Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 10 The Other Side of Zero Class 6 Question Answer to understand textbook questions step by step.
Class 6 Maths Chapter 10 The Other Side of Zero Solutions
Ganita Prakash Class 6 Chapter 10 Solutions
Class 6 Maths Ganita Prakash Chapter 10 Solutions The Other Side of Zero
Question 1.
Evaluate these expressions:
(a) (+ 1) + (+ 4) = _______
(b) (+ 4) + (+ 1) = _______
(c) (+ 4) + (- 3) = _______
(d) (- 1) + (+ 2) = _______
(e) (- 1) + (+ 1) = _______
(f) 0 + (+ 2) = _______
(g) 0 + (- 2) = _______
Solutions:
(a) (+ 1) + (+ 4) = + 5
(b) (+ 4) + (+ 1) = + 5
(c) (+ 4) + (- 3) = + 1
(d) (- 1) + (+ 2) = + 1
(e) (- 1) + (+ 1) = 0
(f) 0 + (+ 2) = + 2
(g) 0 + (- 2) = – 2
Question 2.
Try to subtract: – 3 – (+ 5). How many zero pairs will you have to put in? What is the result?
Solution:
– 3 – (+ 5) = – 8
We want to take away 5 positive tokens when we have 3 negative tokens. So, we add 5 zero pairs. After removing the 5 positive tokens, we have 8 negative tokens left.
Question 3.
Suppose you start with 0 rupees in your bank account and then you have debits of ₹ 1, ₹ 2, ₹ 4, ₹ 8, ₹ 16, ₹ 32, ₹ 64 and ₹ 128 and then a single credit of ₹ 256. What is your bank account balance now?
Solution:
Total debits = ₹ 1 + ₹ 2 + ₹ 4 + ₹ 8 + ₹ 16 + ₹ 32 + ₹ 64 + ₹ 128 = ₹ 255
There is a single credit of ₹ 256.
∴ Final balance = Total credit – Total debit = ₹ 256 – ₹ 255 = ₹ 1
Question 4.
Looking at the geographical cross section, fill in the respective heights:
Solution:
A = + 1500 m
B = – 500 m
C = + 300 m
D = – 1200 m
E = + 1200 m
F = – 200 m
G = + 100 m
Question 5.
Leh in Ladakh gets very cold during winter. The following is a table of temperature readings taken during different times of the day and night in Leh on a day in November. Match the temperature with the appropriate time of the day and night.
Solution:
Usually, the coldest time of the day is very early in the morning, before the sun rises, and it starts to get warmer after sunrise, reaching the warmest in the afternoon. At night, the temperature starts to drop again.
So, we can match the temperatures like this:
– 4°C → 2:00 a.m. (This is the coldest time when it is still dark and freezing.)
– 2°C → 11:00 p.m. (It is night time, so it is also very cold, but not as cold as early morning.)
8°C → 11:00 a.m. (It is late morning, the sun is up and it is getting warmer.)
14°C → 2:00 p.m. (This is the afternoon and the warmest time of the day.)
This shows us how the temperature rises as the sun comes up and falls again after sunset.
Question 6.
Complete the grids to make the required border sum
Solution:
Question 7.
There are two dice whose faces have these numbers: – 1, 2, – 3, 4, – 5, 6. The smallest possible sum upon rolling these dice is – 10 = (- 5) + (- 5) and the largest possible sum is 12 = (6) + (6). Some numbers between (- 10) and (+ 12) are not possible to get by adding numbers on these two dice. Find those numbers.
Solution:
Let’s find the sums that are not possible when rolling these two dice.
The faces of the dice are:
– 1, 2, -3, 4, -5, and 6.
First, let’s list all possible sums:
The sum of two negative numbers:
(- 1) + (- 1) = -2; (- 1) + (- 3) = – 4; (- 1) + (- 5) = – 6
(- 3) + (- 3) = – 6; (- 3) + (- 5) = – 8 (- 5) + (- 5) = – 10
The sum of one negative and one positive number:
(-1) + 2 = 1; (-1)+ 4 = 3; (-1) + 6 = 5; (-3) + 2 = – 1; (-3)+ 4=1;
(- 3) + 6 = 3; (- 5) + 2 = – 3; (- 5) + 4 = – 1; (- 5) + 6 = 1
The sum of two positive numbers:
2 + 2 = 4; 2 + 4 = 6; 2 + 6 = 8; 4 + 4 = 8; 4 + 6=10; 6 + 6=12
Now, let’s list all the possible sums in ascending order:
– 10, -8, -6, -4, -3,-2,- 1, 1, 3, 4, 5, 6, 8, 10, 12
The sum of numbers between – 10 and 12 that are not possible to get are: – 9, – 7, – 5, 0, 2, 7, 9, 11 .
Question 8.
This string has a total of 100 tokens arranged in a particular pattern. What is the value of the string?
Solution:
Total tokens =100
Total sets of 5 tokens = \(\frac{100}{5}\) = 20
Value of one set of 5 tokens = 3 + (- 2) = 1
Hence, total value of string = 1 × 20 = 20
InText Questions
Question 1.
Can there be a number less than 0? Can you think of any ways to have less than 0 of something?
Solution:
Yes, there can be numbers less than 0.
They are called negative numbers and written with a minus sign, like – 1, – 2, – 3, etc.
Temperature: In cold places, temperature can go below 0 °C, like – 5 °C or – 10 °C.
Money: If you have ₹ 10 and will have – ₹ 10. you spend ₹ 20, you
Question 2.
Connect the inverses by drawing lines.
Solution:
The inverses of + 5, – 7, – 8 and + 9 are – 5, + 7, + 8 and – 9 respectively.
Question 3.
Should we write – 3 <-4or-4 <-3?
Solution:
We should write – 4 < – 3, because on the number line, – 4 is to the left of – 3, which makes it smaller.
So, – 4 < -3 or -3 > -4.
Question 4.
Evaluate 15 – 5, 100 – 10 and 74 – 34.
Solution:
(i) What should we add to 5 to make 15? i.e.
5 + _______ = 15.
The missing number is 10. So, 15 – 5 = 10.
(ii) What should we add to 10 to make 100? i.e.
10 + ______ = 100.
The missing number is 90. So, 100 – 10 = 90.
(iii) What should we add to 34 to make 74? i.e.
34 + _______ = 74.
The missing number is 40. So, 74 – 34 = 40.
Question 5.
Use unmarked number lines to evaluate these expressions:
(a) _______ – 125 + (- 30) = _______
(b) _______ + 105 – (- 55) = _______
(c) _______ + 80 – (- 150) = _______
(d) _______ – 99 – (- 200) = _______
Solution:
(a) – 125 + (- 30) = – 155
We start at – 125 and move 30 steps left on the number line.
(b) + 105 – (- 55) = + 105 + ( + 55) = + 160
We start at + 105 and move 55 steps right on the number line.
(c) + 80 – (- 150) = + 80 + (+ 150) = + 230
We start at + 80 and move 150 steps right on the number line.
(d) -99-(-200) = -99 + 200 = + 101
We start at – 99 and move 200 steps right on the number line.
The Other Side of Zero Class 6 Extra Questions
The Other Side of Zero Class 6 Very Short Question Answer
Question 1.
In a game, Rishi scored + 10,-5, + 3,-8 and + 4 in five rounds. What is his total score?
Solution:
Given, the scores of Rishi in five rounds are +10, – 5, + 3, – 8 and + 4.
Total score of Rishi
= + 10 + (- 5) + (+ 3) + (- 8) + (+ 4)
= (+ 10 + 3 + 4) + (-5-8)
[Grouping positive and negative integers]
= (+ 17) + (- 13)
= (+ 17)-(+ 13)
[The number that is being added can be replaced by . its additive inverse and then subtracted.]
= + 4
∴ The total score of Rishi is + 4.
Question 2.
Arrange the following integers in ascending order:
9, -7, -4, 0, 3
Solution:
Given integers are: 9, – 7, – 4, 0, 3
For positive integers 3 and 9, we have 3 < 9.
For negative integers – 7 and – 4, we have – 7 < – 4.
We know that on the number line, the numbers right to 0 are greater than 0 and the numbers left to 0 are less than 0. Also, every positive integer is greater than every negative integer.
Hence, the given integers in ascending order: – 7, -4, 0, 3, 9
Question 3.
A boat sailed 40 km to the east of a harbour and then 70 km to the west from there. How far from the harbour is the boat finally?
Solution:
The boat starts from point A at harbour. It sailed 40 km to the east to reach point B. From point B, it sailed 70 km to the west to reach point C, as shown in figure.
Now, the position of boat = (+ 40) + (- 70) = – 30
Hence, the boat is 30 km to the west of the harbour.
Question 4.
Write 5 distinct integers whose sum is 6.
Solution:
We know that the sum of an integer with its additive inverse is zero.
The additive inverse of – 1 is 1 and the additive inverse of – 2 is 2.
∴ 1 + (- 1) = 0 and 2 + (- 2) = 0
Now, [1 + (- 1)] + [2 + (- 2)] + 6 = 0 + 0 + 6 = 6
∴ 5 distinct integers whose sum is 6 are – 1, 1, -2, 2, 6.
Question 5.
Arrange the following integers in descending order:
-8, -3, 5, 1, -1
Solution:
Given integers are – 8, – 3, 5, 1 and – 1.
We know that a number is greater than every number that is to its left on the number line.
For positive integers 1 and 5, we have 5 > 1.
For negative integers -1,-3 and – 8, we have
– 1 > – 3 > – 8.
Also, every positive integer is greater than every negative integer.
Hence, the given integers in descending order:
5, 1, – 1, -3, -8
The Other Side of Zero Class 6 Short Question Answer
Question 1.
Complete the additions using tokens:
(i) (+ 4) + (- 8)
(ii) (- 3) + (+ 4)
(iii) (- 10) + ( + 6)
(iv) (+ 7) + (- 5)
Solution:
(i) (+ 4) + (- 8) = – 4
(ii) (- 3) + (+ 4) = + 1
(iii) (- 10) + (+ 6) = – 4
(iv) (+ 7) + (- 5) = + 2
Question 2.
Complete the subtractions using tokens:
(i) (+ 6) – (+ 3)
(ii) (+ 8) – (+ 7)
(iii) (- 9) – (- 5)
(iv) (- 7) – (- 4)
Solution:
(i) (+ 6) – (+ 3) = + 3
(ii) (+ 8) – (+ 7) = + 1
(iii) (- 9) – (- 5) = – 4
(iv) (- 7) – (- 4) = – 3
Question 3.
Complete the following subtractions using tokens:
(i) (+ 9) – (+ 2)
(ii) (+ 5) – (4- 3)
(iii) (- 8) – (- 3)
(iv) (- 6) – (- 1)
Solution:
(i) (+ 9) – (+ 2) = + 7
(ii) (+ 5) – (+ 3) = + 2
(iii) (- 8) – (- 3) = – 5
(iv) (- 6) – (- 1) = – 5
Question 4.
Complete the following subtractions using tokens:
(i) (+ 3) – (+ 8)
(ii) (+ 7) – (+ 9)
(iii) (+ 4) – (- 4)
(iv) (+ 6) – (- 5)
(v) (- 4) – (+ 3)
(vi) (- 6) – (+ 6)
Solution:
(i) (+ 3) – (+ 8) = – 5
(ii) (+ 7) – (+ 9)
(iii) (+ 4) – (- 4)
(iv) (+ 6) – (- 5)
(v) (- 4) – (+ 3)
(vi) (- 6) – (+ 6)
The Other Side of Zero Class 6 Long Question Answer
Question 1.
There are two dice whose faces have the numbers 1, – 2, 3, – 4, 5 and – 6. If these dice are rolled together, find the smallest possible sum of the numbers so obtained.
Solution:
Given, there are two dice whose faces have the numbers 1, – 2, 3, – 4, 5 and – 6.
To get the smallest possible sum, we need to add the smallest number from each die.
A number is always greater than any number to its left on the number line.
For positive integers 1, 3 and 5, we have 1 < 3 < 5.
For negative integers -2,-4 and – 6, we have -6 < -4 < -2.
We know that every positive integer is greater than every negative integer.
Hence, the given integers in ascending order: -6, -4, -2, 1, 3, 5
The smallest number on each die is – 6.
Hence, smallest possible sum = (- 6) + (- 6) = – 12
Question 2.
Using the number line write the integer which is
(i) 3 more than – 1
(ii) 4 less than – 1
(iii) 4 more than – 9
Solution:
(i) To find the integer which is 3 more than – 1, we start at – 1 on the number line and move 3 steps forward. We reach 2. Hence, the required integer is 2.
(ii) To find the integer which is 4 less than – 1, we start at – 1 on the number line and move 4 steps backward. We reach – 5. Hence, the required integer is -5.
(iii) To find the integer which is 4 more than – 9, we start at – 9 on the number line and move 4 steps forward. We reach – 5. Hence, the required integer is – 5.
Question 3.
There are two dice whose faces have the numbers – 8, – 3, – 1, 2, 4 and 6. If these dice are rolled together, find the largest possible sum of the numbers so obtained.
Solution:
Given, there are two dice whose faces have the numbers – 8, – 3, – 1, 2, 4 and 6.
To get the largest possible sum, we need to add the largest number from each die.
We know that a number is greater than every number that is to its left on the number line.
For positive integers 2, 4 and 6, we have 2 < 4 < 6.
For negative integers -1,-3 and – 8, we have – 8 < – 3 < – 1.
We know that every positive integer is greater than every negative integer.
Hence, the given integers in ascending order: – 8, -3, – 1, 2, 4, 6
The largest number on each die is 6.
So, largest possible sum = (6) + (6) = 12
Question 4.
Complete the subtractions using tokens:
(i) (+ 4) – (+ 7)
(ii) (+ 5) – (+ 8)
(iii) (+ 5) – (- 3)
(iv) (+ 6) – (- 2)
(v) (- 4) – (+ 2)
Solution:
(i) (+ 4) – (+ 7) = – 3
[Adding 3 zero pairs]
(ii) (+ 5) – (+ 8) = – 3
[Adding 3 zero pairs]
(iii) (+ 5) – (- 3) = + 8
[Adding 3 zero pairs]
(iv) (+ 6) – (- 2) = + 8
[Adding 2 zero pairs]
(v) (- 4) – (+ 2) = – 6
[Adding 2 zero pairs]
Question 5.
Complete the grids to make the required border
(i)
| -7 | ||
| 4 | ||
| -3 |
Border sum = + 5
(ii)
| -12 | ||
| -1 | ||
| 7 |
Border sum = – 4
(iii)
| 3 | 11 | |
| -2 | -3 |
Border sum = 0
(iv)
| -15 | 5 | |
| 10 | ||
| 4 |
Border sum = 3
Solution:
We have to take that number which makes the sum of all the numbers in a row or column equal to the border sum. So, the complete grids are given below.
(i)
| -7 | 3 | 9 |
| -4 | 4 | |
| 16 | -3 | -8 |
Border sum = + 5
(ii)
| 2 | -12 | 6 |
| -1 | -17 | |
| -5 | -6 | 7 |
Border sum = – 4
(iii)
| 3 | -14 | 11 |
| -1 | -8 | |
| -2 | 5 | -3 |
Border sum = 0
(iv)
| -15 | 13 | 5 |
| 10 | -6 | |
| 8 | -9 | 4 |
Border sum = 3
Question 6.
Complete the following additions using tokens:
(i) (+ 7) + (- 2)
(ii) (- 12) + (+ 4)
(iii) (- 9) + (+ 5)
(iv) (-7) + (+ 11)
(v) (- 3) + (+ 8)
(vi) (- 6) + (+ 12)
Solution:
(i) (+ 7) + (-2) = + 5
(ii) (- 12) + (+ 4) = – 8
(iii) (- 9) + (+ 5) = – 4
(iv) (-7) + (+ 11) = + 4
(v) (- 3) + (+ 8) = + 5
(vi) (- 6) + (+ 12) = + 6
Question 7.
Complete the grids to make the required border
(i)
| 0 | ||
| 5 | ||
| -2 | ||
| Border sum = – 2 | ||
(ii)
| 6 | ||
| -5 | ||
| -3 | ||
| Border sum = + 3 | ||
(iii)
| 0 | ||
| 3 | 5 | |
| 8 | ||
| Border sum = 4 | ||
(iv)
| 3 | ||
| -7 | 6 | |
| -8 | ||
| Border sum = 5 | ||
Solution:
We have to take that number which makes the sum of all the numbers in a row or column equal to the border sum. So, the complete grids are given below.
(i)
| 1 | 0 | -3 |
| -1 | 5 | |
| -2 | 4 | -4 |
(ii)
| 5 | -8 | 6 |
| -5 | -6 | |
| 3 | -3 | 3 |
(iii)
| 3 | 0 | 1 |
| 3 | 5 | |
| -2 | 8 | -2 |
(iv)
| 3 | -5 | 7 |
| -7 | 6 | |
| -2 | 4 | -8 |
The Other Side of Zero Class 6 Case Based Questions
Question 1.
In Shimla, the temperature in the morning was – 2°C. By afternoon, it rose by 5°C. In another city, Manali, the temperature was – 5°C in the morning and increased by 3°C in the afternoon.
Based on the above information, answer the following questions:
(i) What was the afternoon temperature in Shimla?
(ii) What was the afternoon temperature in Manali?
(iii) What is the difference in afternoon temperatures of both cities?
Solution:
(i) Given, morning temperature in Shimla = – 2 °C
Increase in temperature by afternoon = + 5 °C
To find afternoon temperature, we add the change to the morning temperature.
∴ Afternoon temperature in Shimla = (- 2 °C) + (+ 5°C) = + 3°C
(ii) Given, morning temperature in Manali = – 5°C
Increase in temperature by afternoon = + 3 °C
To find afternoon temperature, we add the change to the morning temperature.
∴ Afternoon temperature in Manali = (- 5 °C) + (+ 3 °C)= – 2 °C
(iii) Afternoon temperature in Shimla = + 3 °C
Afternoon temperature in Manali = – 2 °C
Difference in afternoon temperatures of both cities
= + 3 °C – (- 2 °C)
= + 3 °C + (+ 2 °C) = 5 °C
Question 2.
In a geographical cross-section, some points appear above sea level (marked with positive height values) and others lie below sea level (represented by negative values), using sea level defined as zero as the reference point.
Based on the given information, answer the following questions:
(i) Write the highest and lowest points in the given geographical cross section?
(ii) What is the difference between heights of the highest and lowest points?
(iii) Write the points A, B, …, G in a sequence of increasing order of heights.
OR
(iii) Write the points A, B, …, G in a sequence of decreasing order of heights.
Solution:
(i) E(500 m) is the highest point and C (- 400 m) is the lowest point.
(ii) Difference between heights of E and C
= 500 – (- 400) = 500 + (+ 400) = 900 m
(iii) Points in increasing order are as:
C(- 400 m), B(- 300 m), G(- 200 m), D(0 m), A(200 m), F(300 m), E(500 m)
OR
(iii) Points in decreasing order, are as:
E(500 m), F(300 m), A(200 m), D(0 m), G(- 200 m), B(- 300 m), C(- 400 m)