Management Aptitude Test 2014 Paper

(b) Let the time taken by Ram to finish the race = t min
Case I
Ram gives Shyam a start of 200 m and beats him by 30 s, i.e., $\frac{1}{2}$ min.
∴ Speed of Ram = $\frac{2000}{\mathit{t}}$
and, speed of Shyam = $\frac{1800}{\mathit{t}+\frac{1}{2}}$
Case II
Here, Ram gives Shyam a start of 3 min and losses to Shyam by 1000 m.
∴ Time taken by Shyam - Time taken by Ram = 3 m
$\frac{2000}{\frac{1800}{\mathit{t}+\frac{1}{2}}}-\frac{1000}{\frac{2000}{\mathit{t}}}$ = 3
⇒ $\frac{2000}{\frac{1800\times 2}{2\mathit{t}+1}}-\frac{\mathit{t}}{2}$ = 3
⇒ $\frac{2000\times \left(2\mathit{t}+1\right)}{1800\times 2}-\frac{\mathit{t}}{2}$ = 3
⇒ $\frac{5}{9}\left(2\mathit{t}+1\right)-\frac{\mathit{t}}{2}$= 3
⇒ $\frac{10\left(2\mathit{t}+1\right)-9\mathit{t}}{18}$ = 3
⇒ 20t + 10 - 9t = 18 x 3
⇒ 11t = 54 - 10
∴ t = $\frac{44}{10}$ = 4min
So, time taken by Ram to finish the race is 4 min.
Again, speed of Shyam
= $\frac{1800}{\mathit{t}+\frac{1}{2}}$ ⇒ $\frac{1800}{4+\frac{1}{2}}$
= $\frac{1800}{8+1}\times 2$
= 400 m/min
So, to cover 2000 m Shyam will take 5 min.