574 views

0 votes

A boat travels upstream from $P$ to $Q$ and downstream from $Q$ to $P$ in $4$ hours. If the speed of the boat in still water is $12$ km/hr and the speed of the current is $4$ km/hr, then what is the distance from $P$ to $Q$?

- $31\dfrac{1}{3}$ km
- $41\dfrac{1}{3}$ km
- $21\dfrac{1}{3}$ km
- $11\dfrac{1}{3}$km

1 vote

Best answer

Given that, the speed of the boat in still water is $S_{b} = 12\;\text{km/hr}$ and the speed of the current is $S_{c} = 4\;\text{km/hr}.$

In upstream : The speed $S_{1} = S_{b} – S_{c} = 12-4 = 8\;\text{km/hr}$

In downstream : The speed $S_{2} = S_{b} + S_{c} = 12+4 = 16\;\text{km/hr}$

And, the total time $T = 4\;\text{hrs}$

We know that, $\text{Speed} = \dfrac{\text{Distance}}{\text{Time}}$

Let’s distance from $P$ to $Q$ is $D.$

Now, $\dfrac{D}{S_{1}} + \dfrac{D}{S_{2}} = T$

$\implies \dfrac{D}{8} + \dfrac{D}{16} = 4$

$\implies \dfrac{2D+D}{16} = 4$

$\implies 3D = 64$

$\implies D = \dfrac{64}{3}\;\text{km} = 21\dfrac{1}{3}\;\text{km}.$

So, the correct answer is $(C).$

In upstream : The speed $S_{1} = S_{b} – S_{c} = 12-4 = 8\;\text{km/hr}$

In downstream : The speed $S_{2} = S_{b} + S_{c} = 12+4 = 16\;\text{km/hr}$

And, the total time $T = 4\;\text{hrs}$

We know that, $\text{Speed} = \dfrac{\text{Distance}}{\text{Time}}$

Let’s distance from $P$ to $Q$ is $D.$

Now, $\dfrac{D}{S_{1}} + \dfrac{D}{S_{2}} = T$

$\implies \dfrac{D}{8} + \dfrac{D}{16} = 4$

$\implies \dfrac{2D+D}{16} = 4$

$\implies 3D = 64$

$\implies D = \dfrac{64}{3}\;\text{km} = 21\dfrac{1}{3}\;\text{km}.$

So, the correct answer is $(C).$