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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