Question

A boat takes $7$ hours to travel $30$ km upstream and $28$ km downstream. It takes $5$ hours to travel $21$ km upstream and to return back. Find the speed of the boat in still water.

• A. $10km/hr$
• B. $20km/hr$
• C. $14km/hr$
• D. $6km/hr$

Answer 1

The correct option is A

$10\mathrm{km}/\mathrm{hr}$

Step 1: Assumption

Let the speed of the boat in still water be $\mathrm{x}\mathrm{km}/\mathrm{hr}$.

Let the speed of the stream be $\mathrm{y}\mathrm{km}/\mathrm{hr}$.

The speed upstream will be $(\mathrm{x}-\mathrm{y})\mathrm{km}/\mathrm{hr}$.

The speed downstream will be $(\mathrm{x}+\mathrm{y})\mathrm{km}/\mathrm{hr}$.

Step 2: Use the given information.

The time taken to travel $30$ km upstream and $28$ km downstream is $7$ hours.

Since, $\frac{\mathrm{Distance}}{\mathrm{Speed}}=\mathit{\text{Time}}$. Therefore:

$\begin{array}{rcl}{T}_{Upstream}+T_{Downstream}& =& T_{Total}\\ \frac{30}{\mathrm{x}-\mathrm{y}}+\frac{28}{x+y}& =& 7.......\left(1\right)\end{array}$.

and the time taken to travel $21$ km upstream and to return is $5$ hours.

So, $\frac{21}{\mathrm{x}-\mathrm{y}}+\frac{21}{x+y}=5......\left(2\right)$

Step 3: Solve obtained system of equations using the elimination method.

Let $\frac{1}{\mathrm{x}-\mathrm{y}}=pand\frac{1}{\mathrm{x}+\mathrm{y}}=\mathrm{q}$.

So, the equations (1) and (2) become:

$30p+28q=7...\left(3\right)$

$21p+21q=5...\left(4\right)$

Solve as:

Multiply equation (3) by 3 and (4) by 4 to get:

$\left[30p+28q=7\right]\times 3$

$\left[21p+21q=5\right]\times 4$

$$\begin{gathered} 90p+84q=21 \\ 84p+84q=20 \end{gathered}$$

On subtracting, we get

$\begin{array}{rcl}\left(90p+84q\right)-\left(84p+84q\right)& =& 21-20\\ 6p& =& 1\\ p& =& \frac{1}{6}\end{array}$

On substitution, we get:

$\begin{array}{rcl}30\left(\frac{1}{6}\right)+28q& =& 7\\ 5+28q& =& 7\\ q& =& \frac{2}{28}\\ q& =& \frac{1}{14}\end{array}$.

Now substitute back the values to get:

$\frac{1}{\mathrm{x}-\mathrm{y}}=\frac{1}{6}and\frac{1}{\mathrm{x}+\mathrm{y}}=\frac{1}{14}$

$$\begin{gathered} x-y=6 \\ x+y=14 \end{gathered}$$

In adding, we get

$\begin{array}{rcl}\left(x-y\right)+\left(x+y\right)& =& 14+6\\ 2x& =& 20\\ x& =& 10\end{array}$.

The speed of the boat in still water is $10km/hr$.

Hence, option (a) is correct.