A cylinder (r = 4 cm, h = 10 cm) is bored through by a hole (r = 2 cm, full height). What percentage of original volume is removed?
- (a)25%
- (b)50%
- (c)75%
- (d)37.5%
Answer
Why
Correct — A. The hole is a cylinder of the same height, so compare the two volumes directly.
Original: π × 4² × 10 = 160π cm³
Hole: π × 2² × 10 = 40π cm³
Divide: 40π ÷ 160π = 1⁄4 = 25% → option (a)
Why the others are wrong
- (b)50% — 50% is the ratio of the radii, 2 ÷ 4. Volume depends on radius squared, so the hole holds one quarter of the volume, not half.
- (c)75% — 75% is the share that stays: 120π of 160π cm³. The question asks for the share removed, which is 40π.
- (d)37.5% — 37.5% = 3⁄8 would need a hole with r² = 6, a radius of about 2.45 cm. A 2 cm hole gives r² = 4 against 16.
Concept
Volume of a cylinder is πr²h. When the hole runs the full height, both solids share h, so the fraction removed is (hole radius ÷ outer radius)².
That square is why halving the radius removes a quarter of the volume, not a half. The 10 cm height does not affect the percentage.
Key facts
- Volume of a cylinder = πr²h.
- Volume of a hollow cylinder = πh(R² − r²).
- Here 160π − 40π = 120π cm³ remains, 75% of the original.
Study next
Common traps
- Comparing the radii (2 : 4) instead of their squares (4 : 16).
- Answering with what remains (75%) when the question asks what is removed.
17 Sep 2025, 16:00, Quant Q.13 also turns on radius entering volume as a square: a cone's radius rises 20% while its height falls 10%, keyed a 29.6% increase.
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