NCERT Solutions Class 7 Maths Chapter 1 Integers Ex 1.3

NCERT Solutions Class 7 Maths
Chapter – 1 (Integers)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 1 Integers Exercise 1.3 has been provided here to help the students in solving the questions from this exercise.

Chapter 1: Integers

Exercise – 1.3

1. Find each of the following products:
(a) 3 × (-1)
(b) (-1) × 225
(c) (-21) × (-30)
(d) (-316) × (-1)
(e) (-15) × 0 × (-18)
(f) (-12) × (-11) × (10)
(g) 9 × (-3) × (-6)
(h) (-18) × (-5) × (-4)
(i) (-1) ×(-2) × (-3) × 4
(j) (-3) × (-6) × (-2) × (-1)

Solution –
(a) 3 × (-1)
By the rule of Multiplication of integers,
= 3 × (-1)
= -3 [∵ (+ × – = -)]

(b) (–1) × 225
By the rule of Multiplication of integers,
= (-1) × 225
= -225 [∵ (- × + = -)]

(c) (–21) × (–30)
By the rule of Multiplication of integers,
= (-21) × (-30)
= 630 [∵ (- × – = +)]

(d) (–316) × (–1)
By the rule of Multiplication of integers,
= (-316) × (-1)
= 316 [∵ (- × – = +)]

(e) (–15) × 0 × (–18)
By the rule of Multiplication of integers,
= (–15) × 0 × (–18)
= 0
∵ Any integer is multiplied with zero and the answer is zero itself.

(f) (–12) × (–11) × (10)
By the rule of Multiplication of integers,
= (–12) × (-11) × (10)
First multiply the two numbers having same sign,
= 132 × 10 [∵ (- × – = +)]
= 1320

(g) 9 × (–3) × (– 6)
By the rule of Multiplication of integers,
= 9 × (-3) × (-6)
First multiply the two numbers having same sign,
= 9 × 18 [∵ (- × – = +)]
= 162

(h) (–18) × (–5) × (– 4)
By the rule of Multiplication of integers,
= (-18) × (-5) × (-4)
First multiply the two numbers having same sign,
= 90 × – 4 [∵ (- × – = +)]
= – 360 [∵ (+ × – = -)]

(i) (–1) × (–2) × (–3) × 4
By the rule of Multiplication of integers,
= [(–1) × (–2)] × [(–3) × 4]
= 2 × (-12) [∵ (- × – = +), (- × + = -)]
= – 24

(j) (–3) × (–6) × (–2) × (–1)
By the rule of Multiplication of integers,
= [(–3) × (–6)] × [(–2) × (–1)]
First multiply the two numbers having same sign,
= 18 × 2 [∵ (- × – = +)
= 36

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

Solution –
(a) 18 × [7 + (–3)] = [18 × 7] + [18 × (–3)]
Let us consider the Left Hand Side (LHS) first = 18 × [7 + (–3)]
= 18 × [7 – 3]
= 18 × 4
= 72
Now, consider the Right Hand Side (RHS) = [18 × 7] + [18 × (–3)]
= [126] + [-54]
= 126 – 54
= 72
By comparing LHS and RHS,
72 = 72
LHS = RHS
Hence, the given equation is verified.

(b) (–21) × [(– 4) + (– 6)] = [(–21) × (– 4)] + [(–21) × (– 6)]
Let us consider the Left Hand Side (LHS) first = (–21) × [(– 4) + (– 6)]
= (-21) × [-4 – 6]
= (-21) × [-10]
= 210
Now, consider the Right Hand Side (RHS) = [(–21) × (– 4)] + [(–21) × (– 6)]
= [84] + [126]
= 210
By comparing LHS and RHS,
210 = 210
LHS = RHS
Hence, the given equation is verified.

3.
(i) For any integer a, what is (–1) × a equal to?
(ii) Determine the integer whose product with (-1) is 0.
(a) -22
(b) 37
(c) 0

Solution –
(i) For any integer a, what is (–1) × a equal to?
= (-1) × a = -a
Because, when we multiplied any integer a with -1, then we get additive inverse of that integer.

(ii). Determine the integer whose product with (–1) is
(a) –22
Now, multiply -22 with (-1), we get
= -22 × (-1)
= 22
Because, when we multiplied integer -22 with -1, then we get additive inverse of that integer.

(b) 37
Now, multiply 37 with (-1), we get
= 37 × (-1)
= -37
Because, when we multiplied integer 37 with -1, then we get additive inverse of that integer.

(c) 0
Now, multiply 0 with (-1), we get
= 0 × (-1)
= 0
Because, the product of negative integers and zero give zero only.

4. Starting from (–1) × 5, write various products showing some pattern to show (–1) × (–1) = 1.
Solution –
The various products are,
= -1 × 5 = -5
= -1 × 4 = -4
= -1 × 3 = -3
= -1 × 2 = -2
= -1 × 1 = -1
= -1 × 0 = 0
= -1 × -1 = 1
We concluded that the product of one negative integer and one positive integer is negative integer. Then, the product of two negative integers is a positive integer.

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) 26 × (-48) + (-48) × (-36)
The given equation is in the form of Distributive law of Multiplication over Addition.
= a × (b + c) = (a × b) + (a × c)
Let, a = -48, b = 26, c = -36
Now,
= 26 × (– 48) + (– 48) × (–36)
= -48 × (26 + (-36)
= -48 × (26 – 36)
= -48 × (-10)
= 480 [∵ (- × – = +)]

(b) 8 × 53 × (–125)
The given equation is in the form of Commutative law of Multiplication.
= a × b = b × a
Then,
= 8 × [53 × (-125)]
= 8 × [(-125) × 53]
= [8 × (-125)] × 53
= [-1000] × 53
= – 53000 [∵ (- × + = -)]

(c) 15 × (–25) × (– 4) × (–10)
The given equation is in the form of Commutative law of Multiplication.
= a × b = b × a
Then,
= 15 × [(–25) × (– 4)] × (–10)
= 15 × [100] × (–10)
= 15 × [-1000]
= – 15000

(d) (– 41) × 102
The given equation is in the form of Distributive law of Multiplication over Addition.
= a × (b + c) = (a × b) + (a × c)
= (-41) × (100 + 2)
= (-41) × 100 + (-41) × 2
= – 4100 – 82
= – 4182

(e) 625 × (–35) + (– 625) × 65
The given equation is in the form of Distributive law of Multiplication over Addition.
= a × (b + c) = (a × b) + (a × c)
= 625 × [(-35) + (-65)]
= 625 × [-100]
= – 62500

(f) 7 × (50 – 2)
The given equation is in the form of Distributive law of Multiplication over Subtraction.
= a × (b – c) = (a × b) – (a × c)
= (7 × 50) – (7 × 2)
= 350 – 14
= 336

(g) (–17) × (–29)
The given equation is in the form of Distributive law of Multiplication over Addition.
= a × (b + c) = (a × b) + (a × c)
= (-17) × [-30 + 1]
= [(-17) × (-30)] + [(-17) × 1]
= [510] + [-17]
= 493

(h) (–57) × (–19) + 57
The given equation is in the form of Distributive law of Multiplication over Addition.
= a × (b + c) = (a × b) + (a × c)
= (57 × 19) + (57 × 1)
= 57 [19 + 1]
= 57 × 20
= 1140

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 –
Temperature of the room in the beginning = 40°C
Temperature after 1 hour
= 40°C – 1 × 5°C
= 40°C – 5°C – 35°C
Similarly, temperature of the room after 10 hours
= 40°C – 10 × 5°C
= 40°C – 50°C
= -10°C

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?
Solution –
Marks awarded to Mohan = 4 × 5
= 20 for correct answers
Marks awarded to Mohan = 6 × (-2)
= – 12 for incorrect answers.
∴ Total marks obtained by Mohan = 20 + (-12) = 20 – 12 = 8

(ii) Reshma gets five correct answers and five incorrect answers, what is her score?
Solution –
Marks awarded to Reshma for correct answers = 5 × 5
= 25
Marks awarded to Reshma for incorrect answers = 5 × (-2)
= -10
∴ Total marks obtained by Reshma = 25 + (-10)
= 25 – 10
= 15

(iii) Heena gets two correct and five incorrect answers out of seven questions she attempts. What is her score?
Solution –
Marks awarded to Heena for correct answers = 2 × 5 = 10
Marks awarded to Heena for incorrect answers = 5 × (-2) = -10
Number of question not attempted by Heena = 10 – (2 + 5) = 10 – 7 = 3
Marks awarded to Heena for non-attempted questions = 3 × 0 = 0
∴ Total marks obtained by Heena
= 10 + (-10) + 0
= 10 – 10 + 0
= 0

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 one white cement bag = ₹ 8
loss on one grey cement bag = ₹ –5
Profit on 3,000 bags of white cement = ₹ (8 × 3,000)
= ₹ 24,000
Loss on 5,000 bags of grey cement = ₹ (-5 × 5000)
= – ₹ 25,000
Total loss = – ₹ 25,000 + ₹ 24,000
= – ₹ 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 –
Selling price of grey bags at a loss of ₹ 5 = ₹ (5 × 6,400) = ₹ 32,000
For no profit and no loss, the selling price of white bags = ₹ 32,000
Rate of selling price of white bags at a profit of ₹ 8 per bag.
∴ Number of white cement bags sold = 32,000/8 = 4000
Hence, the 4000 bags of white cement have neither profit nor loss.

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) × __ = 27
Let us assume the missing integer be x,
Then,
= (–3) × (x) = 27
= x = – (27/3)
= x = -9
Let us substitute the value of x in the place of blank,
= (–3) × (-9) = 27 [∵ (- × – = +)]

(b) 5 × __ = –35
Let us assume the missing integer be x,
Then,
= (5) × (x) = -35
= x = – (-35/5)
= x = -7
Let us substitute the value of x in the place of blank,
= (5) × (-7) = – 35 [∵ (+ × – = -)]

(c) __ × (– 8) = –56
Let us assume the missing integer be x,
Then,
= (x) × (-8) = -56
= x = (-56/-8)
= x = 7
Let us substitute the value of x in the place of blank,
= (7) × (-8) = -56 … [∵ (+ × – = -)]

(d) _____ × (–12) = 132
Let us assume the missing integer be x,
Then,
= (x) × (-12) = 132
= x = – (132/12)
= x = – 11
Let us substitute the value of x in the place of blank,
= (–11) × (-12) = 132 … [∵ (- × – = +)]

NCERT Solutions Class 7 Maths Chapter 1 Integers Ex 1.3