# A motorboat whose speed in still water is $18km/hr$, takes $1$ an hour more to go $24$ km upstream than to return downstream to the same spot. Find the speed of the stream.

Last updated date: 23rd Mar 2023

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Answer

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Hint- For solving such types of questions, we will use the basic concepts and formula of speed and time.

Upstream means against the direction of water flow.

Downstream means in the direction of water flow.

Given:

Speed of boat in still water$ = 18km/hr$

Let speed of the stream $ = s$

Speed of boat upstream = Speed of boat in still water $ - $Speed of stream$ = 18 - s$

Speed of boat downstream= Speed of boat in still water + Speed of stream $18 + s$

Time taken for upstream = Time taken to cover downstream +1

$

\Rightarrow \dfrac{{Dis\tan c{e_{upstream}}}}{{Spee{d_{upstream}}}} = \dfrac{{Dis\tan c{e_{downstream}}}}{{Spee{d_{downstream}}}} + 1 \\

\Rightarrow \dfrac{{24}}{{18 - s}} = \dfrac{{24}}{{18 + s}} + 1 \\

\Rightarrow 24(18 + s) = 24(18 - s) + (18 - s)(18 + s) \\

\Rightarrow {s^2} + 48s - 324 = 0 \\

\Rightarrow {s^2} + 54s - 6s - 324 = 0 \\

\Rightarrow (s + 54)(s - 6) = 0 \\

\Rightarrow s = 6, - 54 \\

\Rightarrow s \ne - 54 \\

\\

$

Neglecting the negative value of s because speed of stream cannot be negative.

Thus, $s = 6km/hr$

Note- In this type of problem formulate the algebraic equation form the given conditions and then proceed further. Neglect the negative part as speed can’t be negative. As seen above the time taken by boat in order to reach upstream is always greater than the time taken by the boat to cover the same distance downstream as in downstream the speed of the boat increases due to the support from the stream.

Upstream means against the direction of water flow.

Downstream means in the direction of water flow.

Given:

Speed of boat in still water$ = 18km/hr$

Let speed of the stream $ = s$

Speed of boat upstream = Speed of boat in still water $ - $Speed of stream$ = 18 - s$

Speed of boat downstream= Speed of boat in still water + Speed of stream $18 + s$

Time taken for upstream = Time taken to cover downstream +1

$

\Rightarrow \dfrac{{Dis\tan c{e_{upstream}}}}{{Spee{d_{upstream}}}} = \dfrac{{Dis\tan c{e_{downstream}}}}{{Spee{d_{downstream}}}} + 1 \\

\Rightarrow \dfrac{{24}}{{18 - s}} = \dfrac{{24}}{{18 + s}} + 1 \\

\Rightarrow 24(18 + s) = 24(18 - s) + (18 - s)(18 + s) \\

\Rightarrow {s^2} + 48s - 324 = 0 \\

\Rightarrow {s^2} + 54s - 6s - 324 = 0 \\

\Rightarrow (s + 54)(s - 6) = 0 \\

\Rightarrow s = 6, - 54 \\

\Rightarrow s \ne - 54 \\

\\

$

Neglecting the negative value of s because speed of stream cannot be negative.

Thus, $s = 6km/hr$

Note- In this type of problem formulate the algebraic equation form the given conditions and then proceed further. Neglect the negative part as speed can’t be negative. As seen above the time taken by boat in order to reach upstream is always greater than the time taken by the boat to cover the same distance downstream as in downstream the speed of the boat increases due to the support from the stream.

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