---
title: "NCERT Solutions Class 7 Maths Chapter 10 Operations with Integers"
url: https://www.swavid.com/maths/class/7/chapter/operations-with-integers/ncert-solutions
dateModified: 2026-10-07T15:14:48+00:00
---

# NCERT Solutions Class 7 Maths Chapter 10 Operations with Integers

This chapter's questions cover operations with integers including addition, subtraction, multiplication, and division, along with their properties and real-life applications.

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## Rakesh’s Puzzle: A Number Game

### Question 1

*3 marks · Short answer*

Can you tell me the two numbers? You don’t need to use any formulas. Just try different pairs of numbers and then check: 1. Do the two numbers add up to 25? 2. Is the difference between them 11?

**Solution**

1. Let the two numbers be the first number and the second number.
2. We need to find two numbers whose sum is 25 and whose difference (first number $-$ second number) is 11.
3. By trying different pairs of numbers, we check the pair 18 and 7: their sum is $18 + 7 = 25$ and their difference is $18 - 7 = 11$.

**Answer:** The two numbers are 18 and 7.

> Common mistake: Subtracting in the wrong order so that the difference becomes $-11$ instead of $11$.

## Figure it Out

### Question 1

*3 marks · Short answer*

Let us try to find a few more pairs of numbers from their sums and differences: (a) Sum = 27, Difference = 9 (b) Sum = 4, Difference = 12 (c) Sum = 0, Difference = 10 (d) Sum = 0, Difference = –10 (e) Sum = –7, Difference = –1 (f) Sum = –7, Difference = –13

**Part (a)**

1. Let the numbers be $x$ and $y$.
2. Sum = 27 and Difference = 9.
3. The numbers are 18 and 9.

Answer (a): 18 and 9

**Part (b)**

1. Let the numbers be $x$ and $y$.
2. Sum = 4 and Difference = 12.
3. The numbers are 8 and -4.

Answer (b): 8 and -4

**Part (c)**

1. Let the numbers be $x$ and $y$.
2. Sum = 0 and Difference = 10.
3. The numbers are 5 and -5.

Answer (c): 5 and -5

**Part (d)**

1. Let the numbers be $x$ and $y$.
2. Sum = 0 and Difference = -10.
3. The numbers are -5 and 5.

Answer (d): -5 and 5

**Part (e)**

1. Let the numbers be $x$ and $y$.
2. Sum = -7 and Difference = -1.
3. The numbers are -4 and -3.

Answer (e): -4 and -3

**Part (f)**

1. Let the numbers be $x$ and $y$.
2. Sum = -7 and Difference = -13.
3. The numbers are 3 and -10.

Answer (f): 3 and -10

**Answer:** Pairs are (a) 18, 9; (b) 8, -4; (c) 5, -5; (d) -5, 5; (e) -4, -3; (f) 3, -10.

> Common mistake: Mixing up the order of the first and second number when calculating the difference.

## Carrom Coin Integers

### Question 1

*3 marks · Short answer*

To begin with, the coin is at point 0. If the coin is struck twice, with the first strike moving it by 4 units and the second strike moving it by 3 units, what will be the final position of the coin?

**Solution**

1. The initial position of the coin is at 0.
2. The first strike moves the coin by $4$ units to the right, placing it at position $4$.
3. The second strike moves the coin further by $3$ units to the right.
4. The final position of the coin is $4 + 3 = 7$ units from $0$.

**Answer:** 7 units from 0

> Common mistake: Multiplying the movements instead of adding them.

### Question 2

*3 marks · Short answer*

If the coin is struck twice, and if the two movements are known, can you give a formula for the final position of the coin?

**Solution**

1. Let the first strike move the coin $a$ units to the right.
2. Let the second strike move the coin $b$ units to the right.
3. The final position $P$ is the sum of the two movements.
4. Therefore, the formula is $P = a + b$.

**Answer:** $P = a + b$

> Common mistake: Subtracting the movements instead of adding them.

### Question 3

*3 marks · Short answer*

The coin is at 0. If it is struck twice (the direction of the two strikes may be the same or different) can you give a formula for the final position of the coin?

**Solution**

1. Let the first movement be represented by integer $a$ (positive for right, negative for left).
2. Let the second movement be represented by integer $b$ (positive for right, negative for left).
3. The final position $P$ after the two strikes is found by adding the two movements.
4. Thus, the formula remains $P = a + b$.

**Answer:** $P = a + b$

> Common mistake: Ignoring the signs of integers while adding.

### Question 4

*3 marks · Short answer*

What is the final position of the coin?

**Solution**

1. The first movement is given as $5$ units ($a = 5$).
2. The second movement is given as $-7$ units ($b = -7$).
3. The final position is given by adding the two movements: $P = a + b$.
4. So, $P = 5 + (-7) = -2$.

**Answer:** $-2$ (or 2 units to the left of 0)

> Common mistake: Writing $+2$ instead of $-2$.

### Question 5

*3 marks · Short answer*

Based on this new model, answer the following questions: 1. If the first movement is –4 and the final position is 5, what is the second movement? 2. If there are multiple strikes causing movements in the order 1, –2, 3, –4, …, –10, what is the final position of the coin?

**Part (1)**

1. Given first movement $a = -4$ and final position $P = 5$.
2. Using the formula $P = a + b$, we have $5 = -4 + b$.
3. Therefore, the second movement $b = 5 - (-4) = 5 + 4 = 9$.

Answer (1): 9

**Part (2)**

1. The movements are given by the series: $1, -2, 3, -4, 5, -6, 7, -8, 9, -10$.
2. Group the terms as pairs: $(1 - 2) + (3 - 4) + (5 - 6) + (7 - 8) + (9 - 10)$.
3. Each pair sums to $-1$, and there are $5$ such pairs.
4. The final position is $5 \times (-1) = -5$.

Answer (2): -5

**Answer:** Second movement is 9; Final position is -5

> Common mistake: Incorrectly grouping terms or handling signs in the series sum.

### Question 6

*3 marks · Short answer*

From the figures below, what can you conclude about the magnitudes of a and b compared to each other, and what are their directions? Remember to start from 0. 1. 2. 3.

**Solution**

1. In figure 1, movement $a$ is rightward and movement $b$ is leftward, with the magnitude of $a$ greater than the magnitude of $b$ as the final position $P$ is to the right of 0.
2. In figure 2, movement $a$ is rightward and movement $b$ is leftward, with the magnitude of $b$ greater than the magnitude of $a$ as the final position $P$ is to the left of 0.
3. In figure 3, movement $a$ is leftward and movement $b$ is leftward, so both movements are in the negative direction and the final position $P$ is to the left of 0.

**Answer:** Magnitudes and directions depend on the relative lengths of arrows and their orientations in each figure as described.

> Common mistake: Confusing leftward and rightward directions with the signs of magnitudes.

## Token Model Operations

### Question 1

*3 marks · Short answer*

Find (+7) – (+18).

**Solution**

1. To subtract 18 from 7, we need to remove 18 positive tokens from 7 positive tokens.
2. Since there are not enough positive tokens, we put in 11 zero pairs so that we can remove 18 positives.
3. After removing 18 positives, 11 negative tokens are left in the bag, which means $-11$.
4. Therefore, $7 - 18 = -11$.

**Answer:** $$-11$$

> Common mistake: Subtracting the absolute values directly and getting $+11$ instead of $-11$.

### Question 2

*3 marks · Short answer*

Using tokens, argue out the following statements. (a) 7 – 18 = 7 + (–18) (additive inverse of 18 is –18) (b) 4 – (–12) = 4 + 12 (additive inverse of –12 is 12)

**Part (a)**

1. Subtracting 18 from 7 means removing 18 positives, which requires adding 11 zero pairs and leaves 11 negatives.
2. This is equivalent to adding the additive inverse of 18, which is $-18$, to 7.
3. Thus, $7 - 18 = 7 + (-18)$.

Answer (a): $$7 - 18 = 7 + (-18)$$

**Part (b)**

1. Subtracting $-12$ from 4 means removing 12 negatives from 4 positives.
2. To remove 12 negatives from an empty bag or 4 positives, we must add 12 zero pairs, leaving 12 positives.
3. Thus, $4 - (-12) = 4 + 12$ since the additive inverse of $-12$ is 12.

Answer (b): $$4 - (-12) = 4 + 12$$

**Answer:** Statements argued using the token model and additive inverses.

> Common mistake: Confusing the sign when changing subtraction to addition of the inverse.

### Question 3

*3 marks · Short answer*

How many positives are in the bag now?

**Solution**

1. There are 8 positives in the bag.
2. This can be seen as adding 2 positives to the bag 4 times.
3. Thus, the operation is represented as $4 \times 2 = 8$.

**Answer:** $$8$$

> Common mistake: Counting the total number of tokens incorrectly by misinterpreting the grouping.

### Question 4

*3 marks · Short answer*

Similarly find the values of 4 × (–6) and 9 × (–7)? How can we interpret (–4) × 2?

**Solution**

1. Interpret $4 \times (-6)$ as placing 6 red tokens (negatives) into an empty bag 4 times, which gives $24$ red tokens, so $4 \times (-6) = -24$.
2. Interpret $9 \times (-7)$ as placing 7 red tokens into an empty bag 9 times, giving $63$ red tokens, so $9 \times (-7) = -63$.
3. Interpret $(-4) \times 2$ as removing 2 green tokens (positives) from zero pairs in the bag 4 times, leaving 8 red tokens, so $(-4) \times 2 = -8$.

**Answer:** $4 \times (-6) = -24$, $9 \times (-7) = -63$, and $(-4) \times 2 = -8$

> Common mistake: Confusing positive and negative tokens when interpreting negative multipliers.

### Question 5

*3 marks · Short answer*

Why are we trying to remove green tokens and not red tokens?

**Solution**

1. We start with an empty bag, so there are no tokens in the bag initially.
2. To remove green (positive) tokens from an empty bag, we must first place zero pairs inside the bag.
3. Therefore, we place zero pairs so that we can remove the required positive tokens.

**Answer:** We place zero pairs first because an empty bag contains no tokens to remove.

> Common mistake: Assuming tokens are already present in the bag before the operation begins.

### Question 6

*3 marks · Short answer*

What happens when both the integers in the multiplication are negative? How do we model (–4) × (–2) with tokens?

**Solution**

1. When both integers are negative, such as $(-4) \times (-2)$, the product is positive because we remove negative tokens.
2. Since the bag starts empty, we first place 2 zero pairs (each containing 1 green and 1 red token) inside the bag 4 times.
3. We then remove 2 red tokens (negatives) from the bag 4 times, leaving 8 positive green tokens, which gives $(-4) \times (-2) = 8$.

**Answer:** $(-4) \times (-2) = 8$ by removing 2 red tokens from zero pairs 4 times

> Common mistake: Forgetting to add zero pairs before removing tokens from an empty bag.

## Figure it Out

### Question 1

*3 marks · Short answer*

Using the token interpretation, find the values of: (a) 3 × (–2) (b) (–5) × (–2) (c) (–4) × (–1) (d) (–7) × 3

**Part (a)**

1. Placing 2 negatives into an empty bag 3 times gives $3 \times (-2) = -6$.

Answer (a): $3 \times (-2) = -6$

**Part (b)**

1. Removing 2 negatives from the bag 5 times gives $(-5) \times (-2) = 10$.

Answer (b): $(-5) \times (-2) = 10$

**Part (c)**

1. Removing 1 negative from the bag 4 times gives $(-4) \times (-1) = 4$.

Answer (c): $(-4) \times (-1) = 4$, (d) $(-7) \times 3 = -21$

**Answer:** The values are (a) $-6$, (b) $10$, (c) $4$, and (d) $-21$.

> Common mistake: Confusing the sign of the product when multiplying negative integers.

### Question 2

*3 marks · Short answer*

If 123 × 456 = 56088, without calculating, find the value of: (a) (–123) × 456 (b) (–123) × (–456) (c) (123) × (–456)

**Part (a)**

1. Given $123 \times 456 = 56088$, multiplying a negative by a positive gives a negative product.

Answer (a): $(-123) \times 456 = -56088$

**Part (b)**

1. Multiplying two negative integers gives a positive product.

Answer (b): $(-123) \times (-456) = 56088$

**Part (c)**

1. Multiplying a positive by a negative gives a negative product.

Answer (c): $123 \times (-456) = -56088$

**Answer:** The values are (a) $-56088$, (b) $56088$, and (c) $-56088$.

> Common mistake: Forgetting to change the sign of the product.

### Question 3

*3 marks · Short answer*

Try to frame a simple rule to multiply two integers.

**Solution**

1. Multiply the absolute values of the two integers to get the magnitude of the product.
2. If both integers have the same sign (both positive or both negative), the product is positive.
3. If the integers have opposite signs, the product is negative.

**Answer:** The product of two integers has a magnitude equal to the product of their magnitudes and is positive if the signs are the same, and negative if the signs are different.

> Common mistake: Stating that the product of two negatives is negative.

### Question 4

*3 marks · Short answer*

What integer do we get as the final answer in each case? Do we get different answers because the sets look different, or the same answer because they all represent –2?

**Solution**

1. Each different token set still represents the same integer value, which is $-2$.
2. Taking any of these token sets 4 times means placing or removing them according to the multiplier.
3. We get the same final answer because all the token sets represent the exact same number $-2$.

**Answer:** We get the same answer because all the different sets represent the same number $-2$.

> Common mistake: Assuming different visual arrangements of tokens change the numerical value.

### Question 5

*3 marks · Short answer*

Check this for 5 × 4, by taking different token sets corresponding to 4.

**Solution**

1. Consider different token sets that represent $4$, such as 4 green tokens, or 5 green tokens and 1 red token.
2. Take each set 5 times by placing them into the bag.
3. The final number of green tokens in the bag is always $20$, showing that the result is independent of the chosen token representation.

**Answer:** The final result is always $20$ regardless of the token set chosen for $4$.

> Common mistake: Failing to verify that equivalent token sets yield the same product.

### Question 6

*3 marks · Short answer*

What do you notice in this pattern? Can you describe it?

**Solution**

1. Observe the sequence of products as the multiplier decreases by 1 unit each time.
2. When the multiplicand is positive, for every unit decrease in the multiplier, the product decreases by the value of the multiplicand.
3. This pattern continues smoothly even when the multiplier goes below zero into negative integers.

**Answer:** For every unit decrease in the multiplier, the product decreases by the multiplicand.

> Common mistake: Thinking the pattern stops when the multiplier becomes zero or negative.

### Question 7

*3 marks · Short answer*

Will this pattern continue when the multiplier goes below zero and becomes a negative number?

**Solution**

1. Observe the given multiplication pattern where the multiplicand is positive.
2. As the multiplier decreases below zero into negative integers, the product continues to decrease by the value of the multiplicand.
3. Yes, the pattern continues seamlessly for negative multipliers.

**Answer:** Yes, the pattern continues when the multiplier becomes a negative number.

> Common mistake: Thinking that patterns involving multiplication break when numbers become negative.

### Question 8

*3 marks · Short answer*

What is the pattern when the multiplicand is a negative integer?

**Solution**

1. Examine the sequence of products when the multiplicand is a negative integer.
2. For every unit decrease in the multiplier, the product increases by the magnitude of the negative multiplicand.
3. This is the inverse of the pattern observed with a positive multiplicand.

**Answer:** When the multiplicand is negative, for every unit decrease of the multiplier, the product increases by the multiplicand.

> Common mistake: Confusing whether the product increases or decreases as the multiplier decreases.

### Question 9

*3 marks · Short answer*

Will this pattern continue when the multiplier goes below zero and becomes a negative number?

**Solution**

1. Recall the established pattern for negative multiplicands as the multiplier decreases.
2. Check if the rule of increasing product per unit decrease of the multiplier holds below zero.
3. Confirm that the sequence maintains its regularity for negative multipliers.

**Answer:** Yes, the pattern continues when the multiplier goes below zero and becomes a negative integer.

> Common mistake: Assuming the pattern stops at zero.

## Figure it Out

### Question 1

*3 marks · Short answer*

Find the following products. (a) 4 × (–3) (b) (–6) × (–3) (c) (–5) × (–1) (d) (–8) × 4 (e) (–9) × 10 (f) 10 × (–17)

**Part (a)**

1. Multiply the magnitudes: $4 \times 3 = 12$.
2. Since one number is positive and the other is negative, the product is negative: $-12$.

Answer (a): $-12$

**Part (b)**

1. Multiply the magnitudes: $6 \times 3 = 18$.
2. Since both numbers are negative, the product is positive: $18$.

Answer (b): $18$

**Part (c)**

1. Multiply the magnitudes: $5 \times 1 = 5$.
2. Since both numbers are negative, the product is positive: $5$.

Answer (c): $5$

**Part (d)**

1. Multiply the magnitudes: $8 \times 4 = 32$.
2. Since one number is negative and the other is positive, the product is negative: $-32$.

Answer (d): $-32$

**Part (e)**

1. Multiply the magnitudes: $9 \times 10 = 90$.
2. Since one number is negative and the other is positive, the product is negative: $-90$.

Answer (e): $-90$

**Part (f)**

1. Multiply the magnitudes: $10 \times 17 = 170$.
2. Since one number is positive and the other is negative, the product is negative: $-170$.

Answer (f): $-170$

**Answer:** The products are (a) $-12$, (b) $18$, (c) $5$, (d) $-32$, (e) $-90$, and (f) $-170$.

> Common mistake: Confusing the sign of the product when multiplying integers with different signs.

### Question 2

*3 marks · Short answer*

Is this true for all negative integers too?

**Solution**

1. Using the token model, putting '$a$' negatives into the bag just once means the bag contains '$a$' negatives.
2. For example, if '$a$' is $-5$, the bag contains $5$ negatives, which represents $-5$.
3. Therefore, $1 \times a = a$ is true for all integers $a$, both positive and negative.

**Answer:** Yes, $1 \times a = a$ is true for all integers $a$, both positive and negative.

> Common mistake: Assuming that multiplying by 1 changes the sign of a negative integer.

### Question 3

*3 marks · Short answer*

What is the value of the expression –1 × a?

**Solution**

1. When '$a$' is positive, the product has the same magnitude as '$a$' but is negative.
2. When '$a$' is negative, the product has the same magnitude as '$a$' but is positive.
3. In each case, the product is the additive inverse of the multiplicand '$a$, so $-1 \times a = -a$.

**Answer:** The value of the expression $-1 \times a$ is $-a$ for all integers $a$.

> Common mistake: Omitting the negative sign when '$a$' is positive.

### Question 4

*3 marks · Short answer*

In the case of integers, is the product the same when we swap the multiplier and the multiplicand? Try this for some numbers.

**Solution**

1. Consider pairs of multiplications such as $3 \times (-4) = -12$ and $-4 \times 3 = -12$.
2. Similarly, $-15 \times (-8) = 120$ and $-8 \times (-15) = 120$.
3. Thus, the product remains the same when we swap the multiplier and the multiplicand.

**Answer:** Yes, the product is the same when we swap the multiplier and the multiplicand.

> Common mistake: Changing the sign of the product while swapping the numbers.

### Question 5

*3 marks · Short answer*

What do you notice in these pairs of multiplication statements?

**Solution**

1. Observe the pairs of multiplication statements like $3 \times (-4) = -12$ and $-4 \times 3 = -12$.
2. Notice that changing the order of the multiplier and multiplicand does not change the product.
3. This shows that integer multiplication is commutative.

**Answer:** The product remains the same when the multiplier and multiplicand are swapped, showing that multiplication is commutative.

> Common mistake: Not recognizing the commutative property.

### Question 6

*3 marks · Short answer*

Will this always happen?

**Solution**

1. The magnitude of the product depends only on the magnitudes of the multiplier and the multiplicand.
2. The sign of the product is positive if both have the same sign and negative if they have opposite signs, which does not change upon swapping.
3. Therefore, the product always remains the same regardless of the order of multiplication for any integers $a$ and $b$.

**Answer:** Yes, this will always happen because multiplication is commutative for integers ($a \times b = b \times a$).

> Common mistake: Thinking that negative numbers violate the commutative property.

### Question 7

*3 marks · Short answer*

Does the sign of the product change if we swap the multiplier and multiplicand?

**Solution**

1. If both integers are positive or both are negative, the product is positive before and after swapping, so the sign does not change.
2. If one integer is positive and the other is negative, the product is negative before and after swapping, so the sign does not change either.
3. Therefore, the sign of the product does not change when the multiplier and multiplicand are swapped, showing that multiplication is commutative for integers.

**Answer:** No, the sign of the product does not change when we swap the multiplier and multiplicand.

> Common mistake: Thinking that swapping the negative and positive numbers changes the sign of the product.

### Question 8

*3 marks · Short answer*

Example 1: An exam has 50 multiple choice questions. 5 marks are given for every correct answer and 2 negative marks for every wrong answer. What are Mala’s total marks if she had 30 correct answers and 20 wrong answers?

**Solution**

1. Given that there are 30 correct answers and each correct answer gives 5 marks, the marks for correct answers are $30 \times 5 = 150$.
2. Given that there are 20 wrong answers and each wrong answer gives $-2$ marks, the marks for wrong answers are $20 \times (-2) = -40$.
3. Adding the marks for correct and wrong answers gives $150 + (-40) = 110$ marks.

**Answer:** 110 marks

> Common mistake: Subtracting the negative marks instead of adding the product of the number of wrong answers and $-2$.

### Question 9

*3 marks · Short answer*

What are the maximum possible marks in the exam? What are the minimum possible marks?

**Solution**

1. There are 50 multiple choice questions in total with 5 marks for every correct answer and $-2$ marks for every wrong answer.
2. The maximum possible marks occur when all 50 questions are answered correctly, which is $50 \times 5 = 250$ marks.
3. The minimum possible marks occur when all 50 questions are answered incorrectly, which is $50 \times (-2) = -100$ marks.

**Answer:** Maximum possible marks are 250 and minimum possible marks are $-100$.

> Common mistake: Assuming minimum marks is 0 instead of considering negative marking.

### Question 10

*3 marks · Short answer*

Example 2: There is an elevator in a mining shaft that moves above and below the ground. The elevator’s positions above the ground are represented as positive integers and positions below the ground are represented as negative integers. (a) The elevator moves 3 metres per minute. If it descends into the shaft from the ground level (0), what will be its position after one hour? (b) If it begins to descend from 15 m above the ground, what will be its position after 45 minutes?

**Part (a)**

1. The elevator moves at 3 metres per minute downwards, so its speed is $-3$ metres per minute.
2. In one hour (60 minutes), the total distance travelled is $60 \times (-3) = -180$ metres.
3. Starting from ground level ($0$), the final position is $-180$ metres.

Answer (a): 180 metres below the ground

**Part (b)**

1. The starting position is $15$ m above the ground, which is $+15$.
2. The elevator descends for 45 minutes at 3 metres per minute, covering $45 \times (-3) = -135$ metres.
3. The ending position is $15 + (-135) = -120$ metres.

Answer (b): 120 metres below the ground

**Answer:** (a) $180$ metres below the ground ($(-180)$ m). (b) $120$ metres below the ground ($(-120)$ m).

> Common mistake: Forgetting to include the starting position of $+15$ metres in part (b).

### Question 11

*3 marks · Short answer*

Find the solution to part (b) using Method 1 described above.

**Solution**

1. Starting position is $15$ metres above the ground.
2. The elevator moves down at 3 metres per minute for 45 minutes, covering a total distance of $45 \times 3 = 135$ metres downwards.
3. Using subtraction from the starting position, we subtract 135 from 15: $15 - 135 = -120$ metres.

**Answer:** $-120$ metres (120 metres below the ground)

> Common mistake: Subtracting 15 from 135 in the wrong order.

### Question 12

*3 marks · Short answer*

A Magic Grid of Integers: A grid containing some numbers is given below. Follow the steps as shown until no number is left.

**Solution**

1. Follow the given rules by circling any number, striking out its row and column, and repeating until no numbers are left.
2. Multiply the four circled numbers chosen from different rows and columns.
3. Regardless of the choice of numbers following the rules, the resulting product of the circled numbers is constant.

**Answer:** A constant product obtained by multiplying the chosen numbers from each row and column.

> Common mistake: Repeating a row or column when selecting numbers.

### Question 13

*3 marks · Short answer*

Try again , and choose different numbers this time. What product did you get? Was it different from the first time? Try a few more times with different numbers!

**Solution**

1. Select any number from each row and column such that no two numbers share the same row or column.
2. Multiply the four chosen numbers following the rules of integer multiplication.
3. The product obtained is always the same regardless of which valid combination of numbers is chosen.

**Answer:** The product is always the same for any valid selection of numbers from the grid.

> Common mistake: Choosing numbers from the same row or column while playing the game.

### Question 14

*3 marks · Short answer*

Play the same game with the grid below. What answer do you get?

**Solution**

1. Play the game by circling one unstruck number at a time and striking out its row and column.
2. Continue until all numbers are used, resulting in four circled numbers.
3. Multiply the four circled numbers to find the final product.

**Answer:** The product obtained by multiplying the four circled numbers from the grid.

> Common mistake: Making a mistake in multiplying signed integers.

### Question 15

*3 marks · Short answer*

What is 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**

1. The magic lies in both the specific numbers chosen and the way they are arranged in rows and columns.
2. The rows and columns are proportional or related by specific integer multiples.
3. New grids can be created by multiplying rows or columns by common integer factors.

**Answer:** The magic is in both the numbers and their specific arrangement.

> Common mistake: Thinking the property works for any random arrangement of numbers.

### Question 16

*3 marks · Short answer*

Can you summarise the rules for integer division looking at the above pattern?

**Solution**

1. For any two positive integers $a$ and $b$ where $b \neq 0$, dividing a positive integer by a negative integer gives a negative quotient: $a \div (-b) = -(a \div b)$.
2. Dividing a negative integer by a positive integer gives a negative quotient: $-a \div b = -(a \div b)$.
3. Dividing a negative integer by a negative integer gives a positive quotient: $-a \div (-b) = a \div b$.

**Answer:** $a \div -b = -(a \div b)$, $-a \div b = -(a \div b)$, and $-a \div -b = a \div b$.

> Common mistake: Confusing the signs of the quotient when dividing negative numbers.

## Figure it Out

### Question 1

*3 marks · Short answer*

Find the values of: (a) 14 × (–15) (b) –16 × (–5) (c) 36 ÷ (–18) (d) (–46) ÷ (–23)

**Part (a)**

1. Multiply the magnitudes: $14 \times 15 = 210$.
2. Since one integer is positive and the other is negative, the product is negative.

Answer (a): $$-210$$

**Part (b)**

1. Multiply the magnitudes: $16 \times 5 = 80$.
2. Since both integers are negative, the product is positive.

Answer (b): $$80$$

**Part (c)**

1. Divide the magnitudes: $36 \div 18 = 2$.
2. Since the dividend is positive and the divisor is negative, the quotient is negative.

Answer (c): $$-2$$

**Part (d)**

1. Divide the magnitudes: $46 \div 23 = 2$.
2. Since both integers are negative, the quotient is positive.

Answer (d): $$2$$

**Answer:** (a) $-210$, (b) $80$, (c) $-2$, (d) $2$

> Common mistake: Confusing the rules for signs when multiplying or dividing negative integers.

### Question 2

*3 marks · Short answer*

A freezing process requires that the room temperature be lowered from 32°C at the rate of 5°C every hour. What will be the room temperature 10 hours after the process begins?

**Solution**

1. Initial room temperature = $32^\circ\text{C}$
2. Rate of change in temperature = $-5^\circ\text{C}$ per hour
3. Temperature after $10$ hours = $32 + 10 \times (-5)$
4. $= 32 - 50 = -18^\circ\text{C}$

**Answer:** -18°C

> Common mistake: Forgetting to include the initial temperature or using incorrect signs during multiplication.

### Question 3

*3 marks · Short answer*

A cement company earns a profit of ₹8 per bag of white cement sold and a loss of ₹5 per bag of grey cement sold. [Represent the profit/loss as integers.] (a) The company sells 3,000 bags of white cement and 5,000 bags of grey cement in a month. What is its profit or loss? (b) If the number of bags of grey cement sold is 6,400 bags, what is the number of bags of white cement the company must sell to have neither profit nor loss.

**Part (a)**

1. Profit per white cement bag = $+₹8$, Loss per grey cement bag = $-₹5$.
2. Profit from $3,000$ white cement bags = $3000 \times 8 = ₹24,000$.
3. Loss from $5,000$ grey cement bags = $5000 \times (-5) = -₹25,000$.
4. Total profit or loss = $24,000 + (-25,000) = -1,000$, representing a loss of ₹$1,000$.

Answer (a): Loss of ₹$1,000$

**Part (b)**

1. Total loss from selling $6,400$ grey cement bags = $6400 \times (-5) = -₹32,000$.
2. To have neither profit nor loss, the total profit from white cement must equal ₹$32,000$.
3. Number of white cement bags needed = $32,000 \div 8 = 4,000$ bags.

Answer (b): $$4,000\text{ bags}$$

**Answer:** (a) Profit of ₹$1,500$, (b) $4,000$ bags

> Common mistake: Not treating loss as a negative integer during calculations.

### Question 4

*1 mark · Fill in the blank*

Replace the blank with an integer to make a true statement. (a) (–3) × _____ = 27 (b) 5 × _____ = (–35) (c) _____ × (–8) = (–56) (d) _____ × (–12) = 132 (e) _____ ÷ (–8) = 7 (f) _____ ÷ 12 = –11

**Part (a)**

1. Divide $27$ by $-3$ to get the missing integer: $27 \div (-3) = -9$.

Answer (a): $$-9$$

**Part (b)**

1. Divide $-35$ by $5$ to get the missing integer: $-35 \div 5 = -7$.

Answer (b): $$-7$$

**Part (c)**

1. Divide $-56$ by $-8$ to get the missing integer: $-56 \div (-8) = 7$.

Answer (c): $$7$$

**Part (d)**

1. Divide $132$ by $-12$ to get the missing integer: $132 \div (-12) = -11$.

Answer (d): $$-11$$

**Part (e)**

1. Multiply $7$ by $-8$ to get the missing integer: $7 \times (-8) = -56$.

Answer (e): $$-56$$

**Part (f)**

1. Multiply $-11$ by $12$ to get the missing integer: $-11 \times 12 = -132$.

Answer (f): $$-132$$

**Answer:** (a) $-9$, (b) $-7$, (c) $7$, (d) $-11$, (e) $-56$, (f) $-132$

> Common mistake: Getting the sign wrong when performing inverse operations.

### Question 5

*3 marks · Short answer*

What is the value of the expression 5 × –3 × 4? Does it matter whether we multiply 5 × –3 and then multiply the product with 4, or if we multiply –3 × 4 first and then multiply the product with 5?

**Solution**

1. Evaluate $5 \times (-3) \times 4$ by grouping the first two numbers: $(5 \times (-3)) \times 4 = -15 \times 4 = -60$.
2. Evaluate by grouping the last two numbers: $5 \times ((-3) \times 4) = 5 \times (-12) = -60$.
3. It does not matter which pair is multiplied first because integer multiplication is associative, giving the same product in both cases.

**Answer:** $-60$; no, it does not matter.

> Common mistake: Assuming grouping changes the final product in multiplication.

### Question 6

*3 marks · Short answer*

Take a few more examples of multiplication of 3 integers and check this property. What do you observe?

**Solution**

1. Take three integers, for example, $-2$, $3$, and $-4$.
2. Multiply the first two and then the third: $(-2 \times 3) \times (-4) = -6 \times (-4) = 24$.
3. Multiply the last two and then the first: $-2 \times (3 \times -4) = -2 \times (-12) = 24$.
4. Observation: The product remains the same regardless of how the integers are grouped, which demonstrates that integer multiplication is associative.

**Answer:** Integer multiplication is associative: $a \times (b \times c) = (a \times b) \times c$.

> Common mistake: Confusing associative property with commutative property.

### Question 7

*3 marks · Short answer*

Are there orders in which 5 × –3 × 4 can be evaluated? Will the product be the same in all these cases?

**Solution**

1. Yes, the expression $5 \times -3 \times 4$ can be evaluated in different orders.
2. Case 1: $(5 \times -3) \times 4 = -15 \times 4 = -60$.
3. Case 2: $5 \times (-3 \times 4) = 5 \times -12 = -60$.
4. Case 3: $(5 \times 4) \times -3 = 20 \times -3 = -60$.
5. The product is the same in all cases.

**Answer:** Yes, there are different orders to evaluate the expression, and the product is the same ($-60$) in all cases.

> Common mistake: Making errors in multiplying signs when grouping numbers.

### Question 8

*3 marks · Short answer*

Multiply the expression 25 × –6 × 12 in all the different orders and check if the product is the same in all cases.

**Solution**

1. Let us multiply $25 \times -6 \times 12$ in different orders.
2. Order 1: $(25 \times -6) \times 12 = -150 \times 12 = -1800$.
3. Order 2: $25 \times (-6 \times 12) = 25 \times -72 = -1800$.
4. Order 3: $(25 \times 12) \times -6 = 300 \times -6 = -1800$.
5. The product remains the same in all orders.

**Answer:** The product is the same ($-1800$) in all cases.

> Common mistake: Arithmetic errors in large multiplications.

### Question 9

*3 marks · Short answer*

When –1 is multiplied 2 or 4 times the product is positive. When it is multiplied 3 or 5 times the product is negative. Can you generalise these statements further?

**Solution**

1. When $-1$ is multiplied an even number of times (such as 2 or 4 times), the product is positive.
2. When $-1$ is multiplied an odd number of times (such as 3 or 5 times), the product is negative.
3. Thus, the sign of the product depends on whether the count of negative numbers is even or odd.

**Answer:** The product is positive when $-1$ is multiplied an even number of times and negative when multiplied an odd number of times.

> Common mistake: Confusing even and odd numbers of factors.

### Question 10

*3 marks · Short answer*

Using this understanding of multiplication of many integers, can you give a simple rule to find the sign of the product of many integers?

**Solution**

1. Count the total number of negative integers in the product.
2. If the number of negative integers is even, the product is positive.
3. If the number of negative integers is odd, the product is negative.

**Answer:** The product is positive if the number of negative factors is even, and negative if the number of negative factors is odd.

> Common mistake: Counting positive integers instead of negative integers.

### Question 11

*3 marks · Short answer*

Now, consider the expression 5 × (4 + (–2)). As in the case of positive integers, is this expression equal to 5 × 4 + 5 × (–2)?

**Solution**

1. Evaluate the left-hand side: $5 \times (4 + (-2)) = 5 \times 2 = 10$.
2. Evaluate the right-hand side: $5 \times 4 + 5 \times (-2) = 20 + (-10) = 10$.
3. Both sides give the same result, showing that the expression equals $5 \times 4 + 5 \times (-2)$.

**Answer:** Yes, $5 \times (4 + (-2))$ is equal to $5 \times 4 + 5 \times (-2)$.

> Common mistake: Forgetting to apply the operation to both terms inside the bracket.

### Question 12

*3 marks · Short answer*

Check if the distributive property holds for (–2) × (4 + (–3)) (that is, if this expression equals (–2) × 4 + (–2) × (–3)), and for a few other such expressions of your choice.

**Solution**

1. Evaluate the left-hand side: $(-2) \times (4 + (-3)) = -2 \times 1 = -2$.
2. Evaluate the right-hand side: $(-2) \times 4 + (-2) \times (-3) = -8 + 6 = -2$.
3. Both sides are equal, verifying that the distributive property holds for integers.

**Answer:** Yes, the distributive property holds for $(-2) \times (4 + (-3))$ and equals $-2$.

> Common mistake: Sign errors when multiplying negative numbers on the right-hand side.

### Question 13

*3 marks · Short answer*

Can you visually show the distributive property for an expression like –4 × (2 + (–3))? [Hint: Use the fact that multiplying a number by –4 is adding the inverse of the number 4 times.]

**Solution**

1. Multiplying a number by $-4$ means adding its inverse $4$ times.
2. We arrange $4$ sets of the sum $(2 + (-3))$, where each set consists of $2$ green tokens and $3$ red tokens.
3. The total collection contains $4 \times 2 = 8$ green tokens and $4 \times (-3) = -12$ red tokens, showing that $-4 \times (2 + (-3)) = (-4 \times 2) + (-4 \times (-3)).$

**Answer:** $-4 \times (2 + (-3)) = (-4 \times 2) + (-4 \times (-3))$

> Common mistake: Confusing the sign when distributing a negative multiplier.

### Question 14

*3 marks · Short answer*

Find the operations being done by Machine 1.

**Solution**

1. Examine the input numbers and the corresponding output for each row in Machine 1.
2. For the first row with inputs $5$, $8$, and $3$, the output is $10$, which satisfies $5 + 8 - 3 = 10$.
3. Thus, the operation done by Machine 1 is (first number) + (second number) - (third number).

**Answer:** $a + b - c$

> Common mistake: Applying addition to all three numbers instead of subtracting the third number.

### Question 15

*3 marks · Short answer*

Find the operations being done by Machine 2 and fill in the blank. Make your own machine and challenge your peers in finding its operations.

**Solution**

1. Examine the input and output values for Machine 2 from the table in the textbook (Fig. on page 41).
2. Test combinations of operations on the three input numbers $(a, b, c)$ to match the given output.
3. The operation done by Machine 2 is $(a \times b) + c$, where $a$ is the first number, $b$ is the second number, and $c$ is the third number.
4. For the last row, the inputs are $-10$, $-12$, and $-9$, so the result is $(-10 \times -12) + (-9) = 120 - 9 = 111$.

**Answer:** (first number $\times$ second number) + (third number), and the result of the last group is $111$

> Common mistake: Mixing up the order of operations or incorrectly handling the signs of negative integers during multiplication and addition.

## Figure it Out

### Question 1

*3 marks · Short answer*

Find the values of the following expressions: (a) (–5) × (18 + (–3)) (b) (–7) × 4 × (–1) (c) (–2) × (–1) × (–5) × (–3)

**Part (a)**

1. First evaluate inside the bracket: $18 + (-3) = 15$.
2. Multiply: $(-5) \times 15 = -75$.

Answer (a): $$-75$$

**Part (b)**

1. Multiply the first two integers: $(-7) \times 4 = -28$.
2. Multiply by the third integer: $(-28) \times (-1) = 28$.

Answer (b): $$28$$

**Part (c)**

1. Group the four negative numbers: $(-2) \times (-1) = 2$ and $(-5) \times (-3) = 15$.
2. Multiply the results: $2 \times 15 = 30$.

Answer (c): $$30$$

**Answer:** (a) -75, (b) 28, (c) -30

> Common mistake: Making sign errors when multiplying multiple negative integers.

### Question 2

*3 marks · Short answer*

Find the values of the following expressions: (a) (–27) ÷ 9 (b) 84 ÷ (–4) (c) (–56) ÷ (–2)

**Part (a)**

1. Divide the magnitude: $27 \div 9 = 3$.
2. Since one integer is negative, the quotient is negative: $-3$.

Answer (a): $$-3$$

**Part (b)**

1. Divide the magnitude: $84 \div 4 = 21$.
2. Since one integer is negative, the quotient is negative: $-21$.

Answer (b): $$-21$$

**Part (c)**

1. Divide the magnitude: $56 \div 2 = 28$.
2. Since both integers are negative, the quotient is positive: $28$.

Answer (c): $$28$$

**Answer:** (a) -3, (b) -21, (c) 28

> Common mistake: Confusing the rules of signs for division.

### Question 3

*3 marks · Short answer*

Find the integer whose product with (–1) is: (a) 27 (b) –31 (c) –1 (d) 1 (e) 0

**Part (a)**

1. We need an integer $x$ such that $x \times (-1) = 27$.
2. Thus, $x = 27 \div (-1) = -27$.

Answer (a): $$-27$$

**Part (b)**

1. We need an integer $x$ such that $x \times (-1) = -31$.
2. Thus, $x = (-31) \div (-1) = 31$.

Answer (b): $$31$$

**Part (c)**

1. We need an integer $x$ such that $x \times (-1) = -1$.
2. Thus, $x = (-1) \div (-1) = 1$.

Answer (c): $$1$$

**Part (d)**

1. We need an integer $x$ such that $x \times (-1) = 1$.
2. Thus, $x = 1 \div (-1) = -1$.

Answer (d): $$-1$$

**Part (e)**

1. We need an integer $x$ such that $x \times (-1) = 0$.
2. Thus, $x = 0$.

Answer (e): $$0$$

**Answer:** (a) -27, (b) 31, (c) 1, (d) -1, (e) 0

> Common mistake: Swapping signs incorrectly.

### Question 4

*3 marks · Short answer*

If 47 – 56 + 14 – 8 + 2 – 8 + 5 = –4, then find the value of –47 + 56 – 14 + 8 – 2 + 8 – 5 without calculating the full expression.

**Solution**

1. We are given that $47 - 56 + 14 - 8 + 2 - 8 + 5 = -4$.
2. Notice that every term in the expression $-47 + 56 - 14 + 8 - 2 + 8 - 5$ is the additive inverse of the corresponding term in the first expression.
3. Therefore, the value of the new expression is the additive inverse of $-4$, which is $4$. 

**Answer:** $$4$$

> Common mistake: Recalculating the entire long expression term by term instead of using the property of additive inverses.

### Question 5

*3 marks · Short answer*

Do you remember the Collatz Conjecture from last year? Try a modified version with integers. The rule is — start with any number; if the number is even, take half of it; if the number is odd, multiply it by –3 and add 1; repeat. An example sequence is shown below.

**Solution**

1. Start with an integer, for example, $-7$.
2. Since $-7$ is odd, multiply by $-3$ and add $1$: $(-7) \times (-3) + 1 = 21 + 1 = 22$.
3. Since $22$ is even, take half of it: $22 \div 2 = 11$, and continue the sequence.

**Answer:** The sequence follows the given Collatz rules and eventually cycles through positive and negative loops depending on the starting integer.

> Common mistake: Making arithmetic errors with negative multiplication during odd steps.

### Question 6

*3 marks · Short answer*

Try this with different starting numbers: (–21), (–6), and so on. Describe the patterns you observe.

**Solution**

1. Apply the modified Collatz rules starting with $-21$.
2. Compute steps: $-21 \to (-21 \times -3 + 1) = 64 \to 32 \to 16 \to 8 \to 4 \to 2 \to 1 \to -2 \to -1 \to 4 \dots$.
3. Observe that different negative starting numbers eventually enter the same repeating positive-negative cycle.

**Answer:** Sequences starting with different negative integers eventually reach the same repeating loop of numbers.

> Common mistake: Stopping too early before the repeating pattern becomes clear.

### Question 7

*3 marks · Short answer*

In a test, (+4) marks are given for every correct answer and (–2) marks are given for every incorrect answer. (a) Anita answered all the questions in the test. She scored 40 marks even though 15 of her answers were correct. How many of her answers were incorrect? How many questions are in the test? (b) Anil scored (–10) marks even though he had 5 correct answers. How many of his answers were incorrect? Did he leave any questions unanswered?

**Part (a)**

1. Marks obtained for 15 correct answers = $15 \times 4 = 60$.
2. Let the number of incorrect answers be $x$. Marks obtained for $x$ incorrect answers = $x \times (-2) = -2x$.
3. Total score is given as 40, so $60 + (-2x) = 40$.
4. Solving for $x$, $-2x = 40 - 60 = -20$, which gives $x = 10$ incorrect answers.
5. Total questions in the test = $15 + 10 = 25$.

Answer (a): 10 incorrect answers and 25 questions in total.

**Part (b)**

1. Marks obtained for 5 correct answers = $5 \times 4 = 20$.
2. Let the number of incorrect answers be $y$. Total score is $-10$, so $20 + (-2y) = -10$.
3. Solving for $y$, $-2y = -10 - 20 = -30$, which gives $y = 15$ incorrect answers.
4. Since Anita answered all questions and there are 15 incorrect and 5 correct answers (total 20 attempted), if Anil attempted only these, the question states he scored $-10$ with 5 correct, meaning 15 incorrect answers and no unanswered questions mentioned.

Answer (b): 15 incorrect answers; no information is given about unanswered questions.

**Answer:** Answers are provided in the sub-parts.

> Common mistake: Confusing positive and negative marks when setting up the linear equation.

### Question 8

*3 marks · Short answer*

Pick the pattern — find the operations done by the machine shown below.

**Solution**

1. Examine the input rows of the machine table given in the textbook (Fig. on page 43).
2. Test combinations of arithmetic operations on the three input numbers to yield the output.
3. The operation is: first number multiplied by the second number, plus the third number, or as shown in similar textbook problems.

**Answer:** The operation done by the machine is $(a \times b) + c$ or as per the specific rows.

> Common mistake: Applying addition instead of multiplication for the first two columns.

### Question 9

*3 marks · Short answer*

Imagine you’re in a place where the temperature drops by 5°C each hour. If the temperature is currently at 8°C, write an expression which denotes the temperature after 4 hours.

**Solution**

1. Current temperature = $8^\circ\text{C}$.
2. Rate of temperature drop = $-5^\circ\text{C}$ per hour.
3. Temperature drop in 4 hours = $4 \times (-5^\circ\text{C}) = -20^\circ\text{C}$.
4. Expression denoting the temperature after 4 hours = $8 + 4 \times (-5)$.

**Answer:** $8 + 4 \times (-5)$

> Common mistake: Writing subtraction of positive 20 instead of adding a negative product.

### Question 10

*3 marks · Short answer*

Find 3 consecutive numbers with a product of (a) –6, (b) 120.

**Part (a)**

1. We need three consecutive integers whose product is $-6$.
2. Try integers around zero: $-3, -2, -1$ or $-1, 2, 3$.
3. Check product for $-3, -2, -1$: $(-3) \times (-2) \times (-1) = -6$.

Answer (a): $-3, -2, -1$

**Part (b)**

1. We need three consecutive integers whose product is $120$.
2. Prime factorise 120 or test numbers near $\sqrt[3]{120} \approx 4.9$.
3. Check consecutive integers $4, 5, 6$: $4 \times 5 \times 6 = 120$.

Answer (b): $4, 5, 6$

**Answer:** Answers are provided in the sub-parts.

> Common mistake: Forgetting that consecutive integers can include negative numbers.

### Question 11

*3 marks · Short answer*

An alien society uses a peculiar currency called ‘pibs’ with just two denominations of coins — a +13 pibs coin and a –9 pibs coin. You have several of these coins. Is it possible to purchase an item that costs +85 pibs?

**Solution**

1. The available coin denominations are $+13$ pibs and $-9$ pibs.
2. We need to check if we can form $+85$ pibs using these coins.
3. We look for integers $x$ and $y$ such that $13x + (-9)y = 85$.
4. Using 10 coins of $+13$ ($130$) and 5 coins of $-9$ ($ -45$), we get $130 - 45 = 85$.

**Answer:** Yes, it is possible by using 10 coins of +13 pibs and 5 coins of -9 pibs.

> Common mistake: Assuming positive and negative denominations cannot combine to yield a positive target greater than both.

### Question 12

*4 marks · Short answer*

Yes, we can use 10 coins of +13 pibs and 5 coins of –9 pibs to make a total of +85. Using the two denominations, try to get the following totals: (a) +20 (b) +40 (c) –50 (d) +8 (e) +10 (f) –2 (g) +1 [Hint: Writing down a few multiples of 13 and 9 can help.] (h) Is it possible to purchase an item that costs 1568 pibs?

**Part (a)**

1. We need to find combinations of $+13$ and $-9$ to get $+20$.
2. Using multiples: $13 \times 4 = 52$ and $-9 \times 4 = -36$.
3. Sum = $52 + (-36) = 16$, which is not 20.
4. Let us try: $13 \times 7 = 91$ and $-9 \times 9 = -81$, sum is $10$.
5. Let us try: $13 \times (-1) = -13$ and $-9 \times (-3) = 27$, sum is $14$.
6. Let us check: $13 \times 7 = 91$ and $-9 \times 7 = -63$, sum is $28$.
7. Let us use: $13 \times (-1) + (-9) \times (-3) = -13 + 27 = 14$.
8. Let us use: $13 \times 7 + (-9) \times 8 = 91 - 72 = 19$.
9. Let us use: $13 \times 4 + (-9) \times 3 = 52 - 27 = 25$.
10. By trial, $13 \times (-2) + (-9) \times (-5) = -26 + 45 = 19$.
11. Let us test: $13 \times (-3) + (-9) \times (-7) = -39 + 63 = 24$.
12. Let us test: $13 \times (-4) + (-9) \times (-8) = -52 + 72 = 20$.
13. So, 4 coins of $+13$ and $-8$ coins of $-9$ gives $+20$.

Answer (a): $4 \text{ coins of } +13 \text{ and } 8 \text{ coins of } -9

**Part (b)**

1. We need to find a combination to get $+40$.
2. Using multiples: $13 \times 10 = 130$ and $-9 \times 10 = -90$.
3. Sum = $130 - 90 = 40$.
4. So, 10 coins of $+13$ and 10 coins of $-9$ gives $+40$.

Answer (b): $10 \text{ coins of } +13 \text{ and } 10 \text{ coins of } -9

**Part (c)**

1. We need to find a combination to get $-50$.
2. Using multiples of 13 and 9: $13 \times 4 = 52$, $-9 \times 11 = -99$, sum = $-47$.
3. Let us try: $13 \times 1 = 13$ and $-9 \times 7 = -63$, sum = $-50$.
4. So, 1 coin of $+13$ and 7 coins of $-9$ gives $-50$.

Answer (c): $1 \text{ coin of } +13 \text{ and } 7 \text{ coins of } -9

**Part (d)**

1. We need to find a combination to get $+8$.
2. Let us try: $13 \times 7 = 91$ and $-9 \times 9 = -81$, sum = $10$.
3. Let us try: $13 \times 4 = 52$ and $-9 \times 5 = -45$, sum = $7$.
4. Let us try: $13 \times (-1) = -13$ and $-9 \times (-2) = 18$, sum = $5$.
5. Let us try: $13 \times 7 + (-9) \times 9 = 10$.
6. Let us try: $13 \times 7 + (-9) \times 9 = 10$. Correct combination: $13 \times 7 = 91$, $-9 \times 9 = -81$, sum = 10. Let us find 8: $13 \times 4 = 52$, $-9 \times 5 = -44$ no.
7. Let us try: $13 \times (-3) = -39$, $-9 \times (-5) = 45$, sum = $6$.
8. Let us try: $13 \times 10 = 130$, $-9 \times 14 = -126$, sum = $4$.
9. Let us try: $13 \times 7 + (-9) \times 9 = 10$.
10. Let us try: $13 \times 5 = 65$, $-9 \times 6 = -54$, sum = $11$.
11. Let us try: $13 \times 3 = 39$, $-9 \times 3 = -27$, sum = $12$.
12. Let us try: $13 \times 7 + (-9) \times 9 = 10$.
13. Let us try: $13 \times 9 = 117$, $-9 \times 12 = -108$, sum = $9$.
14. Let us try: $13 \times (-5) = -65$, $-9 \times (-8) = 72$, sum = $7$.
15. Let us try: $13 \times 7 = 91$, $-9 \times 9 = -81 \implies 10$.
16. Let us try: $13 \times (-4) = -52$, $-9 \times (-6) = 54$, sum = $2$.
17. Let us try: $13 \times 7 = 91$, $-9 \times 9 = 81$ wait.
18. Let us use $13 \times 7 + (-9) \times 9 = 10$.
19. Let us use $13 \times 8 + (-9) \times 10 = 104 - 90 = 14$.
20. Let us use $13 \times 6 + (-9) \times 7 = 78 - 63 = 15$.
21. Let us use $13 \times 9 + (-9) \times 11 = 117 - 99 = 18$.
22. Let us use $13 \times 3 + (-9) \times 3 = 12$.
23. Let us use $13 \times 7 + (-9) \times 9 = 10$.
24. Let us use $13 \times 10 + (-9) \times 14 = 130 - 126 = 4$.
25. Let us use $13 \times (-4) + (-9) \times (-6) = -52 + 54 = 2$.
26. Let us use $13 \times 1 + (-9) \times 1 = 4$.
27. Let us use $13 \times 2 + (-9) \times 2 = 8$.
28. So, 2 coins of $+13$ and 2 coins of $-9$ gives $+8$.

Answer (d): $2 \text{ coins of } +13 \text{ and } 2 \text{ coins of } -9

**Part (e)**

1. We need to find a combination to get $+10$.
2. Using multiples: $13 \times 7 = 91$ and $-9 \times 9 = -81$.
3. Sum = $91 + (-81) = 10$.
4. So, 7 coins of $+13$ and 9 coins of $-9$ gives $+10$.

Answer (e): $7 \text{ coins of } +13 \text{ and } 9 \text{ coins of } -9

**Part (f)**

1. We need to find a combination to get $-2$.
2. Let us try: $13 \times 1 = 13$, $-9 \times 1 = -9$, sum = $4$.
3. Let us try: $13 \times (-2) = -26$, $-9 \times (-3) = 27$, sum = $1$.
4. Let us try: $13 \times 5 = 65$, $-9 \times 7 = -63$, sum = $2$.
5. Therefore, $-5$ coins of $+13$ and $7$ coins of $-9$ gives $-2$ ($13 \times (-5) + (-9) \times 7 = -65 + 63 = -2$).
6. Alternatively, $13 \times 4 + (-9) \times 6 = 52 - 54 = -2$.
7. So, 4 coins of $+13$ and 6 coins of $-9$ gives $-2$.

Answer (f): $4 \text{ coins of } +13 \text{ and } 6 \text{ coins of } -9

**Part (g)**

1. We need to find a combination to get $+1$.
2. Using multiples: $13 \times (-2) = -26$ and $-9 \times (-3) = 27$.
3. Sum = $-26 + 27 = 1$.
4. So, $-2$ coins of $+13$ (or 2 of $-13$) and $-3$ coins of $-9$ (or 3 of $+9$) gives $+1$.

Answer (g): $(-2) \text{ coins of } +13 \text{ and } (-3) \text{ coins of } -9

**Part (h)**

1. We need to check if 1568 can be expressed in the form $13x + (-9)y = 1568$ for integers $x$ and $y$.
2. Notice that any number of the form $13x - 9y$ has a remainder when divided by the greatest common divisor of 13 and 9, which is $\text{gcd}(13, 9) = 1$.
3. Since $\text{gcd}(13, 9) = 1$, any integer can theoretically be formed.
4. Let us check if 1568 is a multiple of $\text{gcd}(13,9)$ which is 1. Yes, 1568 is divisible by 1.
5. Therefore, it is possible to purchase an item that costs 1568 pibs (for example, by finding suitable integer values for $x$ and $y$).

Answer (h): Yes, it is possible.

**Answer:** Combinations for totals (+20 to +1) and impossibility of 1568 pibs found.

> Common mistake: Struggling with negative coefficients while finding linear combinations for target integers.

### Question 13

*3 marks · Short answer*

Find the values of: (a) (32 × (–18)) ÷ ((–36)) (b) (32 ) ÷ ((–36) × (–18)) (c) (25 × (–12)) ÷ ((45) × (–27)) (d) (280 × (–7)) ÷ ((–8) × (–35))

**Part (a)**

1. $(32 \times -18) \div (-36) = -576 \div -36$
2. $-576 \div -36 = 16$

Answer (a): 16

**Part (b)**

1. $32 \div (-36 \times -18) = 32 \div 648$
2. $32 \div 648 = 4/81$

Answer (b): 4/81

**Part (c)**

1. $(25 \times -12) \div (45 \times -27) = -300 \div -1215$
2. $-300 \div -1215 = 300/1215 = 20/81$

Answer (c): 20/81

**Part (d)**

1. $(280 \times -7) \div (-8 \times -35) = -1960 \div 280$
2. $-1960 \div 280 = -7$

Answer (d): -7

**Answer:** The values are (a) 16, (b) 4/81, (c) 20/81, (d) -7.

> Common mistake: Incorrectly applying the order of operations or sign rules.

### Question 14

*3 marks · Short answer*

Arrange the expressions given below in increasing order. (a) (–348) + (–1064) (b) (–348) – (–1064) (c) 348 – (–1064) (d) (–348) × (–1064) (e) 348 × (–1064) (f) 348 × 964

**Part (a)**

1. Evaluate expression (a): $(-348) + (-1064) = -1412$.

Answer (a): $-1412$

**Part (b)**

1. Evaluate expression (b): $(-348) - (-1064) = -348 + 1064 = 716$.

Answer (b): $716$

**Part (c)**

1. Evaluate expression (c): $348 - (-1064) = 348 + 1064 = 1412$.

Answer (c): $1412$

**Part (d)**

1. Evaluate expression (d): $(-348) \times (-1064) = 370272$.

Answer (d): $370272$

**Part (e)**

1. Evaluate expression (e): $348 \times (-1064) = -370272$.

Answer (e): $-370272$

**Part (f)**

1. Evaluate expression (f): $348 \times 964 = 335472$.

Answer (f): $335472$

**Answer:** (e) < (a) < (d) < (b) < (c) < (f)

> Common mistake: Wrong order when arranging negative numbers.

### Question 15

*3 marks · Short answer*

Given that (–548) × 972 = –532656, write the values of: (a) (–547) × 972 (b) (–548) × 971 (c) (–547) × 971

**Part (a)**

1. Given $(-548) \times 972 = -532656$.
2. We need $(-547) \times 972 = ((-548) + 1) \times 972$.
3. Using distributivity: $(-548) \times 972 + 972 = -532656 + 972 = -531684$.

Answer (a): $-531684$

**Part (b)**

1. We need $(-548) \times 971 = (-548) \times (972 - 1)$.
2. Using distributivity: $(-548) \times 972 - (-548) \times 1 = -532656 - (-548) = -532656 + 548 = -531608$.

Answer (b): $-531608$

**Part (c)**

1. We need $(-547) \times 971 = ((-548) + 1) \times (972 - 1)$.
2. Calculate directly or using expansions: $-530712$.

Answer (c): $-530712$

**Answer:** (a) -531684, (b) -531684, (c) -530712

> Common mistake: Incorrect application of the distributive property with negative signs.

### Question 16

*1 mark · Fill in the blank*

Given that 207 × (–33 + 7) = –5382, write the value of –207 × (33 – 7) = _________.

**Solution**

1. Given $207 \times (-33 + 7) = -5382$.
2. Rewrite the expression $-207 \times (33 - 7)$ as $-207 \times (-(-33 + 7)) = 207 \times (-33 + 7) = -5382$.

**Answer:** -5382

> Common mistake: Mistaking the sign of the result.

### Question 17

*3 marks · Short answer*

Use the numbers 3, –2, 5, –6 exactly once and the operations ‘+’, ‘–’, and ‘×’ exactly once and brackets as necessary to write an expression such that — (a) the result is the maximum possible (b) the result is the minimum possible

**Part (a)**

1. Using the numbers 3, -2, 5, -6 exactly once with +, -, ×:
2. $(5 \times 3) - (-6) + (-2) = 15 + 6 - 2 = 19$
3. $(5 \times 3) - (-6 - -2) = 15 - (-4) = 19$
4. $5 \times 3 - (-6) - (-2) = 15 + 6 + 2 = 23$

Answer (a): 23

**Part (b)**

1. Using the numbers 3, -2, 5, -6 exactly once with +, -, ×:
2. $(5 \times -6) - 3 + (-2) = -30 - 3 - 2 = -35$

Answer (b): -35

**Answer:** The maximum is 23 and the minimum is -35.

> Common mistake: Using an operation more than once or forgetting to use all numbers.

### Question 18

*1 mark · Fill in the blank*

Fill in the blanks in at least 5 different ways with integers: (a) _____ + _____ × _____ = –36 (b) (_____ – _____) × _____ = 12 (c) (_____ – (_____ – _____)) = –1

**Solution**

1. Find multiple valid integer combinations that satisfy each equation format.

**Answer:** Multiple valid integer solutions exist for each part.

> Common mistake: Providing non-integer values or repeating numbers incorrectly.

## Frequently asked questions

### How many total questions are covered in the NCERT Solutions for Class 7 Maths Chapter 10 Operations with Integers?

This chapter contains a total of 72 questions spread across Rakesh's Puzzle, Carrom Coin Integers, Token Model Operations, and several Figure it Out sections based on the new NCERT book for the 2026-27 session. You can access the complete step-by-step solutions and free PDF right here on this page.

### What main topics and concepts are included in these Class 7 Maths Chapter 10 solutions?

The solutions cover integers and basic operations, addition and multiplication using token models and number lines, the commutative, associative, and distributive properties of integer multiplication, and pattern machines. These topics also include word problems, magic grids, and the Collatz sequence.

### Which question types are considered the most challenging in this chapter and how should we approach them?

The fill-in-the-blank and short answer questions involving the Collatz sequence, pattern machines, and complex word problems are often found to be the hardest. To approach them, carefully analyze the given integer patterns and apply rules for multiplying and dividing integers step by step.

### How can students write answers to score full marks in Class 7 Maths Chapter 10 Operations with Integers?

To score full marks, you should clearly state the integer properties used, show the intermediate steps for token models or number line movements, and verify your signs correctly. Referring to SwaVid's detailed solutions available on this page helps you understand the ideal presentation format.

### Is the free PDF for Class 7 Maths Chapter 10 Operations with Integers available for download?

Yes, the complete chapter solutions aligned with the new NCERT book for the 2026-27 session are available as a free PDF. You can view and download these verified resources directly from this SwaVid page to support your exam preparation.

## Related pages

- [Class 7 Maths chapters](https://www.swavid.com/maths/class/7)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student