---
title: "NCERT Solutions Class 7 Maths Chapter 2 Arithmetic Expressions"
url: https://www.swavid.com/maths/class/7/chapter/arithmetic-expressions/ncert-solutions
dateModified: 2026-10-07T14:58:42+00:00
---

# NCERT Solutions Class 7 Maths Chapter 2 Arithmetic Expressions

This chapter's questions cover arithmetic expressions, including evaluating and comparing expressions, applying operations, brackets, and properties of addition and multiplication.

Free PDF (32 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-7/swavid-ncert-solutions-class-7-maths-chapter-2-arithmetic-expressions-1ca43c1d71.pdf

## Page 24

### Question 1

*3 marks · Short answer*

Choose your favourite number and write as many expressions as you can having that value.

**Solution**

1. Let us choose the number 24.
2. The arithmetic expressions for the number can be written using different operations.
3. Some expressions having the value 24 are $12 + 12$, $4 \times 6$, $48 \div 2$, $34 - 10$, and $20 + 4$.

**Answer:** Expressions for the value 24 are $12 + 12$, $4 \times 6$, $48 \div 2$, $34 - 10$, and $20 + 4$.

> Common mistake: Using operations incorrectly so that the expression does not evaluate to the chosen number.

## Figure it Out

### Question 1

*1 mark · Fill in the blank*

Fill in the blanks to make the expressions equal on both sides of the $=$ sign:
(a) $13 + 4 = \_\_\_\_ + 6$
(b) $22 + \_\_\_\_ = 6 \times 5$
(c) $8 \times \_\_\_\_ = 64 \div 2$
(d) $34 - \_\_\_\_ = 25$

**Part (a) (0.25 marks)**

1. Evaluate the LHS: $13 + 4 = 17$.
2. Find the missing number by solving $\text{blank} + 6 = 17$, which gives $17 - 6 = 11$.

Answer (a): 11

**Part (b) (0.25 marks)**

1. Evaluate the RHS: $6 \times 5 = 30$.
2. Find the missing number by solving $22 + \text{blank} = 30$, which gives $30 - 22 = 8$.

Answer (b): 8

**Part (c) (0.25 marks)**

1. Evaluate the RHS: $64 \div 2 = 32$.
2. Find the missing number by solving $8 \times \text{blank} = 32$, which gives $32 \div 8 = 4$.

Answer (c): 4

**Part (d) (0.25 marks)**

1. Consider the equation $34 - \text{blank} = 25$.
2. Find the missing number by subtracting $25$ from $34$, which gives $34 - 25 = 9$.

Answer (d): 9

**Answer:** (a) 11, (b) 8, (c) 4, (d) 9

> Common mistake: Confusing addition and subtraction operations when balancing the two sides of the equals sign.

### Question 2

*3 marks · Short answer*

Arrange the following expressions in ascending (increasing) order of their values.
(a) $67 - 19$
(b) $67 - 20$
(c) $35 + 25$
(d) $5 \times 11$
(e) $120 \div 3$

**Solution**

1. Evaluate expression (a): $67 - 19 = 48$.
2. Evaluate expression (b): $67 - 20 = 47$.
3. Evaluate expression (c): $35 + 25 = 60$.
4. Evaluate expression (d): $5 \times 11 = 55$.
5. Evaluate expression (e): $120 \div 3 = 40$.
6. Arrange the values in ascending order: $40 < 47 < 48 < 55 < 60$.

**Answer:** $120 \div 3 < 67 - 20 < 67 - 19 < 5 \times 11 < 35 + 25$

> Common mistake: Confusing ascending order with descending order.

## Page 26

### Question 1

*3 marks · Short answer*

Use '>' or '<' or '=' in each of the following expressions to compare them. Can you do it without complicated calculations? Explain your thinking in each case.
(a) $245 + 289 \quad \_\_\_\_ \quad 246 + 285$
(b) $273 - 145 \quad \_\_\_\_ \quad 272 - 144$
(c) $364 + 587 \quad \_\_\_\_ \quad 363 + 589$
(d) $124 + 245 \quad \_\_\_\_ \quad 129 + 245$
(e) $213 - 77 \quad \_\_\_\_ \quad 214 - 76$

**Part (a) (0.6 marks)**

1. Compare the left-hand side and right-hand side expressions without full calculation.
2. Notice that $245 + 289 = 245 + 1 + 288 = 246 + 288$, whereas the right side is $246 + 285$.
3. Since $288 > 285$, the left side is greater.

Answer (a): $>$

**Part (b) (0.6 marks)**

1. Compare $273 - 145$ and $272 - 144$.
2. Raja had 273 marbles and lost 145, which is the same as losing 1 more than Joy who had 272 and lost 144.
3. Since one lost 1 more from a starting number that was 1 more, both expressions evaluate to the same value.

Answer (b): $=$

**Part (c) (0.6 marks)**

1. Compare $364 + 587$ and $363 + 589$.
2. The left side has 364 (1 more than 363) and 587 (2 less than 589).
3. The net change makes the right side larger.

Answer (c): $<$

**Part (d) (0.6 marks)**

1. Compare $124 + 245$ and $129 + 245$.
2. Both sides share the term $245$.
3. Since $124 < 129$, the left side is less than the right side.

Answer (d): $<$

**Part (e) (0.6 marks)**

1. Compare $213 - 77$ and $214 - 76$.
2. On the left, 213 is decreased by 77. On the right, 214 (1 more) is decreased by 76 (1 less).
3. Subtracting a larger amount from a smaller starting number gives a smaller value.

Answer (e): $<$

**Answer:** (a) >, (b) =, (c) <, (d) <, (e) <

> Common mistake: Evaluating each expression fully using arithmetic instead of using reasoning about terms.

## Page 28

### Question 1

*3 marks · Short answer*

Check if replacing subtraction by addition in this way does not change the value of the expression, by taking different examples.

**Solution**

1. Let us take two numbers, 18 and 10.
2. Evaluating the subtraction expression: $18 - 10 = 8$.
3. Replacing subtraction with the addition of the inverse: $18 + (-10) = 8$.
4. Since both expressions evaluate to 8, replacing subtraction by addition does not change the value.

**Answer:** Replacing subtraction with addition does not change the value of the expression.

> Common mistake: Forgetting to change the sign of the number to its inverse when converting subtraction to addition.

### Question 2

*3 marks · Short answer*

Can you explain why subtracting a number is the same as adding its inverse, using the Token Model of integers that we saw in the Class 6 textbook of mathematics?

**Solution**

1. In the Token Model, positive integers are represented by positive tokens and negative integers by negative tokens, where a positive and a negative token together make zero.
2. Subtracting a positive number is equivalent to removing positive tokens, which leaves the same result as adding negative tokens (its inverse).
3. Therefore, subtracting a number gives the same result as adding its inverse integer.

**Answer:** Refer to Chapter 10 of the Class 6 Mathematics Textbook for the Token Model explanation.

> Common mistake: Confusing the rules of positive and negative tokens.

## Page 29

### Question 1

*1 mark · True or false*

Does changing the order in which the terms are added give different values?

**Solution**

1. Changing the order in which the terms are added does not change the value of the expression.

**Answer:** False

> Common mistake: Thinking that changing the order of terms alters the sum.

### Question 2

*3 marks · Short answer*

Will this also hold when there are terms having negative numbers as well? Take some more expressions and check.

**Solution**

1. Consider the expression with negative numbers: $(-4) + (-2) = -6$.
2. Swapping the terms gives: $(-2) + (-4) = -6$.
3. Thus, swapping terms does not change the value even when negative numbers are involved.

**Answer:** Yes, swapping terms does not change the value of the expression even with negative numbers.

> Common mistake: Incorrectly applying integer sign rules when adding negative terms.

## Page 30

### Question 1

*3 marks · Short answer*

Can you explain why this is happening using the Token Model of integers that we saw in the Class 6 textbook of mathematics?

**Solution**

1. In the Token Model, positive numbers are represented by yellow tokens and negative numbers by red tokens.
2. Combining terms in an expression means combining collections of tokens together.
3. Since counting or combining tokens together does not depend on the order or grouping in which they are put together, changing the order or grouping of terms does not change the total value of the expression.

**Answer:** Combining tokens in any order or grouping yields the same total number of positive or negative tokens, showing that addition is commutative and associative.

> Common mistake: Confusing positive and negative tokens when combining them.

### Question 2

*3 marks · Short answer*

Will this also hold when there are terms having negative numbers as well? Take some more expressions and check. Consider expressions with more than 3 terms also.

**Solution**

1. Yes, swapping or grouping terms holds true even when the terms include negative numbers.
2. Consider an expression with four terms: $15 + (-8) + 5 + 12$.
3. Adding them in any order, such as grouping positive numbers together: $(15 + 5 + 12) + (-8) = 32 + (-8) = 24$, gives the same result.

**Answer:** Yes, swapping and grouping terms works for negative numbers as well, giving the same value regardless of the order.

> Common mistake: Making errors with signs when adding negative numbers in different orders.

### Question 3

*3 marks · Short answer*

Can you explain why this is happening using the Token Model of integers that we saw in the Class 6 textbook of mathematics?

**Solution**

1. In the Token Model, each term represents a collection of yellow (positive) or red (negative) tokens.
2. When grouping three terms, such as $(-7) + 10 + (-11)$, we are simply gathering these sets of tokens into one combined collection.
3. The total number and type of resulting tokens remain exactly the same whether we group the first two terms first or the last two terms first.

**Answer:** Grouping tokens in different ways results in the same final collection of tokens, explaining why associative property holds true.

> Common mistake: Forgetting that addition of integers can be regrouped in any manner.

## Page 31

### Question 1

*3 marks · Short answer*

Manasa is adding a long list of numbers. It took her five minutes to add them all and she got the answer 11749. Then she realised that she had forgotten to include the fourth number 9055. Does she have to start all over again?

**Solution**

1. No, Manasa does not have to start all over again.
2. By the commutative and associative properties of addition, the order and grouping of numbers in addition do not change the sum.
3. She can simply add the forgotten number 9055 to her obtained sum of 11749 to get the correct total: $11749 + 9055 = 20804$.

**Answer:** No, she does not have to start all over again. She can just add 9055 to 11749 to get the correct sum of 20804.

> Common mistake: Thinking that a missed number requires re-adding the entire list from the beginning.

## Page 32

### Question 1

*3 marks · Short answer*

If the total number of friends goes up to 7 and the tip remains the same, how much will they have to pay? Write an expression for this situation and identify its terms.

**Solution**

1. The cost of one dosa is ₹23 and the tip is ₹5.
2. For 7 friends, the total cost of 7 dosas is $7 \times 23$ and the added tip is $5$.
3. The expression describing the total amount to be paid is $7 \times 23 + 5$, and its terms are $7 \times 23$ and $5$.

**Answer:** Expression: $7 \times 23 + 5$; Terms: $7 \times 23, 5$

> Common mistake: Listing individual numbers inside the product term as separate terms instead of treating the multiplication as a single term.

### Question 2

*3 marks · Short answer*

Think and discuss why she wrote this.

**Solution**

1. There are 6 groups of 5 students each, which can be represented by the product $6 \times 5$.
2. Ruby, who is out of the game, sits separately, representing 3 more students or an addition of $3$.
3. Thus, the expression $6 \times 5 + 3$ describes 3 more than 6 times 5.

**Answer:** Ruby wrote $6 \times 5 + 3$ because there are 6 groups of 5 students and 3 remaining students sitting separately.

> Common mistake: Writing addition before multiplication without understanding terms.

## Page 33

### Question 1

*3 marks · Short answer*

For each of the cases below, write the expression and identify its terms:
If the teacher had called out '4', Ruby would write ____________
If the teacher had called out '7', Ruby would write ____________
Write expressions like the above for your class size.

**Part (a) (1.5 marks)**

1. The expression for group size 4 with 33 students and Ruby sitting aside is $8 \times 4 + 1$.
2. The terms of this expression are $8 \times 4$ and $1$.
3. Thus, Ruby would write $8 \times 4 + 1$ with terms $8 \times 4, 1$.

Answer (a): $8 \times 4 + 1$; Terms: $8 \times 4, 1$

**Part (b) (1.5 marks)**

1. The expression for group size 7 with 33 students and Ruby sitting aside is $4 \times 7 + 5$.
2. The terms of this expression are $4 \times 7$ and $5$.
3. Thus, Ruby would write $4 \times 7 + 5$ with terms $4 \times 7, 5$.

Answer (b): $4 \times 7 + 5$; Terms: $4 \times 7, 5$

**Answer:** Expressions and terms for group sizes 4 and 7.

> Common mistake: Confusing the number of full groups with the remaining students.

### Question 2

*3 marks · Short answer*

Identify the terms in the two expressions above.

**Part (a) (1.5 marks)**

1. Consider the first expression: $432 = 4 \times 100 + 1 \times 20 + 1 \times 10 + 2 \times 1$.
2. Terms are the parts separated by a plus sign.
3. The terms are $4 \times 100, 1 \times 20, 1 \times 10, \text{ and } 2 \times 1$.

Answer (a): Terms: $4 \times 100, 1 \times 20, 1 \times 10, 2 \times 1$

**Part (b) (1.5 marks)**

1. Consider the second expression: $432 = 8 \times 50 + 1 \times 10 + 4 \times 5 + 2 \times 1$.
2. Terms are the parts separated by a plus sign.
3. The terms are $8 \times 50, 1 \times 10, 4 \times 5, \text{ and } 2 \times 1$.

Answer (b): Terms: $8 \times 50, 1 \times 10, 4 \times 5, 2 \times 1$

**Answer:** Terms identified for both ways of making ₹432.

> Common mistake: Including multiplication signs inside terms as separate additions.

### Question 3

*3 marks · Short answer*

Can you think of some more ways of giving ₹432 to someone?

**Solution**

1. Write an expression using available notes and coins to make ₹432.
2. One such expression is $40 \times 10 + 3 \times 10 + 2 \times 1$.
3. Identify the terms as $40 \times 10, 3 \times 10, \text{ and } 2 \times 1$.

**Answer:** Expression: $40 \times 10 + 3 \times 10 + 2 \times 1$; Terms: $40 \times 10, 3 \times 10, 2 \times 1$

> Common mistake: Writing incorrect multipliers that do not sum up to 432.

## Page 34

### Question 1

*3 marks · Short answer*

What is the expression for the arrangement in the right making use of the number of yellow and blue squares?

**Solution**

1. Observe the arrangement of yellow and blue squares in the right picture on page 34 of the textbook.
2. The arrangement shows two groups, each containing 5 yellow squares and 3 blue squares.
3. Using brackets to describe this grouping, the expression is $2 \times (5 + 3)$.

**Answer:** $2 \times (5 + 3)$

> Common mistake: Writing the expression without brackets as $2 \times 5 + 3$, which changes the value and meaning.

## Figure it Out

### Question 1

*5 marks · Short answer*

Find the values of the following expressions by writing the terms in each case.
(a) $28 - 7 + 8$
(b) $39 - 2 \times 6 + 11$
(c) $40 - 10 + 10 + 10$
(d) $48 - 10 \times 2 + 16 \div 2$
(e) $6 \times 3 - 4 \times 8 \times 5$

**Part (a) (1 mark)**

1. Write the expression by converting subtraction to addition of the inverse: $28 + (-7) + 8$.
2. Identify the terms: $28$, $-7$, and $8$.
3. Evaluate the terms to get the final value: $28 - 7 + 8 = 29$.

Answer (a): $29$

**Part (b) (1 mark)**

1. Write the expression as a sum of terms: $39 + (-2 \times 6) + 11$.
2. Identify the terms: $39$, $-2 \times 6$, and $11$.
3. Evaluate the terms and add: $39 - 12 + 11 = 38$.

Answer (b): $38$

**Part (c) (1 mark)**

1. Write the expression as a sum of terms: $40 + (-10) + 10 + 10$.
2. Identify the terms: $40$, $-10$, $10$, and $10$.
3. Evaluate the terms and add: $40 - 10 + 10 + 10 = 50$.

Answer (c): $50$

**Part (d) (1 mark)**

1. Write the expression as a sum of terms: $48 + (-10 \times 2) + (16 \div 2)$.
2. Identify the terms: $48$, $-10 \times 2$, and $16 \div 2$.
3. Evaluate the terms and add: $48 - 20 + 8 = 36$.

Answer (d): $36$

**Part (e) (1 mark)**

1. Write the expression as a sum of terms: $(6 \times 3) + (-4 \times 8 \times 5)$.
2. Identify the terms: $6 \times 3$ and $-4 \times 8 \times 5$.
3. Evaluate the terms and add: $18 - 160 = -142$.

Answer (e): $-142$

**Answer:** (a) $29$, (b) $38$, (c) $50$, (d) $36$, (e) $-142$

> Common mistake: Forgetting to convert subtraction into the addition of the negative inverse before identifying terms, or evaluating operations from left to right instead of evaluating terms first.

### Question 2

*3 marks · Short answer*

Write a story/situation for each of the following expressions and find their values.
(a) $89 + 21 - 10$
(b) $5 \times 12 - 6$
(c) $4 \times 9 + 2 \times 6$

**Part (a) (1 mark)**

1. Situation: A library had 89 books, bought 21 more books, and then 10 books were borrowed.
2. Write the expression: $89 + 21 - 10$.
3. Evaluate the expression: $110 - 10 = 100$.

Answer (a): 100

**Part (b) (1 mark)**

1. Situation: Ravi bought 5 packets containing 12 chocolates each and gave away 6 chocolates.
2. Write the expression: $5 \times 12 - 6$.
3. Evaluate the expression: $60 - 6 = 54$.

Answer (b): 54

**Part (c) (1 mark)**

1. Situation: There are 4 boxes with 9 pencils each and 2 boxes with 6 pens each. Find the total number of items.
2. Write the expression: $4 \times 9 + 2 \times 6$.
3. Evaluate the expression: $36 + 12 = 48$.

Answer (c): 48

**Answer:** Situations created and values evaluated.

> Common mistake: Writing stories that do not match the order of operations in the given expression.

### Question 3

*3 marks · Short answer*

For each of the following situations, write the expression describing the situation, identify its terms and find the value of the expression.
(a) Queen Alia gave 100 gold coins to Princess Elsa and 100 gold coins to Princess Anna last year. Princess Elsa used the coins to start a business and doubled her coins. Princess Anna bought jewellery and has only half of the coins left. Write an expression describing how many gold coins Princess Elsa and Princess Anna together have.
(b) A metro train ticket between two stations is ₹40 for an adult and ₹20 for a child. What is the total cost of tickets:
(i) for four adults and three children?
(ii) for two groups having three adults each?
(c) Find the total height of the window by writing an expression describing the relationship among the measurements shown in the picture.

**Part (a) (1 mark)**

1. Princess Elsa's coins: $2 \times 100$; Princess Anna's coins: $100 \div 2$.
2. Write the expression: $2 \times 100 + \frac{100}{2}$.
3. Identify terms: $2 \times 100$ and $\frac{100}{2}$. Evaluate to get $200 + 50 = 250$.

Answer (a): Expression: $2 \times 100 + \frac{100}{2}$, Terms: $2 \times 100, \frac{100}{2}$, Value: 250 gold coins

**Part (i) (1 mark)**

1. Write the expression for four adults and three children: $4 \times 40 + 3 \times 20$.
2. Identify terms: $4 \times 40$ and $3 \times 20$.
3. Evaluate the expression: $160 + 60 = ₹220$.

Answer (i): Expression: $4 \times 40 + 3 \times 20$, Terms: $4 \times 40, 3 \times 20$, Value: ₹220

**Part (ii) (0.5 marks)**

1. Write the expression for two groups having three adults each: $2 \times (3 \times 40)$.
2. Identify terms: $2 \times (3 \times 40)$.
3. Evaluate the expression: $2 \times 120 = ₹240$.

Answer (ii): Expression: $2 \times (3 \times 40)$, Terms: $2 \times (3 \times 40)$, Value: ₹240

**Part (c) (0.5 marks)**

1. From the figure in the textbook (Fig. p. 29/34), use the measurements: 7 sections of grill of height 5 cm, 6 sections of border/spacers of height 2 cm, and 2 gaps of height 3 cm.
2. Write the expression for total height: $7 \times 5 + 6 \times 2 + 2 \times 3$.
3. Identify terms: $7 \times 5$, $6 \times 2$, $2 \times 3$. Evaluate to get $35 + 12 + 6 = 53$ cm.

Answer (c): Expression: $7 \times 5 + 6 \times 2 + 2 \times 3$, Terms: $7 \times 5, 6 \times 2, 2 \times 3$, Value: 53 cm

**Answer:** Expressions, terms, and values determined for all parts.

> Common mistake: Misidentifying the terms in expressions involving products and sums.

## Figure it Out

### Question 1

*1 mark · Fill in the blank*

Fill in the blanks with numbers, and boxes with operation signs such that the expressions on both sides are equal.
(a) $24 + (6 - 4) = 24 + 6 \square $
(b) $38 + (\square \square ) = 38 + 9 - 4$
(c) $24 - (6 + 4) = 24 \square 6 - 4$
(d) $24 - 6 - 4 = 24 - 6 \square $
(e) $27 - (8 + 3) = 27 \square 8 \square 3$
(f) $27 - (\square \square ) = 27 - 8 + 3$

**Part (a)**

1. Using the rule for removing brackets preceded by a negative sign, $24 + (6 - 4) = 24 + 6 - 4$.

Answer (a): $\ominus 4$

**Part (b)**

1. Comparing the RHS $38 + 9 - 4$ with $38 + (\dots)$, the missing terms inside the bracket are $9 - 4$.

Answer (b): $9 - 4$

**Part (c)**

1. Removing the brackets preceded by a minus sign, $24 - (6 + 4) = 24 - 6 - 4$.

Answer (c): $-$

**Part (d)**

1. Simplifying by combining terms, $24 - 6 - 4 = 24 - 6 \ominus 4$.

Answer (d): $4$

**Answer:** (a) $\ominus 4$, (b) $9 - 4$, (c) $-$, (d) $4$, (e) $- 8 - 3$, (f) $8 - 3$

> Common mistake: Forgetting to change the sign of terms inside brackets when preceded by a minus sign.

### Question 2

*3 marks · Short answer*

Remove the brackets and write the expression having the same value.
(a) $14 + (12 + 10)$
(b) $14 - (12 + 10)$
(c) $14 + (12 - 10)$
(d) $14 - (12 - 10)$
(e) $-14 + (12 - 10)$
(f) $14 - (-12 - 10)$

**Part (a) (0.5 marks)**

1. Remove the brackets preceded by a plus sign without changing signs: $14 + 12 + 10$.

Answer (a): $14 + 12 + 10$

**Part (b) (0.5 marks)**

1. Remove the brackets preceded by a minus sign, changing inner signs: $14 - 12 - 10$.

Answer (b): $14 - 12 - 10$

**Part (c) (0.5 marks)**

1. Remove the brackets preceded by a plus sign: $14 + 12 - 10$.

Answer (c): $14 + 12 - 10$

**Part (d) (0.5 marks)**

1. Remove the brackets preceded by a minus sign: $14 - 12 + 10$.

Answer (d): $14 - 12 + 10$

**Part (e) (0.5 marks)**

1. Remove the brackets preceded by a plus sign: $-14 + 12 - 10$.

Answer (e): $-14 + 12 - 10$

**Part (f) (0.5 marks)**

1. Remove the brackets preceded by a minus sign: $14 + 12 + 10$.

Answer (f): $14 + 12 + 10$

**Answer:** (a) $14 + 12 + 10$, (b) $14 - 12 - 10$, (c) $14 + 12 - 10$, (d) $14 - 12 + 10$, (e) $-14 + 12 - 10$, (f) $14 + 12 + 10$

> Common mistake: Failing to change all signs inside a bracket when a negative sign precedes it.

### Question 3

*3 marks · Short answer*

Find the values of the following expressions. For each pair, first try to guess whether they have the same value. When are the two expressions equal?
(a) $(6 + 10) - 2$ and $6 + (10 - 2)$
(b) $16 - (8 - 3)$ and $(16 - 8) - 3$
(c) $27 - (18 + 4)$ and $27 + (-18 - 4)$

**Part (a) (1 mark)**

1. Evaluate $(6 + 10) - 2 = 16 - 2 = 14$.
2. Evaluate $6 + (10 - 2) = 6 + 8 = 14$.
3. Both expressions are equal.

Answer (a): Equal (value 14)

**Part (b) (1 mark)**

1. Evaluate $16 - (8 - 3) = 16 - 5 = 11$.
2. Evaluate $(16 - 8) - 3 = 8 - 3 = 5$.
3. The expressions are not equal.

Answer (b): Not equal ($11 \neq 5$)

**Part (c) (1 mark)**

1. Evaluate $27 - (18 + 4) = 27 - 22 = 5$.
2. Evaluate $27 + (-18 - 4) = 27 - 18 - 4 = 5$.
3. Both expressions are equal.

Answer (c): Equal (value 5)

**Answer:** (a) Both equal to $14$, (b) $11 \neq 5$, (c) Both equal to $5$

> Common mistake: Assuming subtraction is associative.

### Question 4

*3 marks · Short answer*

In each of the sets of expressions below, identify those that have the same value. Do not evaluate them, but rather use your understanding of terms.
(a) $319 + 537$, $319 - 537$, $-537 + 319$, $537 - 319$
(b) $87 + 46 - 109$, $87 + 46 - 109$, $87 + 46 - 109$, $87 - 46 + 109$, $87 - (46 + 109)$, $(87 - 46) + 109$

**Part (a) (1.5 marks)**

1. Identify the terms of $319 - 537$ as $319$ and $-537$.
2. Identify the terms of $-537 + 319$ as $-537$ and $319$.
3. By the commutative property of addition, these expressions have the same value.

Answer (a): $319 - 537$ and $-537 + 319$

**Part (b) (1.5 marks)**

1. Express all forms with their respective terms.
2. The expressions with terms $87, 46, -109$ or equivalent matching terms have equal values.

Answer (b): $87 + 46 - 109$, $87 - 46 + 109$ and $(87 - 46) + 109$

**Answer:** (a) $319 - 537$ and $-537 + 319$, (b) $87 + 46 - 109$, $87 - 46 + 109$ and $(87 - 46) + 109$

> Common mistake: Ignoring the sign attached to a number when determining its term.

### Question 5

*3 marks · Short answer*

Add brackets at appropriate places in the expressions such that they lead to the values indicated.
(a) $34 - 9 + 12 = 13$
(b) $56 - 14 - 8 = 34$
(c) $-22 - 12 + 10 + 22 = -22$

**Part (a) (1 mark)**

1. Place brackets around $9 + 12$ so that it is subtracted together: $34 - (9 + 12) = 34 - 21 = 13$.

Answer (a): $34 - (9 + 12) = 13$

**Part (b) (1 mark)**

1. Place brackets around $56 - 14$: $(56 - 14) - 8 = 42 - 8 = 34$.

Answer (b): $(56 - 14) - 8 = 34$

**Part (c) (1 mark)**

1. Place brackets around $12 + 10$: $-22 - (12 + 10) + 22 = -22 - 22 + 22 = -22$.

Answer (c): $-22 - (12 + 10) + 22 = -22$

**Answer:** (a) $34 - (9 + 12) = 13$, (b) $(56 - 14) - 8 = 34$, (c) $-22 - (12 + 10) + 22 = -22$

> Common mistake: Placing brackets incorrectly leading to an altered sign upon expansion.

### Question 6

*1 mark · Fill in the blank*

Using only reasoning of how terms change their values, fill the blanks to make the expressions on either side of the equality ($=$) equal.
(a) $423 + \_\_\_\_\_\_ = 419 + \_\_\_\_\_\_$
(b) $207 - 68 = 210 - \_\_\_\_\_\_$

**Part (a)**

1. By the commutative property of addition, $423 + 419 = 419 + 423$.

Answer (a): $419$, $423$

**Part (b)**

1. Rewrite $207 - 68$ as $207 + 3 - (3 + 68) = 210 - 71$.

Answer (b): $71$

**Answer:** (a) $419$, $423$, (b) $71$

> Common mistake: Not accounting for the change in value when adjusting numbers across an equality sign.

### Question 7

*3 marks · Short answer*

Using the numbers 2, 3 and 5, and the operators '+' and '-', and brackets, as necessary, generate expressions to give as many different values as possible. For example, $2 - 3 + 5 = 4$ and $3 - (5 - 2) = 0$.

**Solution**

1. Using the numbers 2, 3, and 5 with operators '+' and '-' and brackets, different expressions can be formed.
2. Expression 1: $2 + 3 + 5 = 10$
3. Expression 2: $5 - 3 - 2 = 0$
4. Expression 3: $(5 - 3) + 2 = 4$
5. Expression 4: $5 - (3 + 2) = 0$

**Answer:** Multiple expressions can be formed, such as $2 + 3 + 5 = 10$, $5 - 3 - 2 = 0$, and $(5 - 3) + 2 = 4$.

> Common mistake: Forgetting to include brackets or using numbers other than 2, 3, and 5.

### Question 8

*3 marks · Short answer*

Whenever Jasoda has to subtract 9 from a number, she subtracts 10 and adds 1 to it. For example, $36 - 9 = 26 + 1$.
(a) Do you think she always gets the correct answer? Why?
(b) Can you think of other similar strategies? Give some examples.

**Part (a) (2 marks)**

1. Subtracting 9 is equivalent to subtracting 10 and adding 1 since $-9 = -10 + 1$.
2. For example, $36 - 9 = 36 - 10 + 1 = 26 + 1 = 27$.

Answer (a): Yes, because subtracting 10 and adding 1 has the same net effect as subtracting 9.

**Part (b) (1 mark)**

1. To subtract 8, we can subtract 10 and add 2 because $-8 = -10 + 2$.
2. For example, $55 - 8 = 55 - 10 + 2 = 45 + 2 = 47$.

Answer (b): Yes, e.g., to subtract 8, subtract 10 and add 2.

**Answer:** (a) Yes, she always gets the correct answer because subtracting 9 is the same as subtracting 10 and adding 1. (b) Yes, to subtract 8, we can subtract 10 and add 2.

> Common mistake: Adding instead of subtracting during the adjustment step.

### Question 9

*3 marks · Short answer*

Consider the two expressions: a) $73 - 14 + 1$, b) $73 - 14 - 1$. For each of these expressions, identify the expressions from the following collection that are equal to it.
(a) $73 - (14 + 1)$
(b) $73 - (14 - 1)$
(c) $73 + (-14 + 1)$
(d) $73 + (-14 - 1)$

**Part a (1.5 marks)**

1. The given expression is $73 - 14 + 1$.
2. Removing brackets in $73 - (14 + 1)$ gives $73 - 14 - 1$.
3. Converting subtraction to addition of inverse, $73 + (-14 + 1) = 73 - 14 + 1$.
4. Therefore, expression (c) is equal to $73 - 14 + 1$.

Answer a: Expression (c) is equal to $73 - 14 + 1$.

**Part b (1.5 marks)**

1. The given expression is $73 - 14 - 1$.
2. Removing brackets in $73 - (14 + 1)$ gives $73 - 14 - 1$, which matches expression (a).
3. Converting subtraction to addition, $73 + (-14 - 1) = 73 - 14 - 1$, which matches expression (d).
4. Therefore, expressions (a) and (d) are equal to $73 - 14 - 1$.

Answer b: Expressions (a) and (d) are equal to $73 - 14 - 1$.

**Answer:** Expressions (b) and (c) are equal to $73 - 14 + 1$, and expressions (a) and (d) are equal to $73 - 14 - 1$.

> Common mistake: Confusing the sign changes when removing brackets preceded by a negative sign.

## Page 39

### Question 1

*3 marks · Short answer*

If another friend, Sangmu, joins them and orders the same items, what will be the expression for the total amount to be paid?

**Solution**

1. Lhamo and Norbu each ordered a vegetable cutlet costing ₹43 and a rasgulla costing ₹24, which can be represented as $(43 + 24)$ for one person.
2. When another friend Sangmu joins them and orders the same items, the total number of friends becomes 3.
3. Therefore, the expression for the total amount to be paid by all 3 friends is $3 \times (43 + 24)$.

**Answer:** $3 \times (43 + 24)$

> Common mistake: Writing $3 \times 43 + 24$, which incorrectly applies the tip or only multiplies the cutlet cost by 3 instead of the entire combined cost.

## Page 40

### Question 1

*3 marks · Short answer*

$5 \times 4 + 3 \neq 5 \times (4 + 3)$. Can you explain why?

**Solution**

1. First, evaluate the expression on the left side, $5 \times 4 + 3 = 20 + 3 = 23$.
2. Next, evaluate the expression inside the brackets on the right side, $4 + 3 = 7$, and then multiply by $5$, giving $5 \times 7 = 35$.
3. Since $23 \neq 35$, the two expressions are not equal because brackets change the order of operations.

**Answer:** $5 \times 4 + 3 = 23$ while $5 \times (4 + 3) = 35$, so they are not equal.

> Common mistake: Performing addition before multiplication without brackets.

### Question 2

*1 mark · True or false*

Is $5 \times (4 + 3) = 5 \times (3 + 4) = (3 + 4) \times 5$?

**Solution**

1. Evaluate $5 \times (4 + 3) = 5 \times 7 = 35$.
2. Evaluate $5 \times (3 + 4) = 5 \times 7 = 35$ and $(3 + 4) \times 5 = 7 \times 5 = 35$.
3. Since all three expressions evaluate to 35, the statement is True.

**Answer:** True

> Common mistake: Assuming that changing the order of numbers inside brackets changes the value of the product.

## Page 41

### Question 1

*3 marks · Short answer*

Use this method to find the following products:
(a) $95 \times 8$
(b) $104 \times 15$
(c) $49 \times 50$
Is this quicker than the multiplication procedure you use generally?

**Part (a) (1 mark)**

1. $95 \times 8 = (100 - 5) \times 8$
2. $= 100 \times 8 - 5 \times 8$
3. $= 800 - 40 = 760$

Answer (a): 760

**Part (b) (1 mark)**

1. $104 \times 15 = (100 + 4) \times 15$
2. $= 100 \times 15 + 4 \times 15$
3. $= 1500 + 60 = 1560$

Answer (b): 1560

**Part (c) (1 mark)**

1. $49 \times 50 = (50 - 1) \times 50$
2. $= 50 \times 50 - 1 \times 50$
3. $= 2500 - 50 = 2450$

Answer (c): 2450 and it is quicker than the general multiplication procedure

**Answer:** (a) 760, (b) 1560, (c) 2450; yes, this method is quicker.

> Common mistake: Multiplying incorrectly when splitting the numbers or missing the subtraction sign.

### Question 2

*3 marks · Short answer*

Which other products might be quicker to find like the ones above?

**Solution**

1. This method is useful when one of the numbers is close to a multiple of 10, 50, 100, 1000, etc.
2. Examples of such products include $98 \times 7$, $103 \times 9$, and $49 \times 5$.
3. These can be easily computed by splitting the number into a sum or difference from a nearby base number.

**Answer:** Products where one of the numbers is close to a multiple of 10, 50, 100, 1000, etc.

> Common mistake: Choosing numbers that are far from multiples of 10 or 100, which makes the splitting method more complicated than standard multiplication.

## Figure it Out

### Question 1

*1 mark · Fill in the blank*

Fill in the blanks with numbers, and boxes by signs, so that the expressions on both sides are equal.
(a) $3 \times (6 + 7) = 3 \times 6 + 3 \times 7$
(b) $(8 + 3) \times 4 = 8 \times 4 + 3 \times 4$
(c) $3 \times (5 + 8) = 3 \times 5 \square 3 \times \_\_\_\_$
(d) $(9 + 2) \times 4 = 9 \times 4 \square 2 \times \_\_\_\_$
(e) $3 \times (\_\_\_\_ + 4) = 3 \square + \_\_\_\_$
(f) (\_\_\_\_ $+ 6) \times 4 = 13 \times 4 + \_\_\_\_$
(g) $3 \times (\_\_\_\_ + \_\_\_\_) = 3 \times 5 + 3 \times 2$
(h) (\_\_\_\_ $+ \_\_\_\_) \times \_\_\_\_ = 2 \times 4 + 3 \times 4$
(i) $5 \times (9 - 2) = 5 \times 9 - 5 \times \_\_\_\_$
(j) $(5 - 2) \times 7 = 5 \times 7 - 2 \times \_\_\_\_$
(k) $5 \times (8 - 3) = 5 \times 8 \square 5 \times \_\_\_\_$
(l) $(8 - 3) \times 7 = 8 \times 7 \square 3 \times 7$
(m) $5 \times (12 - \_\_\_\_) = \square 5 \times \_\_\_\_$
(n) $(15 - \_\_\_\_) \times 7 = \square 6 \times 7$
(o) $5 \times (\_\_\_\_ - \_\_\_\_) = 5 \times 9 - 5 \times 4$
(p) (\_\_\_\_ $- \_\_\_\_) \times \_\_\_\_ = 17 \times 7 - 9 \times 7$

**Part (a)**

1. Using the distributive property of multiplication over addition.

Answer (a): $3 \times (6 + 7) = 3 \times 6 + 3 \times 7$

**Part (b)**

1. Using the distributive property of multiplication over addition.

Answer (b): $(8 + 3) \times 4 = 8 \times 4 + 3 \times 4$

**Part (c)**

1. Multiply 3 with each term inside the bracket.

Answer (c): $3 \times (5 + 8) = 3 \times 5 + 3 \times 8$

**Part (d)**

1. Multiply each term inside the bracket by 4.

Answer (d): $(9 + 2) \times 4 = 9 \times 4 + 2 \times 4$

**Part (e)**

1. Distribute 3 to the terms inside the bracket.

Answer (e): $3 \times (10 + 4) = 3 \times 10 + 3 \times 4$

**Part (f)**

1. Distribute 4 to the terms inside the bracket.

Answer (f): $(7 + 6) \times 4 = 13 \times 4 + 6 \times 4$

**Part (g)**

1. Compare with the right side expression $3 \times 5 + 3 \times 2$.

Answer (g): $3 \times (5 + 2) = 3 \times 5 + 3 \times 2$

**Part (h)**

1. Compare with the right side expression $2 \times 4 + 3 \times 4$.

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

**Part (i)**

1. Use the distributive property for subtraction.

Answer (i): $5 \times (9 - 2) = 5 \times 9 - 5 \times 2$

**Part (j)**

1. Use the distributive property for subtraction.

Answer (j): $(5 - 2) \times 7 = 5 \times 7 - 2 \times 7$

**Part (k)**

1. Apply the distributive property with a minus sign.

Answer (k): $5 \times (8 - 3) = 5 \times 8 - 5 \times 3$

**Part (l)**

1. Apply the distributive property with a minus sign.

Answer (l): $(8 - 3) \times 7 = 8 \times 7 - 3 \times 7$

**Part (m)**

1. Match terms with the right side.

Answer (m): $5 \times (12 - 3) = 5 \times 12 - 5 \times 3$

**Part (n)**

1. Match terms with the right side.

Answer (n): $(15 - 6) \times 7 = 15 \times 7 - 6 \times 7$

**Part (o)**

1. Match terms with the right side.

Answer (o): $5 \times (9 - 4) = 5 \times 9 - 5 \times 4$

**Part (p)**

1. Match terms with the right side.

Answer (p): $(17 - 9) \times 7 = 17 \times 7 - 9 \times 7$

**Answer:** (c) $3 \times (5 + 8) = 3 \times 5 + 3 \times 8$

> Common mistake: Forgetting to multiply the number outside with every term inside the brackets.

### Question 2

*3 marks · Short answer*

In the boxes below, fill '<', '>' or '=' after analysing the expressions on the LHS and RHS. Use reasoning and understanding of terms and brackets to figure this out and not by evaluating the expressions.
(a) $(8 - 3) \times 29 \quad \square \quad (3 - 8) \times 29$
(b) $15 + 9 \times 18 \quad \square \quad (15 + 9) \times 18$
(c) $23 \times (17 - 9) \quad \square \quad 23 \times 17 + 23 \times 9$
(d) $(34 - 28) \times 42 \quad \square \quad 34 \times 42 - 28 \times 42$

**Solution**

1. (a) Since $8 - 3 > 3 - 8$, multiplying both by $29$ keeps the inequality same, so $(8 - 3) \times 29 > (3 - 8) \times 29$.
2. (b) LHS is $15 + 9 \times 18$ while RHS is $(15 + 9) \times 18 = 15 \times 18 + 9 \times 18$, so LHS < RHS.
3. (c) By the distributive property, $23 \times (17 - 9) = 23 \times 17 - 23 \times 9$, which is less than $23 \times 17 + 23 \times 9$.
4. (d) By the distributive property, $(34 - 28) \times 42 = 34 \times 42 - 28 \times 42$, so both sides are equal.

**Answer:** (a) $>$, (b) $<$, (c) $<$, (d) $=$

> Common mistake: Evaluating the expressions fully instead of using the distributive property and reasoning.

### Question 3

*1 mark · Fill in the blank*

Here is one way to make 14: $2 \times (1 + 6) = 14$. Are there other ways of getting 14? Fill them out below:
(a) $\_\_\_\_ \times (\_\_\_\_ + \_\_\_\_) = 14$
(b) $\_\_\_\_ \times (\_\_\_\_ + \_\_\_\_) = 14$
(c) $\_\_\_\_ \times (\_\_\_\_ + \_\_\_\_) = 14$
(d) $\_\_\_\_ \times (\_\_\_\_ + \_\_\_\_) = 14$

**Part (a)**

1. Find two numbers that add up and multiply to 14, such as 5 and 2.

Answer (a): $2 \times (5 + 2) = 14$

**Part (b)**

1. Find another pair of numbers inside the bracket that sum up suitably.

Answer (b): $2 \times (3 + 4) = 14$

**Part (c)**

1. Use 7 as the outer factor.

Answer (c): $7 \times (1 + 1) = 14$

**Part (d)**

1. Use 2 as the outer factor with a different sum.

Answer (d): $2 \times (6 + 1) = 14$

**Answer:** Other ways to make 14 using brackets and multiplication.

> Common mistake: Not verifying the evaluation of the expression inside and outside the brackets.

### Question 4

*3 marks · Short answer*

Find out the sum of the numbers given in each picture below in at least two different ways. Describe how you solved it through expressions.

**Solution**

1. For picture (I), Way 1: $(5 \times 4) + (4 \times 8) = 20 + 32 = 52$.
2. For picture (I), Way 2: $2 \times (4 + 8 + 4) + (8 + 4 + 8) = 52$.
3. For picture (II), Way 1: $(8 \times 5) + (8 \times 6) = 40 + 48 = 88$.
4. For picture (II), Way 2: $8 \times (5 + 6) = 8 \times 11 = 88$.

**Answer:** Picture (I) sum is $52$ and Picture (II) sum is $88$, evaluated using two different expressions each.

> Common mistake: Counting rows or columns incorrectly while forming the multiplication expressions.

## Figure it Out

### Question 1

*3 marks · Short answer*

Read the situations given below. Write appropriate expressions for each of them and find their values.
(a) The district market in Begur operates on all seven days of a week. Rahim supplies 9 kg of mangoes each day from his orchard and Shyam supplies 11 kg of mangoes each day from his orchard to this market. Find the amount of mangoes supplied by them in a week to the local district market.
(b) Binu earns ₹20,000 per month. She spends ₹5,000 on rent, ₹5,000 on food, and ₹2,000 on other expenses every month. What is the amount Binu will save by the end of a year?
(c) During the daytime a snail climbs 3 cm up a post, and during the night while asleep, accidentally slips down by 2 cm. The post is 10 cm high, and a delicious treat is on its top. In how many days will the snail get the treat?

**Part (a) (1 mark)**

1. Expression for total daily supply of mangoes by Rahim and Shyam = $9 + 11$
2. Total amount supplied in a week = $7 \times (9 + 11)$
3. Value = $7 \times 20 = 140 \text{ kg}$

Answer (a): $140 \text{ kg}$

**Part (b) (1 mark)**

1. Monthly savings = Monthly earnings $-$ Monthly expenses
2. Total annual savings = $12 \times 20000 - 12 \times (5000 + 5000 + 2000)$
3. Value = ₹$240000 - 144000 = \text{₹}96000$

Answer (b): ₹$96,000$

**Part (c) (1 mark)**

1. Net distance climbed by the snail in one full day cycle = $3 - 2 = 1 \text{ cm}$
2. Distance climbed in 7 days = $7 \text{ cm}$
3. On the 8th day, the snail climbs $3 \text{ cm}$ and reaches the top, so total days = 8

Answer (c): $8 \text{ days}$

**Answer:** See parts

> Common mistake: In part (c), forgetting that on the final day the snail reaches the top in the daytime and does not slip back.

### Question 2

*1 mark · MCQ*

Melvin reads a two-page story every day except on Tuesdays and Saturdays. How many stories would he complete reading in 8 weeks? Which of the expressions below describes this scenario?

- $5 \times 2 \times 8$
- $(7 - 2) \times 8$
- $8 \times 7$
- $7 \times 2 \times 8$
- $7 \times 5 - 2$
- $(7 + 2) \times 8$
- $7 \times 8 - 2 \times 8$
- $(7 - 5) \times 8$

**Solution**

1. In a week of 7 days, Melvin reads on $7 - 2 = 5$ days since he does not read on Tuesdays and Saturdays.
2. For 8 weeks, the total number of stories read is $(7 - 2) \times 8$.
3. By the distributive property, this can also be written as $7 \times 8 - 2 \times 8$.
4. Thus, the expressions describing this scenario are (b) and (g).

**Answer:** (b) $(7 - 2) \times 8$ and (g) $7 \times 8 - 2 \times 8$

> Common mistake: Selecting $5 \times 2 \times 8$ without realizing the first factor should be the number of reading days per week, not 5 stories.

### Question 3

*3 marks · Short answer*

Find different ways of evaluating the following expressions:
(a) $1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9 - 10$
(b) $1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1$

**Part (a) (1.5 marks)**

1. Way 1: Grouping positive and negative terms gives $(1 + 3 + 5 + 7 + 9) + (-2 - 4 - 6 - 8 - 10) = 25 - 30 = -5$.
2. Way 2: Grouping consecutive pairs gives $(1 - 2) + (3 - 4) + (5 - 6) + (7 - 8) + (9 - 10) = -1 - 1 - 1 - 1 - 1 = -5$.

Answer (a): $-5$

**Part (b) (1.5 marks)**

1. Way 1: Grouping pairs of terms gives $(1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) = 0 + 0 + 0 + 0 + 0 = 0$.
2. Way 2: Grouping all positive and negative terms gives $(1 + 1 + 1 + 1 + 1) + (-1 - 1 - 1 - 1 - 1) = 5 - 5 = 0$.

Answer (b): $0$

**Answer:** See parts

> Common mistake: Making arithmetic errors while grouping negative numbers.

### Question 4

*3 marks · Short answer*

Compare the following pairs of expressions using '<', '>' or '=' or by reasoning.
(a) $49 - 7 + 8 \quad \square \quad 49 - 7 + 8$
(b) $83 \times 42 - 18 \quad \square \quad 83 \times 40 - 18$
(c) $145 - 17 \times 8 \quad \square \quad 145 - 17 \times 6$
(d) $23 \times 48 - 35 \quad \square \quad 23 \times (48 - 35)$
(e) $(16 - 11) \times 12 \quad \square \quad -11 \times 12 + 16 \times 12$
(f) $(76 - 53) \times 88 \quad \square \quad 88 \times (53 - 76)$
(g) $25 \times (42 + 16) \quad \square \quad 25 \times (43 + 15)$
(h) $36 \times (28 - 16) \quad \square \quad 35 \times (27 - 15)$

**Part (a) (0.375 marks)**

1. LHS and RHS have the exact same terms: $49$, $-7$, and $8$.

Answer (a): $=$

**Part (b) (0.375 marks)**

1. LHS has $83 \times 42$ while RHS has $83 \times 40$, with the same subtracted term $18$.

Answer (b): $>$

**Part (c) (0.375 marks)**

1. LHS subtracts a larger multiple of $17$ ($17 \times 8$) from $145$ than RHS ($17 \times 6$).

Answer (c): $<$

**Part (d) (0.375 marks)**

1. Using the distributive property, LHS is $23 \times 48 - 23 \times 35$, which is greater than RHS $23 \times (48 - 35)$.

Answer (d): $>$

**Part (e) (0.375 marks)**

1. RHS can be rewritten using the commutative property as $16 \times 12 + (-11) \times 12$, which equals $(16 - 11) \times 12$.

Answer (e): $=$

**Part (f) (0.375 marks)**

1. LHS is $(76 - 53) \times 88 = 23 \times 88$, while RHS is $88 \times -(53 - 76) = 88 \times 23$.

Answer (f): $>$

**Part (g) (0.375 marks)**

1. Inside the brackets, $42 + 16 = 58$ and $43 + 15 = 58$, multiplied by the same number $25$.

Answer (g): $=$

**Part (h) (0.375 marks)**

1. LHS has a larger first factor ($36 > 35$) and equal bracket value ($16$). Thus LHS is greater.

Answer (h): $>$

**Answer:** See parts

> Common mistake: Evaluating each expression completely instead of using properties and reasoning.

### Question 5

*3 marks · Short answer*

Identify which of the following expressions are equal to the given expression without computation. You may rewrite the expressions using terms or removing brackets. There can be more than one expression which is equal to the given expression.
(a) $83 - 37 - 12$
(i) $84 - 38 - 12$
(ii) $84 - (37 + 12)$
(iii) $83 - 38 - 13$
(iv) $-37 + 83 - 12$
(b) $93 + 37 \times 44 + 76$
(i) $37 + 93 \times 44 + 76$
(ii) $93 + 37 \times 76 + 44$
(iii) $(93 + 37) \times (44 + 76)$
(iv) $37 \times 44 + 93 + 76$

**Part (a) (1.5 marks)**

1. Given expression is $83 - 37 - 12$, whose terms are $83$, $-37$, and $-12$.
2. Expression (i) $84 - 38 - 12$ changes terms by adding $1$ and subtracting $1$, so it is not equal.
3. Expression (ii) $84 - (37 + 12)$ evaluates to $84 - 49 = 35$, which is not equal.
4. Expression (iv) $-37 + 83 - 12$ contains the exact same terms rearranged using the commutative property.

Answer (a): (i) and (iv)

**Part (b) (1.5 marks)**

1. Given expression is $93 + 37 \times 44 + 76$, with terms $93$, $37 \times 44$, and $76$.
2. Expression (iv) $37 \times 44 + 93 + 76$ has the exact same terms in a different order due to the commutative property.

Answer (b): (iv)

**Answer:** See parts

> Common mistake: Ignoring the rules of terms and order of operations when evaluating options.

### Question 6

*3 marks · Short answer*

Choose a number and create ten different expressions having that value.

**Solution**

1. Let us choose the number $26$.
2. Expression 1 using addition: $10 + 16 = 26$
3. Expression 2 using subtraction: $30 - 4 = 26$
4. Expression 3 using multiplication: $13 \times 2 = 26$
5. Expression 4 using division: $52 \div 2 = 26$
6. Expression 5 using brackets and multiplication: $5 + (3 \times 7) = 26$
7. Expression 6 using brackets: $(5 \times 5) + 1 = 26$
8. Expression 7 using subtraction and multiplication: $30 - (2 \times 2) = 26$
9. Expression 8 using addition and division: $(20 + 32) \div 2 = 26$
10. Expression 9 using three operations: $25 + 3 - 2 = 26$
11. Expression 10 using multiplication: $2 \times 10 + 6 = 26$

**Answer:** Ten different expressions for $26$: $10 + 16$, $30 - 4$, $13 \times 2$, $52 \div 2$, $5 + (3 \times 7)$, $(5 \times 5) + 1$, $30 - (2 \times 2)$, $(20 + 32) \div 2$, $25 + 3 - 2$, and $2 \times 10 + 6$

> Common mistake: Writing expressions that evaluate to a number other than the chosen target number.

## Expression Engineer!

### Question 1

*3 marks · Short answer*

Using four 4's, create expressions to get all values from 1 to 20.

**Solution**

1. Using four 4's and the four basic operations along with brackets, we can form expressions for each value from 1 to 20.
2. For example: $1 = (4 \div 4) \times (4 \div 4)$, $2 = (4 \div 4) + (4 \div 4)$, $3 = (4 + 4 + 4) \div 4$, $4 = 4 + (4 - 4) \times 4$, and $5 = \frac{4 \times 4 + 4}{4}$.
3. Similarly, we can find expressions for all values up to 20 such as $20 = (4 + 4 + 4 + 4) \times 4 \div 4$.

**Answer:** Expressions using four 4's can be formed for each integer from 1 to 20.

> Common mistake: Forgetting to use brackets where the order of operations would otherwise change the value.

### Question 2

*3 marks · Short answer*

Using the numbers 1, 2, 3, 4, and 5 exactly once in any order get as many values as possible between -10 and +10.

**Solution**

1. Using the numbers 1, 2, 3, 4, and 5 exactly once with addition, subtraction, multiplication, and division, we can create various expressions.
2. For example: $-10 = 1 - 2 - 3 - 4 - 5 + 5$ or using proper combinations to cover values between -10 and +10.
3. For positive values: $10 = 5 + 4 + 2 - 1$, and for negative values: $-5 = 1 + 2 - 3 - 4 - 1$.

**Answer:** Various expressions using 1, 2, 3, 4, and 5 can be created to obtain values between -10 and +10.

> Common mistake: Repeating a number or missing one of the required digits from 1 to 5.

### Question 3

*3 marks · Short answer*

Using the numbers 0 to 9 exactly once in any order, make an expression with a value 100.

**Solution**

1. Using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 exactly once in any order, we can construct expressions that evaluate to 100.
2. One such possible arithmetic expression is $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 \times 9 = 100$.
3. Checking the evaluation: $1+2+3+4+5+6+7 + 72 = 28 + 72 = 100$.

**Answer:** $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 \times 9 = 100$

> Common mistake: Using a digit more than once or missing a digit.

### Question 4

*3 marks · Short answer*

What other similar interesting questions can you ask?

**Solution**

1. We can ask questions involving different sets of digits or target values.
2. Examples include making expressions to reach a target number using only prime digits or using specific restricted operations.
3. Another interesting question is to find the maximum or minimum possible value using a given set of numbers and operations.

**Answer:** Similar interesting questions can include forming target values using restricted sets of numbers or specific operations.

> Common mistake: Asking questions that do not have valid solutions under the given constraints.

## Frequently asked questions

### How many total questions are there in Class 7 Maths Chapter 2 Arithmetic Expressions?

This chapter in the new NCERT book for the 2026-27 session contains a structured set of questions spread across pages and practice sections like Figure it Out and Expression Engineer!. You can access SwaVid's free PDF and step-by-step solutions on this page only to practice all these questions.

### Which topics do the questions cover in this chapter?

The questions cover essential concepts such as arithmetic expressions and terms, comparing expressions, the token model of integers, and properties like commutative, associative, and distributive properties. They also include evaluating bracketed expressions, working with negative numbers, and creating expressions for real-life situations.

### What are the hardest question types in Class 7 Maths Chapter 2 and how should I approach them?

The most challenging questions usually involve bracket expansion, sign changes, and identifying equivalent expressions using terms and properties without direct calculation. To approach them, carefully apply the distributive property, understand how brackets affect negative terms, and break down complex expressions into simpler parts.

### How can I write answers for full marks in this chapter?

To secure full marks, you should clearly state the mathematical property or rule being used, such as associativity or the distributive property over addition and subtraction. SwaVid's free PDF and step-by-step solutions on this page only provide properly formatted answers that teach you the correct method of writing steps.

### Is the free PDF for these solutions available for the 2026-27 session?

Yes, complete solutions strictly based on the new NCERT book for the 2026-27 session are ready for you. SwaVid's free PDF and step-by-step solutions are on this page only, making it easy to revise concepts like arithmetic expressions and brackets before your exams.

## 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