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Question

A man travels three-fifth of a distance AB at a speed 3a, and the remaining at a speed 2b. If he goes from B to A and returns at a speed 5c in the same time, then :

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Solution

The correct option is **C**

**Approach 1:** Conventional Approach :

option (c) time taken to cover AC =3x5×3a=x5a hr.

Time taken to cover CB =x5b hr.

Time taken to cover BA and back AB =2x5c

Given x5a+x5b=2x5c→1a+1b=2c

**Approach 2: **Shortcut using Assumption !

As the questions in terms of variables we can assume values for all the given variables .

Let the distance AB be 5 kms:

The man travels three-fifth of a distance AB at a speed 3a. If we take a=1 we see that he covers a distance of 3 km. The remaining distance of 2 kms is covered in 2b time we get b=1. Also, he goes from B to A and returns at a speed 5c in the same time, we see that c=0.5. Sub these values into the options we find that only (c) fits in. (all the other options get eliminated)

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