CBSE students can refer to NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3 Textbook Questions and Answers are provided by experts in order to help students secure good marks in exams.

## Class 7 Maths NCERT Solutions Chapter 1 Integers Ex 1.3

**Ex 1.3 Class 7 Maths Question 1.**

Find each of the following products :

**(a)** 3 x (- 1)

**(b) **(- 1) x 225**
(c)** (- 21) x (- 30)

**(d)**(- 316)

**x (- 1)**

**(e)**(- 15) x O x (- 18)

**(f)**(- 12) x (- 11) x (10)

**(g)**9 x (- 3) x (- 6)

**(h)**(- 18) x (-5) x (- 4)

**(i)**(- 1) x (- 2) x (- 3) x 4 Sol. (a) 3 x (- 1) = – (3 x 1) = – 3

**(j)**(- 3) x (- 6) x (- 2) x (- 1)

**Solution.**

**(a) 3 × (- 1)**

3 × (- 1) = – (3 × 1) = – 3.

**(b) (- 1) × 225**

(- 1) × 225 = – (1 × 225) = – 225.

**(c) (- 21) × (- 30)**

(- 21) × (- 30) = 21 × 30 = 630.

**(d) (- 316) × (- 1)**

(- 316) × (- 1) = 316 × 1 = 316.

**(e) (- 15) × 0 × (- 18)**

(- 15) × 0 × (- 18) = (- 15) × 0) × (- 18) = 0 × (-18) = 0.

**(f) (- 12) × (-11) × 10**

(- 12) × (- 11) × 10

= [(- 12) × (- 11)] × 10

= (12 × 11) × 10

= 132 × 10 = 1320.

**(g) 9 × (- 3) × (- 6)**

9 × (- 3) × (- 6) = 9 × [(- 3) × (- 6)]

= 9 × (3 × 6)

= 9 × 18 = 162.

**(h) (- 18) × (- 5) × (- 4)**

(- 18) × (- 5) × (- 4)

= [(- 18) × (- 5)] × (- 4)

= (18 × 5) × (- 4)

= – (90 × 4) = – 360.

**(i) (- 1) × (- 2) × (- 3) × 4**

(- 1) × (- 2) × (- 3) × 4

= [(- 1) × (- 2)] × [(- 3) × 4]

= (1 × 2) × [- (3 × 4)]

= 2 × [-(12)] = 2 × ( – 12)

= – ( 2 × 12) = -24.

**(j) (- 3) × (- 6) × (- 2) × (- 1)**

(- 3) × (- 6) × (- 2) × (- 1)

= [(- 3) × (- 6)] × [(- 2) × (- 1)]

= (3 × 6) × (2 × 1) = 18 × 2 = 36.

**Ex 1.3 Class 7 Maths Question 2.**

Verify the following:

**(a)** 18 x [7 + (- 3)] = [18 x 7] + [18 x (- 3)]

**(b)** (- 21) x [(- 4) + (-6)] = [(- 21) x (- 4)] + [(- 21) x (- 6)]

**Solution.**

**(a)** We have,

18 x [7 + (- 3)] = 18 x 4 = 72

and, [18 x 7] + [18 x (- 3)] = 126 – 54 = 72

18 x [7 + (- 3)] = [18 x 7] + [18 x (- 3)]

**(b)** We have,

(- 21) x [(- 4) + (- 6)] = (- 21) x (- 4 – 6)

= (-21)(-10) = 210 and, [(- 21) x (- 4) + [(- 21) x (- 6)]

= 84 + 126 =210

∴ (- 21) x [(- 4) + (- 6)] = [(- 21) x (- 4)] + [(- 21) x (- 6)]

**Ex 1.3 Class 7 Maths Question 3.**

**(i)** For any integer a, what is (-1) x an equal to?

**(ii)** Determine the integer whose product with (-1) is

**(a)** – 22

**(b)** 37

**(c)** 0

**Solution.
**

**(i)**(-1) × a = – a

**(ii)**(a) 22 (b)-37 (c) 0.

**Ex 1.3 Class 7 Maths Question 4.
**Starting from (- 1) x 5, write various products showing some pattern to show (- 1) x (- 1) = 1.

**Solution.**

(- 1) x 5 = – 5

(- 1) x 4 = -4 = [-5 – (-1)] = – 5 +1

(- 1) x 3 = -3 = [-4 – (-1)] = – 4 +1

(- 1) x 2 = -2 = [-3 – (-1)] = – 3 +1

(- 1) x 1 = -1 = [-2 – (-1)] = – 2 +1

(- 1) x 0 = 0 = [-1 – (-1)] = – 1 + 1

(- 1) x (- 1) =[0 – (-1)] = 0 + 1 = 1

**Ex 1.3 Class 7 Maths Question 5.
**Find the product, using suitable properties :

**(a)**26 x (- 48) + (- 48) x (- 36)

**(b)**8 x 53 x (-125)

**(c)**15 x (- 25) x (-4) x (- 10)

**(d)**(- 41) x 102

**(e)**625 x (- 35) +(- 625) x 65

**(f)**7 x (50 – 2)

**(g)**(- 17) x (- 29)

**(h)**(- 57) x (- 19) + 57

**Solution.**

**(a) 26 × (- 48) + (- 48) × (- 36)**

26 × (- 48) + (- 48) × (- 36)

= [- (26 × 48)] + (48 × 36)

a × (- b) = – (a × b) = (- 1248) + 1728

= 480. [(-a) × (-b) = a × b]

**Aliter.**26 × (- 48) + (- 48) × (- 36)

= 26 ×(- 48) + (-36) × (-48)

= [26 + (-36)] × (-48)

[Distributive property] = (-10) × (-48) = 480.

**(b) 8 × 53 × (- 125)**

8 × 53 × (- 125) = (8 × 53) × (- 125)

= 424 × (- 125)

= – (424 × 125)

∝ × (- b) = – (a × b) = – [424 × (100 + 25)] = – [424 × 100 + 424 × 25]

Distributivity of multiplication over addition = – [42400 + 10600] = – 53000

**Aliter.** 8 × 53 × (- 125)

= 8 × (- 125) × 53

Commutativity of multiplication

= [8 × (- 125)] × 53 = [- (8 × 125)] × 53

α × (- b) = – (a × b) = (- 1000) × 53 = – (1000 × 53)

(- α) × b = – (a × b) = – 53000.

**(c) 15 × (- 25) × (- 4) × (- 10)**

15 × (- 25) × (- 4) × (- 10)

= 15 × (- 25) × (- 10) × (-4)

Commutativity of multiplication = 15 × (- 10) × (- 25) × (- 4)

Commutativity of multiplication = [(15) × (- 10)] × [(- 25) × (- 4)]

= [-(15 × 10)] × [25 × 4]

α × (- b) = -(a × b) (- a) × (- b) = a × b

= (- 150) × 100 = -(150 × 100) = – 15000.

**(d) (- 41) × 102**

(- 41) × 102 = – (41 × 102)

(- α) × b = – (a × b) = – [41 × (100 + 2)] = – [41 × 100 + 41×2]

Distributivity of multipli-cation over addition

= – [4100 + 82] = – 4182.

**(e) 625 × (- 35) + (- 625) × 65**

625 × (- 35) + (- 625) × 65

= 625 × (- 35) + 625 × (- 65)

(- α) × b = a × (- b) = 625 × [(-35) + (- 65)]

Distributivity of multipli-cation over addition

= 625 × (- 100) = – (625 × 100)

α × (- b) = – (a × b) = – 62500.

**(f) 7 × [50 – 2]**

7 × [50 – 2] = 7 × 50 – 7 × 2

Distributivity of multiplication over subtraction = 350 – 14 = 336.

**(g) (- 17) × (- 29)**

(- 17) × (- 29) = 17 × 29

(-α) × (- b) = a × b = 17 × (30-1) = 17 × 30- 17 × 1

Distributivity of multiplication over subtraction = 510 – 17 = 493.

**(h) (- 57) × (- 19) + 57**

(- 57) × (- 19) + 57 = 57 × 19 + 57

(-α) × (- b) = a × b = 57 × 19 + 57 ×1 I a × 1 = a = 57 × (19 + 1)

Distributivity of multipli-cation over addition = 57 × 20 = 1140.

**Ex 1.3 Class 7 Maths Question 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?

**Solution.**

Initial room temperature = 40 X

Temperature lowered every hour = (- 5) °C

Temperature lowered in 10 hours = (- 5) x 10 °C = – 50 °C

∴ Room temperature after 10 hours = 40 X – 50 X = – 10 °C

**Ex 1.3 Class 7 Maths Question 7.**

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?

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

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

**Solution.**

**(i)** Marks awarded for one correct answer = 5

Marks scored for 4 correct answer = 5 x 4 = 20

Marks awarded for one incorrect answer = (- 2)

Marks scored for 6 incorrect answer = (- 2) x 6 = – 12

Hence, Mohan’s score = 20 – 12 = 8 marks.

**(ii)** Reshma’s score for 5 correct answers = 5 x 5 = 25 marks

Reshma’s score for 5 incorrect answers = (-2) x 5 = – 10 marks

Hence, Reshma’s score = 25 – 10 = 15 marks

**(iii)** Heena’s score for 2 correct and 5 incorrect answers

= (5 x 2) + {(- 2) x 5}

= 10+ (- 10) = 10 – 10 = 0.

**Ex 1.3 Class 7 Maths Question 8.**

Does a cement company earn a profit? 8 per bag of white cement sold and a loss of? 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?

**(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?

**Solution.**

Profit on sale of 1 bag of white cement = ₹ 8

Loss on sale of 1 bag of grey cement = – ₹ 5

**(a)** Profit on sale of 3000 bags of white cement

= ₹ (3000 x 8)

= ₹ 24,000

Loss on sale of 5000 bags of grey cement = ₹ (5000 x – 5)

= – ₹ 25,000

Difference between the two = ₹ 24,000 – ₹ 25,000 = – ₹ 1,000

Hence, there is a loss of ₹ 1000.

**(b)** Loss on the sale of 6400 bags of grey cement = ₹ (6400 x 5) = ₹ 32,000

In order to have neither profit nor loss, the profit on the sale of white cement should be? 32,000.

Number of white cement bags sold

\(=\frac { Totalprofit }{ Profitperbag }\)

\(=\frac { 32000 }{ 8 }\)

Hence, 4000 bags of white cement should be sold to have neither profit nor loss.

**Question 9.**

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

**(a)** (- 3) x = 27

**(b)** 5 x = – 35

**(c)** 7 x (- 8) = – 56

**(d)** (- 11) x (- 12) = 132

**Solution.
**

**(a)**(- 3) ×

**(- 9)**= 27

**(b)**5 ×

**(- 7)**= – 35

**(c)**

**7**× (- 8) = – 56

**(d)**

**(- 11)**× (- 12) = 132.