**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.

**NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1****NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.2****NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3****NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4**

**Ex 1.3 Class 7
Maths Question 1.**

Find each of the following products:**(a)**3 × (-1)

**(b)**(-1) × 225

**(-21) × (-30)**

**(c)****(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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