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
Integers Questions For Class 7 Question 1.
Fill in the blanks using < or >.
(a) -3 …… -4
(b) 6 ……. -20
(c) -8 …… -2
(d) 5 …… -7
Solution:
(a) -3 > -4
(b) 6 > -20
(c) -8 < -2
(d) 5 > -7
Extra Questions For Class 7 Maths Integers With Answers 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
Integers Class 7 Extra Questions 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]
Class 7 Maths Chapter 1 Extra Questions 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
Integers Class 7 Questions 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
Class 7 Integers Extra Questions 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
Integer Questions For Class 7 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
Extra Questions For Class 7 Maths Integers 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.
Integers Extra Questions Class 7 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)
Problems On Integers For Class 7 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
Questions On Integers For Class 7 With Answers 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.
Class 7 Maths Ch 1 Extra Questions 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