A girl can row her boat at $5 km/h$ in still water. If she takes $1$ hour more to row the boat $5.2 km$ upstream then to return downstream, find the speed of the stream.

What I had done so far:


  • Velocity of water be $x km/h$

We know,

Velocity of boat is $5 km/h$

Velocity of boat in upstream = $(5-x) km/h$

Velocity of boat in downstream = $(5+x) km/h$

Distance = $5.2 km$

Time taken to cover distance in upstream = $\frac{5.2}{5-x} hours$

Time taken in downstream = $\frac{5.2}{5+x} hours$

Now how can I create a quadratic equation along with this collected observations?

  • $\begingroup$ 5.3, or 5.2? You've used both in the question. $\endgroup$ – Simon Hayward Oct 30 '12 at 14:19
  • $\begingroup$ Oh! Sorry its $5.2 km$ $\endgroup$ – Saharsh Oct 30 '12 at 14:20
  • $\begingroup$ Also, do you really need to form a quadratic to solve this? Have you been instructed to? Aren't you actually solving two simultaneous equations? $\endgroup$ – Simon Hayward Oct 30 '12 at 14:22
  • $\begingroup$ Yep! It is compulsory to have a quadratic equation. And we have to solve the derived quadratic equation by factorisation or Discriminant formula $\endgroup$ – Saharsh Oct 30 '12 at 14:23
  • $\begingroup$ Misread the first sentence. Have posted an appropriate answer now. $\endgroup$ – Simon Hayward Oct 30 '12 at 14:31

Ok, so we have $2 \text{(hours)} = \frac{5.2}{5-x}+\frac{5.2}{5+x}$, which rearranges simply enough into:- $2(5+x)(5-x)= 5.2(5+x)+5.2(5-x)$ from which you can gather terms to form a neat quadratic and solve for x.

Is that enough for you to go on? I can add more if needed.

  • $\begingroup$ Can you explain, How $2 \text{(hours)} = \frac{5.2}{5-x}+\frac{5.2}{5+x}$, Rest are understood by me. $\endgroup$ – Saharsh Oct 30 '12 at 14:34
  • $\begingroup$ It takes two hours to go 5.2km upstream and them back downstream if I've understood you correctly. $\endgroup$ – Simon Hayward Oct 30 '12 at 14:36
  • $\begingroup$ @Simon I'd read this as $t_{downstream} = \frac{5.2}{5+x}$, $t_{upstream} = t_{downstream} + 1 = \frac{5.2}{5-x}$ which yields $\frac{5.2}{5-x} = \frac{5.2}{5+x} +1$ $\endgroup$ – roman Oct 30 '12 at 14:47
  • $\begingroup$ Ah ok. Yes the original question is lacking punctuation! @Alpha, is Roman's reading of the question more correct? $\endgroup$ – Simon Hayward Oct 30 '12 at 14:50
  • $\begingroup$ Also, did you mean "then" or "than" in the original question? It alters the meaning! $\endgroup$ – Simon Hayward Oct 30 '12 at 14:52

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