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RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions

March 18, 2026 by Prasanna Leave a Comment

Practicing RBSE Class 6 Maths Solutions and Class 6 Maths Chapter 10 The Other Side of Zero Solutions Question Answer helps develop logical thinking and accuracy.

The Other Side of Zero Class 6 Solutions

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

Figure it Out (Page 245)

Question 1.
You start from Floor +2 and press -3 in the lift. Where will you reach? Write an expression for this movement.
Solution:
Here, starting floor =+2
The number of button presses = -3
∴ Target floor = Starting floor + Movement
= (+ 2) + (- 3) = -1

Question 2.
Evaluate these expressions (you may think of them as Starting Floor + Movement by referring to the Building of Fun).
(a) (+ 1) + (+ 4) = ……………………
(b) (+ 4) + (+ 1) = ……………………
(c) (+ 4) + (- 3) = ……………………
(d) (- 1) + (+ 2) = ……………………
(e) (- 1) + (+ 1) = ……………………
(f) (0) + (+ 2) = ……………………
(g) (0) + (- 2) = ……………………
Solution:
(a) +5
(b) +5
(c) +1
(d) +1
(e) 0
(f) +2
(g) -2

Question 3.
Starting from different floors, find the movements required to reach Floor – 5. For example, if I start at Floor +2, 1 must press – 7 to reach Floor – 5. The expression is (+ 2) + (- 7) = -5.
Find more such starting positions and the movements needed to reach Floor – 5 and write the expressions.
Solution:
(a) If I start at floor +1,1 must press -6 to reach floor -5.
The expression is : (+1) + (-6) = -5.

(b) If I start at floor +3, I must press -8 to reach floor -5.
The expression is : (+3) + (-8) = -5.

RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions

Figure it Out (Page 246)

Question 1.
Evaluate these expressions by thinking of them as the resulting movement of combin¬ing button presses:
(a) (+ 1) + (+ 4) = ……………………
(b) (+ 4) + (+ 1) = ……………………
(c) (+ 4) + (-  3) + (- 2) = ……………………
(d) (- 1) + (+ 2) + (- 3) = ……………………
Solution:
(a) +5
(b) +5
(c) (+ 4) + (- 5) = -1
(d) (- 4) + (+ 2) = -2

Figure it Out (Page 247)

Question 1.
Compare the following numbers using the Building of Fun and fill in the boxes with < or >.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 1
Notice that all negative number floors are below Floor 0. So, all negative numbers are less than 0. All the positive number floors are above Floor 0. So, all positive numbers are greater than 0.
Solution:
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 2

Question 2.
Imagine the Building of Fun with more floors. Compare the numbers and fill in the boxes with < or >:
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 3
Solution:
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 4

Question 3.
If Floor A = – 12, Floor D = – 1 and Floor E = + 1 in the building shown on the right as a line, find the numbers of Floors B, C, F, G and H.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 5
Solution:
B = -9
C = -6
F = +2
G = +6
H = +ll

Question 4.
Mark the following floors building shown on the right,
a. -7
b. -4
c. +3
d. -10
Solution:
To be done by the students.

RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions

Figure it Out (Page 249)

Question 1.
Complete these expressions. You may think of them as finding the movement needed to reach the Target Floor from the Starting Floor.
(a) (+ 1) – (+ 4) = ……………………
(b) (0) – (+ 2) = ……………………
(c) (+ 4) – (+ 1) = ……………………
(d) (0) – (- 2) = ……………………
(e) (+ 4) – (- 3) = ……………………
(f) (- 4) – (- 3) = ……………………
(g) (- 1) – (+ 2) = ……………………
(h) (- 2) – (- 2) = ……………………
(i) (- 1) – (+ 1) = ……………………
(j) (+3) – (-3) = ……………………
Solution:
(a) (+ 1) – (+ 4) = -3
(b) (0) – (+ 2) = -2
(c) (+ 4) – (+ 1) = +3
(d) (0) – (- 2) = +2
(e) (+ 4) – (- 3) = +7
(f) (- 4) – (- 3) = -1
(g) (- 1) – (+ 2) = -3
(h) (- 2) – (- 2) = 0
(i) (- 1) – (+ 1) = -2
(J) (+ 3) – (- 3) = +6

Figure it Out (Page 251)

Question 1.
Complete these expressions.
(a) (+ 40) + …………………… = +200
(b) (+ 40) + …………………… = -200
(c) (- 50) + …………………… = +200
(d) (- 50) + …………………… = -200
(e) (- 200) – (- 40) = ……………………
(f) (+ 200) – (+ 40) = ……………………
(g) (- 200) – (+ 40) = ……………………
Solution:
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 6

Figure it Out (Page 253)

RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 7
Question 1.
Mark 3 positive numbers and 3 negative numbers on the number line above.
Solution:
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 8
Positive numbers : 1, 3, 5
Negative numbers : -2, -5, -6

Question 2.
Write down the above 3 marked negative numbers in the following boxes: RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 9
Solution:
3 marked negative numbers : RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 10

Question 3.
Is 2 > -3? Why? Is – 2 < 3? Why?
Solution:
Yes. 2 is a positive number and -3 is a negative number. As positive numbers are always greater than negative numbers, therefore 2 > -3 and 3 > -2.

Question 4.
What are
(i) – 5 + 0
(ii) 7 + (- 7)
(iii) – 10 + 20
(iv) 10 – 20
(v) 7 – (- 7)
(vi) – 8 – (- 10)?
Solution:
(i) – 5 + 0 = -5
(ii) 7 + (- 7) = 0
(iii) – 10 + 20 = 10
(iv) 10 – 20 = -10
(v) 7 – (- 7) =14
(vi) – 8 – (- 10)= – 8 + 10 = 2

RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions

Figure it Out (Page 257)

Question 1.
Complete the additions using tokens,
(a) (+ 6) + (+ 4)
(b) (- 3) + (- 2)
(c) (+ 5) + (- 7)
(d) (- 2) + (+ 0)
Solution:
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 11

Question 2.
Cancel the zero pairs in the following two sets of tokens. On what floor is the lift attendant in each case? What is the corresponding addition statement in each case?
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 12
Solution:
a) RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 13
Therefore, the lift attendent is on the second floor below the ground floor.
The corresponding addition statement: (+ 3) + (- 5) = (-2)

b) RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 14
Therefore, the lift attendent is on the third floor above the ground floor.
The corresponding addition statement : (+6) + (-3) = (+3)

Figure it Out (Page 258)

Question 1.
Evaluate the following differences using tokens. Check that you get the same result as with other methods you now know:
(a) (+ 10) – (+ 7)
(b) (- 8) – (-4)
(c) (- 9) – (- 4)
(d) (+ 9) – (+ 12)
(e) (- 5) – (- 7)
(f) (- 2) – (- 6)
Solution:
(a) Here, from 10 positives we take away 7 positives.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 15
∴ (+10) – (+7) = +3

(b) Here, from 8 negatives we take away 4 negatives.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 16
∴ (-8) – (-4) = -4

(c) Here from 9 negatives take away 4 negatives.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 17
∴ (-9) – (-4) = -5

(d) There are not enough tokens to take out 12 positives from 9 positives so we put down 3 zero pairs.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 18
Therefore, (+9) – (+12) = -3

(e) There are not enough tokens to take out 7 negatives from 5 negatives so we put down 2 zero pairs.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 19
Therefore, (-5) – (-7) = +2

(f) There are not enough tokens to take out 6 negatives from 2 negatives so we put down 4 zero pairs.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 20
Therefore, (-2) – (-6) = +4

Question 2.
Complete the subtractions :
(a) (- 5) – (- 7)
(b) (+ 10) – (+ 13)
(c) (- 7) – (- 9)
(d) (+ 3) – (+ 8)
(e) (- 2) – (- 7)
(f) (+ 3) – (+ 15)
Solution:
(a) There are not enough tokens to take out 7 negatives from 5 negatives so we put an extra two zero pairs.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 21
∴ (-5) – (-7) = +2

(b) There are not enough tokens to take out 13 positives from 10 positives, so we put an extra 3 zero pairs.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 22
∴ (+10) – (+13) = -3

(c) There are not enough tokens to take out 9 negatives from 7 negative, so we put an extra two zero pairs.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 23
Therefore, (-7) – (-9) = +2

(d) There are not enough tokens to take out 8 positives from 3 positives, so we put an extra 5 zero pairs.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 24
Therefore, (+3) – (+8) = -5

(e) There are not enough tokens to take out 7 negatives from 2 negatives, therefore we put an extra 5 zero pairs.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 25
Therefore, (-2) – (-7) = +5

(f) There are not enough tokens to take 15 positives from 3 positives, therefore we put an extra 12 zero pairs.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 26
Therefore, (+3) – (+15) = -12

RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions

Figure it Out (Page 259)

Question 1.
Try to subtract : – 3 – (+ 5).
How many zero pairs will you have to put in? What is the result?
Solution:
We have to take out 5 positives from 3 negatives. But there are not enough positives, so we have to put in 5 zero pairs.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 27
Therefore, -3 – (+5) = -8

Question 2.
Evaluate the following using tokens.
(a) (- 3) – (+ 10)
(b) (+ 8) – (- 7)
(c) (- 5) – (+ 9).
(d) (- 9) – (+ 10)
(e) (+ 6) – (- 4)
(f) (- 2) – (+ 7)
Solution:
(a) (- 3) – (+ 10)
We have to take out 10 positives from 3 negatives, but there are not enough positives. Therefore, we have to put in 10 zero pairs. Now we can take out 10 positives.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 28
Therefore, (-3) – (+10) = -13

(b) (+ 8) – (- 7)
We have to take out 7 negatives from 8 positives. But there are not enough negatives, therefore we have to put in 7 zero pairs. Now we can take out 7 negatives.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 29
Therefore, (+ 8) – (- 7) = 15

(c) (- 5) – (+ 9)
We have to take out 9 positives from 5 negatives. There are not enough positives, therefore we have to put in 9 zero pairs. Now we can take out 9 positives.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 30
Therefore, (- 5) – (+ 9) = -14

(d) (- 9) – (+ 10)
We have to take out 10 positives from 9 negatives, But there are not enough positives. Therefore, we have to put in 10 zero pairs. Now we can take out 10 positives.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 31
Therefore, (-9) – (+ 10) = -19

(e) (+ 6) – (- 4)
We have to take out 4 negatives from 6 positives. But there are not enough negatives. Therefore, we put down 4 zero pairs. Now we can take away 4 negatives.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 32
Therefore, (+ 6) – (- 4) = +10

(f) (- 2) – (+ 7)
We have to take out 7 positives from 2 negatives. But there are not enough positives. Therefore, we put down 7 zero pairs. Now we can take away 7 positives.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 33
Therefore, – 2 – (+ 7) = -9

Figure it Out (Page 260)

Question 1.
Suppose you start with 0 rupees in your bank account, and then you have credits of ₹30, ₹40, and ₹50, and debits of ₹40, ₹50, and ?60. What is your bank account balance now?
Solution:
Here, Initial account balance = ₹0
Credits = ₹30 + ₹40 + ₹50 = ₹120
Debits – ₹40 + ₹50 + ₹60 = ₹150
and Balance = credits – debits
∴ = ₹120 – ₹150 = -₹30
Therefore, the bank account balance is – ₹30.

Question 2.
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:
Here, Initial account balance = ₹0
Debits = ₹1 + ₹2 + ₹4 + ₹8 + ₹16 + ₹32 + ₹64 + ₹128 = ₹255
Credits = ₹256
∴ Balance = Credits – Debits = ₹256 – ₹255 = ₹1
Therefore, the bank account balance is ₹1.

Question 3.
Why is it generally better to try and maintain a positive balance in your bank account? What are circumstances under which it may be worthwhile to temporarily have a negative balance?
Solution:
To be done by the students.

RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions

Figure it Out (Page 261)

Question 1.
Looking at the geographical cross section fill in the respective heights :
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 34
Solution:
(a) A = +1500 m.
(b) B = -500 m.
(c) C = +300 m.
(d) D = -1200 m.
(e) E = +1200 m.
(f) F = -200 m.
(g) G = +100 m.

Question 2.
Which is the highest point in this geographical cross-section? Which is the lowest point?
Solution:
The highest point = A
The lowest point = D

Question 3.
Can you write the points A, B, …, G in a sequence of decreasing order of heights? Can you write the points in a sequence of increasing order of heights?
Solution:
Decreasing order of heights : A > E > C > G > F > B > D
And, increasing order of heights : D < B < F < G < C < E < A

Question 4.
What is the highest point above sea level on Earth? What is its height?
Solution:
The highest point above sea level on Earth is Mount Everest. Its height is 8848 m above the sea level.

Question 5.
What is the lowest point with respect to sea level on land or on the ocean floor? What is its height? (This height should be negative).
Solution:
The lowest known point on the Earth is Marina Trench in the Pacific Ocean. Its depth is 11034 m below sea level. Which is written as -11034 m.

Figure it Out (Page 262)

Question 1.
Do you know that there are some places in India where temperatures can go below 0°C? Find out the places in India where temperatures sometimes go below 0°C. What is common among these places? Why does it become colder there and not in other places?
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 35
Solution:
The places in India where temperature can go below 0°C—
(1) Ladakh
(2) Himachal Pradesh
(3) Jammu and Kashmir
(4) Sikkim
(5) Arunachal Pradesh

These places arc located in Himalayan region at high altitudes. These regions arc farther from the equator receiving less direct sunlight, so it becomes colder there. Also higher altitude leading to colder temperature.

Question 2.
Leh in Ladakh gets very cold during winter. The following is a table of temperature readings taken during different times of the day/night in Leh on a day in November. Match the temperature with the appropriate time of the day/night.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 36
Solution:
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 37

RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions

Figure it Out (Page 263)
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 38
Question 1.
Do the calculations for the second grid above and find the border sum.
Solution:
On analyzing the given grid—
The above grid is a 3 × 3 arrangement of numbers.
The sum of numbers in each row and column should be the same.
The border sum is as follows:
Top row : 5 + (-3) + (-5) = -3
Bottom row : -8 + (-2) + 7 = -3
Left column : 5 + 0 + (-8) = -3
Right column : (-5) + (-5) + 7 = -3
Therefore, the border sum of the above grid is -3.

Question 2.
Complete the grids to make the required border sum :
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 39
Solution:
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 40
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 41

Question 3.
For the last grid above, find more than one way of filling the numbers to get border sum – 4.
Solution:
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 42

Question 4.
Which other grids can be filled in multiple ways? What could be the reason?
Solution:
Grid with a larger size (more rows and column) can be filled in multiple ways. Because there are more degrees of freedom to distribute numbers while maintaining the border sum.

Question 5.
Make a border integer square puzzle and challenge your classmates.
Solution:
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 43

Figure it Out (Page 264)

Question 1.
Try afresh, choose different numbers this time. What sum did you get? Was it different from the first time? Try a few more times!
Solution:
Let’s circle the number -5 as per the game. Now let’s strike out the row and column with the number -5.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 44
Again let’s circle the number 3 and strike out the row and column with the number 3.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 45
Again let’s circle the number -1 and strike out the row and clolumn with number -1
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 46
Again let’s circle the balance number 2 and strike out the row and column with the number 2.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 47
Now let’s add the circled numbers = (- 5) + 3 + (-1) + 2 = – 6 + 5 = -1
Here, we get Name value (-1).

RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions

Question 2.
Play the same game with the grids below. What answer did you get?
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 48
Solution:
(i) Let’s circle the number 1.
Now, according to the game let’s cross out the row and column with number 1.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 49
Let’s circle the number 13.
Now cross out the row and column with number 13.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 50
Again, let’s circle the number – 20.
Now we will cross out the row and column with number – 20.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 51
Let’s circle the number -2.
Now cross out the row and column with number -2.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 52
Now let’s add the circled numbers—
1 + 13 – 20 – 2 = 14 – 22 = -8
Thus, we arrive at the value -8.

(ii) Let’s circle the number 0.
Now as per game, we will cross out the row and column with number 0.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 53
Now, let’s circle the number -5.
We will cross out the row and column with number -5.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 54
Let’s circle the number 1.
We cross out the row and column with number 1.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 55
Again, let’s circle the number -10.
We cross out the row and column with number -10.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 56
Now we will add the circled numbers—
= 0 + (-5) + 1 + (-10) = -14
Which is the required answer.

Question 3.
What could be so special about these grids? Is the magic in the numbers or the way they are arranged or both? Can you make more such grids?
Solution:
Grids can be seem like magic due to both the numbers involved and their arrangement. The numbers in a grid can exibit specific patterns or sequences. Yes, we can make more such grids.

Figure it Out (Page 265)

Question 1.
Write all the integers between the given pairs, in increasing order.
(a) 0 and -7
(b) -4 and 4
(c) -8 and -15
(d) -30 and -23
Solution:
(a) The integers between 0 and -7 in increasing order are : -6, -5, -4, -3, -2, -1
(b) The integers between -4 and 4 in increasing order:-3, -2, -1, 0, 1, 2, 3
(c) The integers between -8 and -15 in increasing order : -14, -13, -12, -11, -10, -9
(d) The integers between -30 and -23 in
increasing order: -29, -28, -27, -26, -25, -24

Question 2.
Give three numbers such that their sum is -8.
Solution:
Three numbers such that their sum is -8 are : -11, 2 and 1
When we add them together we get :
– 11 + 2 + 1 = -8

Question 3.
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 upon rolling these two dice.
The faces of dice have numbers : -1, 2, -3, 4, -5, 6.
First we find all possible sums.

(i) The sum of two negative numbers :

  • (-1) + (-1) = -2
  • (-1) + (-3) = -4
  • (-1) + (-5) = -6
  • (-3) + (-3) = -6
  • (-3) + (-5) = -8
  • (-5) + (-5) = -10

(ii) The sum of one negative and one positive numbers:

  • (-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

(iii) The sum of two positive numbers :

  • 2 + 2 = 4
  • 2 + 4 = 6
  • 2 + 6 = 8
  • 4 + 4 = 8
  • 4 + 6 = 10
  • 6 + 6 = 12

Let’s arrange these sums in ascending order : -10, -8, -6, -4, -3, -2, -1, 1, 3, 4, 5, 6, 8, 10, 12
∴ The numbers between (-10) and (+12) that are not possible to get the adding numbers on these two dice :
-9, -7, -5, 0, 2, 7, 9, 11

Question 4.
Solve these :

(a) 8 – 13 (b) (- 8) – (13)
(c) (- 13) – (- 8) (d) (- 13) + (- 8)
(e) 8 + (- 13) (f) (- 8) – (- 13)
(g) 13 – 8 (h) 13 – (- 8)

Solution:
(a) 8 – 13 = -5
(b) (-8) – (13) = – 8 – 13 = -21
(c) (- 13) – (- 8) = -13 + 8 = -5
(d) (- 13) + (- 8) = -13 – 8 = -21
(e) 8 + (- 13) = 8 – 13 = -5
(f) (- 8) – (- 13) = -8 + 13 = 5
(g) (13) – 8 = 13 – 8 = 5
(h) 13 – (-8) = 13 + 8 = 21

RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions

Question 5.
Find the years below.
(a) From the present year, which year was it 150 years ago? ____________
(b) From the present year, which year was it 2200 years ago? ____________
(Hint: Recall that there was no year 0.)
(c) What will be the year 320 years after 680 BCE? ____________
Solution:
(a) 150 years ago from the present year (2025):
2025 – 150 = 1875

(b) 2200 years ago from the present year :
Since there was no year 0, we need to account for it in our calculation:
2025 – 2200 = -175
The year -175 corresponds to 176 BCE, so 2200 years ago, it was the year 176 BCE.

(c) As BCE is before Christ, hence let’s write 680 BCE = -680.
Hence, 320 years after 680 BCE
= – 680 + 320
= -360 = 360 BCE.

Question 6.
Complete the following sequences :
(a) (-40), (-34), (-28), (-22), …………………, …………………, …………………
(b) 3, 4, 2, 5, 1, 6, 0, 7, …………………, …………………, …………………
(c) …………………, …………………, 12, 6, 1, (-3), (-6), …………………, …………………, …………………
Solution:
(a) -40, -34, -28, -22, …………………, …………………, …………………
∵ (-34) – (-40) = – 34 + 40 = 6
= (-28) – (-34) = – 28 + 34 = 6
(-22) – (-28) = -22 + 28 = 6
Clearly, this is a sequence where each term increases by 6. Each term is obtained by adding 6 to the previous term.
Hence, the sequence is : (-40), (-34), (-28), (-22), (-16), (-10), (-4).

(b) Subtracting the last number of the sequence with the preceding number :
7 – 0 = 7
0 – 6 = -6
6 – 1 = 5
1 – 5 = -4
5 – 2 = 3
2 – 4 = -2
4 – 3 = 1
Therefore in the sequence, numbers are decreasing by 1 with alternate positive and nega¬tive integers.
Hence, the next number :
7 + (- 8) = -1
– 1 + 9 = 8
8 – 10 = -2
– 2 + 11 = 9
and so on.

∴ Complete sequence : 3, 4, 2, 5, 1, 6, 0, 7, -1, 8, -2, 9 …………………..
Let’s check : – 1 – 7 = -8
8 + 1 = 9
– 2 – 8 = -10

(c) Subtracting the last number of the sequence from the preceding number :
(- 6) – (- 3) = – 6 + 3 = -3
(- 3) – (1) = – 3 – 1 = -4
1 – 6 = -5
6 – 12 = -6
Clearly, on subtracting the number before 12 from 12, we get -7, hence on adding 7 to 12 we will get the number before 12.
∴ Number before 12 : 12 + 7 = 19
Similarly, Number before 19 : 19 + 8 = 27
Number after – 6 : – 6 + (- 2) = -8
Number after – 8 : = 8 + (- 1) = -9
Number after – 9 : -9 + 0 = -9
Hence, required sequence is :
27, 19, 12, 6, 1, (-3), (-6), (-8), (-9), (-9)

Question 7.
Here are six integer cards: (+1), (+7), (+18), (-5), (-2), (-9).
You can pick any of these and make an ex-pression using addition(s) and subtraction(s). Here is an expression: (+18) + (+ 1) – (+ 7) – (- 2) which gives a value (+14). Now, pick cards and make an expression such that its value is closer to (-30).
Solution:
Using the given cards, one possible expression such that its value is closer to (-30) is:
(- 9) + (-5) + (- 2) + (- 18) + (+ 1).
Let’s calculate the value step by step:
(1) – 9 + (- 5) = -14
(2) – 14 + (- 2) = -16
(3) – 16 +(- 18) = -34
(4) – 34 + (+ 1) = -33
Hence, the value of the expression is (-33) which is closer to (-30).

Question 8.
The sum of two positive integers is always positive but a (positive integer) – (positive integer) can be positive or negative. What about
(a) (positive) – (negative)
(b) (positive) + (negative)
(c) (negative) + (negative)
(d) (negative) – (negative)
(e) (negative) – (positive)
(f) (negative) + (positive)
Solution:
(a) (positive) – (negative) : Subtracting a negative number is the same as adding the corresponding positive number. Therefore, it is always positive, e.g., 6 – (- 3) = 6 + 3 = 9

(b) (positive) + (negative) : The result depends on the magnitudes of the numbers. If the positive number is greater the result is positive, if the negative number is greater in magnitude the result is negative.
e.g., 8 + (- 4) = 4 (positive)
4 + (- 7) = -3 (negative)

(c) (negative) + (negative) : The sum of two negatives is always negative.
e.g., -3 + (-6) = -9

(d) (negative) – (negative) : If the first negative number is greater in magnitude the result will be negative, if first negative number is smaller in magnitude, the result will be positive.
i. e., -5 – (-3) = – 5 + 3 = -2 (negative)
-5 – (-8) = – 5 + 8 = 3 (positive)

(e) (negative) – (positive) : On subtracting a positive number from a negative number, the result will always be negative.
e.g., – 4 – 2 = -6

(f) (negative) + (positive) : It depends on the magnitudes of the numbers. If the positive number is greater, the result is positive. If the negative number is greater in magnitude the result is negative.
e.g., – 4 + 6 = 2 (positive)
– 4 + 2 = -2 (negative)

Question 9.
This string has a total of 100 tokens arranged in a particular pattern. What is the value of the string?
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 57
Solution:
Let us analyze the pattern :
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 58
Here, one set of 5 tokens has 3 positives and 2 negatives. Since there are 100 tokens in the string,
∴ Total sets of 5 tokens = 100 ÷ 5 = 20 sets
Value of 1 set = 3 – 2 = 1 Hence, the value of the string = 20 × 1 = 20

RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions

Figure it Out (Page 268)

Question 1.
Can you explain each of Brahmagupta’s rules in terms of Bela’s Building of Fun, or in terms of a number line?
Solution:
Brahmagupta gave explicit rules for operations on all numbers — positive, negative and zero. We can explain Brahmagupta’s rules in terms of Bela’s Building of Fun and a number line:
Brahmagupta’s Rules—
1. Addition of positive numbers :
Rule : The sum of two positives is positive.

  • Bela’s Building : If Bela starts on floor +3 and moves up 2 floors she reaches on the floor +5.
  • Number line : On a number line, moving from 3 to 5 by adding 2. Example : 3 + 2 = 5

2. Addition of negative numbers :
Rule : The sum of two negati\es is negative.

  • Bela’s Building : If Bela starts 3 floors below ground level (-3) and moves down 2 more floors, she reaches 5 floors below ground (-5).
  • Number line : Moving from -3 to -5 by adding -2. ‘ Example : (- 3) + (- 2) = (-5)

3. Addition of a positive and a negative number :
Rule : Subtract the smaller number (without the sign) from the greater number (without the sign) and place the sign of the greater number to obtain the result.

  • Bela’s Building : If Bela starts on the floor + 3 and moves down 5 floors, she reaches 2 floors below ground (-2).
  • Number line : Moving from 3 to -2 by adding -5.
    Example : 3 + (- 5) = (-2)

4. Subtraction of a positive number from a negative number :
Rule : Subtracting a positive number from a negative number is like adding the two numbers and keeping the negative sign.

  • • Bela’s Building : If Bela starts 3 floors below ground (-3) and moves down 2 more floors, she reaches 5 floors below ground floor (-5).
    • Number line : moving from -3 to -5 by subtracting 2.
    Example : – 3 – 2 = -5

5. Subtraction of a negative number from a positive number :
Rule: Subtracting a negative number is the same as adding the corresponding positive number.

  • Bela’s Building : If Bela starts on the floor +3 and moves up 2 floors, she reaches floor +5.
  • Number line: Moving from 3 to 5 by subtracting -2.
    Example : 3 – (- 2) = 5

6. Subtraction of a negative number from a negative number :
Rule : Subtracting a negative number from a negative number is like adding the absolute values and keeping the negative sign.
• Bela’s Building: If Bela is 3 floor below ground (-3) and moves up 2 floors, she reaches first floor below ground (-1).
Number line : Moving from -3 to -1 by subtracting -2.
Example : (- 3) – (- 2) = (-1)

Question 2.
Give your own examples of each rule.
Solution:
(1) Addition of positive numbers :
Example : 8 + 5 = 13
Bela’s Building : Starting on the 8th floor and moving up 5 floors to reach 13th floor.

(2) Addition of negative numbers :
Example : (- 5) + (- 3) = (-8)
Bela s Building : Starting 5 floors below ground and moving down 3 more floors to reach 8 floors below ground.

(3) Addition of a positive and a negative number :
Example : 6 + (- 2) = 4
Bela’s Building : Starting on the 6th floor and moving down 2 floors to reach 4th floor.

(4) Subtraction of a positive number from a negative number :
Example : – 4 – 3 = – 7
Bela’s Building: Starting 4th floor below ground and moving down 3 more floors to reach 7 floors below ground.

(5) Subtraction of a negative number from a positive number :
Example : 6 – (- 2) = 8
Bela’s Building : Starting on the 6th floor and moving up 2 floors to reach 8th floor.

(6) Subtraction of a negative number from a negative number :
Example : – 5 (- 3) = -2
Bela’s Building : Starting 5 floors below ground and moving up 3 floors to reach 2 floors below ground.

The Other Side of Zero Class 6 Question Answer

The Other Side of Zero Class 6 Extra Questions

Multiple Choice Questions—

Question 1.
Which of the following is a set of integers?
(a) {1, 3, 5 ……………….}
(b) {1, 2, 3, 4 ……………….}
(c) ( ……………….-2, -1, 0, 1, 2, 3,)
(d) none of the above
Answer:
(c) ( ……………….-2, -1, 0, 1, 2, 3,)

Question 2.
By subtracting (-7) from (-15) we get :
(a) -8
(b) 8
(c) -22
(d) -10
Answer:
(a) -8

Question 3.
Fourth integer to the left of the integer 3 will be :
(a) 1
(b) -1
(c) 2
(d) -3
Answer:
(b) -1

Question 4.
The value of (- 5) + (- 13) will be :
(a) +8
(b) -18
(c) -13
(d) (-5)
Answer:
(b) -18

RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions

Question 5.
The additive inverse of 14 is :
(a) -14
(b) zero
(c) \(\frac{1}{14}\)
(d) none of these
Answer:
(a) -14

Question 6.
The integer just before -18 is :
(a) -19
(b) -17
(c) 19
(d) 17
Answer:
(a) -19

Question 7.
Which of the following statements is true?
(a) -20 > -21
(b) -20 < -21 (c) 20 > 21
(d) 20 = 21
Answer:
(a) -20 > -21

Question 8.
(- 4) + (- 5) = ?
(a) 9
(b) -9
(c) 1
(d) -1
Answer:
(b) -9

Question 9.
0 + (- 2) + 3 = ?
(a) -5
(b) -6
(c) 0
(d) 1
Answer:
(d) 1

Question 10.
The integer between -2 and -1 is :
(a) 0
(b) 1
(c) -1.5
(d) none of these
Answer:
(d) none of these

Fill in the blanks—

1. Every positive integer is greater than every ………………………. .
2. …………………….. is smaller than every positive integer.
3. Zero is ………………………. than every negative integer.
4. 8 – (……….) = 0
5. The value of 5 – (- 11) will be …………………….. .
6. The value of (- 2) – (3) + 5 will be …………………….. .
Answer:
1. negative integer
2. 0 (zero)
3. greater
4. 8
5. 16
6. 0

RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions

Write True/False for the following statements—

1. The sum of two positives is always positive. (True/False)
2. A positive number can also be obtained by adding two negative numbers. (True/False)
3. The additive inverse of 5 is -5. (True/False)
4. The sum of three different integers can not be zero. (True/False)
Answer:
1. True
2. False
3. True
4. False

Make the right match—

Question 1.

Part (A) Part (B)
(i) 5 – (-3) (a) 1
(ii) – 7 + (- 8) (b) 0
(iii) 4 + (- 3) (c) -15
(iv) – 8 + (8) (d) 8

Answer:
(i) – (d), (ii) – (c), (iii) – (a), (iv) – (b).

Part (A) Part (B)
(i) 5 – (-3) (d) 8
(ii) – 7 + (- 8) (c) -15
(iii) 4 + (- 3) (a) 1
(iv) – 8 + (8) (b) 0

Question 2.

Part (A) Part (B)
(i) -9 RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 59 -12 (a) >
(ii) -18 RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 59 -10 (b) –
(iii) 9 + (- 9) RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 59 0 (c) <
(iv) 8 RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 59 8 = 0 (d) –

Answer:
(i) – (a), (ii) – (c), (iii) – (d), (iv) – (b).

Part (A) Part (B)
(i) -9 RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 59 -12 (a) >
(ii) -18 RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 59 -10 (c) <
(iii) 9 + (- 9) RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 59 0 (d) –
(iv) 8 RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 59 8 = 0 (b) –

Very Short Answer Type Questions—

Question 1.
Find the sum of -50, -200 and 300.
Solution:
(- 50) + (- 200) + (300)
= – 250 + 300
= – 250 + 250 + 50
= 0 + 50 = 50

Question 2.
Find the value of: (- 7) + (- 9) + 4 + 6.
Solution:
(- 7) + (- 9) + (4) + (6)
= -16 + 4 + 6
= -16 + 10
= -6

Question 3.
Find the value of: 37 + (- 2) + (- 65) + (- 8)
Solution:
(37) + (- 2) + (- 65) + (- 8)
= 37 + (- 75)
= 37 + (- 37) + (- 38)
= 0 + (- 38) = -38

Question 4.
Find the value of : (- 217) + (- 100) + (- 50)
Solution:
(- 217) + (- 100) + (- 50)
= – 217 + (- 150)
= -367

Question 5.
Write four negative integers greater than -20.
Solution:
Four negative integers greater than -20 are:
-19, -18,-17 and -16.

RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions

Question 6.
Write four negative integers smaller than -10.
Solution:
-11, -12, -13, -14

Question 7.
Find the value : (- 15) – (- 18)
Solution:
(- 15) – (- 18) = – 15 + 18
= – 15 + 15 + 3
= 0 + 3 = 3

Question 8.
Find the value : (- 10) + (- 20) – 10
Solution:
– 10 + (- 20) – 10 = -20 + (- 20) = -40

Question 9.
Draw a number line and show which number will we reach if we move 4 numbers to the right of -2.
Solution:
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 60

Question 10.
Represent (- 1) + (- 7) on a number line.
Solution:
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 61
= (- 1) + (- 7) = -8

Short Answer Type Questions—

Question 1.
Write all the integers between the given pairs (write them in the increasing order) :
(a) 0 and -7
(b) -4 and 4
(c) -8 and -15
(d) -30 and -23
Solution:
(a) The integers between 0 and -7 are (increasing order):
-6, -5, -4, -3, -2, -1.

(b) The integers between -4 and 4 in increasing order are :
-3, -2, -1, 0, 1, 2, 3.

(c) The integers between -8 and -15 in increasing order are :
-14, -13, -12, -11, -10, -9.

(d) The integers between -30 and -23 in increasing order are :
-29, -28, -27, -26, -25, -24.

Question 2.
Write the following numbers with appropriate signs:
(a) 100 m below sea level.
(b) 25°C above 0°C temperature.
(c) 15°C below 0°C temperature.
(d) Any five numbers less than 0.
Solution:
(a) -100 m,
(b) 25°C,
(c) -15°C
(d) -1, -2, -3, -4, -5

Question 3.
Represent the following positions with + or – sign :
(a) 8 steps to the left of zero
(b) 7 steps to the right of zero
(c) 11 steps to the right of zero
(d) 6 steps to the left of zero
Solution:
(a) -8
(b) +7
(c) +11
(d) -6

Question 4.
Find the solution of the following addition using number line : – 3 + 7 + (- 5)
Solution:
First we move three steps to the left of 0 to reach -3. From here we move seven steps to the right of -3 to reach 4. From here we move 5 steps to the left of 4 to reach -1 as shown on number line.
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 62

RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions

Question 5.
Mark -3, 7, -4, -8, -1 and 3 on the number line—
Solution:
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 63
Clearly the integers -3, 7, -4, -8, -1 and 3 are represented by respective points A, B, C, D, E and F on the number line.

Essay Type Questions—

Question 1.
Following is the list of temperatures of five places in India, on a particular day of the year :
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 64
(a) Write the temperatures of these places in the form of integers in the blank column.
(b) Following is the number line representing the temperature in degree Celsius :
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 65
Plot tbe of the city against its temperature.
(c) Which is the coolest place?
(d) Write the names of the places where temperatures are above 10°C.
Solution:
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 66
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 67
(b) Number line
RBSE Class 6 Maths Chapter 10 The Other Side of Zero Solutions 68
(c) Siachen
(d) Ahmedabad and Delhi

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