Management Aptitude Test 2017 Paper

(d) 10 kmph
In the time taken by the boat in travelling 20+ (20 -12) km, the log travels 12 km.
Let, ${\mathit{S}}_{\mathit{B}}$ be the speed of the boat in still water and ${\mathit{S}}_{\mathit{s}}$ be the speed of the stream.
Then, $\frac{12}{{\mathit{S}}_{\mathit{S}}}$ = $\frac{20}{\left({\mathit{S}}_{\mathit{B}}+{\mathit{S}}_{\mathit{S}}\right)}$ + $\frac{8}{\left({\mathit{S}}_{\mathit{B}}-{\mathit{S}}_{\mathit{S}}\right)}$ ........(i)
It is also known that
$\frac{20}{\left({\mathit{S}}_{\mathit{B}}+{\mathit{S}}_{\mathit{S}}\right)}$ + $\frac{20}{\left({\mathit{S}}_{\mathit{B}}-{\mathit{S}}_{\mathit{S}}\right)}$ = 7 hours ........(ii)
From Eq. (i)
12 = $\frac{20{\mathit{S}}_{\mathit{S}}}{\left({\mathit{S}}_{\mathit{B}}+{\mathit{S}}_{\mathit{S}}\right)}$ + $\frac{8{\mathit{S}}_{\mathit{S}}}{\left({\mathit{S}}_{\mathit{B}}-{\mathit{S}}_{\mathit{S}}\right)}$
⇒ 12 $\left({\mathit{S}}_{\mathit{B}}^2-{\mathit{S}}_{\mathit{S}}^2\right)$ = 20${\mathit{S}}_{\mathit{S}}{\mathit{S}}_{\mathit{B}}-20{\mathit{S}}_{\mathit{S}}^2$ + $8{\mathit{S}}_{\mathit{B}}{\mathit{S}}_{\mathit{S}}+5{\mathit{S}}_{\mathit{S}}^2$
⇒ $12{\mathit{S}}_{\mathit{B}}^2-12{\mathit{S}}_{\mathit{S}}^2$ = $28{\mathit{S}}_{\mathit{S}}{\mathit{S}}_{\mathit{B}}-12{\mathit{S}}_{\mathit{S}}^2$
⇒ 12${\mathit{S}}_{\mathit{B}}=28{\mathit{S}}_{\mathit{S}}$
⇒ ${\mathit{S}}_{\mathit{B}}=\frac{7}{3}{\mathit{S}}_{\mathit{S}}$
Now, putting the value of ${\mathit{S}}_{\mathit{B}}$ in Eq. (ii) we get
$\frac{20}{\left(\frac{7}{3}{\mathit{S}}_{\mathit{S}}+{\mathit{S}}_{\mathit{S}}\right)}$ + $\frac{20}{\left(\frac{7}{3}{\mathit{S}}_{\mathit{S}}-{\mathit{S}}_{\mathit{S}}\right)}$ = 7
⇒ $\frac{60}{10{\mathit{S}}_{\mathit{S}}}+\frac{60}{4{\mathit{S}}_{\mathit{S}}}$ = 7
⇒ $\frac{120+300}{20{\mathit{S}}_{\mathit{S}}}$ = 7
⇒ ${\mathit{S}}_{\mathit{S}}=\frac{420}{7\times 20}$ = 3
∴ ${\mathit{S}}_{\mathit{B}}+{\mathit{S}}_{\mathit{S}}$ = $\frac{7}{3}{\mathit{S}}_{\mathit{S}}+{\mathit{S}}_{\mathit{S}}=\frac{10}{3}$ × 3 = 10 kmph