NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 2 PDF Download

NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 2 help students understand Arithmetic Expressions. This chapter teaches how to make and solve expressions using addition, subtraction, and brackets.

The solutions are based on the Class 7 Maths Chapter 2 syllabus and explain each step clearly so students can learn with confidence.

Class 7 Maths Ganita Prakash Chapter 2 Arithmetic Expressions

Chapter 2 of Ganita Prakash Class 7 is about Arithmetic Expressions. In this chapter, students learn how to form and solve expressions using numbers and basic operations like addition and subtraction. 

The chapter also explains how brackets and the order of operations are important in solving problems correctly. These concepts help students understand how to handle expressions in a better way. The Ganita Prakash Class 7 Chapter 2 Solutions Arithmetic Expressions are designed to make learning easy with step-by-step answers and simple examples.

Class 7 Maths Ganita Prakash Chapter 2 Solutions

Below are the Class 7 Maths Ganita Prakash Chapter 2 Question Answers. This chapter is based on Arithmetic Expressions and helps students understand how to use numbers, brackets, and basic operations in expressions.

 NCERT In-Text Questions (Page 24)

Q.1. Choose your favourite number and write as many expressions as you can having that value.
Solution:
Let us choose the number 20.
We can write the arithmetic expressions for the number as follows:
12 + 8 = 20; 4 × 5 = 20; 40 ÷ 2 = 20, etc.

Figure it Out (Page 25)

Question 1. Fill in the blanks to make the expressions equal on both sides of the ‘=’ sign:
(a) 13 + 4 = _________ + 6
(b) 22 + _________ = 6 × 5
(c) 8 × _________ = 64 ÷ 2
(d) 34 – _________ = 25
Solution:
(a) 13 + 4 = 17
11 + 6 = 17
Therefore, 13 + 4 = 11 + 6

(b) Since 6 × 5 = 30
22 + 8 = 30
Therefore, 22 + 8 = 6 × 5

(c) Since 64 ÷ 2 = 32
8 × 4 = 32
Therefore, 8 × 4 = 64 ÷ 2

(d) Since 34 – 25 = 9
Therefore, 34 – 9 = 25

Question 2. Arrange the following expressions in ascending (increasing) order of their values.
(a) 67 – 19
(b) 67 – 20
(c) 35 + 25
(d) 5 × 11
(e) 120 ÷ 3
Solution:
(a) 67 – 19 = 48
(b) 67 – 20 = 47
(c) 35 + 25 = 60
(d) 5 × 11 = 55
(e) 120 ÷ 3 = 40
Clearly, 40 < 47 < 48 < 55 < 60
Therefore, 120 ÷ 3 < 67 – 20 < 67 – 19 < 5 × 11 < 35 + 25
Thus, (e) < (b) < (a) < (d) < (c).

Comparing Expressions

NCERT In-Text Questions (Page 26)

Q.1. 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 + 289 246 + 285
(b) 273 – 145 272 – 144
(c) 364 + 587 363 + 589
(d) 124 + 245 129 + 245
(e) 213 – 77 214 – 76
Solution:

(a) 245 + 289 < 246 + 285
Explanation:
245 is 1 less than 246, but 289 is 4 more than 285.
So the second sum is larger by 3.

(b) 273 – 145 = 272 – 144
Explanation:
273 is 1 more than 272, and 145 is also 1 more than 144.
So the difference stays the same.

(c) 364 + 587 < 363 + 589
Explanation:
364 is 1 more than 363, but 587 is 2 less than 589.
So the second sum is greater by 1.

(d) 124 + 245 < 129 + 245
Explanation:
Only the first numbers are different.
129 is 5 more than 124, so the second sum is greater by 5.

(e) 213 – 77 < 214 – 76
Explanation:
214 is 1 more than 213, and 76 is 1 less than 77.
So the second difference is greater by 2.

NCERT In-Text Questions (Pages 29-31)

Q.1. Will this also hold when there are terms having negative numbers as well? Take some more expressions and check.
Solution:
Yes, swapping the terms having negative numbers does not change the sum.
As (-3) + (-2) = -5 or (-2) + (-3) = -5
(Answer may vary)

Q.2. Can you explain why this is happening using the Token Model of integers that we saw in the Class 6 textbook of mathematics?
Solution:
In the Token Model of integers, we use positive (+) and negative (–) tokens to show numbers.

• A positive token means +1

• A negative token means –1

• When we put one positive and one negative token together, they cancel each other (because +1 and –1 = 0)

Let’s take an example like 5 – 3.
This means we have 5 positive tokens, and we take away 3 positive tokens, so we are left with 2 positive tokens.

Now let’s say we have –2 + 5.
We start with 2 negative tokens, and we add 5 positive tokens.
Pairing 2 positives with 2 negatives, we are left with 3 positive tokens (because 2 pairs cancel out).

So, using the Token Model helps us understand how adding and subtracting positive and negative numbers works by canceling tokens.

Q.3. Will this also hold when there are terms having negative numbers as well? Take some more expressions and check.
Solution:
Yes, this will also work for negative numbers. Let’s check a few examples:

1. –5 + 3 and 3 + (–5)
   → Both give the same result: –2
     So, changing the order works here too.

2. –4 + (–6) and –6 + (–4)
   → Both give the same result: –10
   Works again.

3. 7 + (–2) and (–2) + 7
   → Both give the result: 5
   Still true.

This shows that the rule holds even when we use negative numbers.

NCERT In-Text Questions (Pages 32-33)

Q.1. 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:
Since the total number of friends = 7
and the cost of each dosa = ₹ 23
Therefore, the total cost of 7 dosas = 7 × 23
As the tip remains the same, that is ₹ 5.
So, the expression for describing the total cost is 7 × 23 + 5 = 7 × 23 + 5 = 161 + 5 = ₹ 166.
The terms in the expression 7 × 23 + 5 are 7 × 23, 5.

Q.2. 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 _________

Solution:
If the teacher had called our ‘4’, Ruby would write 8 × 4 + 1
Terms: 8 × 4, 1
If the teacher had called our ‘7’, Ruby would write 4 × 7 + 5
Terms: 4 × 7, 5

Figure it Out (Pages 34-35)

Question 1. Find the values of the following expressions by writing the terms in each case.
(a) 28 – 7 + 8
(b) 39 – 2 × 6 + 11
(c) 40 – 10 + 10 + 10
(d) 48 – 10 × 2 + 16 + 2
(e) 6 × 3 – 4 × 8 × 5
Solution:
(a) 28 – 7 + 8 = 28 + (-7) + 8
Terms: 28, -7, and 8
28 – 7 + 8
= 28 + (-7) + 8
= 21 + 8 = 29

(b) 39 – 2 × 6 + 11 = 39 + (-2 × 6) + 11
Terms: 39, -2 × 6, and 11
39 – 2 × 6 + 11
= 39 + (-2 × 6) + 11
= 39 + (-12) + 11
= 27 + 11
= 38

(c) 40 – 10 + 10 + 10 = 40 + (-10) + 10 + 10
Terms: 40, -10, 10, and 10
40 – 10 + 10 + 10
= 40 + (-10) + 10 + 10
= 30 + 10 + 10
= 40 + 10
= 50

(d) 48 – 10 × 2 + 16 + 2 = 48 + (-10 × 2) + (16 + 2)
Terms: 48, -10 × 2, 16 + 2
48 – 10 × 2 + 16 + 2
= 48 + (-10 × 2) + (16 + 2)
= 48 + (-20) + (8)
= 28 + 8 = 36

(e) 6 × 3 – 4 × 8 × 5 = (6 × 3) + (-4 × 8 × 5)
Terms: 6 × 3, 4 × 8 × 5
6 × 3 – 4 × 8 × 5
= (6 × 3) + (-4 × 8 × 5)
= 18 + (-160)
= -142

(a) 89 + 21 – 10
Riya and Siya are cousins. They went to the beach, from where Riya collected 89 stones and Siya collected 21 stones. When they reached home, their other younger sister asked them for some stones. They give her 10 stones from their collection. How many stones were left with both Riya and Siya?
89 + 21 – 10 = 89 + 21 + (-10) = 89 + 11 = 100

(b) 5 × 12 – 6
Radha bought 5 pens from a stationery shop. The cost of each pen is ₹ 12. The shopkeeper also gave her a discount of ₹ 6 on the total cost. Find the total amount that she has to pay to the shopkeeper.
5 × 12 – 6 = 5 × 12 + (-6) = 60 + (-6) = 54

(c) 4 × 9 + 2 × 6
Sumit bought 4 pencils, the cost of each pencil is ₹ 9, and he also bought 2 erasers, which cost ₹ 6 each. How much money does he have to spend to buy these items?
4 × 9 + 2 × 6 = 36 + 12 = 48

Figure it Out (Pages 37-38)

Question 1. 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 _________
(b) 38 + (_________ ________) = 38 + 9 – 4
(c) 24 – (6 + 4) = 24 6 – 4
(d) 24 – 6 – 4 = 24 – 6 _________
(e) 27 – (8 + 3) = 27 _________ 8 _________ 3
(f) 27 – (_________ ________) = 27 – 8 + 3

Solution:
(a) 24 + (6 – 4) = 24 + 6 – 4
(b) 38 + (9 – 4) = 38 + 9 – 4
(c) 24 – (6 + 4) = 24 – 6 – 4
(d) 24 – 6 – 4 = 24 – 6 – 4
(e) 27 – (8 + 3) = 27 – 8 – 3
(f) 27 – (8 – 3) = 27 – 8 + 3

Question 2. 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)

Solution:
(a) 14 + (12 + 10)
= 14 + 12 + 10
= 14 + 22
= 36

(b) 14 – (12 + 10)
= 14 – 12 – 10
= 14 – 22
= -8

(c) 14 + (12 – 10)
= 14 + 12 – 10
= 14 + 2
= 16

(d) 14 – (12 – 10)
= 14 – 12 + 10
= 14 – 2
= 12

(e) -14 + 12 – 10
= -14 + 2
= -12

(f) 14 – (-12 – 10)
= 14 + 12 + 10
= 14 + 22
= 36

Question 3. 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)

Solution:
(a) (6 + 10) – 2 and 6 + (10 – 2)
(6 + 10) – 2 = 16 – 2 = 14
and 6 + (10 – 2) = 6 + 8 = 14
Clearly, (6 + 10) – 2 = 6 + (10 – 2)
Hence, the expressions in part (a) have the same value.

(b) 16 – (8 – 3) and (16 – 8) – 3
16 – (8 – 3) = 16 – 5 = 11
and (16 – 8) – 3 = 8 – 3 = 5
16 – (8 – 3) ≠ (16 – 8) – 3
Hence, the expressions in part (b) do not have the same value.

(c) 27 – (18 + 4) and 27 + (-18 – 4)
27 – (18 + 4) = 27 – 22 = 5
and 27 + (-18 – 4) = 27 + (-22) = 5
Clearly, 27 – (18 + 4) = 27 + (-18 – 4)
Hence, the expressions in part (c) have the same value.

NCERT In-Text Questions (Pages 39-40)

Q.1. If another friend, Sangmu, joins them and orders the same items, what will be the expression for the total amount to be paid?
Solution:
If another friend, Sangmu, joins them (Lhamo and Norbu) and orders the same items, then the expression for the total amount will be 3 × (43 + 24).

5 × 4 + 3 ≠ 5 × (4 + 3). Can you explain why?
Is 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5?
Solution:
Expression 5 × 4 + 3 means 3 more than 5 × 4, which is equal to 23, but 5 × (4 + 3) means 5 times (4 + 3), which is equal to 35.
Hence, 5 × 4 + 3 ≠ 5 × (4 + 3)
5 × (4 + 3), 5 × (3 + 4), and (3 + 4) × 5 have the same meaning, which is 5 times the sum of 3 and 4, and give the same value.
Hence, 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5

Tinker the Terms II

NCERT In-Text Questions (Page 41)

Q.1. Use this method to find the following products:
(а) 95 × 8
(b) 104 × 15
(c) 49 × 50
Is this quicker than the multiplication procedure you use generally?

Solution:
(a) 95 × 8 = (100 – 5) × 8
= 100 × 8 – 5 × 8
= 800 – 40
= 760

(b) 104 × 15 = (100 + 4) × 15
= 100 × 15 + 4 × 15
= 1500 + 60
= 1560

(c) 49 × 50 = (50 – 1) × 50
= 50 × 50 – 50 × 1
= 2500 – 50
= 2450
Yes, this procedure is quicker than the general multiplication procedure.

Figure it Out (Pages 41-42)

Question 1. Fill in the blanks with numbers and boxes by signs, so that the expressions on both sides are equal.
(а) 3 × (6 + 7) = 3 × 6 + 3 × 7
(b) (8 + 3) × 4 = 8 × 4 + 3 × 4
(c) 3 × (5 + 8) = 3 × 5 3 × ________
(d) (9 + 2) × 4 = 9 × 4 2 × ________
(e) 3 × ( ________ + 4) = 3 ________ + ________
(f) (________ + 6) × 4 = 13 × 4 + ________
(g) 3 × (________ + ________) = 3 × 5 + 3 × 2
(h) (________ + ________) × ________ = 2 × 4 + 3 × 4
(i) 5 × (9 – 2) = 5 × 9 – 5 × ________
(j) (5 – 2) × 7 = 5 × 7 – 2 × ________
(k) 5 × (8 – 3) = 5 × 8 5 × ________
(l) (8 – 3) × 7 = 8 × 7 3 × 7
(m) 5 × (12 – ________) = ________ 5 × ________
(n) (15 – ________) × 7 = ________ 6 × 7
(o) 5 × (________ – ________) = 5 × 9 – 5 × 4
(p) (________ – ________) × ________ = 17 × 7 – 9 × 7
Solution:
(а) 3 × (6 + 7) = 3 × 6 + 3 × 7
(b) (8 + 3) × 4 = 8 × 4 + 3 × 4
(c) 3 × (5 + 8) = 3 × 5 + 3 × 8
(d) (9 + 2) × 4 = 9 × 4 + 2 × 4
(e) 3 × (10 + 4) = 30 + 12
(f) (13 + 6) × 4 = 13 × 4 + 24
(g) 3 × (5 + 2) = 3 × 5 + 3 × 2
(h) (2 + 3) × 4 = 2 × 4 + 3 × 4
(i) 5 × (9 – 2) = 5 × 9 – 5 × 2
(j) (5 – 2) × 7 = 5 × 7 – 2 × 7
(k) 5 × (8 – 3) = 5 × 8 – 5 × 3
(l) (8 – 3) × 7 = 8 × 7 – 3 × 7
(m) 5 × (12 – 3) = 60 – 5 × 3
(n) (15 – 6) × 7 = 105 – 6 × 7
(o) 5 × (9 – 4) = 5 × 9 – 5 × 4
(p) (17 – 9) × 7 = 17 × 7 – 9 × 7

Question 2. In the boxes below, fill in ‘<’, ‘>’ 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) × 29 (3 – 8) × 29
(b) 15 + 9 × 18 (15 + 9) × 18
(c) 23 × (17 – 9) 23 × 17 + 23 × 9
(d) (34 – 28) × 42 34 × 42 – 28 × 42
Solution:
(a) (8 – 3) × 29 > (3 – 8) × 29
Because, (3 – 8) × 29 = -(8 – 3) × 29
⇒ (8 – 3) × 29 > (3 – 8) × 29

(b) 15 + 9 × 18 < (15 + 9) × 18
Because, (15 + 9) × 18 = 15 × 18 + 9 × 18 and 15 × 18 > 15
So, 15 + 9 × 18 < (15 + 9) × 18

(c) 23 × (17 – 9) < 23 × 17 + 23 × 9
Because, 23 × (17 – 9) = 23 × 17 – 23 × 9
Clearly, 23 × 17 > 23 × 17 – 23 × 9
⇒ 23 × (17 – 9) < 23 × 17 + 23 × 9

(d) (34 – 28) × 42 = 34 × 42 – 28 × 42

Question 3. Here is one way to make 14: 2 × (1 + 6) = 14. Are there other ways of getting 14? Fill them out below:
(a) ________ × (________ + ________) = 14
(b) ________ × (________ + ________) = 14
(c) ________ × (________ + ________) = 14
(d) ________ × (________ + ________) = 14
Solution:
(a) 2 × (5 + 2) = 14
(b) 2 × (3 + 4) = 14
(c) 2 × (4 + 3) = 14
(d) 2 × (6 + 1) = 14

Figure it Out (Pages 42-44)

Question 1. 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 the 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 number 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 the 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 top. In how many days will the snail get the treat?
Solution:
(a) Supplies of mangoes by Rahim in the market each day = 9 kg
Supplies of mangoes by Shyam in the market each day = 11 kg
Total supplies of mangoes in the market on each day = (9 + 11) kg
Therefore, total supplies of mangoes in the market in all 7 days = 7 × (9 + 11) kg
= 7 × 20
= 140 kg

(b) Binu’s per month earning = ₹ 20,000
Binu’s total monthly expenditures = ₹ 5,000 on rent + ₹ 5,000 on food + ₹ 2,000 on other expenses
= 5,000 + 5,000 + 2,000
Therefore, Binu’s monthly savings = ₹ 20,000 – ₹(5,000 + 5,000 + 2,000)
= ₹ 20,000 – ₹ 12,000
= ₹ 8,000
Thus, Binu’s total yearly savings = 12 × 8000 = 96000
Hence, Binu will save ₹ 96000 by the end of the year.

(c) Since the snail climbs 3 cm up the post in daytime and slips down by 2 cm at night.
So, the distance climbed by the snail of the post = 3 – 2 = 1 cm in a day.
∴ The distance climbed in 7 days = 7 cm
The height of the post is 10 cm.
The distance climbed on the 8th day before slipping = 7 + 3 = 10 cm
So, the snail will take 8 days to reach the top of the post and get the delicious treat.

Question 2. 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?
(a) 5 × 2 × 8
(b) (7 – 2) × 8
(c) 8 × 7
(d) 7 × 2 × 8
(e) 7 × 5 – 2
(f) (7 + 2) × 8
(g) 7 × 8 – 2 × 8
(h) (7 – 5) × 8
Solution:
Number of days in a week except Tuesday and Saturday = 7 – 2
Since Melvin reads a two-page story every day except Tuesday and Saturday.
Therefore, number of stories read in a week = 1 × (7 – 2)
So, number of stories read in 8 weeks = 8 × 1 × (7 – 2)
= 8 × (7 – 2) or (7 – 2) × 8 [Expression (b)]
or 7 × 8 – 2 × 8 [Expression (g)]
Only expressions (b) and (g) describe this scenario.

Question 3. Find different ways of evaluating the following expressions:
(а) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
(b) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
Solution:
(a) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
= (1 + 3 + 5 + 7 + 9) + (-2 – 4 – 6 – 8 – 10)
= 25 + (-30)
= -5
or
1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
= (1 – 2) + (3 – 4) + (5 – 6) + (7 – 8) +(9 – 10)
= (-1) + (-1) + (-1) + (-1) + (-1)
= -5

(b) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
= (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1)
= 0 + 0 + 0 + 0 + 0
= 0
or
1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
= (1 + 1 + 1 + 1 + 1) + (-1 – 1 – 1 – 1 – 1)
= 5 + (-5)
= 0

Question 4. Compare the following pairs of expressions using ‘<’, ‘>’, or ‘=,’ or by reasoning.
(a) 49 – 7 + 8 49 – 7 + 8
(b) 83 × 42 – 18 83 × 40 – 18
(c) 145 – 17 × 8 145 – 17 × 6
(d) 23 × 48 – 35 23 × (48 – 35)
(e) (16 – 11) × 12 -11 × 12 + 16 × 12
(f) (76 – 53) × 88 88 × (53 – 76)
(g) 25 × (42 + 16) 25 × (43 + 15)
(h) 36 × (28 – 16) 35 × (27 – 15)

Solution:
(a) 49 – 7 + 8 = 49 – 7 + 8
(∵ All the terms on both sides are the same)

(b) 83 × 42 > 83 × 40
∴ 83 × 42 – 18 > 83 × 40 – 18

(c) 17 × 8 > 17 × 6
⇒ -17 × 8 < -17 × 6
∴ 145 – 17 × 8 < 145 – 17 × 6

(d) 23 × (48 – 35) = 23 × 48 – 23 × 35
and 35 < 23 × 35 23 × 48 – 35 > 23 × (48 – 35)

(e) (16 – 11) × 12 = 16 × 12 – 11 × 12 = -11 × 12 + 16 × 12
∴ (16 – 11) × 12 = -11 × 12 + 16 × 12

(f) (76 – 53) × 88 = 76 × 88 – 53 × 88 = -(53 – 76) × 88
∴ (76 – 53) × 88 > 88 × (53 – 76)

(g) 43 + 15 = 42 + 1 + 15 = 42 + 16
⇒ 25 × (43 + 15) = 25 × (42 + 16)
∴ 25 × (42 + 16) = 25 × (43 + 15)

(h) 35 × (27 – 15) = 35 × (28 – 16)
∴ 36 × (28 – 16) > 35 × (27 – 15)

Question 5. 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 that 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 × 44 + 76
(i) 37 + 93 × 44 + 76
(ii) 93 + 37 × 76 + 44
(iii) (93 + 37) × (44 + 76)
(iv) 37 × 44 + 93 + 76

Solution:
(a) 83 – 37 – 12 = 83 – 37 – 12 + (1 – 1)
= (83 + 1) – 37 – 1 – 12
= 84 – 38 – 12 (option (i))
= 34
Or
83 – 37 – 12 = -37 + 83 – 12
= 46 – 12
= 34 (option (iv))
Hence, (i) and (iv) are equal to the given expression 83 – 37 – 12.

(b) (iv) 37 × 44 + 93 + 76
Rearrange the terms, and we get 93 + 37 × 44 + 76, which is equal to the given expression.
Hence, (iv) is equal to the given expression 93 + 37 × 44 + 76.

Question 6. Choose a number and create ten different expressions having that value.
Solution:
Let’s choose the number 24.
Here are ten different expressions that give the value 24:

1. 20 + 4 = 24

2. 30 – 6 = 24

3. 6 × 4 = 24

4. 48 ÷ 2 = 24

5. (3 × 8) = 24

6. 25 – 1 = 24

7. (12 × 2) = 24

8. 100 – 76 = 24

9. 8 + 8 + 8 = 24

10. 6 × (10 – 6) = 24

NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 2 PDF Download

NCERT Class 7 Maths Chapter 2 helps students understand arithmetic expressions in a clear and easy way. These solutions include step-by-step answers to all textbook questions and follow the class 7 maths syllabus. 

They are useful for quick revision and better exam preparation. Students can download the PDF form of these solutions from the link given below.

Benefits of Using NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 2

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•  Covers all questions from the chapter for better exam preparation.

• Follows the latest syllabus and exam pattern.

•  Step-by-step explanations help in understanding concepts clearly.

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