Question

A motor boat can travel $30km$ upstream and $28km$ downstream in $7$ hours. It can travel $21km$ upstream and return in $5$ hours. Find the speed of the boat in still water in $km/h$.

• A. $3$
• B. $15$
• C. $10$
• D. $11$

Answer 1

Answer: (C)

$10$

Let the speed of the boat in still water be $x$ km/h and speed of the stream is $y$ km/h.

Therefore, speed of the boat while upstream is $(x-y)$ km/h and speed of the boat while downstream is $(x+y)$ km/h

As we know that $speed=\frac{distance}{time}$, therefore, $time=\frac{distance}{speed}$

It is given that the motor boat can travel $30$ km upstream and $28$ km downstream in $7$ hours and also it can travel $21$ km upstream and return in $5$ hours, thus,

$\frac{30}{x+y}+\frac{28}{x-y}=7.............\left(1\right)\frac{21}{x+y}+\frac{21}{x-y}=5.............\left(2\right)$

Let $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$, then the equations $\left(1\right)$ and $\left(2\right)$ becomes:

$30u+28v=7..........\left(3\right)$

$21u+21v=5..........\left(4\right)$

Multiplying equation $\left(3\right)$ by $21$ and equation $\left(4\right)$ by $30$ we get,

$630u+588v=147..........\left(5\right)$

$630u+630v=150..........\left(6\right)$

Now subtracting equation $\left(5\right)$ from equation $\left(6\right)$, we get

$42v=3$

$\Rightarrow v=\frac{1}{14}$

Substitute the value of $v$ in equation $\left(4\right)$ then, $u=\frac{1}{6}$

Since $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$, therefore,

$x+y=6..........\left(7\right)$

$x-y=14..........\left(8\right)$

Adding equations $\left(7\right)$ and $\left(8\right)$, we get:

$2x=20$

$\Rightarrow x=10$

Hence, the speed of the boat in still water is $10$ km/h.