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## Karnataka State Syllabus Class 7 Maths Chapter 1 Integers Ex 1.3

Question 1.

Find each of the following products

a) 3 × (- 1)

3 × (- 1) = – 3

b) (- 1) × 225

(- 1) × 225 = – 225

c) (- 21) × (- 30)

(- 21) × (- 30) = 630

d) (- 316) × (- 1)

(- 316) × (- 1) = 316

e) (- 15) × 0 × (- 18)

(- 15) × 0 × (- 18)

f) (- 12) × (- 11) × (10)

(- 12) × (- 11) × (10) = 1320

g) 9 × (- 3) × (- 6)

9 × (- 3) × (- 6) = 162

h) (- 18) × (- 5) × (- 4)

(- 18) × (- 5) × (- 4) = – 360

i) (- 1) × (2) × (- 3) × 4

(- 1) × (- 2) × (- 3) × 4 = – 24

j) (- 3) × (- 6) × (- 2) × (- 1)

(- 3) × (- 6) × (- 2) × (- 1) = 36

Question 2.

Verify the following :

a) 18 × [7 + (- 3)] = [18 × 7] + [18 × (- 3)]

Solution:

18 × [7 – 3] = [126] + [18 × – 3]

18 × [4] = 126 + [- 54]

72 = 72 (verified)

b) (- 21) × [(- 4) + (- 6)] = [(- 21) × (- 4)] + [(- 21) × (- 6)]

Solution:

– 21 × [- 4 + – 6] = [- 21 × – 4] + [- 21 × – 6]

– 21 × – 10 = [84] + [126]

+ 210 = 210

LHS = RHS (verified)

Question 3.

i) For any integer a, what is (- 1) × a equal to?

– 1 × a = – a

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

a) – 22

– 1 × – 22 = + 22

b) 37

– 1 × 37 = – 37

c) 0

– 1 × 0 = 0

Question 4.

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

Solution:

(- 1) × 5 = – 5

(- 1) × 4 = – 4 [= (- 5) + 1]

(- 1) × 3 = – 3 [= (- 4) + 1]

(- 1) × 2 = – 2 [= (- 3) + 1]

(- 1) × 1 = – 1 [= (- 2) + 1]

(- 1) × 0 = 0 [= (- 1) + 1]

(- 1) × (- 1) = 1 [= 0 + 1]

Question 5.

End the product, using suitable properties:

a) 26 × (- 48) + (- 48) × (- 36)

= 26 × (- 48)+ (- 36) × (- 48)

= [26 + (- 36)] × (-48) ∵ Distributive Property = (- 10) × (- 48)

= + 480

b) 8 × 53 × (- 125)

= 8 × (- 125) × 53

= [8 × (- 125)] × 53

= [- (8 × 125)] × 53 ∵ a × (- b) = – (a × b)

= (- 1000) × 53 = – (1000 x 53) ∵ (- a) × (b) = – (a × b)

= – 53000

c) 15 × (- 25) × (- 4) × (- 10)

= 15 × (- 25) × (- 10) × (- 4)

∵ commutativity of multiplication

= 15 × (- 10) × (- 25) × (- 4)

= [15 × (- 10)] × [(- 25) × (- 4)]

= [- (15 × 10)] × [25 × 4] (∵ a × (- b) = – (a × b)

= – 150 × 100 (- a) × (- b) = a × b

= – 15000

d) (- 41) × 102

= -(41 × 102) ∵ (- a) × b = – (a × b)

=-[41 × (100 + 2)]

= -[41 × 100 + 41 × 2]

Distributivity of multiplication over addition.

= – [4100 + 82]

= – 4182

e) 625 × (- 35) + (- 625) × 65

= 625 × (- 35) + 625 × (- 65) ∵ (- a) × b = a × (- b)

= 625 × [(- 35) + (- 65)] ∵ Distributivity of multiplication over addition

= 625 × (- 100)

= – (625 × 100)

f) 7 × (50 – 2)

= 7 × 50 – 7 × 2

∵ Distributivity of multiplication over subtraction

= 350 – 14 = 336

g) (-17) × (- 29)

= 17 × 29 ( ∵ (- a) × (- 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 ∵ (- a) × (- b) = a × b

= 57 × 19 + 57 × 1 ∵ (a × 1 = a)

= 57 × (19 + 1)

∴ Distributivity of multiplication over subtraction.

= 57 × 20 = 1140

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 temperature = 40°C

Room temperature decreases at 5°C per hour

Temperature change in 1 hour = -5°C

Temperature change in 10 hour = -5 × 10°C = -50°C

Room temperature after 10 hours = Initial temperature + Temperature change in 10 hour

= 40 + (-50)

= 40 – 50

∴ Room temperature after 10 hours = -10°C

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 question not attempted.

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

Solution:

Total questions = 10

Marks for correct answer = +5

Marks for incorrect answer = -2

Marks for no attempt = 0

Mohan marks for 4 correct answers = 4 × 5 = 20

Mohan marks for 6 incorrect answers= 6 × (-2) = 12

∴ Total marks = 20 + (-12) = 20 – 12 = 8

Total marks scored by Mohan is = 8

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

Solution:

Reshma marks for 5 correct answers = 5 × 5 = 25

Reshma marks for 5 incorrect answers

= 5 × (- 2) = – 10

∴ Total marks = 25 + (- 10) = 25 – 10 = 15

∴ Total marks scored by Reshma = 15

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

Solution:

Heena marks for 2 correct answers = 2 × 5 = 10

Heena marks for 5 incorrect answers = 5 × (-2) = -10

∴ Heena total marks = 10 + (-10) = 10 – 10 = 0

∴ Total marks scored by Heena = 0

Question 8.

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.

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?

Solution:

Profit on white cement = Rs +8

Loss on grey cement = Rs -5

∴ Profit on 1 bag of white cement = Rs 8

Profit on 3,000 bags of white cement

= Rs 8 × 3000 = Rs 24,000

Loss on 1 bag cement = Rs -5

Loss on 5000 bags grey cement

= Rs -5 × 5000 = Rs -25000

Total profit / loss = Total profit + Total loss

= 24000 + (-2500)

= 24000 – 25000

∴ Total loss = -1000

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:

Loss on 1 bag grey cement bag = Rs – 5

Loss on 6400 bag grey cement =Rs – 5 × 6400

= Rs – 32000

Let number of bags of white cement = n

profit on 1 bag white cement = Rs 8

Profit on n bags white cement = Rs 8n

∴ Total profit / loss = 0

Total profit + Total loss = 0

8n + (- 32000) = 0

8n – 32000 = 0

⇒ 8n = 32000

n = 4000

∴ Number of white cement bags = n = 4000

Question 9.

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

a) (- 3) × ___ = 27

Solution:

(- 3) × __– 9__ = 27

b) 5 × ___ = – 35

Solution:

5 × __– 7__ = – 35

c) __ × (- 8) = – 56

Solution:

__7__ × (- 8) = – 56

d) __ × (- 12) = 132

Solution:

__– 11__ × (- 12) = 132