NCERT Solutions for Class 7 Maths Chapter 1 Integers

Edited By Ramraj Saini | Updated on May 21, 2023 - 4:56 p.m. IST

NCERT Solutions for Class 7 Maths Chapter 1 Integers

NCERT Solutions for Class 7 Maths Chapter 1 Integers are discussed here. These NCERT solutions are created by expert team at Careers360 keeping in mind the latest CBSE syllabus 2023. These solutions are simple, comprehensive and cover step by step solutions of each problem. The first chapter of class 7 starts with recollecting the concepts of integers studied in the previous class of NCERT. The first exercise explained in the NCERT class 7 maths chapter 1 solutions are based on these recollected concepts. In NCERT solutions for class 7 maths chapter 1, you will study more about the integers, their properties, and operations. There are a total of 4 exercises with 30 questions in this NCER book chapter. All the questions are explained in the CBSE NCERT solutions for class 7 maths chapter 1.

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The NCERT solutions are prepared by expert teachers to help students. NCERT Class 7 maths chapter 1 solutions are important for a student. All the topics included in NCERT Class 7 Maths chapter 1 solutions are extremely important for higher classes as well. In this NCERT Solutions for Class 7 Maths Chapter 1, we will study important topics of NCERT solutions for class 7 like properties of addition, subtraction, multiplication, and division of Integers. Here you will get NCERT Class 7 Maths chapter 1 solutions to all four exercises.

NCERT Solutions for Class 7 Maths Chapter 1 Integers Topic 1.2

1. A number line representing integers is given below

1643863676143

–3 and –2 are marked by E and F respectively. Which integers are marked by B, D, H, J, M and O?

Answer: First, we complete the number line.

1643863701109

Now, the integers marked by:

B = -6

D = -4

H = 0

J = 2

M = 5

O = 7

2. Arrange 7, –5, 4, 0 and – 4 in ascending order and then mark them on a number line to check your answer.

Answer: The given number are: 7, –5, 4, 0 and – 4

Arranging them in ascending order (increasing order)

$-5,-4, 0 , 4, 7$

On a number line, as we move towards the right, the number increases.

3 . We have done various patterns with numbers in our previous class.
Can you find a pattern for each of the following? If yes, complete them:
(a) 7, 3, – 1, – 5, _____, _____, _____.
(b) – 2, – 4, – 6, – 8, _____, _____, _____.
(c) 15, 10, 5, 0, _____, _____, _____.
(d) – 11, – 8, – 5, – 2, _____, _____, _____.
Make some more such patterns and ask your friends to complete them.

Answer:

(a) 7, 3, – 1, – 5, $\underline{-9} ,\underline{ -13},\underline{ -17},\underline{ -21}$

The pattern is : $7-4 =3; 3-4 =-1\ and\ so\ on$

(b) – 2, – 4, – 6, – 8, $\underline{-10},\underline{ -12},\underline{ -14},\underline{ -16}$

The pattern is : $-2-2 =-4; -4-2 =-6\ and\ so\ on$

The pattern is : $15-10 =5; 10-5=5\ and\ so\ on$

(d) – 11, – 8, – 5, – 2, $\underline{1},\underline{ 4},\underline{ 7}, \underline{10}$

The pattern is : $-11+3 =-8; -8+3=-5\ and\ so\ on$

NCERT Solutions for Class 7 Maths Chapter 1 Integers Topic 1.3.5

1. Write a pair of integers whose sum gives
(a) a negative integer (b) zero
(c) an integer smaller than both the integers. (d) an integer smaller than only one of the integers.
(e) an integer greater than both the integers.

Answer:

(a) a negative integer : $-8 \ \&\ 3$
(b) zero : $-8 \ \&\ 8$
(c) an integer smaller than both the integers. $-8 \ \&\ -3$
(d) an integer smaller than only one of the integers. $-8 \ \&\ 3$
(e) an integer greater than both the integers. $8 \ \&\ 3$

2. Write a pair of integers whose difference gives
(a) a negative integer. (b) zero.
(c) an integer smaller than both the integers. (d) an integer greater than only one of the integers.
(e) an integer greater than both the integers.

Answer:

(a) a negative integer : $-8 \ \&\ 3 : -8 - 3 = -11$
(b) zero : $8 \ \&\ 8$
(c) an integer smaller than both the integers. $-8 \ \&\ 3$
(d) an integer smaller than only one of the integers. $-8 \ \&\ -3$
(e) an integer greater than both the integers. $8 \ \&\ -3$

NCERT Solutions for Class 7 Maths Chapter 1 Integers Topic 1.4.1

1. Find:
(i) 6 × (–19)
(ii) 12 × (–32)
(iii) 7 × (–22)

Answer:

Multiplying a positive integer and a negative integer, we multiply them as whole numbers and put a minus sign (–) before the product. We thus get a negative integer

(i) $6 \times (-19) = -(6\times19) = -114$

(ii) $12 \times (-32) = -(12\times32) = -384$

(iii) $7 \times (-22) = -(7\times22) = -154$

1. Find:
(a) 15 × (–16) (b) 21 × (–32)
(c) (– 42) × 12 (d) –55 × 15

Answer: We know, multiplication of a positive and negative integer is given by:

$a \times (- b) = (- a) \times b = - (a \times b)$

(a) $15 \times (-16) = -(15\times16) = -240$

(b) $21 \times (-32) = -(21\times32) = -672$

(d) $ï¿½55 \times 15 = -(55\times15) = -825$ $(-55)\times 15=-(55\times 15)=825$

2. Check if (a) 25 × (–21) = (–25) × 21 (b) (–23) × 20 = 23 × (–20)

Write five more such examples.

Answer: We know, when multiplying a positive integer and a negative integer, we multiply them as whole numbers and put a minus sign (–) before the product. We thus get a negative integer.

(a) $25 \times (-21) = (-25) \times 21$

L.H.S = $25 \times (-21) = -(25\times21) = -525$

R.H.S = $(-25) \times 21 = -(25\times21) = -525$

Therefore, L.H.S = R.H.S

(b) $(-23) \times 20 = 23 \times (-20)$

L.H.S = $(-23) \times 20 = -(23\times20) = -460$

R.H.S = $23 \times (-20) = -(23\times20) = -460$

Therefore, L.H.S = R.H.S

Five more examples:

$\\ 20 \times (-1) = (-20) \times 1 \\ 15 \times (-20) = (-15) \times 20 \\ 3 \times (-2) = (-3) \times 2 \\ (-10) \times 5= 10\times (-5) \\ (-50) \times 110= 50\times (-110)$

(i) Starting from (–5) × 4, find (–5) × (– 6)

Answer: Given,

$\\ (-5) \times 4 = -20$

$\\ (-5) \times 3 = -15 = -20+5 \\ (-5) \times 2 = -10 = -20 +10 \\ (-5) \times 1 = -5 = -20+15 \\ (-5) \times 0 = 0 = -20 + 20 \\ (-5) \times (-1) = 5 = -20+25 \\ (-5) \times (-2) = 10 = -20+30$

Therefore,

$\\ (-5) \times (-6) = -20+50=30$

(ii) Starting from (– 6) × 3, find (– 6) × (–7)

Answer: Given,

$(- 6) \times 3 = -18$

$\\ (-6) \times 2 = -12 = -18 +6 \\ (-6) \times 1 = -6 = -18+12 \\ (-6) \times 0 = 0 = -18 + 18 \\ (-6) \times (-1) = 6 = -18+24 \\ (-6) \times (-2) = 12 = -18+30 \\ \\ (-6) \times (-3) = 18 = -18+36$

Therefore,

$\\ (-6) \times (-7) = -18+60=42$

Q. Find: (–31) × (–100), (–25) × (–72), (–83) × (–28)

Answer: We know,

Multiplication of two negative integers : $(- a) \times (- b) = a \times b$

$\\ (-31) \times (-100) = 31 \times100 = 3100 \\ \\ (-25) \times (-72) = 25\times 72 = 1800 \\ \\ (-83) \times (-28) = 83\times 28 = 2324$

NCERT Soltions for Class 7 Maths Chapter 1 Integers Topic 1.4.3

(i) The product (–9) × (–5) × (– 6)×(–3) is positive whereas the product (–9) × ( –5) × 6 × (–3) is negative. Why?

Answer: We know,

If the number of negative integers in a product is even, then the product is a positive integer and if the number of negative integers in a product is odd, then the product is a negative integer.

Hence, the product $(-9) \times (-5) \times (- 6)\times(-3)$ is positive whereas the product $(-9) \times (-5) \times 6\times(-3)$ is negative

(ii) What will be the sign of the product if we multiply together:

(a) 8 negative integers and 3 positive integers?

(b) 5 negative integers and 4 positive integers?

(d) (–1), 2m times, m is a natural number?

Answer: We know,

And, any number of positive integers will always give a positive integer.

So, the sign of the product will be decided by the number of negative integers.

(a) 8 negative integers and 3 positive integers: Even number of negative integers, hence product is positive.

(b) 5 negative integers and 4 positive integers: Odd number of negative integers, hence product is negative.

(d) (–1), 2m times, m is a natural number: Even number of negative integers, hence product is positive.

NCERT Solutions for Class 7 Maths Chapter 1 Integers Topic 1.5.6

(i) Is 10 × [(6 + (–2)] = 10 × 6 + 10 × (–2)?

Answer:

L.H.S = $10 \times [(6 + (-2)] = 10\times4 =40$

R.H.S = $10 \times 6 + 10 \times (-2) = 60 + (-20) = 40$

Therefore,L.H.S = R.H.S

Hence, $10 \times [(6 + (-2)] = 10 \times [(6 + (-2)]$

(ii) Is (–15) × [(–7) + (–1)] = (–15) × (–7) + (–15) × (–1)?

Answer:

L.H.S = $(-15) \times [(-7) + (-1)] = (-15)\times(-8) = 120$

R.H.S = $(-15) \times (-7) + (-15) \times (-1) = [-(15 \times 7)] + [-(15 \times 1) ]=(105) + 15 = 120$

Therefore, L.H.S = R.H.S

Hence, $(-15) \times [(-7) + (-1)] = (-15) \times (-7) + (-15) \times (-1)$

(i) Is 10 × (6 – (–2)] = 10 × 6 – 10 × (–2)?

Answer:

L.H.S = $10 \times [6 - (-2)] = 10\times[6+2] = 10\times8 = 80$

R.H.S = $10 \times 6- 10 \times (-2) = 60 - (-20) = 60+20 = 80$

$\therefore$ L.H.S = R.H.S

(ii) Is (–15) × [(–7) – (–1)] = (–15) × (–7) – (–15) × (–1)?

Answer:

L.H.S = $(-15) \times [(-7) - (-1)] =(-15) \times [-7 +1]$

$= (-15) \times (-6) = -(15\times6) = -90$

R.H.S = $(-15) \times (-7) - (-15) \times (-1) = -(15\times7)-[-(15\times1)]$

$= -105+15 = -90$

Therefore, L.H.S = R.H.S

NCERT Solutions for Class 7 Maths Chapter 1 Integers Topic 1.5.7

Q. Find (– 49) × 18; (–25) × (–31); 70 × (–19) + (–1) × 70 using distributive property.

Answer: We know,

Distributive law: for any integers a, b and c, $a \times (b + c) = a \times b + a \times c$

$\\ (- 49) \times 18 = (-50+1)\times18 = (-50)\times18 + 1\times18 \\ = -900 + 18 = -782 \\ \\ (-25) \times (-31) = (-25)\times((-30) + (-1)) \\ = (-25)\times(-30) + (-25)\times(-1) = 750+25 = 725 \\ \\ 70 \times (-19) + (-1) \times 70 = 70\times((-19)+ (-1)) \\ = 70\times(-20) = -1400$

NCERT Solutions for Class 7 Maths Chapter 1 Integers Topic 1.6

Find:
(a) (–100) ÷ 5 (b) (–81) ÷ 9

Answer: When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put a minus sign (–) before the quotient

(a) $(-100) \div 5 = -\frac{100}{5} = -20$

(b) $(-81) \div 9 = -(\frac{81}{9} )= -9$

(d) $(-32) \div 2 = -(\frac{32}{2} )= -16$

Find: (a) 125 ÷ (–25) (b) 80 ÷ (–5) (c) 64 ÷ (–16)

Answer: When we divide a positive integer by a negative integer, we first divide them as whole numbers and then put a minus sign (–) before the quotient.

(a) $125 \div (-25) = -\left (\frac{125}{25} \right ) = -5$

(b) $80\div (-5) = -\left (\frac{80}{5} \right ) = -16$

Find: (a) (–36) ÷ (– 4) (b) (–201) ÷ (–3) (c) (–325) ÷ (–13)

Answer: When we divide a negative integer by a negative integer, we first divide them as whole numbers and then put a positive sign (+).

(a)

$\\ (-36) \div (- 4) \\ = +\left (\frac{36}{4} \right ) \\ = 9$

(b)

$\\ (-201) \div (- 3) \\ = +\left (\frac{201}{3} \right ) \\ = 67$

(c)

$\\ (-325) \div (- 13) \\ = +\left (\frac{325}{13} \right ) \\ = 25$

NCERT Solutions for Class 7 Maths Chapter 1 Integers Topic 1.7

Q. Is (i) 1 ÷ a = 1?
(ii) a ÷ (–1) = – a? for any integer a.
Take different values of a and check.

Answer: (i) $1 \div a = 1$ is true only for $a =1$ .

(ii) $a \div (-1) = - a$

Taking, $a =1; 1\div (-1) = -1$

$a =2; 2\div (-1) = -2$

$a =-1; (-1)\div (-1) = 1$

Hence, this is true for every integer.

NCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.1

1. Following number, line shows the temperature in degree celsius (°C) at different places on a particular day.

1643863844004

(a) Observe this number line and write the temperature of the places marked on it.

(b) What is the temperature difference between the hottest and the coldest places among the above?

(d) Can we say the temperature of Srinagar and Shimla taken together is less than the temperature at Shimla? Is it also less than the temperature at Srinagar?

Answer:

(a) The temperature of the places in degree Celsius (°C) marked on it are :

1643863874372

Places	Temperature
Lahaul Spiti	$-8^{\circ} C$
Srinagar	$-2^{\circ} C$
Shimla	$5^{\circ} C$
Ooty	$14^{\circ} C$
Bangalore	$22^{\circ} C$

(b) The hottest temperature is $22^{\circ} C$ and coldest is $-8^{\circ} C$

The temperature difference between them = $[22-(-8)]^{\circ} C = (22+8)^{\circ} C = 30^{\circ} C$

The temperature difference between Lahulspiti and Srinagar is = $[(-2)-(-8)]^{\circ} C = (-2+8)^{\circ} C = 6^{\circ} C$

(d) The temperatures at Srinagar is $-2^{\circ} C$ and Shimla is $5^{\circ} C$

The temperature of Srinagar and Shimla took together = $[(-2)+5]^{\circ} C = 3^{\circ} C$

which is less than the temperature of Shimla.

But, it is not less than the temperature at Srinagar.

2. In a quiz, positive marks are given for correct answers and negative marks are given for incorrect answers. If Jack’s scores in five successive rounds were 25, – 5, – 10, 15 and 10, what was his total at the end?

Answer: Given, Jack’s scores in five successive rounds are $25, - 5, - 10, 15\ and\ 10$

Therefore, his total score = $\\ 25 +(- 5) +(- 10)+ 15+10$

$\\ = 25 - 5 - 10+ 15+10 \\ = 35$

His total score at the end was $35$

3. At Srinagar temperature was – 5°C on Monday and then it dropped by 2°C on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4°C. What was the temperature on this day?

Answer: The temperature of Srinagar on Monday = $-5^{\circ} C$

According to question,

The temperature of Tuesday = $[(-5)-2]^{\circ} C$

$\therefore$ The temperature of Srinagar on Tuesday = $[(-5)-2]^{\circ} C = -8^{\circ} C$

Again, according to question

The temperature of Wednesday= $[(-8)+4]^{\circ} C$

$\therefore$ The temperature of Srinagar on Wednesday = $[(-8)+4]^{\circ} C = -4^{\circ} C$

4. A plane is flying at the height of 5000 m above the sea level. At a particular point, it is exactly above a submarine floating 1200 m below the sea level. What is the vertical distance between them?

Answer: The plane is flying at the height of 5000 m above the sea level

The distance of the plane from sea level = $5000\ m$

Also, the submarine floating 1200 m below the sea level

The distance of the submarine from sea level = $-1200\ m$

Distance between them = $[5000-(-1200)]\ m$

$6200\ m$

5. Mohan deposits Rs 2,000 in his bank account and withdraws Rs 1,642 from it, the next day. If withdrawal of amount from the account is represented by a negative integer, then how will you represent the amount deposited? Find the balance in Mohan’s account after the withdrawal.

Answer: Given,

The amount deposited in the bank account = $Rs.\ 2000$

Amount withdrawn from bank account is= $Rs.\ 1642$

Balance in account = Amount deposited in bank account + Amount withdrawn from the bank account

$\\ = 2000 + (-1642) \\ = 2000 - 1642 \\ = 358$

Therefore, the balance in Mohan’s account after the withdrawal is $Rs.\ 358$

6. Rita goes 20 km towards east from a point A to the point B. From B, she moves 30 km towards west along the same road. If the distance towards east is represented by a positive integer then, how will you represent the distance travelled towards west? By which integer will you represent her final position from A? 1643863911868

Answer: We will represent the distance travelled towards west with negative integer.

Distance travelled towards east = $20\ km$

Distance travelled towards west = $30\ km$

Distance travelled from A = $[20+(-30)]\ km$

$-10\ km$

As the distance is negative, so final position of Rita from A is towards the west direction.

7(i). In a magic square each row, column and diagonal have the same sum. Check if the following is a magic square.

5	-1	-4
-5	-2	7
0	3	-3

Answer: Taking Rows-

$5-1-4 = 0$

$-5-2+7 = 0$

$0+3-3 = 0$

Taking Columns-

$5-5+0 = 0$

$-1-1+3 = 0$

$-4+7-3 = 0$

Taking Diagonals-

$-4-2+0 = -6$

$5-2-3 = 0$

As the sum of one of the diagonals is not equal to 0, it is not a magic square.

7 (ii). In a magic square each row, column and diagonal have the same sum. Check if the following is a magic square.

1	-10	0
-4	-3	-2
-6	4	-7

Answer: Taking Rows-

$1-10+0 = -9$

$-4-3-2 = -9$

$-6+4-7= -9$

Taking Columns-

$1-4-6=-9$

$-10-3+4 = -9$

$0-2-7= -9$

Taking Diagonals-

$0-3-6= -9$

$1-3-7=-9$

As the sum of all the rows, columns and diagonals are equal, this is a magic square.

8. Verify $a - (-b) = a + b$ for the following values of a and b.

(i) $a = 21, b = 18$

(ii) $a = 118, b = 125$

(iii) $a = 75, b = 84$

(iv) $a = 28, b = 11$

Answer:

(i) $a = 21, b = 18$

L.H.S = $a-(-b) = 21-(-18)$

$= 21+18 = 39$

R.H.S = $a+b= 21+18 = 39$

$\therefore$ L.H.S = R.H.S

Hence verified.

(ii) $a = 118, b = 125$

L.H.S = $a-(-b) = 118-(-125)$

$= 118+125 = 243$

R.H.S = $a+b= 118+125 = 243$

$\therefore$ L.H.S = R.H.S

Hence verified.

(iii) $a = 75, b = 84$

L.H.S = $a-(-b) = 75-(-84)$

$= 75+84 = 159$

R.H.S = $a+b= 75+84 = 159$

$\therefore$ L.H.S = R.H.S

Hence verified.

(iv) $a = 28, b = 11$

L.H.S = $a-(-b) = 28-(-11)$

$= 28+11 = 39$

R.H.S = $a+b= 28+11 = 39$

$\therefore$ L.H.S = R.H.S

Hence verified.

9. Use the sign of >, < or = in the box to make the statements true.

(a) (– 8) + (– 4) 1643864041199 (–8) – (– 4)

(b) (– 3) + 7 – (19) 1643864040488 15 – 8 + (– 9)

(d) 39 + (– 24) – (15) 1643864040883 36 + (– 52) – (– 36)

(e) – 231 + 79 + 51 1643864040124 –399 + 159 + 81

Answer:

(Note: On a number line, the number on the right is greater than a number towards its left.)

(a) (– 8) + (– 4) 1643864219651 (–8) – (– 4)

L.H.S = (- 8) + (- 4) = -8-4 = -12

R.H.S = (- 8) - (- 4) = -8+4 = -4

Clearly, R.H.S > L.H.S (b) (– 3) + 7 – (19) 1643864219312 15 – 8 + (– 9)

L.H.S = (- 3) + 7 - (19) = 4-19 = -15

R.H.S = 15 - 8 + (- 9) = 7-9= -2

Clearly, R.H.S > L.H.S

L.H.S = 23 - 41 + 11 = -18+11= -7

R.H.S = 23 - 41 - 11 = -18-11= -29

Clearly, L.H.S > R.H.S

(d) 39 + (– 24) – (15) 1643864218313 36 + (– 52) – (– 36)

L.H.S = 39 + (- 24) - (15) = 39-24-15 = 15-15 = 0

R.H.S = 36 + (- 52) - (- 36) = 36-52+36 = -16+36 = 20

Clearly, R.H.S > L.H.S

(e) – 231 + 79 + 51 1643864218659 –399 + 159 + 81

L.H.S = - 231 + 79 + 51 = -152+51 =-101

R.H.S = -399 + 159 + 81=-240+81 = -159

Clearly, L.H.S > R.H.S

10.(i) A water tank has steps inside it. A monkey is sitting on the topmost step (i.e., the first step). The water level is at the ninth step.

He jumps 3 steps down and then jumps back 2 steps up. In how many jumps will he reach the water level?

Answer: Let the number of steps moved down be represented by positive integers and the number of steps moved up by negative integers.

Given,

Monkey is at the step = 1

Now,

The step after:

1 ^st jump = $1 + 3 = 4$

2 ^nd jump = $4 + (-2) = 2$

3 ^rd jump = $2 + 3 = 5$

4 ^th jump = $5 + (-2) = 3$

5 ^th jump = $3 + 3 = 6$

6 ^th jump = $6 + (-2) = 4$

7 ^th jump = $4 + 3 = 7$

8 ^th jump = $7 + (-2) = 5$

9 ^th jump = $5 + 3 = 8$

10 ^th jump = $8 + (-2) = 6$

11 ^th jump = $6 + 3 = 9$

Therefore, after 11 jumps, the monkey reaches the water level which is at the ninth step.

10.(ii) A water tank has steps inside it. A monkey is sitting on the topmost step (i.e., the first step). The water level is at the ninth step.

After drinking water, he wants to go back. For this, he jumps 4 steps up and then jumps back 2 steps down in every move. In how many jumps will he reach back the top step?

Answer: Let the number of steps moved down be represented by positive integers and the number of steps moved up by negative integers.

Given, Monkey is at the step = 9

Now,

The step after:

1 ^st jump = $9 + (-4) = 5$

2 ^nd jump = $5 + 2 = 7$

3 ^rd jump = $7 + (-4) = 3$

4 ^th jump = $3 + 2 = 5$

5 ^th jump = $5 + (-4) = 1$

Therefore, after 5 jumps, the monkey will reach back at the top.

10.(iii) A water tank has steps inside it. A monkey is sitting on the topmost step (i.e., the first step). The water level is at the ninth step.

If the number of steps moved down is represented by negative integers and the number of steps moved up by positive integers, represent his moves in part (i) and (ii) by completing the following; (a) – 3 + 2 – ... = – 8 (b) 4 – 2 + ... = 8. In (a) the sum (– 8) represents going down by eight steps. So, what will the sum 8 in (b) represent?

Answer: If we represent the down steps by negative and up steps by positive then:

Moves in (i):

$- 3 + 2 - 3 + 2 - 3 + 2 - 3 + 2 - 3 + 2 - 3 = -8$

Also,

Moves in part (ii):

$4 - 2 + 4 - 2 + 4 = 8$

Therefore, the sum +8 represents going up by eight steps.

NCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.2

1. Write down a pair of integers whose:
(a) sum is –7 (b) difference is –10 (c) sum is 0

Answer: A pair of integers whose:

(a) sum is –7 : $-8\ \&\ 1$

(b) difference is –10 : $-8\ \&\ 2 : -8-2= -10$

2. (a) Write a pair of negative integers whose difference gives 8.
(b) Write a negative integer and a positive integer whose sum is –5.
(c) Write a negative integer and a positive integer whose difference is –3.

Answer:

(a) A pair of negative integers whose difference gives 8 :

$-8 \ \& -16 :$

$= -8-(-16) = -8+16 = 8$

(b) A negative integer and a positive integer whose sum is –5:

$-8\ \&\ 3$ :

$-8 + 3 = -5$

$-1\ \&\ 2$

$-1 -(2) = -1-2 = -3$

3. In a quiz, team A scored – 40, 10, 0 and team B scored 10, 0, – 40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?

Answer: Given,

Team A scored : $-40, 10, 0$

$\therefore$ Team A's total = $-40+ 10+ 0 = -30$

Team B scored : $10, 0, - 40$

$\therefore$ Team B's total = $10+ 0+ (- 40)= 10-40 =-30$

Therefore, both the team scored same.

Yes, we can add integers in any order.

4. Fill in the blanks to make the following statements true:
(i) (–5) + (– 8) = (– 8) + (............)
(ii) –53 + ............ = –53
(iii) 17 + ............ = 0
(iv) [13 + (– 12)] + (............) = 13 + [(–12) + (–7)]
(v) (– 4) + [15 + (–3)] = [– 4 + 15] + ............

Answer:

(i) (–5) + (– 8) = (– 8) + ( $\underline{-5}$ )

By the commutative property $a+b=b+a$

(ii) –53 + $\underline{0}$ = –53

$0$ is the additive identity. The number added to $0$ gives the same number.
(iii) 17 + $\underline{(-17)}$ = 0

By additive identity property.
(iv) [13 + (– 12)] + ( $\underline{-7}$ ) = 13 + [(–12) + (–7)]

By associative property $(a+b)+c=a+(b+c)$
(v) (– 4) + [15 + (–3)] = [– 4 + 15] + $\underline{(-3)}$

By associative property $(a+b)+c=a+(b+c)$

NCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.3

1. Find each of the following products:

(a) 3 × (–1) (b) (–1) × 225
(c) (–21) × (–30) (d) (–316) × (–1)
(e) (–15) × 0 × (–18)

Answer:

(a) $3 \times (-1) = -3$

(b) $(-1) \times 225 = -225$

Product of two negative integers is a positive integer

(d) $(-316) \times (-1) = (316\times1) = 316$

(e) $(-15) \times 0 \times (-18)= [(-15) \times 0] \times (-18) = 0\times (-18) = 0$

1. Find each of the following products:

(f) (–12) × (–11) × (10) (g) 9 × (–3) × (– 6)

(h) (–18) × (–5) × (– 4) (i) (–1) × (–2) × (–3) × 4

(j) (–3) × (–6) × (–2) × (–1)

Answer:

Product of even number of negative integers is an even integer.

(f) $\\ (-12) \times (-11) \times (10) =[ -(12\times11)]\times10 = (-132)\times10 = -1320$

(g) $9 \times (-3) \times (- 6) =9 \times [(-3) \times (- 6)] = 9 \times18 = 162$

(h) $(-18) \times (-5) \times (-4) = (-18) \times [(-5) \times (-4)] = (-18) \times20 = -360$

(i)

$\\ (- 1)\times (-2) \times (-3) \times 4 = [(-1)\times (-2) ]\times [(-3) \times 4] \\ = 2\times (-12) = -24$

(j)

$\\ (-3) \times (-6) \times (-2) \times (-1) = [(-3) \times (-6) ]\times [(-2) \times (-1)] \\ = 18 \times2 = 36$

2 (a). Verify the following:

18 × [7 + (–3)] = [18 × 7] + [18 × (–3)]

Answer:

L.H.S = $18 \times [7 + (-3)] = 18\times4 = 72$

R.H.S = $[18 \times 7] + [18 \times (-3)] = 126 + (-54) = 72$

L.H.S = R.H.S

Hence, verified.

2 (b). Verify the following:

[(–21) × [(– 4) + (– 6)] = [(–21) × (– 4)] + [(–21) × (– 6)]

Answer:

L.H.S =

$(-21) \times [(- 4) + (- 6)] = (-21)\times(-10) = 210$

R.H.S

$\\ \ [(-21) \times (-4)] + [(-21) \times (- 6)] = (21\times 4) + (21\times 6) \\ =84+126 = 210$

Therefore, L.H.S = R.H.S

Hence verified.

3. (i) For any integer a, what is (–1) × a equal to?
(ii) Determine the integer whose product with (–1) is
(a) –22 (b) 37 (c) 0

Answer:

(i) For any integer a, $(-1) \times a = -a$ , i.e, the additive inverse of the given integer.

(ii) The integer whose product with $(-1)$ gives the following are:
$\\ (a)\ -22 : 22 \\ (b)\ 37 : -37 \\ (c)\ 0 : 0$

4. Starting from (–1) × 5, write various products showing some pattern to show (–1) × (–1) = 1.

Answer:

Given,

$\\ - 1 \times 5 = - 5$

$\\ - 1 \times 4 = - 4 = - 5 + 1 \\ - 1 \times 3 = -3 = - 4 + 1 \\ - 1 \times 2 =\ \ 2 = - 3 + 1 \\ - 1 \times 1 = - 1 = - 2 + 1 \\ - 1 \times 0 = \ \ 0 = - 1 + 1$

Therefore,

$- 1 \times (-1) = 0 + 1 = 1$

5. Find the product, using suitable properties:
(a) 26 × (– 48) + (– 48) × (–36) (b) 8 × 53 × (–125)
(c) 15 × (–25) × (– 4) × (–10) (d) (– 41) × 102
(e) 625 × (–35) + (– 625) × 65 (f) 7 × (50 – 2)
(g) (–17) × (–29) (h) (–57) × (–19) + 57

Answer: Using suitable properties of multiplication:

(a)

$\\ 26 \times (- 48) + (- 48) \times (-36) \\ = (- 48) \times 26 + (- 48) \times (-36)\ \ [a\times b = b\times a] \\ = (- 48) \times [26 + (-36)] \ \ [a\times b + a\times c = a\times(b+c)] \\ = (-48)\times(-10) = 480$

(b)

$\\ 8 \times 53 \times (-125) \\ = [8 \times 53 ]\times (-125) \ \[a\times b = b\times a] \\ = 53 \times[ 8 \times (-125)] \ \ a\times(b\times c) =( a\times b)\times c \\ = 53\times(-1000) = -53000$

(c)

$\\ 15 \times (-25) \times (- 4) \times (-10) \\ = 15 \times [(-25) \times (- 4)] \times (-10) \\ = 15 \times 100\times (-10) \\ = -15000$

(d)

$\\ (- 41) \times 102 \\ =(- 41) \times(100+2) \\ =(- 41) \times100+(- 41) \times2\ \ [a\times(b+c)=(a\times b) + (a\times c)] \\ = (-4100) + (-82) \\ = -4100-82 = -4182$

(e)

$\\625 \times (-35) + (- 625) \times 65 \\ = (-625) \times 35 + (- 625) \times 65 \ \[a\times b = b\times a] \\ = (-625) \times (35 +65) \ \ [(a\times b) + (a\times c) = a\times(b+c)] \\ = (-625) \times (100) = -62500$

(f)

$\\ 7 \times (50 - 2) \\ = (7 \times 50) - (7 \times2) \ \ [a\times(b-c)=(a\times b) - (a\times c)] \\ = 350 -14 = 336$
(g)

$\\ (-17) \times (-29) \\ =(-17) \times (-30 + 1) \\ = (-17) \times (-30) +(-17) \times 1 \ \ [a\times(b+c)=(a\times b) + (a\times c)] \\ = 510 -17 = 493$

(h)

$\\ (-57) \times (-19) + 57 \\ = (-57) \times (-19) + (-57) \times(-1) \\ = (-57) \times [(-19) +(-1)] \ \ [(a\times b) + (a\times c) = a\times(b+c)] \\ = (-57) \times (-20) = 1140$

6. A certain freezing process requires that room temperature be lowered from 40°C at the rate of 5°C every hour. What will be the room temperature 10 hours after the process begins?

Answer: Given,
Initial temperature = $40^{\circ}C$

Also, Change in temperature per hour = $-5^{\circ}C$

Therefore,Change in temperature after 10 hours = $(-5)\times10 = -50^{\circ}C$

Hence, Final temperature = $40^{\circ}C + (-50^{\circ}C) = -10^{\circ}C$

7 (i). In a class test containing 10 questions, 5 marks are awarded for every correct answer and (–2) marks are awarded for every incorrect answer and 0 for questions not attempted.

(i) Mohan gets four correct and six incorrect answers. What is his score?

Answer: Given,

Total questions = $10$

Marks for correct answer = $+5$

Marks for incorrect answers = $-2$

Marks for unattempted question = $0$

According to question,

Mohan gets four correct and six incorrect answers

Total marks = $4(+5)+6(-2) = 20 -12 = 8$

Therefore, Mohan scored 8 marks.

7(ii). In a class test containing 10 questions, 5 marks are awarded for every correct answer and (–2) marks are awarded for every incorrect answer and 0 for questions not attempted.

(ii) Reshma gets five correct answers and five incorrect answers, what is her score?

Answer: Given,

Total questions = $10$

Marks for correct answer = $+5$

Marks for incorrect answers = $-2$

Marks for unattempted question = $0$

According to question,

Reshma gets five correct answers and five incorrect answers

Total marks = $5(+5)+5(-2) = 25 -10 = 15$

Therefore, Reshma scored 15 marks.

7 (iii). In a class test containing 10 questions, 5 marks are awarded for every correct answer and (–2) marks are awarded for every incorrect answer and 0 for questions not attempted.

(iii) Heena gets two correct and five incorrect answers out of seven questions she attempts. What is her score?

Answer: Given,

Total questions = $10$

Marks for correct answer = $+5$

Marks for incorrect answers = $-2$

Marks for unattempted question = $0$

According to question,

Heena gets two correct and five incorrect answers.

Therefore, number of unattempted questions = $10 -7 =3$

Total marks = $2(+5)+5(-2)+3(0) = 10 -10+0 = 0$

Therefore, Heena scored 0 marks.

8 (a). A cement company earns a profit of Rs. 8 per bag of white cement sold and a loss of Rs. 5 per bag of grey cement sold.

(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: Given,

Profit earned by selling 1 bag of white cement = $Rs.\ 8$

Profit earned by selling 3000 bags of white cement= $Rs.\ (3000\times8) = Rs.\ 24000$

Loss on 1 bag of grey cement = $-Rs.\ 5$

Loss of 5000 bags of grey cement = $-Rs.\( 5000\times5) = -Rs.\ 25000$

Total earnings = $Rs.\ 24000 -Rs.\ 25000 = -Rs.\ 1000$

This means that the company incurred a loss of $Rs.\ 1000$

8 (b). A cement company earns a profit of Rs. 8 per bag of white cement sold and a loss of Rs. 5 per bag of grey cement sold.

(b) What is the number of white cement bags it must sell to have neither profit nor loss, if the number of grey bags sold is 6,400 bags.

Answer: Given,

Profit earned by selling 1 bag of white cement = $Rs.\ 8$

Loss on 1 bag of grey cement = $-Rs.\ 5$

$\therefore$ Loss on 6400 bags of grey cement = $-Rs.\( 6400\times5) = -Rs.\ 32000$

To be neither in profit nor loss, the profit from white cement must be $Rs.\ 32000$

$\therefore$ Number of white cement bags to be sold = $Rs.\ 32000\div Rs.\ 8$

$\\ = \frac{32000}{8} \\ = 4000$

Therefore the company has to sell 4000 white cement bags.

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

Answer:

(a)Given, $(-3) \times \_\_\_\_\ = 27$

The product is positive, therefore there should be even number of negative integers.

$(-3) \times \underline{(-9)}= 27$

(b) Given, $5 \times \_\_\_\_\_ \ = -35$
The product is negative, therefore there should be odd number of negative integers.

$5 \times \underline{(-7)} = -35$

The product is negative, therefore there should be odd number of negative integers.

$\underline{7}\times (- 8) = -56$

(d) $\_\_\_\ \times (-12) = 132$

The product is positive, therefore there should be even number of negative integers.

$\underline{(-11)} \times (-12) = 132$

NCERT Soltions for Class 7 Maths Chapter 1 Integers Exercise 1.4

1. Evaluate each of the following:
(a) (–30) ÷ 10 (b) 50 ÷ (–5) (c) (–36) ÷ (–9)
(d) (– 49) ÷ (49) (e) 13 ÷ [(–2) + 1] (f ) 0 ÷ (–12)
(g) (–31) ÷ [(–30) + (–1)]

(h) [(–36) ÷ 12] ÷ 3 (i) [(– 6) + 5)] ÷ [(–2) + 1]

Answer: Points to keep in mind:

When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put a minus sign (–) before the quotient.

When we divide a positive integer by a negative integer, we first divide them as whole numbers and then put a minus sign (–) before the quotient.

When we divide a negative integer by a negative integer, we first divide them as whole numbers and then put a positive sign (+).

(a) $(-30) \div 10 = -\left (\frac{30}{10} \right ) = -3$

(b) $50 \div (-5) = -\left (\frac{50}{5} \right ) = -10$

(d) $(-49) \div 49 = -\left (\frac{49}{49} \right ) = -1$

(e) $13 \div [(-2) + 1] = 13 \div (-1)= -\left (\frac{13}{1} \right ) = -13$

(f ) $0 \div (-12)= -\left (\frac{0}{12} \right ) = 0$

(g) $(-31) \div [(-30) + (-1)] = (-31) \div (-31)= +\left (\frac{31}{31} \right ) = 1$

(h)

$\\ \ [(-36) \div 12] \div 3 = \left[-\left (\frac{36}{12} \right ) \right ]\div 3 \\ = (-3) \div 3= -\left (\frac{3}{3} \right ) = -1$

(i)

$\\ \ [(- 6) + 5] \div [(-2) + 1] = \\ = (-1) \div (-1)= +\left (\frac{1}{1} \right ) = 1$

2. Verify that $a \div (b + c) \neq (a\div b)+ (a\div c)$ for each of the following values of a , b and c .
(a) a = 12, b = – 4, c = 2 (b) a = (–10), b = 1, c = 1

Answer:

$a \div (b + c) \neq (a\div b)+ (a\div c)$

(a) a = 12, b = – 4, c = 2

L.H.S = $\\ a \div (b + c)$

$\\ = 12 \div [(-4)+2] \\ = 12 \div (-2) \\ = - \left (\frac{12}{2} \right ) \\ = -6$

R.H.S = $(a\div b)+ (a\div c)$

$\\ = [12\div (-4)]+ (12\div 2) \\ = \left [-\left (\frac{12}{4} \right ) \right ] + \left (\frac{12}{2} \right ) \\ = (-3)+6 \\ = 3$

Therefore. $L.H.S \neq R.H.S$

Hence verified.

(b) a = (–10), b = 1, c = 1

L.H.S = $\\ a \div (b + c)$

$\\ = (-10) \div [1+1] \\ = (-10) \div 2 \\ = - \left (\frac{10}{2} \right ) \\ = -5$

R.H.S = $(a\div b)+ (a\div c)$

$\\ = [(-10)\div 1]+ ((-10)\div 1) \\ = \left [-\left (\frac{10}{1} \right ) \right ] + \left [-\left (\frac{10}{1} \right ) \right ] \\ = (-10)+(-10) \\ = -20$

Therefore. $L.H.S \neq R.H.S$

Hence verified.

3. Fill in the blanks:
(a) 369 ÷ _____ = 369 (b) (–75) ÷ _____ = –1
(c) (–206) ÷ _____ = 1 (d) – 87 ÷ _____ = 87
(e) _____ ÷ 1 = – 87 (f) _____ ÷ 48 = –1
(g) 20 ÷ _____ = –2 (h) _____ ÷ (4) = –3

Answer:

(a) Given, $369 \div \underline{\ \ \ \ \ \ } = 369$

A number divided by 1 gives the number itself.

$369 \div \underline{1} = 369$

(b) $(-75) \div \underline{\ \ \ \ \ } = -1$

The product is negative, therefore there must be odd number of negative integers.

$(-75) \div \underline{75} = -1$

(c) $(-206) \div \underline{\ \ \ \ }= 1$

A number divided by itself gives 1

$(-206) \div \underline{(-206)}= 1$

(d) $(-87) \div \underline{\ \ \ \ }= 87$

The product is positive, therefore there must be even number of negative integers.

$(-87) \div \underline{(-1) }= 87$

(e) $\underline{\ \ \ \ } \div 1 = - 87$

A number divided by 1 gives the number itself.

$\underline{(-87)} \div 1 = - 87$

(f) $\underline{\ \ \ \ \ } \div 48 = -1$

The product is negative, therefore there must be odd number of negative integers.

$\underline{(-48)} \div 48 = -1$

(g) $20\div \underline{\ \ \ \ \ } = -2$

The product is negative, therefore there must be odd number of negative integers.

$20\div \underline{(-10)} = -2$

(h) $\underline{\ \ \ \ \ } \div (4) = -3$

The product is negative, therefore there must be odd number of negative integers.

$\underline{\left (\frac{-4}{3} \right ) } \div (4) = -3$

4. Write five pairs of integers (a, b) such that a ÷ b = –3. One such pair is (6, –2) because 6 ÷ (–2) = (–3).

Answer:

Five pairs of integers $(a, b)$ such that $a \div b = -3$ are:

$(-6,2); (9,-3); (-9,3) ; (3,-1) ; (-3,1)$

5. The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until midnight, at what time would the temperature be 8°C below zero? What would be the temperature at mid-night?

Answer: Given,

The temperature at 12 noon = $10^{\circ}C$

Final temperature = $-8^{\circ}C$

$\therefore$ The decrease in temperature = $10 - (-8) = 18^{\circ}C$

Time taken for the temperature to decrease by $2^{\circ}C$ = $1\ hour$

$\therefore$ Time taken for the temperature to decrease by $18^{\circ}C$ =

$=\frac{1}{2}\times18\ hour$

$= 9\ hours$

Now,

Time until midnight = $12\ hours$

$\therefore$ The decrease in temperature in $12\ hours$ = $2\times12 = 24^{\circ}C$

Therefore, the temperature at midnight = $(10 - 24)^{\circ}C = -14^{\circ}C$

Therefore, the temperature at mid-night will be $14^{\circ}C$ below zero.

6. In a class test (+ 3) marks are given for every correct answer and (–2) marks are given for every incorrect answer and no marks for not attempting any question. (i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly? (ii) Mohini scores –5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?

Answer: Given,

Marks for every correct answer = $+3$

Marks for every wrong answer = $- 2$

(i) According to question,

Marks obtained by Radhika = $20$

Number of correct answers = $12$

$\therefore$ Marks obtained for correct answers = $12\times3 = 36$

$\therefore$ Marks obtained for incorrect answers = Total marks – Marks obtained for correct answers

$\\ = 20 - 36 \\= - 16$

$\therefore$ Number of incorrect answers = $(-16)\div (-2)$

$\\ = \frac{16}{2} \\ = 8$

Therefore, Radhika attempted 8 questions incorrectly.

(ii)

According to question,

Marks obtained by Mohini = $-5$

Number of correct answers = $7$

$\therefore$ Marks obtained for correct answers = $7\times3 = 21$

$\therefore$ Marks obtained for incorrect answers = Total marks – Marks obtained for correct answers

$\\ = (-5) - 21 \\= - 26$

$\therefore$ Number of incorrect answers = $(-26)\div (-2)$

$\\ = \frac{26}{2} \\ = 13$

Therefore, Mohini attempted 13 questions incorrectly.

7. An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m.

Answer: Given,

Initial height = $10\ m$

Final depth = $-350\ m$

$\therefore$ Total distance the elevator has to descend = Final position - Initial position

= $|(-350) - 10| = 360 \ m$

Also,

Time the elevator takes to descend $6\ m$ = $1\ min$

$\therefore$ Time the elevator takes to descend $360\ m$ =

$\frac{1}{60}\times360\ min = 60\ min$

$= 1\ hour$

Integers Class 7 Maths Chapter 1- Topics

1.1 Introduction
1.2 Recall
1.3 Properties of Addition and Subtraction of Integers
1.3.1 Closure under Addition
1.3.2 Closure under Subtraction
1.3.3 Commutative Property
1.3.4 Associative Property
1.3.5 Additive Identity
1.4 Multiplication of Integers
1.4.1 Multiplication of a Positive and a Negative Integers
1.4.2 Multiplication of two Negative Integers
1.4.3 Product of three or more Negative Integers
1.5 Properties of Multiplication of Integers
1.5.1 Closure under Multiplication
1.5.2 Commutativity of Multiplication
1.5.3 Multiplication by Zero
1.5.4 Multiplicative Identity
1.5.5 Associativity for Multiplication
1.5.6 Distributive Property
1.5.7 Making Multiplication Easier
1.6 Division of Integers
1.7 Properties of Division of Integers

NCERT Solutions for Class 7 Maths Chapter-Wise

Chapter No.	Chapter Name
Chapter 1	Integers
Chapter 2	Fractions and Decimals
Chapter 3	Data Handling
Chapter 4	Simple Equations
Chapter 5	Lines and Angles
Chapter 6	The Triangle and its Properties
Chapter 7	Congruence of Triangles
Chapter 8	Comparing quantities
Chapter 9	Rational Numbers
Chapter 10	Practical Geometry
Chapter 11	Perimeter and Area
Chapter 12	Algebraic Expressions
Chapter 13	Exponents and Powers
Chapter 14	Symmetry
Chapter 15	Visualising Solid Shapes

NCERT Solutions for Class 7 Subject Wise

NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 Science

Important Points of Class 7 Maths Chapter 1 Integers

Integers are closed for both addition and subtraction. That is, if a and b are any integers then a + b and a – b are also integers.
The addition is commutative for the integers, such that $a + b = b + a$ for all integers 'a' and 'b'.
The addition is associative for the integers, such that, $(a + b) + c = a + (b + c)$ for all the integers 'a', 'b' and 'c'.
Integer 0 is the identity under addition. That is, $a + 0 = 0 + a = a$ for every integer 'a'.
Integers are closed under multiplication. That is, $a \times b$ is an integer for any two integers 'a' and 'b'.
Multiplication is commutative for the integers. That is, $a \times b= b \times a$ for any integers 'a' and 'b.'
The integer 1 is the identity under multiplication, such that, $1 \times a= a \times 1 = a$ for any integer 'a'.
Multiplication is associative for the integers, such that, $(a \times b) \times c = a \times (b \times c)$ for any three integers a, b and c.

If a student is familiar with the CBSE NCERT solutions for Class 7 Maths chapter 1 Integers, then good marks in the exam is not a difficult task. Also, the NCERT solutions for Class 7 Maths chapter 1 Integers help in solving homework problems

Also Check NCERT Books and NCERT Syllabus here:

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions for Class 7 Maths Chapter 1 Integers

Question: How many exercise in NCERT solution Class 7 Maths Chapter Integers?

Answer:

There are 4 exercises in NCERT solution Class 7 Maths Integers

Question: Do I need to solve every problem given in the NCERT Solutions for Class 7 Maths Chapter 1?

Answer:

Indeed, these questions hold significant importance in terms of the exam. Expertly solved, they assist students in easily tackling the exercises. These solutions aid in acquainting students with the concept of integers. You can access the solutions in PDF format on the Careers360 website.

Question: Where can I find the precise answer key for NCERT Solutions for Class 7 Maths Chapter 1?

Answer:

You can obtain the precise PDF format solutions for NCERT Solutions for Class 7 Maths Chapter 1 at BYJU’S. The Mathematics experts at Careers360 have meticulously designed these accurate NCERT Textbook Solutions for the chapter. These solutions are tailored to align with the new pattern of CBSE, ensuring that students acquire comprehensive knowledge to excel in their exams.

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NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 1 Integers

NCERT Solutions for Class 7 Maths Chapter 1 Integers

NCERT Solutions for Class 7 Maths Chapter 1 Integers Topic 1.2

NCERT Solutions for Class 7 Maths Chapter 1 Integers Topic 1.3.5

NCERT Solutions for Class 7 Maths Chapter 1 Integers Topic 1.4.1

NCERT Soltions for Class 7 Maths Chapter 1 Integers Topic 1.4.3

NCERT Solutions for Class 7 Maths Chapter 1 Integers Topic 1.5.6

NCERT Solutions for Class 7 Maths Chapter 1 Integers Topic 1.5.7

NCERT Solutions for Class 7 Maths Chapter 1 Integers Topic 1.6

NCERT Solutions for Class 7 Maths Chapter 1 Integers Topic 1.7

NCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.1

NCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.2

NCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.3

NCERT Soltions for Class 7 Maths Chapter 1 Integers Exercise 1.4

Integers Class 7 Maths Chapter 1- Topics

NCERT Solutions for Class 7 Maths Chapter-Wise

NCERT Solutions for Class 7 Subject Wise