Question #43427

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Question #43427

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Answer 1 · 2015-04-20T04:49:21

It's important to understand what Downstream and Upstream mean here.

Let's assume the Speed of the Boat in Still water to be $S_{b}$ $S_b$ and the speed of the current to be $S_{c}$ $S_c$ (which is 4kmph )

$\cdot D o w n s t r e a m$ $color(green)(*Downstream$ would mean that the current and the boat are moving in the SAME direction . Hence the speed of the current would enhance the speed of the boat and the overall speed would be $S_{b} + S_{c}$ $S_b+S_c$

We know that $T i m e = \frac{D i s tan c e}{S p e e d}$ $Time = (Distance) / (Speed)$

For Downstream it will be
$T i m e (D o w n s t r e a m) = \frac{40}{S_{b} + S_{c}} = \frac{40}{S_{b} + 4}$ $color(green)(Time(Downstream) = 40 / (S_b + S_c) = 40 / (S_b + 4)$

$\cdot U p s t r e a m$ $color(red)(*Upstream$ would mean that the current and the boat are moving in the OPPOSITE direction . Hence the speed of the current would restrict the speed of the boat and the overall speed would be $S_{b} - S_{c}$ $S_b-S_c$

For Upstream it will be
$T i m e (U p s t r e a m) = \frac{24}{S_{b} - S_{c}} = \frac{24}{S_{b} - 4}$ $color(red)(Time(Upstream) = 24 / (S_b - S_c) = 24 / (S_b - 4)$

Based on the data given,

$T i m e (D o w n s t r e a m) = T i m e (U p s t r e a m)$ $color(green)(Time(Downstream) = color(red)(Time(Upstream)$

$\frac{40}{S_{b} + 4} = \frac{24}{S_{b} - 4}$ $40 / (S_b + 4) = 24 / (S_b - 4)$

$40 (S_{b} - 4) = 24 (S_{b} + 4)$ $40(S_b - 4) = 24(S_b + 4)$

$40 S_{b} - 160 = 24 S_{b} + 96$ $40S_b - 160 = 24S_b + 96$

$40 S_{b} - 24 S_{b} = 96 + 160$ $40S_b - 24S_b = 96 + 160$

$16 S_{b} = 256$ $16S_b = 256$

$S_{b} = 16$ $S_b = 16$

The rate of Jai Singh’s boat in still water will be $16 k m p h$ $color(blue)(16 kmph$

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Question #43427

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