equal sides are equal.

Solution:
(a) Both A and R are true and R is the correct explanation of A.
The given triangle is isosceles.
We know, angles opposite to equal sides of an isosceles triangle are equal.
∴ First angle = y, second angle = third angle
= y + 15°
Now, y + (y + 15°) + (y + 15°) = 180°
[The sum of all the angles of a triangle is 180°.]
⇒ y + 2(y + 15°) = 180°
∴ Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Finding the Unknown Class 7 Fill in the Blanks

Question 1.
If the solution is known, the number of equations that can be built are ________ .
Solution: infinite
If the solution is known, the number of equations that can be built are infinite.

Question 2.
Fill in the blanks with integers:
(i) 11 (11 + ____ ) – 98 = 100
(ii) – 3 × (-9 + _____) = 42
Solution: 7, -5
(i) Let y be the required value to fill the blank.
∴ 11(11 + y) – 98 = 100
⇒ 11(11 + y) = 100 + 98
[Adding 98 to both sides]
⇒ 11(11 + y) = 198
⇒ (11 + y) = 198 ÷ 11
[Dividing both sides by 11]
⇒ 11 + y= 18
⇒ y = 18 – 11 = 7
So, 11(11 + 7) – 98 = 100

(ii) Let y be the required value to fill the blank.
∴ -3 × (- 9 + y) = 42
⇒ (- 9 + y) = 42 ÷ (-3)
[Dividing both sides by – 3]
⇒ – 9 + y = -14
⇒ y = – 14 + 9 [Adding 9 to both sides]
⇒ y = – 5
So, – 3 × (- 9 + (- 5)) = 42

Question 3.
In an equation, there is always an ________ sign.
Solution: equals
In an equation, there is always an equals sign.

Question 4.
The value of the variable which satisfies a equation is called a ________ to the equation.
Solution: solution
The value of the variable which satisfies an equation is called a solution to the equation.

Question 5.
The equation representing “one-fifth of a number is 3 more than one-sixth of the same number” is \(\frac{x}{5}\) = ______ .
Solution: \(\frac{x}{6}\) + 3
Let x be the number. Then,
One-fifth of the number = \(\frac{x}{5}\) and one-sixth of the number = \(\frac{x}{6}\)
\(\frac{x}{5}=\frac{x}{6}\) + 3
The equation representing “one-fifth of a number is 3 more than one-sixth of the same number” is \(\frac{x}{5}=\frac{x}{6}\) + 3.

Question 6.
If \(\frac{x}{7}\) of a number is 60. the number is _______ .
Solution: 105
Let the number be x.
∴ \(\frac{4}{7}\) × x = 60
⇒ x = 60 × \(\frac{7}{4}\) = 15 × 7 = 105
If \(\frac{4}{7}\) of a number is 60, the number is 105.

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.]
= \(\frac{99}{12}\) – \(\frac{34}{12}\) + \(\frac{20}{12}\)
= \(\frac{99-34+20}{12}\) = \(\frac{85}{12}\)

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Part 2 Chapter 6 Constructions and Tilings MCQ improves accuracy in objective exams.

MCQ on Constructions and Tilings Class 7

Constructions and Tilings MCQ Class 7

Class 7 Maths Constructions and Tilings MCQ

Question 1.
If we bisect a line segment AB of 9 cm, then what will be the length of each part?
(a) 5 cm
(b) 4 cm
(c) 4.5 cm
(d) 3 cm
Solution:
(c) 4.5 cm
When we bisect a line segment, the length of each part is half of the length of given line segment.
Given, length of the line segment AB = 9 cm After bisecting AB,
Length of each part = (Length of the line segment AB) ÷ 2 = 9 cm ÷ 2 = 4.5 cm.

Question 2.
Two lines are said to be perpendicular if angle between them is:
(a) 45°
(b) 90°
(c) 60°
(d) 30°
Solution:
(b) 90°
We know, two lines are said to be perpendicular if angle between them is 90°.

Question 3.
To construct the perpendicular bisector of a line segment PQ, the compass opening must be:
(a) smaller than half of PQ
(b) more than half of PQ
(c) equal to the length of PQ
(d) equal to half of PQ
Solution:
(b) more than half of PQ
To construct the perpendicular bisector of a line segment PQ, the compass opening must be more than half of PQuestion

Question 4.
If we bisect an angle of a rectangle, we get two angles each of:
(a) 45°
(b) 30°
(c) 60°
(d) 15°
Solution:
(a) 45°
We know that the measure of each interior angle of a rectangle is 90°.
Therefore, if we bisect an angle of a rectangle, we get two angles each of 45°.

Question 5.
What angle would you get if you bisect an exterior angle of an equilateral triangle?
(a) 90°
(b) 60°
(c) 45°
(d) 30°
Solution:
(b) 60°
We know that at each vertex of an equilateral triangle, the measure of exterior angle is 120°.

Therefore, if we bisect an exterior angle of an equilateral triangle, we get two angles measuring 60° each.

Question 6.
If YA = AB and AC = BC, then measure of ∠XYC equals:

(a) 60°
(b) 45°
(c) 30°
(d) None of these
Solution:
(c) 30°
Here, the arc from Y cuts YZ and YX at points A and B respectively and YD bisects ∠XYZ.
As arc radius is same, YA = YB.
∴ YA = YB = AB [Given, YA = AB]
Thus, ∆ABY is an equilateral triangle.
∴ ∠AYB = 60° ⇒ ∠XYC = 60° ÷ 2 = 30°

Question 7.
While constructing the perpendicular bisector of a line segment PQ, if the compass opening is exactly half of PQ, then drawn arcs from P and Q:
(a) do not intersect
(b) intersect at one point
(c) intersect at two points
(d) intersect at four points
Solution:
(b) intersect at one point
While constructing the perpendicular bisector of a line segment PQ, if the compass opening is exactly half of PQ, then drawn arcs from P and Q intersect at one point only.

Question 8.
Which of the following shapes can tile a plane on its own without leaving any gaps or overlaps?
(a) Circle
(b) Oval
(c) Square
(d) Regular pentagon
Solution:
(c) Square
A regular polygon can tile the plane only if its interior angle divides 360° exactly.
Circles and ovals leave gaps, and regular pentagons also fail to tessellate. Squares, however, fit together perfectly and cover the plane without gaps or overlaps.

Question 9.
A 5 × 5 square grid consists of 25 unit squares. Can it be completely covered using 2 × 1 dominoes?
(a) Yes, because 25 squares can be covered by dominoes in different ways.
(b) Yes, because dominoes can be placed both horizontally and vertically.
(c) No, because each domino covers 2 squares, and 25 is an odd number.
(d) None of the above
Solution:
(c) No, because each domino covers 2 squares, and 25 is an odd number.
Each 2 × 1 domino covers exactly 2 unit squares. To completely cover a grid, the total number of squares must be divisible by 2.

Here, the 5 × 5 grid has 25 unit squares. Since 25 is odd, it is not divisible by 2, so it is impossible to cover the entire grid with 2 × 1 dominoes.

Question 10.
Which of the following shapes does not tessellate (tile) a plane on its own?
(a) Equilateral triangle
(b) Square
(c) Hexagon
(d) Circle
Solution:
(d) Circle
Among the given shapes, only equilateral triangles, squares, and regular hexagons can tile the plane by themselves as their interior angle divides 360° exactly.

Question 11.
In a tangram set, which piece has all its sides equal in length?
(a) Square
(b) Parallelogram
(c) Small triangle
(d) Large triangle
Solution:
(a) Square
The tangram set consists of different shapes such as triangles, a square, and a parallelogram. A square has four sides of equal length.

A parallelogram has only opposite sides equal. The small triangles and the large triangles are right-angled isosceles triangles, which have only two sides equal, not all.

Question 12.
The tiling pattern used by bees in honeycombs is based on which of the following shapes?
(a) Squares
(b) Triangles
(c) Hexagons
(d) Pentagons
Solution:
(c) Hexagons
Bees construct honeycombs using hexagonal cells.

Constructions and Tilings Class 7 Fill in the Blanks

Question 1.
If two line segments cut each other at right angles, then they are _________ .
Solution: perpendicular
If two line segments cut each other at right angles, then they are perpendicular.

Question 2.
A straight angle can be bisected into _______ right angles.
Solution: two
We know that a straight angle measures 180°.
When a straight angle is bisected, it is divided into two equal angles, each measuring 90°.
Thus, a straight angle can be bisected into two right angles.

Question 3.
If ray OR bisects ∠POQ of measure 85°, the then they are measure of ∠POR is _______ .
Solution: 42.5°
Given, ∠POQ = 85°
And, ray OR bisects ∠POQ.
Therefore, ∠POR = ∠POQ ÷ 2 = 85° ÷ 2 = 42.5°

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Part 2 Chapter 5 Connecting the Dots MCQ improves accuracy in objective exams.

MCQ on Connecting the Dots Class 7

Connecting the Dots MCQ Class 7

Class 7 Maths Connecting the Dots MCQ

Question 1.
The daily temperatures (in °C) for a week are 30, 27, 29, 35, 34, 32, 30. What is the mean temperature?
(a) 31°C
(b) 32°C
(c) 30°C
(d) 29.5°C
Solution:
(a) 31°C
Given, daily temperatures (in °C) for a week are: 30, 27, 29, 35, 34, 32, 30
We know, Mean = \(\frac{Sum of observations}{Number of observations}\)
∴ Mean temperature (in °C)
= \(\frac{30+27+29+35+34+32+30}{7}\)
= \(\frac{217}{7}\) = 31

Question 2.
The mean of the first 15 natural numbers is:
(a) 5
(b) 7.5
(c) 8
(d) 9.5
Solution:
(c) 8
Sum of first 15 natural numbers
= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 +10 + 11 + 12 + 13 + 14 + 15 = 120
Mean of first 15 natural numbers
= \(\frac{Sum of first 15 natural numbers}{15}\)
= \(\frac{120}{15}\) = 8

Question 3.
The following dot plot shows the hours of sleep taken by a child on different days.

The range of the number of hours of sleep is:
(a) 4
(b) 8
(c) 5
(d) 10
Solution:
(c) 5
We know, range of data = Highest value – Lowest value
In the given dot plot, minimum hours
= 5 and maximum hours = 10
∴ Range = Maximum hours – Minimum hours 10 – 5 = 5

Question 4.
The table shows the number of flowers that bloom in a garden in 7 days.

Day	Mon	Tue	Wed	Thu	Fri	Sat	Sun
No. of flowers	5	4	6	7	3	5	2

What is the median number of flowers?
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
(d) 5
Given, the number of flowers that bloom in garden for 7 days are: 5, 4, 6, 7, 3, 5, 2
Arranging the data in ascending order, we get: 2, 3, 4, 5, 5, 6, 7
Here, number of observations, n = 7, is odd.
∴ Median = \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation
= \(\left(\frac{7+1}{2}\right)^{\text {th }}\) observation = 4^th observation
⇒ Median = 5

Question 5.
The difference between mean and median of the given dot plot showing number of trips taken by different families during an year, is:

(a) 0.33
(b) 3.33
(c) 0.0033
(d) None of these
Solution:
(b) 3.33
Writing down the data from the given dot plot, we get the values:
10, 10, 20, 20, 30, 30, 30, 30, 40, 40, 40, 40, 50, 50, 60
Total number of observations (n) =15 For mean,
Sum = (10 × 2) + (20 × 2) + (30 × 4) + (40 × 4) + (50 × 2) + (60 × 1)
= 20 + 40 + 120 + 160 + 100 + 60 = 500
Mean = \(\frac{500}{15}\) = 33.33
For median, (data is already in ascending order)
Since n = 15 is odd, median = \(\left(\frac{15+1}{2}\right)^{\text {th }}\) = 8^th value of the data
The 8^th value of the data is 30.
∴ Median = 30
Now, Difference = mean – median = 33.33 – 30 = 3.33]

Question 6.
Mean of 10 numbers is 20 and the mean of 5 numbers is 8. Then, the mean of all the 15 numbers is:
(a) 10
(b) 15
(c) 16
(d) 17
Solution:
(c) 16
Given, mean of 10 numbers = 20
∴ Sum of 10 numbers = Mean × 10
= 20 × 10 = 200
Also, mean of 5 numbers = 8
∴ Sum of 5 numbers = Mean × 5
= 8 × 5 = 40
Now, sum of all 15 numbers = Sum of 10 numbers + Sum of 5 numbers
= 200 + 40 = 240
∴ Mean of all 15 numbers
= \(\frac{Sum of 15 numbers}{15}\) = \(\frac{240}{15}\) = 16

Question 7.
If the median of \(\frac{x}{5}, x, \frac{x}{4}, \frac{x}{2} \text { and } \frac{x}{3}\) (where x > 0) is 8, then the value of x would be:
(a) 24
(b) 8
(c) 15
(d) 10
Solution:
(a) 24
Given, the observations are: \(\frac{x}{5}, x, \frac{x}{4}, \frac{x}{2}, \frac{x}{3}\)
To find the median, we first arrange-them in ascending order.
Since x > 0, the larger the denominator, the smaller the value.
Thus arranging the observations in ascending order, we get: \(\frac{x}{5}, \frac{x}{4}, \frac{x}{3}, \frac{x}{2}, x\)
Here, number of observations, n = 5, is odd.
∴ Median = \(\left(\frac{5+1}{2}\right)^{\text {th }}\) observation
= 3^rd observation = \(\frac{x}{3}\)
According to the question,
Median = 8 ⇒ \(\frac{x}{3}\) = 8
⇒ r = 8 × 3 = 24

Question 8.
A cricket player scored 220 runs in 5 matches but did not bat in one of the innings. What is the mean score of the player?
(a) 36.67
(b) 38
(c) 55
(d) 45.67
Solution:
(c) 55
Given, the player scored 220 runs in 5 matches in total and did not bat in one inning.
∴ Mean = \(\frac{Total score}{Number of innings played}\)
= \(\frac{220}{4}\) = 55

Question 9.
The mean of five numbers is 18. If the number 32 is removed, what is the mean of the remaining four numbers?
(a) 12.5
(b) 14.5
(c) 13
(d) 16.5
Solution:
(b) 14.5
Given, mean of 5 numbers = 18
∴ Sum of 5 numbers = 5 × 18 = 90
As, number 32 is removed,
Sum of remaining 4 numbers = 90 – 32 = 58
∴ Mean of remaining four numbers
= \(\frac{58}{4}\) = 14.5

Question 10.
A data set contains 11 values arranged in ascending order. The 6th value is 42. Which of the following must be true?
(a) The mean is 42.
(b) The median is 42.
(c) All values are dose to 42.
(d) There is no outlier in the data.
Solution:
(b) The median is 42.
For an odd number of observations, the median is the middle value.
Here, the number of observations is 11 which is odd.
∴ Median = \(\left(\frac{11+1}{2}\right)^{\mathrm{th}}\) value = 6^th value = 42

Question 11.
The median of prime numbers between 20 and 50 is:
(a) 31
(b) 37
(c) 41
(d) 43
Solution:
(b) 37
The prime numbers between 20 and 50 are:
23, 29, 31, 37, 41, 43, 47 (In ascending order) Here, number of observations, n = 7, is odd.
Median = \(\left(\frac{n+1}{2}\right)^{\mathrm{th}}\) observation
= \(\left(\frac{7+1}{2}\right)^{\mathrm{th}}\) observation = 4^th observation
Thus, median = 37

Question 12.
A data set has mean = 50 and median = 70. Which of the following is most likely to be true?
(a) Presence of higher end outliers
(b) Presence of lower end outliers
(c) Data is balanced
(d) None of these
Solution:
(b) Presence of lower end outliers
Given, for a data set, mean = 50 and median = 70.
In this data set, the mean is much lower than the median (50 < 70).
The median tells us where the middle of the data is.
The mean is sensitive to outliers and gets pulled in their direction.
Because the mean has been pulled down to 50 (below the median), there must be very small numbers (lower-end outliers) dragging the average down.

Question 13.
The following bar graph shows the rainfall at six selected locations (A, B, C, D, E and F) in certain months.

Which of the following statements is correct?
(a) July rainfall exceeds August rainfall by 100 cm at each location.
(b) September rainfall exceeds August rainfall by 50 cm at each location.
(c) July rainfall is lower than August rainfall at each location.
(d) None of the above
Solution:
(c) July rainfall is lower than August rainfall at each location.
From the bar graph, comparing July, August, and September rainfalls at each location A to F, we observe:

• At every location, the August bar is higher than the July bar, i.e. July rainfall never exceeds August rainfall.
• Also, July rainfall is lower than August rainfall in each location.
• September rainfall is not exactly 50 cm more than August at each location.

Question 14.
A bar graph represents the daily water consumption (in litres) of a household.

Which conclusion is most logically sound?
(a) Mean = Median
(b) Mean > Median
(c) Median > Mean
(d) Mean and Median cannot be compared
Solution:
(b) Mean > Median
From the graph, we observe that the water consumption on days 1, 2, 3 and 4 are the same (bars of same height), while the water consumption on day 5 is the highest (longest bar).

As the water consumption on day 5 is higher than the water consumption on other days, it will pull the mean upward.
However, the median water consumption will lie in middle.
Hence, Mean > Median.

Question 15.
A bar graph uses the scale: 1 unit =10 children on y-axis.
If the scale is changed to 1 unit = 20 children, then:
(a) the data values change
(b) the order of bars changes
(c) the height of bars changes
(d) the graph becomes incorrect
Solution:
(c) the height of bars changes
If scale is changed to 1 unit = 20 children, the height of bars will reduce in size, but the data values and order of bars remain unchanged.
So, only the height of bars will change.

Question 16.
The given double bar graph represents the average rainfall of two cities in 5 months.

Which city sees increase in rainfall each month from January to May?
(a) City 1
(b) City 2
(c) Both city 1 and city 2
(d) Neither city 1 nor city 2
Solution:
(a) City 1
From the given bar graph:
Rainfall in City 1 increases steadily from January to May.
Rainfall in City 2 decreases from January to February, then increases from February to March to April and then again decreases in May.

Question 17.
The data given below shows the time (in minutes), taken by nine students to go to school from their homes. Which statement about the data is/are true?
115, 65, 225, 75, 115,65,155,125,115
(i) Median is 115.
(ii) The observation 65 is lowest observation.
(iii) The range of the data is 160.
(iv) The range of the data is 90.
Choose the correct option from the following:
(a) (i), (ii) and (iii)
(b) (i) and (iii) only
(c) (ii) and (iv) only
(d) (iv) only
Solution:
(a) (i), (ii) and (iii)
Arranging the data in ascending order, we get 65, 65, 75, 115, 115, 115, 125, 155, 225
Here, number of observations, n = 9, is odd.
∴ Median = \(\left(\frac{9+1}{2}\right)^{\text {th }}\) observation
= \(\left(\frac{10}{2}\right)^{\text {th }}\) observation = 5^th observation =115
In the given data set, lowest observation = 65 and highest observation = 225.
So, the range of data set
= Highest value- Lowest value
= 225 – 65 = 160
Thus, statements (i), (ii) and (iii) are true.

Question 18.
The dot plots below show the distribution of the number of pockets on clothing for a group of boys and for a group of girls.

Based on the dot plots, which of the following statements are true?
(i) The data varies more for the girls than for the
(ii) The median number of pockets for the boys is more than that for the girls.
(iii) The mean number of pockets for the girls is more than that for the boys.
(iv) The maximum number of pockets for boys is greater than that for the girls.
Choose the correct option from the following:
(a) (i) and (ii)
(b) (iii) and (iv)
(c) (ii) and (iv)
(d) (i) and (iii)
Solution:
(a) (i) and (ii)
For boys: 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6
Range = highest value-lowest value = 6 – 3 = 3
For girls: 0, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 6
Range = highest value-lowest value = 6 – 0 = 6
As the range for girls is higher than the range for boys, the data for girls is more variable than that for boys.
For median:
Number of boys, n = 12, which is even
∴ Median number of pockets for the boys
= \(\frac{\left(\frac{12}{2}\right)^{\text {th }} \text { value }+\left(\frac{12}{2}+1\right)^{\text {th }} \text { value }}{2}\)
= \(\frac{6^{\text {th }} \text { value }+7^{\text {th }} \text { value }}{2}\)
= \(\frac{5+5}{2}=\frac{10}{2}\) = 5
Now, number of girls =13, which is odd.
∴ Median number of pockets for the girls
= \(\left(\frac{13+1}{2}\right)^{\text {th }}\) observation = 7^th observation
= 4
Clearly, the median number of pockets for the boys is more than that for the girls.
For mean:
Total number of pockets for boys
= 3 + 4 + 4 + 4 + 4 + 5 + 5 + 5 + 5 + 5 + 6 + 6
= 3 + 16 + 25 + 12 = 56
∴ Mean number of pockets for boys
= \(\frac{\text { Total number of pockets }}{\text { Number of boys }}=\frac{56}{12}\) ≈ 4.67
Also, total number of pockets for girls
= 0 + 2 + 3 + 3 + 3 + 3 + 4 + 4 + 4 + 4 + 4 + 5 + 6
= 2 + 12 + 20 + 11 = 45
∴ Mean number of pockets for girls
= \(\frac{\text { Total number of pockets }}{\text { Number of girls }}=\frac{45}{13}\) ≈ 3.46
Clearly, the mean number of pockets for the girls is less than that for the boys.
Also, maximum number of pockets for both boys and girls is the same, i.e. 6.
Thus, statements (i) and (ii) are true.

Question 19.
A student’s weekly step counts (in thousands) are: 5, 5, 6, 7, 7, 8, 20
Which of the following will be largely affected by the removal of the outlier?
(i) Mean
(ii) Median
(iii) Range
(iv) None of these
Choose the correct answer from the following:
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (iv) only
Solution:
(c) (i) and (iii)
Observing the given set of data, we see that most numbers are clustered between 5 and 8. The number 20 is much higher than the others and acts as an outlier.
With the presence of an outlier:
Mean steps = \(\frac{5+5+6+7+7+8+20}{7}=\frac{58}{7}\)
= 8.29
The data is already arranged in ascending order.
So, median = 7
Range of data = 20 – 5 = 15
On removing the outlier:
Mean steps: \(\frac{5+5+6+7+7+8}{6}=\frac{38}{6}\) = 6.33
Now, number of observations = 6 is even, so median is average of two middle values.
∴ Median = \(\frac{6+7}{2}=\frac{13}{2}\) = 6.5
Range of data = 8 – 5 = 3
Difference in mean = 8.29 – 6.33 = 1.96
Difference in median = 7 – 6.5 = 0.5
Difference in range = 15 – 3 =12
Therefore, both the mean and the range are largely affected by outliers.

Question 20.
Consider the given dot plot (for 20 students) representing the data of number of pairs of socks owned by a student.

Which of the following statements are correct?
(i) Total 77 pairs of socks are represented in thegiven dot plot.
(ii) Only one student has the highest number of pairs of socks.
(iii) Number of students that own more than 5 pairs is 4.
(iv) Range of the data is 4.
Choose the correct option from the following:
(a) (i), (ii) and (iii)
(b) (i) and (ii) only
(c) (ii) and (iv) only
(d) (iv) only
Solution:
(a)
The total number of pairs of socks
= 0 + 1 + (4 × 2) + (2 × 3) + (5 × 4) + (3 × 5) + (3 × 6) + 9
= 0 + 1 + 8 + 6 + 20 + 15 + 18 + 9 = 77
Since there is only one dot corresponding to 9
i. e. the highest number of pairs of socks, only one student has the highest number of pairs of socks.
The number of students that own more than 5 pairs is 4.
Range of data = Highest value – Lowest value = 9 – 0 = 9
Thus, statements (i), (ii) and (iii) are correct.

Connecting the Dots Class 7 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 ansiyer 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): If two data sets have the same mean, they must have the same sum.
(R): Mean = \(\frac{Sum of observations}{Number of observations}\)
Solution:
(d) A is false but R is true.
We know, Mean = \(\frac{Sum of observations}{Number of observations}\)
Thus, Reason (R) is true.

For two data sets to have the same mean, the ratio of their sum to the number of observations must be equal. However, if the number of observations (n) is different for each set, their sums will also be different even if the mean is the same. For example:

Set A: 10, 10 (Sum = 20, Number of observations = 2, Mean =10)
Set B: 10, 10, 10 (Sum = 30, Number of observations = 3, Mean = 10)
Both have the same mean, but different sums. Thus, the Assertion (A) is false.

Question 2.
(A): If a data set has an outlier at the higher end, then the mean is more likely to be greater than the median.
(R): The outlier at the higher end does not affect the mean.
Solution:
(c) A is true but R is false.
We know that, if in a dataset, the outlier lies at the higher end, the mean is pulled upward.

Because of this, the mean becomes greater than the middle value, i.e. the median.
Thus, Assertion (A) is true, but Reason (R) is false.

Question 3.
(A): An inappropriate scale on a column graph can make the data interpretation misleading.
(R): The scale decides the value represented by each unit length on the axis.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
The height of the bars of column graph is entirely dependent on the chosen scale. Since the scale determines the value represented by each unit on the axis, an inappropriate scale leads to confusion while interpreting the graph.

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

Connecting the Dots Class 7 Fill in the Blanks

Question 1.
If there are even number of observations in a data set, then median is average of _____ middle values.
Solution: two
When number of observations in a data set is even, there exists two middle values.
Thus, if there are even number of observations in a data set, then median is average of two middle values.

Question 2.
Mean of first five multiples of 2 is _________ .
Solution: 6
The first five multiples of 2 are 2, 4, 6, 8, 10.
∴ Mean = \(\frac{Sum of first five multiples of 2}{5}\)
= \(\frac{2+4+6+8+10}{5}=\frac{30}{5}\) = 6
Thus, mean of first five multiples of 2 is 6.

Question 3.
A question is said to be statistical only if its answer is expected to show _________.
Solution: variability
A question is said to be statistical only if its answer is expected to show variability.

Question 4.
The mecpan Qf first 15 odd natural numbers is ________ .
Solution: 15
The first 15 odd numbers in ascending order are: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29 Here, the number of observations, n = 15, is odd.
∴ Median = \(\left(\frac{15+1}{2}\right)^{\text {th }}\)
Observation = 8^th observation =15
Thus, the median of the first 15 odd natural numbers is 15.

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.