Gujarat Board GSEB Textbook Solutions Class 7 Maths Chapter 1 Integers Ex 1.3 Textbook Questions and Answers.
Gujarat Board Textbook Solutions Class 7 Maths Chapter 1 Integers Ex 1.3
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 0 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
(j) (- 3) x (- 6) x (- 2) x (- 1)
Solution:
(a) 3 x (- 1) = – (3 x 1) = – 3
(b) (- 1) x 225 = – (1 x 225) = – 225
(c) (- 21) x (- 30) = +[21 x 30]
= [(20 + 1) x 30]
= 20 x 30 + 1 x 30 = 600 + 30 = 630
(d) (- 316) x (- 1) = + (316 x 1) = 316
(e) (- 15) x 0 x (- 18) = +[(- 15) x 0] x (- 18)
= 0 x (- 18) = 0
(f) (- 12) x (- 11) x 10 = + [12 x 11 x 10]
= [132 x 10] = 1320
[Product of even number of negative integers is positive.]
(g) 9 x (- 3) x (- 6) = + [9 x 3 x 6] = 162
[Product of even number of negative integers is positive.]
(h) (- 18) x (- 5) x (- 4) = – [18 x 5 x 4]
= – [18 x 20] = – 360
[Product of odd number of negative integers is negative.]
(i) (- 1) x (- 2) x (- 3) x 4 = – [1 x 2 x 3 x 4]
= – [2 x 12] = – 24
[Product of odd number of negative integers is negative.]
(j) (- 3) x (- 6) x (- 2) x (- 1) = + [3 x 6 x 2 x 1]
= + [36 x 1] = 36
[Product of even number of negative integers is positive.]
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) L.H.S. = 18 x [7 + (- 3)]
= 18 x [7 – 3] = 18 x [4] = 72
R.H.S. = (18 x 7) + [18 x (- 3)]
= 126 + [ – 54] = 126 – 54 = 72
∵ L.H.S. = R.H.S
18 x [7 + (- 3)]
∴ = [18 x 7] + [18 x (- 3)]
(b) L.H.S. = (- 21) x [(- 4) + (- 6)]
= (- 21) x [- 10] = + (21 x 10) = 210
R.H.S. = [(- 21) x (- 4)] + [(- 21) x (- 6)] = [(+ 84)] + [(+ 126)] = 84 + 126 = 210
∵ L.H.S. = R.H.S.
∴ (- 21) x [(- 4) + (- 6)]
= [(- 21) x (- 4)] + [(- 21) x (- 6)]
Question 3.
(i) For any integer a, what is (- 1) x a equal to?
(ii) Determine the integer whose product with (- 1) is
(a) – 22
(b) 37
(c) 0
Solution:
(i) (- 1) x a = – a
(ii)
(a) ∵ (- 1) x [any integer]
= Additive inverse of the integer
∴ (- 1) x 22 = – 22
(b) (- 1) x (- 37) = 37
(c) (- 1) x (0) = 0 [∵ Product of a negative integer and zero is zero.]
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
(- 1) x 3 = – 3 = (- 4) + 1
(- 1) x 2 = – 2 = (- 3) + 1
(- 1) x 1 = – 1 = (- 2) + 1
(- 1) x 0 = 0 = (- 1) + 1 (- 1)
(- 1) x (- 1) = 1 = 0 + 1
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 x (- 48) + (- 48) x (- 36): [Using distributivity of multiplication over addition]
26 x (- 48) + (- 48) x (- 36) = (- 48)[26 + (- 36)]
= (- 48)[- 10]
= (- 48) x (- 10) = 480
(b) 8 x 53 x (- 125):
8 x 53 x (- 125) = [8 x (- 125)] x 53 [Using associative property for multiplication]
= [- 1000] x 53 = – [1000 x 53]
= – 53000
(c) 15 x (- 25) x (- 4) x (- 10)
Using associative property for multiplication, we have
15 x (- 25) x (- 4) x (- 10)
= [(- 25) x (- 4)] x [(- 10) x 15]
= [100] x [- 150]
= – [100 x 150] = – 15000
(d) (- 41) x 102 = (- 41) x [100 + 2]
= (- 41) x 100 + (- 41) x 2 [∵ a(b + c) = a x b + a x c] = – 4100 + (- 82) = – 4182
(e) 625 x (- 35) + (- 625) x 65
= 625 [(- 35) + (- 65)]
= 625 x [- 100]
= – [625 x 100] = – 62500
(f) 7 x (50 – 2) = 7 x 50 – 7 x 2
= 350 – 14 = 336
(g) (- 17) x (- 29)= + [17 x 29]
= 17 x (30 – 1)
= 17 x 30 – 17 x 1 [∵ a(b – c) = a x b – a x c]
= 510 – 17 = 493
(h) (- 57) x (- 19) + 57
= (- 57) X (- 19) + [- 57 x (- 1)] [∵ 57 = (- 1) x (- 57)]
= (- 57) x [(- 19) + (- 1)]
= (- 57) x (- 20)
= + [57 x 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:
Room temperature = 40°C
Change in temperature per hour = – 5°C
Change in temperature in 10 hours
= 10 x (- 5°C) = – 50°C
Room temperature after 10 hours
= 40°C + (- 50°C) = – 10°C.
Question 7.
In a class test containing 10 question, 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 Jive 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:
Total number of questions =10
Marks for a correct answer = 5
Marks for an incorrect answer = (- 2)
Marks for unanswered question = 0
(i) Mohan’s score = 4 x [5] + 6 x [- 2]
= 20 + [- 12] = 8
(ii) Reshma’s score = 5 x [5] + 5 x [- 2]
= 25 + (- 10) = 15
(iii) Heena’s score = 2 x [5] + 5 x [- 2] + 3 x [0]
= 10 + [- 10] + 0 = 0
Question 8.
A cement company earns a profit of ₹ 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:
(a) ∵ Profit of ₹ 8 is earned on a bag of white cement and a loss of ₹ 5 on a bag of grey cement.
Number of white cement bags sold = 3.000
Number of grey cement bags sold = 5.000
∴ Profit =3000 x ₹ 8 = ₹ 24,000
Loss =5000 x ₹ 5 = ₹ 25,000
Here Loss > Profit
∴ Loss = ₹ (25,000 – 24,000) = ₹ 1000
(b) Number of grey cement bags sold = 6400
∴ Total loss = ₹ 5 x 6400 = ₹ 32000
For no profit and no loss, there should be a profit of ₹ 32000.
∴ Number of white cement bags sold to earn a profit of ₹ 32000 = 32000 ÷ 8 = 4000 bags
Question 9.
Replace the blank with an integer to make it a true statement.
(a) (- 3) x ______ = 27
(b) 5 x ______ = – 35
(c) ______ x (- 8) = – 56
(d) ______ x (- 12) = 132
Solution:
(a) – 3 x ______ = 27
⇒ – 3 x (- 9) = 27 [∵ 3 x 9 = 27]
(b) 5 x ______ = – 35
⇒ 5 x (- 7) = – 35 [∵ 5 x 7 = 35]
(c)______ x (- 8) = – 56
⇒ 7 x (- 8) = – 56 [∵ 7 x 8 = 56]
(d) ______ x (- 12) = 132
⇒ (- 11) x (- 12) = 132 [∵ 11 x 12 = 132]