Xavier Aptitude Test 2013 Paper

(a) Let us take that Ram and Hari initially travels for distance ‘x’ km before Hari gets off and starts walking and Ram gets back in the car to pick up Shyam. Then he picks up Shyam and both of them travel upto the end.
Hari travels in car with the speed of 25 kmph upto ‘x’ km and then he completes the remaining journey of (100 – x) km by walking at a speed of 5 kmph.
time taken by Hari $=\mathrm{}\frac{\mathit{x}}{25}+\mathrm{}\frac{100-\mathit{x}}{5}$
$=\mathrm{}\frac{500-4\mathit{x}}{25}$ ..... (1)
Shyam walks at a speed of 5 kmph for distance (x – y) km,
where he meets Ram and continues the remaining journey of 100 – (x – y) km in the car at a speed of 25 kmph.
$\therefore$ Time taken by Shyam $=\mathrm{}\frac{\mathit{x}-\mathit{y}}{5}+\frac{100-\left(\mathit{x}-\mathit{y}\right)}{25}$
$=\frac{100+4\left(\mathit{x}-\mathit{y}\right)}{25}$ ........ (2)
Ram travels in car with the speed of 25 kmph upto‘x’ km and then he goes back in car a distance of ‘y’ km and picks up Shyam and again travels back ‘y’ km distance to reach the point of ‘x’ km and continues further to complete the remaining (100 – x) km journey.
So, Ram travels in the car a distance of
x + y + y + 100 – x = 100 + 2y
$\therefore$ Time taken by Ram $=\mathrm{}\frac{100+2\mathit{y}}{25}$ ...... (3)
As all three reach the destination at same time
On solving, we get $\mathit{x}=75$ and $\mathit{y}=50$
If we substitute these values in any equation (1) or (2) or (3), we get the answer of 8 hrs.