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A walks at a uniform rate of 2 km/h and 2 h after his start, B cycles after him at the uniform rate of 5 km/h. How far from the starting point will B catch A?
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A) \( \Large 6 \frac{2}{3} min \) |
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B) \( \Large 6 \frac{1}{3} min \) |
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C) \( \Large 6 \frac{5}{7} min \) |
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D) \( \Large 6 \frac{3}{7} min \) |
Correct Answer:
| A) \( \Large 6 \frac{2}{3} min \) |
Description for Correct answer:
Let B catches A after x h.
Then, distance travelled by A in (x+2) h
= Distance travelled by B in x h
According to the question,
\( \Large 2 \left(x+2\right) = 5x \)
\( \Large 2x + 4 = 5x \)
\( \Large 3x = 4 \)
\( \Large x = \frac{4}{3} \)
Distance travelled by B in \( \Large \frac{4}{3} h \)
\( \Large = \frac{4}{3} \times 5 = \frac{20}{3} = 6\frac{2}{3}km \)
Let B catches A after x h.
Then, distance travelled by A in (x+2) h
= Distance travelled by B in x h
According to the question,
\( \Large 2 \left(x+2\right) = 5x \)
\( \Large 2x + 4 = 5x \)
\( \Large 3x = 4 \)
\( \Large x = \frac{4}{3} \)
Distance travelled by B in \( \Large \frac{4}{3} h \)
\( \Large = \frac{4}{3} \times 5 = \frac{20}{3} = 6\frac{2}{3}km \)
Part of solved Time and Distance questions and answers : >> Aptitude >> Time and Distance
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