Distributivity — Class 8 CBSE Mathematics MCQs with Solutions
Free Class 8 CBSE Mathematics Distributivity MCQs with step-by-step solutions (30 questions). Part of Rational Numbers. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Distributivity · easy · theory
The equation (-3/7) × (1/2 + 1/5) = (-3/7) × 1/2 + (-3/7) × 1/5 is an example of which property of rational numbers?
A. Distributivity of multiplication over addition ✓ Correct
B. Commutativity of addition
C. Associativity of multiplication
D. Closure under multiplication
Solution: Multiplying a single number by a sum and spreading it over each term, a×(b+c)=a×b+a×c, is exactly the distributive law.
Q2 — Distributivity · easy · theory
Which of the following statements shows the distributive property?
A. 4 × (5 + 6) = 4 × 5 + 4 × 6 ✓ Correct
B. 4 × 5 = 5 × 4
C. 4 + (5 + 6) = (4 + 5) + 6
D. 4 × 1 = 4
Solution: Only 4 × (5 + 6) = 4 × 5 + 4 × 6 matches the form a×(b+c)=a×b+a×c; the others show commutativity, associativity and the multiplicative identity.
Q3 — Distributivity · easy · theory
For rational numbers a, b and c, the distributive law states that a × (b − c) = ?
A. a×b − a×c ✓ Correct
B. a×b − c
C. a − b×c
D. a×b + a×c
Solution: Distributivity over subtraction spreads the multiplier over both terms, giving a×b − a×c.
Q4 — Distributivity · easy · theory
The property a × (b + c) = a × b + a × c is called the distributivity of multiplication over ______.
A. addition ✓ Correct
B. subtraction
C. division
D. multiplication
Solution: Because the bracket contains a sum (b + c), this is distributivity of multiplication over addition.
Q5 — Distributivity · easy · theory
Using the distributive law, 7 × (10 + 2) = 7 × 10 + 7 × 2 = ?
A. 84 ✓ Correct
B. 72
C. 74
D. 90
Solution: 7 × 10 = 70 and 7 × 2 = 14, and 70 + 14 = 84.
Q6 — Distributivity · easy · theory
Complete using distributivity: 3/5 × (4/7 + 2/9) = 3/5 × 4/7 + 3/5 × ____
A. 2/9 ✓ Correct
B. 3/5
C. 4/7
D. 2/5
Solution: The multiplier 3/5 is distributed over each term of the bracket, so the second term must be 3/5 × 2/9.
Q7 — Distributivity · easy · theory
Taking out the common factor, (2/5 × 3/4) + (2/5 × 1/4) = 2/5 × (3/4 + 1/4). Its value is:
A. 2/5 ✓ Correct
B. 4/5
C. 2/25
D. 5/2
Solution: Inside the bracket 3/4 + 1/4 = 1, so 2/5 × 1 = 2/5.
Q8 — Distributivity · easy · theory
Using distributivity, 6 × (1/2 + 1/3) = 6 × 1/2 + 6 × 1/3 = ?
A. 5 ✓ Correct
B. 6
C. 1
D. 4
Solution: 6 × 1/2 = 3 and 6 × 1/3 = 2, and 3 + 2 = 5.
Q9 — Distributivity · easy · theory
By distributivity, (−2) × (3 + 4) = (−2) × 3 + (−2) × 4 = ?
A. −14 ✓ Correct
B. −2
C. 14
D. −1
Solution: (−2) × 3 = −6 and (−2) × 4 = −8, and −6 + (−8) = −14.
Q10 — Distributivity · easy · theory
Using the distributive law, 2/3 × (1/2 + 1/4) = ?
A. 1/2 ✓ Correct
B. 3/4
C. 1/3
D. 2/9
Solution: Since 1/2 + 1/4 = 3/4, we get 2/3 × 3/4 = 6/12 = 1/2.
Q11 — Distributivity · medium · theory
Using the distributive law, evaluate (2/5)×(3/7) + (2/5)×(4/7).
A. 2/5 ✓ Correct
B. 6/35
C. 8/35
D. 4/7
Solution: Taking out the common factor: (2/5)×(3/7 + 4/7) = (2/5)×(7/7) = (2/5)×1 = 2/5.
Q12 — Distributivity · medium · theory
Which expression correctly applies the distributive law to (-3/4)×(2/3 + 5/6)?
A. (-3/4)×(2/3) + (-3/4)×(5/6) ✓ Correct
B. (-3/4)×(2/3) + (5/6)
C. (-3/4)×(2/3) − (-3/4)×(5/6)
D. (-3/4) + (2/3)×(5/6)
Solution: Distributivity of multiplication over addition gives a×(b+c) = a×b + a×c, so each inner term is multiplied by (-3/4).
Q13 — Distributivity · medium · theory
Evaluate (7/9)×(15/14) − (7/9)×(1/14) using distributivity.
A. 7/9 ✓ Correct
B. 5/6
C. 7/18
D. 8/9
Solution: (7/9)×(15/14 − 1/14) = (7/9)×(14/14) = (7/9)×1 = 7/9.
Q14 — Distributivity · medium · theory
Find the value of (-5/8)×(4/7 + (-4/7)) using the distributive law.
A. 0 ✓ Correct
B. -5/8
C. 5/8
D. -5/14
Solution: Since 4/7 and -4/7 are additive inverses, their sum is 0, and (-5/8)×0 = 0.
Q15 — Distributivity · medium · theory
Use distributivity to compute (6/11)×(11/3 + 11/5).
A. 16/5 ✓ Correct
B. 12/5
C. 6/5
D. 2
Solution: (6/11)×(11/3) + (6/11)×(11/5) = 2 + 6/5 = 10/5 + 6/5 = 16/5.
Q16 — Distributivity · medium · theory
Use distributivity to evaluate (3/4)×(8/9) + (3/4)×(4/9).
A. 1 ✓ Correct
B. 2/3
C. 1/3
D. 4/3
Solution: Taking out the common factor 3/4: (3/4)×(8/9 + 4/9) = (3/4)×(12/9) = (3/4)×(4/3) = 1.
Q17 — Distributivity · medium · theory
Evaluate (3/5)×(5/6 − 5/9) using the distributive law.
A. 1/6 ✓ Correct
B. 5/6
C. 1/2
D. 5/18
Solution: (3/5)×(5/6) − (3/5)×(5/9) = 1/2 − 1/3 = 3/6 − 2/6 = 1/6.
Q18 — Distributivity · medium · theory
Using distributivity, find the value of (-2/3)×(6/5) + (-2/3)×(9/5).
A. -2 ✓ Correct
B. 2
C. -6/5
D. -10/3
Solution: (-2/3)×(6/5 + 9/5) = (-2/3)×(15/5) = (-2/3)×3 = -2.
Q19 — Distributivity · medium · theory
Simplify (4/9)×(3/8) + (5/9)×(3/8) by taking out the common factor.
A. 3/8 ✓ Correct
B. 9/8
C. 1/3
D. 3/16
Solution: Since 3/8 is common, (4/9 + 5/9)×(3/8) = (9/9)×(3/8) = 1×(3/8) = 3/8.
Q20 — Distributivity · medium · theory
Compute (-7/6)×(3/4) + (-7/6)×(1/4) using distributivity.
A. -7/6 ✓ Correct
B. 7/6
C. -7/8
D. -7/24
Solution: (-7/6)×(3/4 + 1/4) = (-7/6)×(4/4) = (-7/6)×1 = -7/6.
Q21 — Distributivity · hard · theory
To simplify (-3/5)×((2/9) − (4/7)) using the distributive law, which expansion is correct?
A. (-3/5)×(2/9) − (-3/5)×(4/7) ✓ Correct
B. (-3/5)×(2/9) + (-3/5)×(4/7)
C. (-3/5)×(2/9) − (4/7)
D. (-3/5)×(2/9) − (2/9)×(4/7)
Solution: By a×(b−c) = a×b − a×c, multiply (-3/5) by BOTH bracket terms and keep the subtraction sign, giving (-3/5)×(2/9) − (-3/5)×(4/7).
Q22 — Distributivity · hard · theory
Evaluate (5/6)×((-2/3) − (4/9)).
A. -25/27 ✓ Correct
B. -5/9
C. 25/27
D. -10/9
Solution: (-2/3) − (4/9) = -6/9 − 4/9 = -10/9, and (5/6)×(-10/9) = -50/54 = -25/27.
Q23 — Distributivity · hard · theory
Using distributivity, find the value of (15/8)×(7/3) + (15/8)×(5/3) − (15/8)×(4/3).
A. 5 ✓ Correct
B. 10
C. 15/8
D. 5/8
Solution: Take out the common factor: (15/8)×(7/3 + 5/3 − 4/3) = (15/8)×(8/3) = 120/24 = 5.
Q24 — Distributivity · hard · theory
Evaluate (-2/3)×((3/5) − (7/10)) + (-2/3)×((1/2) − (1/5)).
A. -2/15 ✓ Correct
B. 2/15
C. -4/15
D. -1/15
Solution: Factor (-2/3): the brackets give (-1/10) + (3/10) = 2/10 = 1/5, so (-2/3)×(1/5) = -2/15.
Q25 — Distributivity · hard · theory
Use the distributive law to compute (-7/8)×(4/5) + (-7/8)×(11/5).
A. -21/8 ✓ Correct
B. 21/8
C. -7/8
D. -7/24
Solution: (-7/8)×(4/5 + 11/5) = (-7/8)×(15/5) = (-7/8)×3 = -21/8.
Q26 — Distributivity · hard · theory
For a = 1/2, b = -3/4 and c = 5/6, the distributive law says a×(b+c) = a×b + a×c. What is this common value?
A. 1/24 ✓ Correct
B. 1/12
C. 1/10
D. -1/24
Solution: b + c = -9/12 + 10/12 = 1/12, so a×(b+c) = (1/2)×(1/12) = 1/24.
Q27 — Distributivity · hard · theory
Evaluate (2/9)×(-3/4) + (5/6)×(-3/4) − (1/3)×(-3/4).
A. -13/24 ✓ Correct
B. 13/24
C. -25/24
D. -13/12
Solution: Here (-3/4) is the common factor: (-3/4)×(2/9 + 5/6 − 1/3) = (-3/4)×(13/18) = -39/72 = -13/24.
Q28 — Distributivity · hard · theory
Find the value of (4/5)×((-7/8) + (2/3)).
A. -1/6 ✓ Correct
B. 1/6
C. -4/11
D. -1/30
Solution: (-7/8) + (2/3) = -21/24 + 16/24 = -5/24, and (4/5)×(-5/24) = -20/120 = -1/6.
Q29 — Distributivity · hard · theory
Using distributivity, evaluate (7/6)×(13/4) − (7/6)×(1/4).
A. 7/2 ✓ Correct
B. 7/6
C. 7/18
D. 7
Solution: (7/6)×(13/4 − 1/4) = (7/6)×(12/4) = (7/6)×3 = 21/6 = 7/2.
Q30 — Distributivity · hard · theory
Evaluate (9/4)×((2/3) − (5/6) + (1/2)).
A. 3/4 ✓ Correct
B. 9/2
C. -3/2
D. 3/8
Solution: The bracket equals 4/6 − 5/6 + 3/6 = 2/6 = 1/3, so (9/4)×(1/3) = 9/12 = 3/4.