NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3 are the part of NCERT Solutions for Class 7 Maths. Here you can find the NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3.



Ex 1.3 Class 7 Maths Question 1.

Find each of the following products:
(a) 3 × (-1)
(b) (-1) × 225
(c)
 (-21) × (-30)
(d) (-316) × (-1)
(e) (-15) × O × (-18)
(f) (-12) × (-11) × (10)
(g) 9 x (-3) × (-6)
(h) (-18) × (-5) × (-4)
(i) (-1) × (-2) × (-3) × 4
(j) (-3) × (-6) × (-2) × (-1)

Solution.
(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) = 0 × (-18) = 0
(f) (-12) × (-11) × (10) = [(-12) × (-11)] × (10)
= (132) × (10) = 1320
(g) 9 × (-3) × (-6) = [9 × (-3)] × (-6) = (-27) × (-6) = 162
(h) (-18) × (-5) × (-4) = [(-18) × (-5)] × (-4)
= 90 × (-4) = -360
(i) (-1) × (-2) × (-3) × 4 = [(-1) × (-2)] × [(-3) × 4]
= (2) × (-12) = -24
(j) (-3) × (-6) × (-2) × (-1) = [(-3) × (-6)] × [(-2) × (-1)] = (18) × (2) = 36

 

Ex 1.3 Class 7 Maths Question 2.

Verify the following:
(a) 18 × [7 + (-3)] = [18 × 7] + [18 × (-3)]
(b) (-21) × [(-4) + (-6)] = [(-21) × (-4)] + [(-21) × (-6)]

Solution.
(a) We have,
LHS = 18 × [7 + (-3)] = 18 × 4 = 72
RHS = [18 × 7] + [18 × (-3)] = 126 – 54 = 72
18 × [7 + (-3)] = [18 × 7] + [18 × (-3)]
(b) We have,
LHS = (-21) × [(-4) + (-6)] = (-21) × (-4 – 6) = (-21) (-10) = 210

RHS = [(-21) × (-4) + [(-21) × (-6)] = 84 + 126 = 210
(-21) × [(-4) + (-6)] = [(-21) × (-4)] + [(-21) × (-6)]

 

Ex 1.3 Class 7 Maths Question 3.

(i) For any integer a, what is (-1) × a equal to?
(ii) Determine the integer whose product with (-1) is
(a) -22
(b) 37
(c) 0

Solution.
(i) For any integer a, (-1) × a = -a.
(ii) We know that the product of any integer and (-1) is the additive inverse of the integer. The integer whose product with (-1) is
(a) -22 is the additive inverse of -22, i.e., 22.
(b) 37 is the additive inverse of 37, i.e., -37.
(c) 0 is the additive inverse of 0, i.e., 0.

 

Ex 1.3 Class 7 Maths 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)] = -5 + 1
(-1) × 3 = -3 = [-4 – (-1)] = -4 + 1
(-1) × 2 = -2 = [-3 – (-1)] = -3 + 1
(-1) × 1 = -1 = [-2 – (-1)] = -2 + 1
(-1) × 0 = 0 = [-1 – (-1)] = -1 + 1
(-1) × (-1) = [0 – (-1)] = 0 + 1 = 1

 

Ex 1.3 Class 7 Maths Question 5.

Find the product, using suitable properties:
(a) 26 × (-48) + (-48) × (-36)
(b) 8 × 53 × (-125)
(c) 15 × (-25) × (-4) × (-10)
(d) (-41) × 102
(e) 625 × (-35) +(-625) × 65
(f) 7 × (50 – 2)
(g) (-17) × (-29)
(h) (-57) × (-19) + 57

Solution.
(a) We have, 26 × (-48) + (-48) × (-36)
= (-48) × 26 + (-48) × (-36)
= (-48) × [26 + (-36)]
= (-48) × (26 – 36)
=(-48) × (-10) = 480
(b) We have, 8 × 53 × (-125) = [8 × (-125)] × 53
= (-1000) × 53 = -53000
(c) We have, 15 × (-25) × (-4) × (-10)
= 15 × [(-25) × (-4)] × (-10)
= 15 × (100) × (-10)
= (15 × 100) × (-10)
= 1500 × (-10) = -15000
(d) We have, (-41) × 102 = (-41) × (100 + 2)
= (-41) × 100 + (-41) × 2 = -4100 – 82 = -4182
(e) We have, 625 × (-35) + (-625) × 65
= 625 × (-35) + (625) × (-65)
= 625 × [(-35) + (-65)]
= 625 × (-100) = -62500
(f) We have, 7 × (50 – 2) = 7 × 50 – 7 × 2
= 350 – 14 = 336
(g) We have, (-17) × (-29) = (-17) × [(-30) + 1]
= (-17) × (-30) + (-17) × 1 = 510 – 17 = 493
(h) We have, (-57) × (-19) + 57 = 57 × 19 + 57 × 1
= 57 × (19 + 1)
= 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°C
Temperature lowered every hour = (-5)°C
Temperature lowered in 10 hours = (-5) × 10°C = -50°C
Room temperature after 10 hours = 40°C – 50°C = -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 awarded for 4 correct answers = 5 × 4 = 20
Marks awarded for one incorrect answer = (-2)
Marks awarded for 6 incorrect answers = (-2) × 6 = -12
Hence, Mohan’s score = 20 – 12 = 8 marks.
(ii) Reshma’s score for 5 correct answers = 5 × 5 = 25 marks
Reshma’s score for 5 incorrect answers = (-2) × 5 = -10 marks
Hence, Reshma’s score = 25 – 10 = 15 marks.
(iii) Heena’s score for 2 correct and 5 incorrect answers
= (5 × 2) + {(-2) × 5}
= 10 + (-10) = 10 – 10 = 0

 

Ex 1.3 Class 7 Maths 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?
(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 the sale of 1 bag of white cement = ₹ 8
Loss on the sale of 1 bag of grey cement = – ₹ 5
(a) Profit on the sale of 3,000 bags of white cement
= ₹ (3000 × 8)
= ₹ 24,000
Loss on the sale of 5,000 bags of grey cement = ₹ (5000 × -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 × 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 = Total profit/Profit per bag
= 32000/8 = 4000
Hence, 4000 bags of white cement should be sold to have neither profit nor loss.


Ex 1.3 Class 7 Maths Question 9.

Replace the blank with an integer to make it a true statement.
(a) (-3) × ______ = 27
(b) 5 × ________ = -35
(c) _______ × (-8) = -56
(d) _______ × (-12) = 132

Solution.
(a) (-3) × (-9) = 27
(b) 5 × (-7) = (-35)
(c) 7 × (-8) = (-56)
(d) (-11) × (-12) = 132



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