NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 1 |

Chapter Name |
Integers |

Exercise |
Ex 1.3 |

Number of Questions Solved |
9 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3

**Question 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 x (- 1) = – (3 x 1) = – 3

(b) (- 1) x 225 = – (1 x 225) = – 225

(c) (- 21) x (- 30) = 21 x 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

(g) 9 x (- 3) x (- 6) = [9 x (- 3)] x (- 6) = (- 27) x (- 6) = 162

(h) (- 18) x (- 5) x (- 4) = [(- 18) x (- 5)] x (- 4) = 90 x (- 4) = – 360

(i) (- 1) x (- 2) x (- 3) x 4 = [(- 1) x (- 2)] x [(- 3) x 4] = (2)x (- 12) = -24

(j) (- 3) x (- 6) x (- 2) x (- 1) = [(- 3) x (- 6)] x [(- 2) x (- 1)] = (18) x (2) = 36

**Question 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)]**

L.H.S. = 18 × [7 + (- 3)]

= 18 × L(7 – 3)] = 18 × (4) = 18 × 4 = 72

R.H.S. = [18 × 7] + [18 × (- 3)]

= 126 + [- (18 × 3)] = 126 + (- 54) = 126 – 54 = 72

So, 18 × [7 + (- 3)]

= [18 × 7] + [18 × (- 3)]

**(b) (- 21) × [(- 4) + (- 6)] = [(- 21) × (- 4)] + [(- 21) × (- 6)]**

L.H.S. = (- 21) × [(- 4) + (- 6)]

= (- 21) × (- 10)

= 21 × 10 = 210

R. H.S. = [(- 21) × (- 4)] + [(- 21) × (- 6)]

= (21 × 4) + (21 × 6)

= 84 + 126 = 210

So, (- 21) × [(- 4) + (- 6)]

= [(- 21) × (- 4)] + [(- 21) × (- 6)].

**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) x a = -a.

**(ii) **We know that the product of any integer and (-1) is the additive inverse of an integer.

The integer whose product with (-1) is

**(a)** additive inverse of -22, t. e., 22.

**(b)** additive inverse of 37, i.e., -37.

**(c)** additive inverse of 0, i.e., 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.**

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 x (-48) + (- 48) x (- 36)

= (- 48) x 26 + (- 48) x (- 36)

= (- 48) x [26 + (- 36)]

= (- 48) x (26 – 36)

=(- 48) x (- 10)= 480

**(b)**We have,

8 x 53 x (- 125) = [8 x (- 125)] x 53

= (- 1000) x 53 = – 53000

**(c)**We have,

15 x (- 25) x (- 4) x (- 10)

=15 x [(- 25) x (-4)] x (- 10)

= 15 x (100) x (- 10)

= (15 x 100) x (- 10)

= 1500 x (- 10) = – 15000

**(d)**We have,

(- 41) x 102 = (- 41) x (100 + 2)

= (- 41) x 100 + (- 41) x 2 = -4100 – 82 = – 4182

**(e)**We have, 625 x (- 35) + (- 625) x 65

= 625 x (- 35) + (625) x (- 65)

= 625 x [(- 35)+ (- 65)]

= 625 x (- 100) = – 62500

**(f)**7 x (50 – 2) = 7 x 50 – 7 x 2

= 350 -14 =336

**(g)**(-17) x (- 29) = (-17) x [(- 30) + 1]

= (- 17) x (- 30) + (- 17) x 1 = 510 – 17 = 493

**(h)**(- 57) x (-19)+ 57 =57 x 19 + 57 x 1

= 57 x (19 +1)

= 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 10 hours after the process begins

= 40°C – 10 × 5°C

= 40°C – 50°C

= – (50 – 40)°C = – 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 si× 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)** Mohan gets for four correct answers 4 × 5 = 20 marks

He also gets for si× incorrect answers. 6 × (- 2) = – 12 marks.

Therefore, Mohan’s score = 20 + (- 12) = 20-12 = 8 marks.

**(ii)** Reshma gets for five correct answers 5 × 5 = 25 marks

She also gets for five incorrect answers 5 × (- 2) = – 10 marks Therefore, Reshma’s score = 25 + (- 10) = 25-10 = 15 marks.

**(iii)** Heena gets for two correct answers

2 × 5 = 10 marks.

She also gets for five incorrect answers 5 × (- 2) = – 10 marks

She didn’t attempt three questions. For these, she gets 3×0 = 0 marks

Therefore, Heena’s score = 10 + (- 10) + 0 = 10 – 10 + 0 = 0 marks.

**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:**

**(a)** The company sells 3,000 bags of white cement. So her profit = 3,000 × 8 = ₹ 24,000

Also, the company sells 5,000 bags of grey cement. So her loss = 5,000 × 5 = ₹ 25,000

Since 25,000 > 24,000

Therefore, the company is at a loss and the loss is = 25000 – 24000 = ₹ 1000

**(b)** Let ‘×’ be the number of white cement bags sold.

According to the question, we get

x × 8 = 6400 × 5

⇒ x = \(\frac { 6400\times 5 }{ 8 }\) = 800 × 5 = 4,000 bags.

Therefore, 4,000 bags of white cement must be sold to have neither profit nor loss.

**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) x (- 9) = 27**(b)**5 x (-7) = (-35)**(c)**7 x (-8) = (-56)**(d)**(-11) x (-12) = 132

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