Quantitative Aptitude Ques 2257

Question: When a person cycled at 10 km/h, he arrived at his office 6 min late. He arrived 6 min early, when he increased his speed by 12 km/h. The distance of his office from the starting place is

Options:

A) 6 km

B) 7 km

C) 12 km

D) 16 km

Show Answer

Answer:

Correct Answer: C

Solution:

  • Let the distance be D km, and actual time taken by man to reached the office at time = x According to the question, $ \frac{D}{10}=x+\frac{6}{60} $ … (i) and $ \frac{D}{12}=x-\frac{6}{60} $ … (ii) From Eqs (i) and (ii), we get $ \frac{D}{10}-\frac{D}{12}=x+\frac{6}{60}-x+\frac{6}{60} $ $ \frac{12D-10D}{120}=\frac{1}{10}+\frac{1}{10} $
    $ \Rightarrow $ $ \frac{2D}{120}=\frac{2}{D}=12,km $ Alternate Method Here, $ b _1=6, $ $ t _2=6, $ $ S _1=10 $ and $ S _2=12 $ Distance $ =\frac{(t _1+t _2)S _1,.,S _2}{(S _2-S _1)\times 60}=\frac{(6+6),10\times 12}{(12-10)\times 60} $ $ =\frac{12\times 120}{2\times 60}=\frac{24}{2}=12,km $
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