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NCERT Solutions Ganita Prakash Chapter 10 & 266Practice with integers — Figure it Out

Book page 265 Updated on2026-09-05

Q1.
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
Answer

‘Between’ means the two given numbers themselves are not written. Increasing order = left to right on the number line.

a. Between 0 and – 7: – 6, – 5, – 4, – 3, – 2, – 1
b. Between – 4 and 4: – 3, – 2, – 1, 0, 1, 2, 3
c. Between – 8 and – 15: – 14, – 13, – 12, – 11, – 10, – 9
d. Between – 30 and – 23: – 29, – 28, – 27, – 26, – 25, – 24
Careful with (c) and (d): the smaller number is written second in the question. – 15 is smaller than – 8, so the list must start at – 14 and climb up to – 9. Do not forget that 0 belongs in the list in part (b).
Q2.
Give three numbers such that their sum is – 8.
Answer

Many answers are possible. Here are a few, each checked:

(– 3) + (– 4) + (– 1) = – 8
5 + (– 6) + (– 7) = – 8
10 + (– 12) + (– 6) = – 8
0 + (– 8) + 0 = – 8
100 + (– 50) + (– 58) = – 8
How to build your own: write down any two numbers, add them, and then choose the third number so that the total moves to – 8. For 5 and – 6 the running total is – 1, and – 1 needs a further – 7 to reach – 8.
Q3.
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.
Answer

Make the addition table of every possible roll. The faces are – 5, – 3, – 1, 2, 4, 6.

+– 5– 3– 1246
– 5– 10– 8– 6– 3– 11
– 3– 8– 6– 4– 113
– 1– 6– 4– 2135
2– 3– 11468
4– 1136810
613581012

The sums that can appear are – 10, – 8, – 6, – 4, – 3, – 2, – 1, 1, 3, 4, 5, 6, 8, 10, 12.

So the numbers between – 10 and + 12 that are not possible are
– 9, – 7, – 5, 0, 2, 7, 9, 11
Why 0 cannot appear: to get 0 the two faces would have to be inverses of each other — like + 5 and – 5, or + 1 and – 1. On these dice the positive faces are 2, 4, 6 and the negative faces are – 1, – 3, – 5, so no pair cancels out.
Check it yourself: that is 8 impossible values out of the 21 whole numbers from – 9 to + 11.
Q4.
Solve these: 8 – 13, (– 8) – (13), (– 13) – (– 8), (– 13) + (– 8), 8 + (– 13), (– 8) – (– 13), (13) – 8, 13 – (– 8)
Answer

Change every subtraction to ‘add the inverse’ first.

ExpressionWritten as an additionAnswer
8 – 138 + (– 13)– 5
(– 8) – (13)– 8 + (– 13)– 21
(– 13) – (– 8)– 13 + 8– 5
(– 13) + (– 8)already an addition– 21
8 + (– 13)already an addition– 5
(– 8) – (– 13)– 8 + 13+ 5
(13) – 813 + (– 8)+ 5
13 – (– 8)13 + 8+ 21
Look at the pattern: only four different answers appear — – 5, – 21, + 5 and + 21. The digits are always 5 or 21; only the signs change. 13 and 8 differ by 5 and add to 21, and the sign depends on which of them ‘wins’.
Q5.
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?
Answer

Take the present year as 2026 CE (do the same working with whatever year it is when you solve this).

a. 150 years back is still in the CE years, so plain subtraction works.

2026 – 150 = 1876 CE

b. 2200 years back takes us past the year 1 CE and into the BCE years. Do it in two steps.

From 2026 CE back to 1 CE: 2026 – 1 = 2025 years
Years still to go back: 2200 – 2025 = 175
Because there is no year 0, the year just before 1 CE is 1 BCE, the one before that is 2 BCE …
So we reach 175 BCE

c. 680 BCE with 320 years added moves us towards the year 1, so the BCE number gets smaller.

680 – 320 = 360 BCE
Why BCE years count backwards: think of BCE years as negative and CE years as positive on a time line, with no 0 in between. Adding years always means moving to the right — 680 BCE + 320 years lands on 360 BCE, which is later in history.
Note: The book's answer key writes ‘– 360 BCE’ for part (c). The minus sign is not needed — ‘BCE’ already says that the year is before the Common Era. The correct answer is 360 BCE.
Q6.
Complete the following sequences: a. (– 40), (– 34), (– 28), (– 22), ___, ___, ___ b. 3, 4, 2, 5, 1, 6, 0, 7, ___, ___, ___ c. ___, ___, 12, 6, 1, (– 3), (– 6), ___, ___, ___
Answer

a. Find the difference between one term and the next.

– 40 + 6→ – 34 + 6→ – 28 + 6→ – 22 → – 16, – 10, – 4

b. This sequence has two threads running through it, one going down and one going up.

Odd places: 3, 2, 1, 0 … each one less → next is – 1, then – 2
Even places: 4, 5, 6, 7 … each one more → next is 8
So the sequence continues – 1, 8, – 2

c. Look at the differences: they grow by 1 each time.

12 – 6→ 6 – 5→ 1 – 4→ – 3 – 3→ – 6
Going backwards the steps were – 7 and – 8, so the two missing numbers in front are 27, 19
Going forward the steps are – 2, – 1, 0, so the sequence continues – 8, – 9, – 9 (and then – 8 again)
Full sequence: 27, 19, 12, 6, 1, – 3, – 6, – 8, – 9, – 9, – 8, – 6 …
Why (c) turns around: the amount being subtracted gets smaller and smaller (6, 5, 4, 3, 2, 1), then becomes 0, and after that the steps turn positive. So the sequence falls, flattens out at – 9, and starts climbing again.
Q7.
Here are six integer cards: (+ 1), (+ 7), (+ 18), (– 5), (– 2), (– 9). You can pick any of these and make an expression 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).
Answer

We can hit exactly – 30. Two neat ways:

Way 1: (– 5) – (+ 7) – (+ 18)
= – 5 – 7 – 18 = – 30 ✔
Way 2: (– 2) + (– 9) – (+ 18) – (+ 1)
= – 2 – 9 – 18 – 1 = – 30 ✔
How to search: to make a very negative value, add the negative cards and subtract the positive ones — subtracting a positive number pushes the value down. The big card (+ 18) is the most useful of all, provided it is subtracted.
Try This: Using the same six cards, can you make + 30? Try (+ 18) + (+ 7) + (+ 1) – (– 5) + … and see how close you get.
Q8.
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)
Answer
Kind of expressionResultExamples
a.(positive) – (negative)always positive7 – (– 2) = 9,   1 – (– 100) = 101
b.(positive) + (negative)positive, negative or zero7 + (– 2) = 5,   2 + (– 7) = – 5,   7 + (– 7) = 0
c.(negative) + (negative)always negative– 3 + (– 4) = – 7
d.(negative) – (negative)positive, negative or zero– 3 – (– 8) = 5,   – 8 – (– 3) = – 5,   – 3 – (– 3) = 0
e.(negative) – (positive)always negative– 3 – 4 = – 7
f.(negative) + (positive)positive, negative or zero– 3 + 8 = 5,   – 8 + 3 = – 5,   – 3 + 3 = 0
Why some are certain and others are not: in (a), (c) and (e) both moves push the same way — in (a) you start above zero and go further up; in (c) and (e) you start below zero and go further down. In (b), (d) and (f) the two numbers pull in opposite directions, so the winner depends on which one has the bigger digits.
Note: (b), (d) and (f) are really the same situation, because d and f can be turned into b by ‘adding the inverse’.
Q9.
This string has a total of 100 tokens arranged in a particular pattern. What is the value of the string?
Answer

Look at the pattern on the string: 3 green tokens then 2 red tokens, repeating again and again.

+ + + – – + + + – – … one block of 5 tokens = + 1
The string repeats a block of 5 tokens — three green and two red. Each block is worth (+ 3) + (– 2) = + 1.
Value of one block = (+ 3) + (– 2) = + 1
Number of blocks in the string = 100 ÷ 5 = 20
Value of the string = 20 × (+ 1) = + 20

Answer: the value of the string is + 20.

Another way to see it: in 20 blocks there are 20 × 3 = 60 green tokens and 20 × 2 = 40 red tokens. Cancel 40 zero pairs and 20 green tokens are left, so the value is + 20. ✔
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