Question

A motor boat whose speed is 20 km/h in still water takes 1 hour more to go 48 km upstream

than to return downstream to the same spot. Find the speed of the stream.

Answer 1

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》To find -

Speed of the stream.

》Given -

Velocity of boat in still water = $20\:km\:{h}^{-1}$

》Solution -

Let the velocity of the stream be $x\:km\:{h}^{-1}$

And we are given that velocity of boat in still water is $20\:km\:{h}^{-1}$

Now,

Velocity upstream = $20-x$

Velocity downstream = $20+x$

According to question,

⇒ $\frac{48}{20-x}-\frac{48}{20+x}$ = $1$

⇒ $48(20+x-20+x)$ = $400-{x}^{2}$

⇒ $48(2x)$ = $400-{x}^{2}$

⇒ ${x}^{2}+96-400$ = $0$

⇒ ${x}^{2}+100x-4x-400$ = $0$

⇒ $x(x+100)-4(x+100)$ = $0$

⇒ $(x-4)(x+100)$ = $0$

⇒ $x=-100$ OR $x=4$

Since, speed can never be negative.

Therefore, speed will be $4\:km\:{h}^{-1}$.

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Answer 2

Answer -

Given that velocity of boat in still water is 20 km/h.

Now, let us assume that speed of stream is x km/h.

So, velocity of water upstream = $\frac{48}{20-x}$

Velocity of water downstream = $\frac{48}{20+x}$

Now ATQ,

==> $\frac{48}{20-x}-\frac{48}{20+x}$ = $1$

==> $48(20+x-20+x)$ = $400-{x}^{2}$

==> $48(2x)$ = $400-{x}^{2}$

==> ${x}^{2}+96-400$ = $0$

==> ${x}^{2}+100x-4x-400$ = $0$

==> $x(x+100)-4(x+100)$ = $0$

==> $(x-4)(x+100)$ = $0$

==> $x=-100$ & $x=4$

We know that speed can't be negative. So, the speed of the stream is 4 km/h.