RBSE Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3

Rajasthan Board RBSE Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3 Textbook Exercise Questions and Answers.

RBSE Class 7 Maths Solutions Chapter 1 Integers Ex 1.3

Question 1.
Find each of the following products:
(a) 3 × (- 1)
Answer:
3 × (- 1) = - (3 × 1) = - 3

(b) (- 1) × 225
Answer:
(- 1) × 225 = - (1 × 225) = - 225

(c) (- 21) × (- 30)
Answer:
(- 21) × (- 30) = + (21 × 30) = 630

(d) (- 316) × (- 1)
Answer:
(- 316) × (- 1) = + (316 × 1) = 316

(e) (- 15) × 0 × (- 18)
Answer:
(- 15) × 0 × (- 18) = 0

(f) (- 12) × (- 11) × (10)
Answer:
(- 12) × (- 11) × (10) = (12 × 11 × 10) = 1320

(g) 9 × (- 3) × (- 6)
Answer:
9 × (- 3) × (- 6) = (9 × 3 × 6) = 162

(h) (-18) × (-5) × (-4)
Answer:
(- 18) × (- 5) × (- 4) = - 18 × (5 × 4) = - 18 × 20 = - 360

(i) (-1) × (- 2) × (- 3) × 4
Answer:
(- 1) × (- 2) × (- 3) × 4 = - (1 × 2 × 3) × 4 = - 24

(j) (- 3) × (- 6) × (- 2) × (- 1)
Answer:
(- 3) × (- 6) × (- 2) × (- 1) = (3 × 6 × 2 × 1) = 36

Question 2.
Verify the following:
(a) 18 × [7 + (- 3)] = [18 × 7] + [18 × (- 3)]
(b) (- 21) × [(- 4) + (- 6)] = [(- 21) × (- 4)] + [(- 21) × (- 6)]
Answer:
(a) LHS = 18 × [7 + (-3)]
= 18 × 4 = 72
RHS = [18 × 7] + [18 × (- 3)J
= 126 + (- 54) = 126 - 54 = 72
∴ LHS = RHS
Hence verified.

(b) LHS = (- 21) × [(- 4) + (- 6)]
= - 21 × (-10) = 210
RHS = [(- 21) × (- 4)] + [(- 21) × (- 6)]
= 84 + 126 = 210
∴ LHS = RHS Hence verified.

Question 3.
(i) For any integer a, what is (- 1) × a equal to?
Answer:
- 1 × a = - (1 × a) = - a, where a is an integer.

(ii) Determine the integer whose product with (- 1) is :
(a) -22
(b) 37
(c) 0
Answer:
(a) 22
(b) - 37
(c) 0

Question 4.
Starting from (- 1) × 5, write various products showing some pattern to show (- 1) × (- 1) = 1.
Answer:
(- 1) × 5 = - 5
(- 1) × 4 = - 4
(- 1) × 3 = - 3
(- 1) × 2 = - 2
(- 1) × 1 = - 1
(- 1) × 0 = 0
(- 1) × (- 1) = 1
We conclude that the product of 1 negative and 1 positive integer is negative whereas the product of 2 negative integers is positive.

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
Answer:
(a) 26 × (- 48) + (- 48) × (- 36)
= (- 48) × [(26) + (- 36)] (By distributive property)
= (- 48) × (- 10) = 480

(b) 8 × 53 × (- 125) =
[(- 125) × 8] × 53 (By commutative property)
= - 1000 × 53 = - 53000

(c) 15 × (-25) × (-4) × (-10)
= [(- 25) × (- 4)] × [15 × (- 10)] (By commutative property)
= 100 × (- 150) = - 15000

(d) (- 41) × 102 = - 41 × (100 + 2) (By distributive property)
= (- 41 × 100) + (- 41 × 2) (By distributive property)
= (- 4100) + (- 82) = - 4182

(e) 625 × (- 35) + (- 625) × 65
= 625 × [(- 35) + (- 65)] (By distributive property)
= 625 × - 100 = - 62500

(f) 7 × (50 - 2)
= (7 × 50) - (7 × 2) (By distributive property)
= 350 - 14 = 336

(g) (- 17) × (- 29)= (- 17) × [- 30 + 1]
= (- 17 × - 30) + (- 17 × 1) (By distributive property)
= 510 - 17 = 493

(h) (- 57) × (- 19) + 57
= (- 57) × (- 19) + 57 × 1 (1 is the multiplicative identity)
= 57 × 19 + 57 × 1 = 57 × (19+ 1) (By distributive property)
= 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?
Answer:
Present room temperature = 40°C
Decrease in temperature per hour = 5°C
Decrease temperature after 10 hours = 5 × 10 = 50°C
Temperature after 10 hours = 40°C + (- 50°C) = - 10°C
So, the room temperature 10 hours after the process will be - 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 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?
Answer:
(i) Marks for 1 correct answer = 5
Marks for 4 correct answers = 5 × 4 = 20
Marks for 1 incorrect answer = (- 2)
Marks for 6 incorrect answers = (- 2) × 6 = -12
∴ Mohan's score = 20 + (- 12) = 20 - 12 = 8
So, Mohan's score is 8 marks.

(ii) Marks for 1 correct answer = 5
Marks for 5 correct answers = 5 × 5 = 25
Marks for 1 incorrect answer = (- 2)
Marks for 5 incorrect answers = (- 2) × 5 = - 10
∴ Reshma's score = 25 + (- 10) = 25 - 10 = 15 marks
So, Reshma's score is 15 marks.

(iii) Marks for 1 correct answer = 5
Marks for 2 correct answers = 5 × 2 = 10
Marks for 1 incorrect answer = (- 2)
Marks for 5 incorrect answers = (- 2) × 5 = - 10
∴ Heena's score = 10 + (- 10) = 10 - 10 = 0
So, Heena's score is 0 marks.

Question 8.
A cement company earns 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?
Answer:
(a) Profit on 1 bag of white cement = ₹ 8
Profit on 3000 bags of white cement = ₹ 8 × 3000 = 24,000
Loss on 1 bag of grey cement = ₹ 5
Loss on 5000 bags of grey cement = ₹5 × 5000 = 25,000
Since, Loss > Profit
Total loss = Loss - Profit
= 25,000 - 24,000
= ₹ 1,000
So, company has loss of ₹ 1,000.

(b) Let the number of white cement bags be x.
Profit = Loss
∴ 5 × 6400 = x × 8
∴ x = \(\frac{5 \times 6400}{8}\) = 4000
So, the company must sell 4000 white cement bags.

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
Answer:
(a) - 9
(b) - 7
(c) 7
(d) - 11