Question

A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. If x represents the speed of the boat in still water in km/hr and y represents the speed of the stream in km/hr, then _________.

• A. $x+y=11,x-y=5$
• B. $x+y=5,x-y=11$
• C. $x+y=6,x-y=10$
• D. $x+y=10,x-y=6$

Answer 1

The correct option is A $x+y=11,x-y=5$

Speed of boat in still water $=x$

Speed of stream $=y$

$\Rightarrow$ Speed of boat upstream $=x-y$

Speed of boat downstream $=x+y$

Thus,

$\frac{30}{x-y}+\frac{44}{x+y}=10$

Let $\frac{1}{x-y}=a$ and $\frac{1}{x+y}=b$

So we have,

$30a+44b=\mathrm{10......}\left(1\right)$

Similarly,

$40a+55b=\mathrm{13......}\left(2\right)$

Multiplying $\left(1\right)$ by $4$ and $\left(2\right)$ by $3$, we get

$120a+176b=\mathrm{40......}\left(3\right)$

$120a+165b=\mathrm{39......}\left(4\right)$

Subtracting $\left(4\right)$ from $\left(3\right)$, we get

$b=\frac{1}{11}$

$\Rightarrow x+y=11$

Putting $b=\frac{1}{11}$ in $\left(1\right)$, we get

$30a+4=10$

$a=\frac{1}{5}$

$\Rightarrow x-y=5$