What is the value of (0.02³ + 0.004³) ÷ (0.2³ + 0.04³)?
- (a)0.002
- (b)0.0045
- (c)0.0102
- (d)0.001
Answer
Why
Correct — D. Each numerator term is the matching denominator term divided by 10.
Pair: 0.02 = 0.1 × 0.2 and 0.004 = 0.1 × 0.04
Cube: 0.02³ = 0.001 × 0.2³ and 0.004³ = 0.001 × 0.04³
Factor: numerator = 0.001 × (0.2³ + 0.04³)
Divide: the bracket cancels, leaving 0.001 → option (d)
Why the others are wrong
- (a)0.002 — 0.002 is double the true value, what splitting the fraction gives: 0.02³ ÷ 0.2³ + 0.004³ ÷ 0.04³ = 0.001 + 0.001. A sum divided by a sum cannot be split into two fractions that way.
- (b)0.0045 — 0.0045 would need every term scaled by about 0.165 (0.165³ ≈ 0.0045). Here each term is scaled by exactly 0.1, so the value is 0.1³ = 0.001.
- (c)0.0102 — 0.0102 is about ten times too large. It would need a scale factor near 0.217 per term; the factor here is 0.1, and 0.1³ = 0.001.
Concept
Common-factor cancellation: if the numerator is (ka)³ + (kb)³ and the denominator is a³ + b³, the numerator equals k³(a³ + b³). The bracket cancels and the value is k³, whatever a and b are.
So pair the terms before any arithmetic. Here 0.02 pairs with 0.2 and 0.004 with 0.04, both with k = 0.1.
Direct arithmetic confirms it: 0.000008 + 0.000000064 = 0.000008064 on top, 0.008 + 0.000064 = 0.008064 below, and the quotient is 0.001.
The shortcut needs the same k for every pair. If one pair scaled by 0.1 and the other by 0.2, nothing would cancel.
Key facts
- (ka)³ + (kb)³ = k³(a³ + b³).
- Dividing a number by 10 divides its cube by 1,000.
- 0.1³ = 0.001.
Study next
Common traps
- Cancelling the factor without cubing it, which gives 0.1.
- Miscounting decimal places: 0.02³ has six, 0.000008.
The same cancellation decides 16 Sep 2025, 12:30, Quant Q.25, (0.05³ + 0.01³) ÷ (0.25³ + 0.05³): every term is divided by 5, so the value is 1⁄125 = 0.008.
18 Sep 2025, 09:00, Quant Q.24 halves every term instead: (0.12³ + 0.06³) ÷ (0.24³ + 0.12³) = 0.5³ = 0.125.
Related PYQs
No directly related past PYQ was found.