Courses

Courses for Kids

Free study material

Talk to our experts

1800-120-456-456

NCERT Solutions for Class 7 Maths Chapter 2 - Operations with Integers

Q: 1. What is Class 7 Maths Chapter 2 Operations with Integers about?

Operations with Integers helps students understand how different mathematical operations are performed with positive and negative integers. The chapter develops calculation skills and helps students apply integer concepts while solving problems.

Q: 2. What should I focus on while studying Class 7 Maths Chapter 2?

Focus on understanding how integers behave under different operations rather than memorising rules. Work through the examples carefully and practise the textbook questions to become comfortable with calculations involving positive and negative numbers.

Q: 3. How should I solve questions from Operations with Integers?

Read the question carefully and identify the operation or concept involved. Apply the appropriate rule step by step and check the signs of the numbers before arriving at the final answer. After solving the question yourself, use the NCERT solution to verify your method.

Q: 4. Are the exercises in Class 7 Maths Chapter 2 important for exams?

Yes. The textbook exercises provide practice with the concepts introduced in the chapter. Solving them regularly can help students improve calculation accuracy and understand how integer operations are used in different types of questions.

Q: 5. Why are integers important in Class 7 Maths?

Integers are used to represent values such as gains and losses, increases and decreases, temperatures above and below zero, and positions on a number line. Understanding integer operations also provides a foundation for many later mathematical concepts.

Q: 6. How can I improve my accuracy while solving integer questions?

Write each step clearly and pay special attention to the signs of positive and negative numbers. Avoid doing too many calculations mentally, especially when a question involves multiple operations. Regular practice can also help reduce sign-related mistakes.

Q: 7. Can I download PDF solutions for Class 7 Maths Chapter 2 Operations with Integers?

Students can use the chapter solutions online and access the PDF option for convenient revision from Vedantu. The solutions can help students review textbook questions, check calculations, and prepare for school assessments.

Class 7 Maths Chapter 2 Operations with Integers Solutions PDF Download

NCERT Solutions for Class 7 Maths Chapter 2 help students understand Operations with Integers through simple explanations and step-by-step answers. As part of Ganita Prakash II, this chapter helps students work confidently with integers and understand the mathematical reasoning behind different operations.

Students can first attempt the textbook questions independently and then use the NCERT Maths Solutions for Class 7 Chapter 2 to check their approach and answers. The solutions can help students understand the steps clearly instead of simply memorising them. Students can also refer to NCERT Solutions for Class 7 Maths for solutions to other chapters.

Chapter 2 Operations with Integers NCERT Question Answers

2.1 A Quick Recap of Integers

Figure It Out (Page 25)

Question 1. 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

(d) Sum = 0, Difference = -10

(e) Sum = -7, Difference = -1

(f) Sum = -7, Difference = -13

Solution:

(a) Sum = 27, Difference = 9

Answer:

Let the two integers be 18 and 9.

First number	Second number	Sum	Difference
20	7	27	13
17	10	27	7
16	11	27	5
18	9	27	9

Sum = 18 + 9 = 27

Difference = 18 - 9 = 9

Therefore, the required pair is (18, 9).

(b) Sum = 4, Difference = 12

Answer:

First number	Second number	Sum	Difference
2	2	4	0
4	0	4	4
10	−6	4	16
8	−4	4	12

Let the two integers be 8 and -4.

Sum = 8 + (-4) = 4

Difference = 8 - (-4) = 12

Therefore, the required pair is (8, -4).

First number	Second number	Sum	Difference
6	−6	0	12
8	−8	0	16
7	−7	0	14
5	−5	0	10

Answer:

Let the two integers be 5 and -5.

Sum = 5 + (-5) = 0

Difference = 5 - (-5) = 10

Therefore, the required pair is (5, -5).

(d) Sum = 0, Difference = -10

First number	Second number	Sum	Difference
4	−4	0	8
3	−3	0	6
−5	5	0	−10
8	−8	0	16

Answer:

Let the two integers be -5 and 5.

Sum = -5 + 5 = 0

Difference = -5 - 5 = -10

Therefore, the required pair is (-5, 5).

(e) Sum = -7, Difference = -1

First number	Second number	Sum	Difference
−9	2	−7	−11
1	−8	−7	9
−6	−1	−7	−5
−4	−3	−7	−1

Answer:

Let the two integers be -4 and -3.

Sum = -4 + (-3) = -7

Difference = -4 - (-3) = -1

Therefore, the required pair is (-4, -3).

(f) Sum = -7, Difference = -13

First number	Second number	Sum	Difference
−4	−3	−7	−1
−8	1	−7	−9
−3	−4	−7	+1
−10	3	−7	−13

Answer:

Let the two integers be -10 and 3.

Sum = -10 + 3 = -7

Difference = -10 - 3 = -13

Therefore, the required pair is (-10, 3).

2.2 Multiplication of Integers

Figure It Out (Page 31)

Question 1. Using the token interpretation, find the values of:

(a) 3 × (-2)

Answer:

Three groups of 2 negative tokens give 6 negative tokens.

Therefore,

3 × (-2) = -6

(b) (-5) × (-2)

Answer:

For (-5) × (-2), we remove 2 negative tokens 5 times.

Using zero pairs, removing 10 negative tokens leaves 10 positive tokens.

Therefore,

(-5) × (-2) = 10

Answer:

For (-4) × (-1), we remove 1 negative token 4 times.

Using zero pairs, removing 4 negative tokens leaves 4 positive tokens.

Therefore,

(-4) × (-1) = 4

(d) (-7) × 3

Answer:

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

7 × 3 = 21

Therefore,

(-7) × 3 = -21

Question 2. If 123 × 456 = 56088, without calculating, find the value of:

(a) (-123) × 456

Answer:

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

Therefore,

(-123) × 456 = -(123 × 456)

= -56088

(b) (-123) × (-456)

Answer:

The product of two negative integers is positive.

Therefore,

(-123) × (-456) = 123 × 456

= 56088

Answer: Multiplying a positive integer by a negative integer gives a negative product.

Therefore,

123 × (-456) = -(123 × 456)

= -56088

Question 3. Try to frame a simple rule to multiply two integers.

Answer:

To multiply two integers:

First, multiply their absolute values.
If both integers have the same sign, the product is positive.
If the integers have different signs, the product is negative.

For example:

(-4) × (-3) = 12

4 × (-3) = -12

Figure It Out (Pages 33-34)

Question 1. Find the following products.

(a) 4 × (-3)

Answer: Multiplying

A positive integer multiplied by a negative integer gives a negative product.

4 × (-3) = -(4 × 3)

= -12

(b) (-6) × (-3)

Answer:

The product of two negative integers is positive.

(-6) × (-3) = 6 × 3

= 18

Answer:

The product of two negative integers is positive.

(-5) × (-1) = 5 × 1

= 5

(d) (-8) × 4

Answer:

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

(-8) × 4 = -(8 × 4)

= -32

(e) (-9) × 10

Answer:

(-9) × 10 = -(9 × 10)

= -90

(f) 10 × (-17)

Answer:

10 × (-17) = -(10 × 17)

= -170

Figure It Out (Page 39)

Question 1. Find the values of:

(a) 14 × (-15)

Answer:

14 × (-15) = -(14 × 15)

= -210

(b) -16 × (-5)

Answer:

The product of two negative integers is positive.

(-16) × (-5) = 80

Answer:

Dividing a positive integer by a negative integer gives a negative quotient.

36 ÷ (-18) = -2

(d) (-46) ÷ (-23)

Answer:Dividing aA negative integer by a negative integer gives a positive quotient.

(-46) ÷ (-23) = 2

Question 2. 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?

Answer:

Initial temperature = 32°C

Temperature decreases every hour = 5°C

Decrease in 10 hours = 5 × 10 = 50°C

Final temperature = 32 - 50

= -18°C

Therefore, the room temperature after 10 hours will be -18°C.

Question 3. 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?

Answer:

Profit on one white cement bag = +₹8

Loss on one grey cement bag = -₹5

Profit from 3,000 white cement bags:

3000 × 8 = ₹24,000

Loss from 5,000 grey cement bags:

5000 × (-5) = -₹25,000

Net amount = 24,000 - 25,000

= -₹1,000

Therefore, the company has a loss of ₹1,000.

(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?

Answer:

Let the number of white cement bags be x.

For neither profit nor loss:

8x + 6400 × (-5) = 0

8x - 32,000 = 0

8x = 32,000

x = 4,000

Therefore, the company must sell 4,000 bags of white cement.

Question 4. Replace the blank with an integer to make a true statement.

(a) (-3) × ________ = 27

Answer:

(-3) × (-9) = 27

Therefore, the required integer is -9.

(b) 5 × ________ = (-35)

Answer:

5 × (-7) = -35

Therefore, the required integer is -7.

Answer:

7 × (-8) = -56

Therefore, the required integer is 7.

(d) ________ × (-12) = 132

Answer:

(-11) × (-12) = 132

Therefore, the required integer is -11.

(e) ________ ÷ (-8) = 7

Answer:

(-56) ÷ (-8) = 7

Therefore, the required integer is -56.

(f) ________ ÷ 12 = -11

Answer:

(-132) ÷ 12 = -11

Therefore, the required integer is -132.

Figure It Out (Pages 42-44)

Question 1. Find the values of the following expressions:

(a) (-5) × (18 + (-3))

Answer:

(-5) × (18 + (-3))

= (-5) × 15

= -75

(b) (-7) × 4 × (-1)

Answer:

(-7) × 4 × (-1)

= (-28) × (-1)

= 28

Answer:

(-2) × (-1) × (-5) × (-3)

= 2 × 15

= 30

Question 2. Find the values of the following expressions:

(a) (-27) ÷ 9

Answer:

(-27) ÷ 9 = -3

(b) 84 ÷ (-4)

Answer:

84 ÷ (-4) = -21

Answer:

(-56) ÷ (-2) = 28

Question 3. Find the integer whose product with (-1) is:

(a) 27

Answer:

(-1) × (-27) = 27

Therefore, the required integer is -27.

(b) -31

Answer:

(-1) × 31 = -31

Therefore, the required integer is 31.

Answer:

(-1) × 1 = -1

Therefore, the required integer is 1.

(d) 1

Answer:

(-1) × (-1) = 1

Therefore, the required integer is -1.

(e) 0

Answer:

(-1) × 0 = 0

Therefore, the required integer is 0.

Question 4. 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.

Answer:

The second expression is the negative of the first expression.

Given:

47 - 56 + 14 - 8 + 2 - 8 + 5 = -4

Therefore:

-47 + 56 - 14 + 8 - 2 + 8 - 5

= -(-4)

= 4

Question 5. 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.

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

Answer:

(a) Following the rules for the sequence starting with -7.

The rule is: if even, take half; if odd, multiply by -3, and add 1.

Start with -7 (odd)

[(-7) × (-3)] + 1 = 21 + 1 = 22 (even)

then 22 ÷ 2 = 11 (odd)

then 11 × (-3) + 1 = -33 + 1 = -32 (even)

then (-32) ÷ 2 = -16 (even)

then (-16) ÷ 2 = -8 (even)

then (-8) ÷ 2 = -4 (even)

then (-4) ÷ 2 = -2 (even)

then (-2) ÷ 2 = – 1 (odd)

then [(-1) × (-3)] + 1 = 4 (even)

then 4 ÷ 2 = 2 (even)

then 2 ÷ 2 = 1 (odd)

(b) (i) Now for the starting number -21 (odd)

then [(-21) × (-3)] + 1 = 63 + 1 = 64 (even)

then 64 ÷ 2 = 32 (even)

then 32 ÷ 2 = 16 (even)

then 16 + 2 = 8 (even)

then 8 ÷ 2 = 4 (even)

then 4 ÷ 2 = 2 (even)

then 2 ÷ 2 = 1 (odd)

then 1 × (-3) + 1 = -2 (even)

then (-2) ÷ 2 = -1 (odd)

then [-1 × (-3)] + 1 = 4 (even)

then 4 ÷ 2 = 2 (even)

then 2 ÷ 2 = 1 (odd)

then 1 × (-3) + 1 = -2 (even)

Hence, the sequence for -21 is

Sequence for the starting number -21 (odd)

(ii) For the sequence, the starting number is -6

-6 is even then -6 + 2 = -3 (odd)

then [(-3) × (-3)] + 1 = 9 + 1 = 10 (even)

then 10 ÷ 2 = 5 (odd)

5 × (-3) + 1 = -15 + 1 = -14 (even)

then -14 ÷ 2 = -7 (odd)

[(-7) × (-3)] + 1 = +21 + 1 = 22 (even)

then 22 ÷ 2 = 11 (odd)

then [11 × (-3)] + 1 = -33 + 1 = -32 (even)

then -32 ÷ 2 = -16 (even)

then -16 ÷ 2 = -8 (even)

then -8 ÷ 2 = -4 (even)

then -4 ÷ 2 = -2 (even)

then (-2) ÷ 2 = -1 (odd)

then [(-1) × (-3)] + 1 = 3 + 1 = 4 (even)

then 4 ÷ 2 = 2 (even)

then 2 ÷ 2 = 1 (odd)

then 1 × (-3) + 1 = -2

Hence, the sequence is

For the sequence, the starting number is -6

Observation: For the starting numbers tested, the sequences eventually reach the repeating cycle -2, -1, 4, 2, 1, -2.

Question 6. 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?

Answer:

Marks for 15 correct answers:

15 × 4 = 60

Anita's final score = 40

Marks lost due to incorrect answers:

60 - 40 = 20

Each incorrect answer reduces the score by 2 marks.

Number of incorrect answers:

20 ÷ 2 = 10

Total number of questions:

15 + 10 = 25

Therefore, Anita gave 10 incorrect answers, and there were 25 questions 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?

Answer:

Marks for 5 correct answers:

5 × 4 = 20

Anil's final score = -10

Therefore, marks lost due to incorrect answers:

20 - (-10) = 30

Each incorrect answer reduces the score by 2 marks.

Number of incorrect answers:

30 ÷ 2 = 15

So, Anil answered:

5 + 15 = 20 questions

From part (a), the test has 25 questions.

Unanswered questions:

25 - 20 = 5

Therefore, Anil had 15 incorrect answers and left 5 questions unanswered.

Question 7. Pick the pattern – find the operations done by the machine shown below:

Answer:

The pattern followed by the machine is:

First number - (Second number × Third number)

Checking the rows:

4 - [8 × (-3)] = 4 - (-24) = 28

6 - (9 × 6) = 6 - 54 = -48

2 - [3 × (-2)] = 2 - (-6) = 8

-9 - [5 × (-8)] = -9 - (-40) = 31

7 - [(-4) × (-6)] = 7 - 24 = -17

For the last row:

-16 - [(-6) × (-9)]

= -16 - 54

= -70

Therefore, the missing value is -70.

Question 8. 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 that denotes the temperature after 4 hours.

Answer:

Current temperature = 8°C

Temperature drop each hour = 5°C

In 4 hours, the total decrease is:

5 × 4

Therefore, the required expression is:

8 - (5 × 4)

Question 9. Find 3 consecutive numbers with a product of (a) -6, (b) 120.

Answer:

(a) Product = -6

Consider the consecutive integers:

-3, -2, -1

Their product is:

(-3) × (-2) × (-1)

= 6 × (-1)

= -6

Therefore, the required integers are -3, -2 and -1.

(b) Product = 120

Consider the consecutive integers:

4, 5, 6

Their product is:

4 × 5 × 6 = 120

Therefore, the required integers are 4, 5 and 6.

Question 10. 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? 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

Answer:

Use 5 coins of +13 pibs and 5 coins of -9 pibs:

5 × 13 + 5 × (-9)

= 65 - 45

= +20 pibs

(b) +40

Answer:

Use 10 coins of +13 pibs and 10 coins of -9 pibs:

10 × 13 + 10 × (-9)

= 130 - 90

= +40 pibs

Answer:

Use 10 coins of +13 pibs and 20 coins of -9 pibs:

10 × 13 + 20 × (-9)

= 130 - 180

= -50 pibs

(d) +8

Answer:

Use 2 coins of +13 pibs and 2 coins of -9 pibs:

2 × 13 + 2 × (-9)

= 26 - 18

= +8 pibs

(e) +10

Answer:

Use 7 coins of +13 pibs and 9 coins of -9 pibs:

7 × 13 + 9 × (-9)

= 91 - 81

= +10 pibs

(f) -2

Answer:

Use 13 coins of +13 pibs and 19 coins of -9 pibs:

13 × 13 + 19 × (-9)

= 169 - 171

= -2 pibs

(g) +1

Answer:

Use 7 coins of +13 pibs and 10 coins of -9 pibs:

7 × 13 + 10 × (-9)

= 91 - 90

= +1 pib

[Hint: Writing down a few multiples of 13 and 9 can help.]

(h) Is it possible to purchase an item that costs 1568 pibs?

Answer:

Yes.

Use 122 coins of +13 pibs and 2 coins of -9 pibs:

122 × 13 + 2 × (-9)

= 1586 - 18

= 1568 pibs

Therefore, it is possible to purchase an item costing 1568 pibs.

Question 11. Find the values of:

(a) (32 × (-18)) ÷ ((-36))

Answer:

(32 × (-18)) ÷ (-36)

= -576 ÷ (-36)

= 16

(b) (32) ÷ ((-36) × (-18))

Answer:

32 ÷ [(-36) × (-18)]

= 32 ÷ 648

= 4/81

Answer:

[25 × (-12)] ÷ [45 × (-27)]

= (-300) ÷ (-1215)

= 300/1215

= 20/81

(d) (280 × (-7)) ÷ ((-8) × (-35))

Answer:

[280 × (-7)] ÷ [(-8) × (-35)]

= -1960 ÷ 280

= -7

Question 12. Arrange the expressions given below in increasing order:

(a) (-348) + (-1064)

(b) (-348) – (-1064)

(d) (-348) × (-1064)

(e) 348 × (-1064)

(f) 348 × 964

Answer:

First, find the value of each expression:

(a) (-348) + (-1064) = -1412

(b) (-348) - (-1064) = -348 + 1064 = 716

(d) (-348) × (-1064) = 370272

(e) 348 × (-1064) = -370272

(f) 348 × 964 = 335472

Therefore, in increasing order:

(e) < (a) < (b) < (c) < (f) < (d)

-370272 < -1412 < 716 < 1412 < 335472 < 370272

Question 13. Given that (-548) × 972 = -532656, write the values of:

(a) (-547) × 972

Answer:

(-547) × 972

= (-548 + 1) × 972

= (-548 × 972) + 972

= -532656 + 972

= -531684

(b) (-548) × 971

Answer:

(-548) × 971

= (-548) × (972 - 1)

= (-548 × 972) - (-548)

= -532656 + 548

= -532108

Answer:

(-547) × 971

= (-548 + 1)(972 - 1)

= (-548 × 972) + 548 + 972 - 1

= -532656 + 1519

= -531137

Question 14. Given that 207 × (-33 + 7) = -5382, write the value of -207 × (33 – 7) = ________

Answer:

33 - 7 = 26

Therefore:

-207 × (33 - 7)

= -207 × 26

= -5382

Hence, the value is -5382.

Question 15. 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

Answer:

One expression that gives the maximum value is:

[(-2) - (3 + 5)] × (-6)

= (-2 - 8) × (-6)

= (-10) × (-6)

= 60

Therefore, the maximum possible value is 60.

(b) The result is the minimum possible value

Answer:

One expression that gives the minimum value is:

[3 - (-2) + 5] × (-6)

= (3 + 2 + 5) × (-6)

= 10 × (-6)

= -60

Therefore, the minimum possible value is -60.

Question 16. Fill in the blanks in at least 5 different ways with integers:

Answer:

(a) Five possible ways are:

(i) 0 + (-6 × 6) = -36

(ii) 4 + (-5 × 8) = -36

(iii) -4 + (-8 × 4) = -36

(iv) 12 + (-8 × 6) = -36

(v) -3 + [3 × (-11)] = -36

(b) Five possible ways are:

(i) (13 - 1) × 1 = 12

(ii) (10 - 4) × 2 = 12

(iii) (1 - 3) × (-6) = 12

(iv) (14 - 10) × 3 = 12

(v) (16 - 13) × 4 = 12

(i) 5 - (10 - 4) = -1

(ii) 0 - (3 - 2) = -1

(iii) -1 - (1 - 1) = -1

(iv) -5 - (0 - 4) = -1

(v) -10 - (-5 - 4) = -1

Learn Operations with Integers with Class 7 Maths Solutions Chapter 2

Understand the basic concepts and operations covered in Operations with Integers through clear explanations.
Use the NCERT Solutions for Class 7 Maths Chapter 2 to check your answers after solving the textbook questions yourself.
Follow each calculation step carefully to understand how integer operations work.
Practise different types of questions regularly to improve accuracy and confidence in Maths.
Use the NCERT Solutions for Class 7 Maths Chapter 2 PDF when revising the chapter before tests and examinations.
The NCERT Solutions for Class 7 Maths Chapter 2 PDF download option can be useful when you want to revisit the solutions during revision.
Refer to CBSE NCERT Solutions for Class 7 Maths Chapter 2 and Maths NCERT Solutions for Class 7 Chapter 2 to review important methods and textbook questions.

Important Study Material Links for Chapter 2: Operations with Integers

S. No	Important Study Material Links for Chapter 2: Operations with Integers
1	Class 7 Chapter 2: Operations with Integers Important Questions
2	Class 7 Chapter 2: Operations with Integers Revision Notes

NCERT Solutions for Class 7 Maths Ganita Prakash I and II (New Syllabus)

S. No	Chapter Name - Ganita Prakash I
1	Chapter 1 Large Numbers Around Us Solutions
2	Chapter 2 Arithmetic Expressions Solutions
3	Chapter 3 A Peek Beyond the Point Solutions
4	Chapter 4 Expressions Using Letter-Numbers Solutions
5	Chapter 5 Parallel and Intersecting Lines Solutions
6	Chapter 6 Number Play Solutions
7	Chapter 7 A Tale of Three Intersecting Lines Solutions
8	Chapter 8 Working with Fractions Solutions
S. No	Chapter Name -Ganita Prakash II
1	Chapter 1: Geometric Twins Solutions
2	Chapter 3: Finding Common Ground Solutions
3	Chapter 4: Another Peek Beyond the Point Solutions
4	Chapter 5: Connecting the Dots… Solutions
5	Chapter 6: Constructions and Tilings Solutions
6	Chapter 7: Finding the Unknown Solutions

Important Related Links for NCERT Class 7 Maths

S. No	Other CBSE Study Materials for Class 7 Maths
1	CBSE Class 7 Maths Revision Notes
2	CBSE Syllabus for Class 7 Maths
3	CBSE Class 7 Maths Important Questions
4	CBSE Class 7 Maths Sample Papers
5	NCERT Books for Class 7 Maths
6.	RS Aggarwal Class 7 Solutions for Maths book
7.	RD Sharma Class 7 Maths Solutions
8.	NCERT Class 7 Maths Exemplar Solutions
9.	NCERT Class 7 Maths Formulas

FAQs on NCERT Solutions for Class 7 Maths Chapter 2 - Operations with Integers

1. What is Class 7 Maths Chapter 2 Operations with Integers about?

Operations with Integers helps students understand how different mathematical operations are performed with positive and negative integers. The chapter develops calculation skills and helps students apply integer concepts while solving problems.

2. What should I focus on while studying Class 7 Maths Chapter 2?

Focus on understanding how integers behave under different operations rather than memorising rules. Work through the examples carefully and practise the textbook questions to become comfortable with calculations involving positive and negative numbers.

3. How should I solve questions from Operations with Integers?

Read the question carefully and identify the operation or concept involved. Apply the appropriate rule step by step and check the signs of the numbers before arriving at the final answer. After solving the question yourself, use the NCERT solution to verify your method.

4. Are the exercises in Class 7 Maths Chapter 2 important for exams?

Yes. The textbook exercises provide practice with the concepts introduced in the chapter. Solving them regularly can help students improve calculation accuracy and understand how integer operations are used in different types of questions.

5. Why are integers important in Class 7 Maths?

Integers are used to represent values such as gains and losses, increases and decreases, temperatures above and below zero, and positions on a number line. Understanding integer operations also provides a foundation for many later mathematical concepts.

6. How can I improve my accuracy while solving integer questions?

Write each step clearly and pay special attention to the signs of positive and negative numbers. Avoid doing too many calculations mentally, especially when a question involves multiple operations. Regular practice can also help reduce sign-related mistakes.

7. Can I download PDF solutions for Class 7 Maths Chapter 2 Operations with Integers?

Students can use the chapter solutions online and access the PDF option for convenient revision from Vedantu. The solutions can help students review textbook questions, check calculations, and prepare for school assessments.