Quantitative Aptitude Ques 804

Question: If $ P=\frac{1}{25}, $ then value of $ 125p^{3}-\frac{1}{64}-\frac{75}{4}p^{2}+\frac{15}{16}p $ is equal to

Options:

A) $ \frac{-1}{8000} $

B) $ \frac{1}{8000} $

C) $ \frac{1}{2000} $

D) $ \frac{-1}{2000} $

Show Answer

Answer:

Correct Answer: A

Solution:

  • $ P=\frac{1}{25} $ Now, $ 125p^{3}-\frac{1}{64}-\frac{75}{4}p^{2}+\frac{15}{16}p $ $ ={{(5p)}^{3}}-{{( \frac{1}{4} )}^{3}}-\frac{3\times 25}{4}p^{2}+\frac{3\times 5}{4^{2}}p $ $ ={{( 5p-\frac{1}{4} )}^{3}} $ On putting $ p=\frac{1}{25}, $ we get $ {{( 5\times \frac{1}{25}-\frac{1}{4} )}^{3}}={{( \frac{1}{5}-\frac{1}{4} )}^{3}} $ $ =( \frac{-1}{20} )=\frac{-1}{8000} $
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