Directions: Answer the questions based on the following information.

A road network (shown figure)connects cities A, B, C and D all road segments are the straight line. D is the mid-point of the road connecting A and C. Roads AB and BC are at right angles to each other with BC shorter than AB, the segment AB is 100km long.

Ms.X and Mr Y leave A at 8.00 a.m .., take different routes to city C and reach at the time. X takes the highway from A to B to C and travels at an average speed of 61.875 km/hr. Y takes the direct route AC and travels at 45 km/hr on segment AD. Y's speed on segment DC is 55 km/hr.

What is the average speed of Y?

47.5 km/hr

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49.5 km/hr

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50 km/hr

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52 km/hr

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Solution

The correct option is B 49.5 km/hr
(b) Time taken by Y to cover AC $= \frac{x}{45} + \frac{x}{55} = \frac{x \times 100}{55 \times 45}$

Average speed of Y between AC.
$= \frac{2 x \times 55 \times 45}{x \times 100} = 49.5$ km/hr.

or just take the harmonic mean of 45 and 55 that will give 49.5 which is the answer.

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A road network (shown figure) connects cities A, B, C and D all road segments are the straight line. D is the mid-point of the road connecting A and C. Roads AB and BC are at right angles to each other with BC shorter than AB, the segment AB is 100km long.

Ms.X and Mr Y leave A at 8.00 a.m .., take different routes to city C and reach at the time. X takes the highway from A to B to C and travels at an average speed of 61.875 km/hr. Y takes the direct route AC and travels at 45 km/hr on segment AD. Y's speed on segment DC is 55 km/hr.

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The correct option is B 49.5 km/hr(b) Time taken by Y to cover AC =x45+x55=x×10055×45 Average speed of Y between AC. =2x×55×45x×100=49.5 km/hr. or just take the harmonic mean of 45 and 55 that will give 49.5 which is the answer.