Management Aptitude Test Sep 2011 Paper

(4) Let upstream speed = x km /h
and downstream speed = y km /h
Then, according to the question,
$\frac{40}{\mathit{x}}+\frac{55}{\mathit{y}}$ = 13 .........(i)
and
$\frac{30}{\mathit{x}}+\frac{44}{\mathit{y}}$ = 10 ....(ii)
Multiplying Eq. (i) by 3 and Eq. (ii) by 4 and then subtracting , we get
$\frac{120}{\mathit{x}}+\frac{165}{\mathit{y}}$ = 39
$\frac{120}{\mathit{x}}+\frac{176}{\mathit{y}}$ = 40
-         -            -

0 - $\frac{11}{\mathit{y}}$ = -1
⇒ y = 11
Putting this value in Eq. (i), we get
$\frac{40}{\mathit{x}}+\frac{55}{11}$ = 13
⇒ $\frac{40}{\mathit{x}}$ = 8
⇒ x = 5
Now , speed of man in still water = $\frac{1}{2}$ ( x + y )
= $\frac{1}{2}$ ( 5 + 11 ) = $\frac{16}{2}$ = 8 km/h