## Integers Class 7 Extra Questions Maths Chapter 1

**Extra Questions for Class 7 Maths Chapter 1 Integers**

### Integers Class 7 Extra Questions Very Short Answer Type

Question 1.

Fill in the blanks using .

(a) -3 …… -4

(b) 6 ……. -20

(c) -8 …… -2

(d) 5 …… -7

Solution:

(a) -3 > -4

(b) 6 > -20

(c) -8

(d) 5 > -7

Question 2.

Solve the following:

(i) (-8) × (-5) + (-6)

(ii) [(-6) × (-3)] + (-4)

(iii) (-10) × [(-13) + (-10)]

(iv) (-5) × [(-6) + 5]

Solution:

(i) (-8) × (-5) + (-6)

= (-) × (-) × [8 × 5] + (-6)

= 40 – 6

= 34

(ii) [(-6) × (-3)] + (-4)

= (-) × (-) × [6 × 3] + (-4)

= 18 – 4

= 14

(iii) (-10) × [(-13) + (-10)]

= (-10) × (-23)

= (-) × (-) × [10 × 23]

= 230

(iv) (-5) × [(-6) + 5]

= (-5) × (-1)

= (-) × (-) × 5 × 1

= 5

Question 3.

Starting from (-7) × 4, find (-7) × (-3)

Solution:

(-7) × 4 = -28

(-7) × 3 = -21 = [-28 + 7]

(-7) × 2 – -14 = [-21 + 7]

(-7) × 1 = -7 = [-14 + 7]

(-7) × 0 = 0 = [-7 + 7]

(-7) × (-1) = 7 = [0 + 7]

(-7) × (-2) = 14 = [7 + 7]

(-7) × (-3) = 21 = [14 + 7]

Question 4.

Using number line, find:

(i) 3 × (-5)

(ii) 8 × (-2)

Solution:

(i) 3 × (-5)

From the number line, we have

(-5) + (-5) + (-5) = 3 × (-5) = -15

(ii) 8 × (-2)

From the number line, we have

(-2) + (-2) + (-2) + (-2) + (-2) + (-2) + (-2) + (-2) = 8 × (-2) = -16

Question 5.

Write five pair of integers (m, n ) such that m ÷ n = -3. One of such pair is (-6, 2).

Solution:

(i) (-3, 1) = (-3) ÷ 1 = -3

(ii) (9, -3) = 9 ÷ (-3) = -3

(iii) (6, -2) = 6 ÷ (-2) = -3

(iv) (-24, 8) = (-24) ÷ 8 = -3

(v) (18, -6) = 18 ÷ (-6) = -3

### Integers Class 7 Extra Questions Short Answer Type

Question 6.

Solve the following:

(i) (-15) × 8 + (-15) × 4

(ii) [32 + 2 × 17 + (-6)] ÷ 15

Solution:

(i) (-15) × 8 + (-15) × 4

= (-15) × [8 + 4]

= (-15) × 12

= -180

(ii) [32 + 2 × 17 + (-6)] ÷ 15

= [32 + 34 – 6] ÷ 15

= [66 – 6] ÷ 15

= 60 ÷ 15

= 4

Question 7.

The sum of two integers is 116. If one of them is -79, find the other integers.

Solution:

Sum of two integers = 116

One integer = -79

Other integer = Sum of integer – One of integer = 116 – (-79) = 116 + 79 = 195

Question 8.

If a = -35, b = 10 cm and c = -5, verify that:

(i) a + (b + c) = (a + b) + c

(ii) a × (b + c) = a × b + a × c

Solution:

(i) Given that a = -35, b = 10, c = -5

LHS = a + (b + c) = (-35) + [10 + (-5)] = (-35) + 5 = -30

RHS = (a + b) + c = [(-35) + 10] + (-5) = (-25) + (-5) = -(25 + 5) = -30

LHS = RHS

Hence, verified.

(ii) a × (b + c) = a × b + a × c

LHS = a × (b + c) = (-35) × [10 + (-5)] = (-35) × 5 = -175

RHS = a × b + a × c = (-35) × 10 + (-35) × (-5) = -350 + (-) × (-) × (35 × 5) = -350 + 175 = -175

LHS = RHS

Hence, verified.

Question 9.

Write down a pair of integers whose

(i) sum is -5

(ii) difference is -7

(iii) difference is -1

(iv) sum is 0

Solution:

(i) (-2) + (-3) = -5

Hence, the required pair of integers = (-2, -3)

(ii) -10 – (-3) = -10 + 3 = -7

Hence, the required pair of integers = (-10, -3)

(iii) (-3) – (-2) = -1

Hence, the required pair of integers = (-3, -2)

(iv) (-4) + (4) = 0

Hence, the required pair of integers = (-4, 4)

Question 10.

You have ₹ 500 in your saving account at the beginning of the month. The record below shows all of your transactions during the month. How much money is in your account after these transactions?

Solution:

Amount in the beginning of the month in the account = ₹ 500

Amount deposited in the account for Jal Board = ₹ 200

Amount paid to Jal Board = ₹ 120

Amount left in the account after the above transactions = ₹ (500 + 200 – 120) = ₹ (700 – 120) = ₹ 580

Amount deposited for LIC India = ₹ 150

Amount paid to LIC India = ₹ 240

Amount left after this transactions = ₹ (580 + 150 – 240) = ₹ (730 – 240) = ₹ 490

Question 11.

The given table shows the freezing points in °F of different gases at sea level. Convert each of these into °C to the nearest integral value using the relations and complete the table

C = (frac { 5 }{ 9 }) [F – 32]

Solution:

Freezing point of Hydrogen = -435°F

C = (frac { 5 }{ 9 }) [-435 – 32]

= (frac { 5 }{ 9 }) [-467]

= 5 × (-51.9)

= 259.5°C or 259°C

For Krypton, freezing point = -251°F

C = (frac { 5 }{ 9 }) [-251 – 32]

= (frac { 5 }{ 9 }) [-283]

= 5 × (-31.4)]

= -157°C

For Oxygen, freezing point = -369°F

C = (frac { 5 }{ 9 }) [-369 – 32]

= (frac { 5 }{ 9 }) [-401]

= 5 × (44.56)

= 222.80° C or 223°C

Hence, the required freezing points at sea level in °C for Hydrogen = -259°C, Krypton = -157°C, Oxygen = -223°C.

Question 12.

Taking today as zero on the number line, if the day before yesterday is 17 January, what is the date on 3 days after tomorrow? [NCERT Exemplar]

Solution:

The date before yesterday = 17 January

The date of yesterday = 17 + 1 = 18 January

Today’s date = 18 + 1 = 19 January

Tomorrow’s date = 19 + 1 = 20 January

Date on 3 days after tomorrow = (20 + 3) = 23rd January