CAT 1999 | Question: 103

Question

Answer the question based on the following information.

A road network (shown in the figure below) connects cities A, B, C and D. All road segments are straight lines. D is the mid-point on 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 $100 \; \text{km}$ long.

Ms X and Mr Y leave A at $8.00 \; \text{a.m.,}$ take different routes to city C and reach at the same time. X takes the highway from A to B to C and travels at an average speed of $61.875$ kmph. Y takes the direct route AC and travels at $45$ kmph on segment AD. Y’s speed on segment DC is $55$ kmph.

What is the average speed of Y?

• A. $47.5$ kmph
• B. $49.5$ kmph
• C. $50$ kmph
• D. $52$ kmph

Answer 1

Firstly kindly have a look on this link http://aptitude.gateoverflow.in/2323/cat1999-105 
here i calculated the value of AD=CD=BD=52.5 km 

theorem used : "Mid point of hypotenous is equidistance from all vertices . "

But for calculating avg speed you dont require the exact value of AD and BD . 
So Average speed of Y =      (AD+BD)
                                     ----------------------------  ; where (AD=BD= x km)
                                      (AD/45 +BD/55)

= 2*45*55/ 100 =49.5 km/h

Option (B) is correct .