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A man rides one third of the distances from A to B at the rate of \( x\)km/hr and the remainder at the rate of \( 2y\) km/hr. If he had travelled at a uniform rate of \( 6z\) km/hr, he could have ridden from A to B and back again in he same time Which one of the following is correct?

A) \( z\) = \( x\) + \( y\)
B) \(3 z\) = \( x\) + \( y\)
C) \( \Large \frac{1}{z} \) = \( \Large \frac{1}{x} \)+\( \Large \frac{1}{y} \)
D) \( \Large \frac{1}{2z} \) = \( \Large \frac{1}{x} \)+\( \Large \frac{1}{y} \)

Correct Answer:



C) \( \Large \frac{1}{z} \) = \( \Large \frac{1}{x} \)+\( \Large \frac{1}{y} \)

Description for Correct answer:

Time taken to travel from A to B with x km/hr speed

\( \Large \frac{d}{3x} hr\)

Time taken to travel from B to C with y km/hr speed

\( \Large = \frac{2d}{3(2y)}hr\)

Total time taken = \( \Large \left(\frac{d}{3x} + \frac{d}{3y}\right)hr\)

\( \Large \frac{d}{3} \left(\frac{1}{x} + \frac{1}{y}\right)hr\)

\( \Large (T) = \frac{d}{3} \left( \frac{y + x}{xy}\right)hr\)

Now according to question,

\( \Large \frac{d}{3} \left( \frac{y + x}{xy}\right) = \frac{2d}{3(6z)} \)

\( \Large \frac{x + y}{xy} = \frac{1}{z} \)

\( \Large \frac{1}{z} = \frac{1}{x} + \frac{1}{y} \)

Option (C)is correct.

Part of solved CDS Maths(1) questions and answers : Exams >> CDSE >> CDS Maths(1)

Comments

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