NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3

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.