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A boat first travels downstream for $21$ km then it turns the opposite direction and travels upstream for $4$ km. It repeats again but now it travels downstream for $12$ km then travels upstream for $7$ km. The time it takes for the first travel and second travel is the same. What is the ratio of the boat's downstream speed to its upstream speed?

It is tempting to do $(21-12):(7-4)=3:1$. But is it correct? How can we just subtract the two downstreams 21 with 12, since the time it takes for the first downstream travel for $21$ km might not be the same as the second downstream travel ($12$ km)? I know that we are assuming the water's speed is constant, but how can we explain this working in more details?

Many thanks!

user71346
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  • What do you mean by the "first travel" and "second travel?" – N. F. Taussig Nov 09 '14 at 12:07
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    @N.F.Taussig, "first travel" means the journey of 21km downstream and 4km upstream. "Second travel" means the journey of 12km downstream and 7km upstream. – Tejas Nov 09 '14 at 12:14

1 Answers1

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Assume $u$ to be the boat's speed and $v$ to be the river's speed.

Let $t_1$,$\;t_2\;$,$\;t_3$,$\;t_4$ be the four time intervals.

We have $$t_1=\frac{21}{u+v}$$ $$t_2=\frac{4}{u-v}$$ $$t_3=\frac{12}{u+v}$$ $$t_4=\frac{7}{u-v}$$

You know that $t_1+t_2=t_3+t_4$ Rearranging, you get $t_1-t_3=t_4-t_2$ From there, you get the required ratio. The time intervals for covering 21km and 12 km are indeed not the same. But since the denominator (that is, the speed) is the same, you can directly subtract the distances. :)

Tejas
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