Quantitative Aptitude Ques 1027

Question: If the radii of the circular ends of a truncated conical bucket which is 45 cm high is 28 cm and 7 cm, then the capacity of the bucket (in cu cm is) (take $ \pi =\frac{22}{7} $ )

Options:

A) 48510

B) 45810

C) 48150

D) 48051

Show Answer

Answer:

Correct Answer: A

Solution:

  • R = 28 cm $ r=7cm $ Volume of bucket $ =\frac{1}{3}\pi h(R^{2}+r^{2}+Rr) $ $ =\frac{1}{3}\times \frac{22}{7}\times 45[28^{2}+7^{2}+(28\times 7)] $ $ =\frac{1}{3}\times \frac{22}{7}\times 45(784+49+196) $ $ =\frac{1}{3}\times \frac{22}{7}\times 45\times 1029=48510cm^{3} $
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