CAT 1999 — QA Question 40

Basics of TSD/ProportinalityEasy

Passage / Data

Directions: Answer the questions 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 km long.
Ms X and Mr Y leave A at 8.00 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?

Answer & solution

A
47.5 kmph
B
49.5 kmph
50 kmph
D
52 kmph

Solution

Since AD = DC, the distance travelled is same for the two stretches.
Hence, the average speed is given by $\frac{2 a b}{(a + b)}$ = $\frac{(2 \times 45 \times 55)}{(45 + 55)} =$ 49.5 kmph