Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Part 2 Chapter 2 Operations with Integers MCQ improves accuracy in objective exams.
MCQ on Operations with Integers Class 7
Operations with Integers MCQ Class 7
Class 7 Maths Operations with Integers MCQ
Question 1.
If sum of two integers is – 5, and one of the integers is 2, then the other integer is:
(a) -2
(b) -3
(c) -7
(d) -8
Solution:
(c) -7
Given, sum of two integers is – 5 and one of the integers is 2.
Let the other integer be 6.
Then, 2 + 6 = – 5 ⇒ b = – 5 – 2 = – 7
Question 2.
A pair of integers that give a product of – 20 is:
(a) -4, -5,
(b) 4, -3
(c) -5, 4
(d) -2, -10
Solution:
(c) -5, 4
We know, ( + ) × (-) = (-) or (-) × (+) = (-)
Thus, one integer must be negative and other must be positive to get product as – 20.
Also, we know that 4 ⇒ 5 = 20.
Taking 5 as negative, we get 4 ⇒ (- 5) = – 20
Thus, the pair of integers is -5 and 4.
Question 3.
For a pair of numbers, if the sum is 10 and the difference is 4, then least number in the pair is:
(a) 5
(b) 6
(c) 7
(d) 3
Solution:
(d) 3
Let a and b be the two numbers. Then,
a + b = 10 …(i)
and a – b = 4 …(ii)
Now, (a + b) + (a – b) = 10 + 4
[On adding (i) and (ii) ]
⇒ 2 × a = 14 ⇒ a = \(\frac{14}{2}\) = 7
Substituting a = 7 in a + b = 10, we get
7 + b = 10 ⇒ b = 10 – 7 = 3
Hence, the numbers are 7 and 3, and the least number in the pair is 3.
Question 4.
The temperature at 12 noon at a certain place was 15°C. If it decreases at the rate of 3°C per hour at what time will it be – 12°C?
(a) 11 am
(b) 11 pm
(c) 10 am
(d) 9 pm
Solution:
(d) 9 pm
Given, temperature at 12 noon, i.e. 12 pm at certain place = 15°C
And, change in temperature in 1 hour = – 3°C
Suppose, – 12°C is the temperature x hours after 12 noon.
∴ x × (-3) = (- 12)- 15
⇒ x × (-3) = – 27 ⇒ x = 9
Thus, time after 9 hours
= 12 pm + 9 hours= 9 pm
Question 5.
What term is used to describe a pair of a positive and a negative token that cancel out each other in the token model?
(a) Additive inverse
(b) Opposite pair
(c) Zero pair
(d) Neutral set
Solution:
(c) Zero pair
A pair of a positive (green) and a negative (red) token is called zero pair.
Question 6.
(-3) × (-6) =
(a) 18
(b) -18
(c) 12
(d) -12
Solution:
(a) 18
We know (-) × (-) = ( + )
Thus, (- 3) × (- 6) = 3 × 6 = 18
Question 7.
If both integers are same, then sum = ____ × integer.
(a) 2
(b) 0
(c) 3
(d) 4
Solution:
(a) 2
Let both integers be a.
Then, sum = a + a = 2 × a
Hence, sum = 2 × integer
Question 8.
If -400 × 3 × 3 + 567 = – 3033, then 43 – 3600 + 567 = ?
(a) – 2990
(b) -2980
(c) -3000
(d) -2970
Solution:
(a) – 2990
Given, – 400 × 3 × 3 + 567 = – 3033
⇒ (- 3600) + 567 = – 3033
Adding 43 to both sides, we get
43 + (- 3600) + 567 = 43 + (- 3033)
⇒ 43 – 3600 + 567 = 43 – 3033 = – 2990
Question 9.
If the sum and difference for a pair of integers is the same, one of the integers, is:
(a) 1
(b) 2
(c) 3
(d) 0
Solution:
(d) 0
Let the first integer be a and the second integer be 6.
We are given that, sum = difference
∴ a + b = a – b
⇒ a + b – a + b = 0 ⇒ 2b = 0 ⇒ b = 0
Thus, one of the integers is 0.
Question 10.
What will be the sign of product if we multiply 37 negative integers and 63 positive integers?
(a) Sometimes positive
(b) Always negative
(c) Never positive
(d) Can’t be determined
Solution:
(b) Always negative
We know, product of odd number of negative integers is negative and product of any number of positive integers is positive.
∴ Product of 37 (odd) negative integers and 63 positive integers is always negative.
Question 11.
The value of (- 30) ÷ 10 is:
(a) 3
(b) -3
(c) 1
(d) 0
Solution:
(b) -3
(-30) ÷ 10 = -(30 + 10)
[∵ (-a) ÷ b = -(a ÷ b)]
= -(3) = -3
Question 12.
The value of 0 ÷ (- 12) is:
(a) 4
(b) -12
(c) 12
(d) 0
Solution:
(d) 0
0 ÷ (- 12) = -(0 ÷ 12)
[∵ a ÷ (-b) = -(a ÷ b)]
= -(0) = 0
Question 13.
An integer which when multiplied by – 9 gives 117, is:
(a) 13
(b) -1053
(c) -13
(d) 1053
Solution:
(c) -13
Let the required integer be x.
Then, x × (- 9) = 117
⇒ x = 117 ÷ (-9) = -(117 ÷ 9) = – 13
Thus, the required integer is – 13.
Question 14.
If 11 × (a + 4) = 11 × (- 3) + 11 × 4, then a is :
(a) -1
(b) -2
(c) – 3
(d) – 4
Solution:
(c) -3
11 × (a + 4) = 11 × a + 11 × 4
[Using distributive property of multiplication over addition]
Given, 11 × (a + 4) = 11 × (- 3) + 11 × 4
⇒ 11 × a + 11 × 4 = 11 × (-3) + 11 × 4
On comparing, we get a = (- 3).
Question 15.
The value of (-36) ÷ (-9) is:
(a) 4
(b) – 9
(c) 6
(d) -2
Solution:
(a) 4
(-36) ÷ (-9) = 36 ÷ 9[∵ (-a) ÷ (-b) = a ÷ b]
= 4
Question 16.
The value of [(- 7) + (- 5)] ÷ [(- 4) + 1] is:
(a) -7
(b) -5
(c) 7
(d) 4
Solution:
(d) 4
[(- 7) + (- 5)] ÷ [(- 4) + 1]
= (-7 – 5) ÷ (-4 + 1)
= (-12) ÷ (-3)
= 12 ÷ 3 [∵ (-a) ÷ (-b) = a ÷ b]
= 4
Question 17.
The value of [8 × (-5) + 15 × 2 + 2] ÷ [1 × (-4)] is:
(a) -15
(b) – 5
(c) 7
(d) 2
Solution:
(d) 2
[8 × (- 5) + 15 × 2 + 2] ÷ [1 × (- 4)]
= (-40 + 30 + 2) ÷ (- 4)
= (-40 + 32) ÷ (- 4)
= (- 8) ÷ (- 4)
= 8 ÷ 4 [∵ (-a) ÷ (-b) = a ÷ b]
= 2
Question 18.
Which of the following expressions are equal to -30?
(i) -20 – (-5 × 2)
(ii) (-6 × 10) + (6 × 5)
(iii) (-2 × 5) + (-4 × 5)
(iv) (-6) × 5
Choose the correct option from the following:
(a) (ii) and (iv) only
(b) (iii) and (iv) only
(c) (ii), (iii) and (iv)
(d) (i), (ii) and (iv)
Solution:
(c) (ii), (iii) and (iv)
(i) -20 – (-5 × 2) = -20 – (-10) = -20 + 10 = -10
(ii) (-6 × 10) + (6 × 5) = -60 + 30 = – 30
(iii) (-2 × 5) + (-4 × 5) = – 10 + (- 20)
= -10 – 20 = -30
(iv) (-6) × 5 = – 30
Thus, expressions (ii), (iii) and (iv) are equal to – 30.
Question 19.
Which of the following expressions result in a negative integer?
(i) (-2) × (-3)
(ii) 10 + 3 × (-7)
(iii) (-5) × 2 + 10
(iv) (-5) × 7
Choose the correct option from the following:
(a) (i) and (iv)
(b) (iii) and (iv)
(c) (i) and (ii)
(d) (ii) and (iv)
Solution:
(d) (ii) and (iv)
(i) (- 2) × (- 3) = 6, which is a positive integer.
(ii) 10 + 3 × (- 7) = 10 + (- 21) = 10 – 21
= -11, which is a negative integer.
(iii) (-5) × 2 + 10 = -10 + 10 = 0, which is neither a positive nor a negative integer.
(iv) (-5) × 7 = -35, which is a negative integer.
Thus, the expressions (ii) and (iv) result in a negative integer.
Question 20.
Which of the following expressions are equal to -25?
(i) – 20 – (5 × 7)
(ii) (-7 × 1) + (- 3 × 6)
(iii) (-5 × 10) + (5 × 5)
(iv) (-5) × 8
Choose the correct option from the following:
(a) (ii) and (iii)
(b) (ii) and (iv)
(c) (iii) and (iv)
(d) (i) and (iv)
Solution:
(a) (ii) and (iii)
(i) -20 – (5 × 7) = – 20 – 35 = – 55 + – 25
(ii) (-7 × 1) + (-3 × 6) = -7 + (-18) = -7 – 18 = -25
(iii) (-5 × 10) + (5 × 5) = -50 + 25 = – 25
(iv) (-5) × 8 = -(5 × 8) = -40 ≠ -25
Thus, expressions (ii) and (iii) are equal to – 25.
Question 21.
Which of the following expressions result in odd integer?
(i) (-7) × (-3)
(ii) 10 + 3 × (- 2)
(iii) (-4) × 2 + 8
(iv) (-3) × 9
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Solution:
(d) (i) and (iv)
(i) (-7) × (-3) = 21,
which is an odd integer.
(ii) 10 + 3 × (- 2) = 10 + (- 6) = 4, which is an even integer.
(iii) (-4) × 2 + 8 = -8 + 8 = 0, which is an even integer.
(iv) (- 3) × 9 = – 27, which is an odd integer.
Thus, the expressions (i) and (iv) result in odd integers.
Question 22.
Which of the following expressions are correct?
(i) 72 ÷ (- 8) = – 9
(ii) 36 ÷ (-12) = – 3
(iii) (-78) ÷ (-2) = 38
Choose the correct option from the following:
(a) (i), (ii) and (iii)
(b) (ii) and (iii) only
(c) (i) and (ii) only
(d) (i) only
Solution:
(c) (i) and (ii) only
(i) 72 ÷ (- 8) = – (72 ÷ 8) = -(9) = – 9
(Correct)
(ii) 36 ÷ (-12) = -(36 ÷ 12) = -3 (Correct)
(iii) (-78) ÷ (-2) = 78 ÷ 2 = 39 (Incorrect)
Operations with Integers Class 7 Assertion and Reason Questions
The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Question 1.
(A): The additive inverse of (-12) × (-3) + 4 is -40.
(R): The additive inverse of a is – a.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
(-12) × (- 3) + 4 = 36 + 4 = 40
We know, the additive inverse of a is – a.
∴ The additive inverse of 40 is – 40.
Hence, the additive inverse of (-12) × (-3) + 4 is – 40.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
Question 2.
(A): The value of (-3) × (-4) × 5 × 2 × (-1) is negative.
(R): The product of odd number of negative integers is positive.
Solution:
(c) A is true but R is false.
We know that the product of odd number of negative integers is always negative.
In product (-3) × (-4) × 5 × 2 × (-1), there are 3 (odd) negative integers.
So, the product is negative.
Thus, Assertion (A) is true, but Reason (R) is false
Operations with Integers Class 7 Fill in the Blanks
Question 1.
60 + 40 + ______ = 0
Solution: -100
Let the required integer be a.
Then, 60 + 40 + a = 0 ⇒ 100 + a = 0
We know that the sum of an integer and its additive inverse is 0.
∴ a = – 100
Thus, 60 + 40 + (-100) = 0.
Question 2.
(-1) × _____ = – 43
Solution: 43
We know, (-1) × a = -a, for all integers a.
Thus, (-1) × 43 = -43
Question 3.
(-9) × (-5) × 6 × (-3) = _______
Solution: -810
The product of odd number of negative integers is negative.
∴ (-9) × (-5) × 6 × (-3)
= -(9 × 5 × 6 × 3) = -810
Question 4.
If we multiply 17 positive integers and 92 negative integers, then the sign of the product is _________ .
Solution: positive
We know that the product of even number of negative integers is always positive.
Final product = (Product of 17 positive integers) × (Product of 92 negative integers)
= (Positive) × (Positive) = Positive
Thus, if we multiply 17 positive integers and 92 negative integers, then the sign of the product is
positive.
Question 5.
When two negative integers are added, we get a ______ integer.
Solution: negative
When two negative integers are added, we get a negative integer.
Question 6.
Find:
(i) -13 + 42 = _____
(ii) 47 + (-26) = ______
(iii) 91 – 19 = ______
(iv) -38 – (-29) = _____
Solution: 29, 21, 72, -9
(i) -13 + 42 = 29
(ii) 47 + (-26) = 21
(iii) 91 – 19 = 72
(iv) -38 – (-29) = – 9
Question 7.
(7 + 3) + (5 + 2) + ( _____ + 6)
= (3 + 6) + (5 + 7) + (2 + 9)
Solution: 9
RHS = (3 + 6) + (5 + 7) + (2 + 9)
= 3 + 6 + 5 + 7 + 2 + 9
= (7 + 3) + (5 + 2) + (6 + 9)
= (7 + 3) + (5 + 2) + (9 + 6) = LHS
∴ (7 + 3) + (5 + 2) + (9 + 6)
= (3 + 6) + (5 + 7) + (2 + 9)
Question 8.
-19 + 23 – 8 + _____ = 12
Solution: 16
Let the required number be x. Then
-19 + 23 – 8 + x = 12 ⇒ 23 – 27 + x = 12
⇒ -4 + x = 12 ⇒ x = 12 + 4= 16
∴ -19 + 23 – 8 + 16 = 12
Question 9.
(-8) × (-9) + (-3) × (-3) × (-3) = _________
Solution: 45, 9, -36
(-8) × (- 9) + (- 3) × (- 3) × (- 3)
= (8 × 9) – (3 × 3 × 3) = 72 – 27 = 45
Question 10.
(-4) × (10 – 5 – 4 + 8) = (-4) × ____ = ______
Solution: 9, -36
(-4) × (10 – 5 – 4 + 8) = (-4) × 9 = – 36
Question 11.
When one of the multiplier or the multiplicand is positive and the other is negative, their product is _______ .
Solution: negative
We know that, when one of the multiplier or the multiplicand is positive and the other is negative, their product is negative.
Question 12.
When both the multiplier and the multiplicand are negative, the product is __________ .
Solution: positive
When both the multiplier and the multiplicand are negative, the product is positive.
Question 13.
Fill in the blanks keeping in view the properties of multiplication and division of integers:
(i) (-2) × 5 = _____ × (-2)
(ii) (-4) × [(____) + (2)] = (-4) × (-5) + (______) × (2)
(iii) 100 × [(____) × (-45)] = [ ____ × (-4)] × (-45)
(iv) (-15) ÷ (_____) = (15) ÷ (-3)
Solution: 5, -5, -4, -4, 100, 3
(i) (-2) × 5 = 5 × (- 2)
[Using commutative law of multiplication]
(ii) (-4) × [(-5) + (2)] = (-4) × (- 5) + (-4) × (2)
[Using distributive law of multiplication]
(iii) 100 × [(-4) × (-45)] = [100 × (-4)] × (-45)
[Using associative law of multiplication]
(iv) (-15) ÷ (3) = (15) ÷ (-3)