Question

A motorboat whose speed in still water is 24 km h takes 1 hour more to go 32 km upstream than to return downstream to the same spot. Find speed of the stream.


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Answer 1

Answer:

• Speed of stream is 8 km/hr.

Given :

• Speed of the boat in still water =24 km/hr

• Distance between the places is 32 km.

• Difference between timings =1 hr.

To find :

• Speed of the stream =?

Step-by-step explanation:

Let the speed of the stream be x km/hr

Speed of the boat in still water =24 km/hr

Speed of the boat in upstream =(24−x) km/hr

Speed of the boat in downstream =(24+x) km/hr

Distance between the places is 32 km.

Time to travel in upstream = d/24 - x

Time to travel in downstream = d/24 + x

Difference between timings =1 hr

Time of upstream journey = Time of downstream journey +1 hr

Therefore, 32/24 - x = 32/24 + x + 1

32/(24 - x) - 32/(24 + x) = 1

768 + 32x - 768 + 32x / (24 - x) (24+ x) = 1

64x = 576 - x²

x² + 64x - 576 = 0

On factoring, we get,

(x + 72) ( x - 8) = 0

So, x = - 72 or 8 (speed of the stream cannot be negative)

Therefore, Speed of stream is 8 km/hr.

Answer 2

AnswEr :

8km/hr.

$\bf{\pink{\underline{\underline{\bf{Given\::}}}}}$

A motorboat whose speed in still water is 24 km/hrs takes 1 hour more to go 32 km upstream than to return downstream to the same spot.

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The speed of the stream.

$\bf{\orange{\underline{\underline{\bf{Explanation\::}}}}}$

Let the speed of the stream be R

Let $\sf{t_{1}}$ and $\sf{t_{2}}$ be the time for the upstream and downstream.

$\bf{\blue{\large{\underline{\underline{\tt{1_st\:\:Case\::}}}}}}$

$\bf{We\:have}\begin{cases}\sf{A\:motorboat\:speed\:for\:upstream=(24-R)km/hr}\\ \sf{Distance\:(d)=32\:km}\\ \sf{Time\:(t)=t_{1}}\end{cases}}$

Formula use :

$\bf{\large{\boxed{\bf{Time=\frac{Distance}{Speed}}}}}}}$

$\dashrightarrow\tt{t_{1}=\dfrac{32}{24-R} }$

$\bf{\blue{\large{\underline{\underline{\tt{2_nd\:\:Case\::}}}}}}$

$\bf{We\:have}\begin{cases}\sf{A\:motorboat\:speed\:for\:downstream=(24+R)km/hr}\\ \sf{Distance\:(d)=32\:km}\\ \sf{Time\:(t)=t_{2}}\end{cases}}$

$\dashrightarrow\tt{t_{2}=\dfrac{32}{24+R} }$

$\bf{\large{\red{\underline{\underline{\tt{A.T.Q.\::}}}}}}$

$\leadsto\sf{t_{1}=t_{2}+1}}$

$\mapsto\tt{\dfrac{32}{24-R} =\dfrac{32}{24+R} +1}\\\\\\\\\mapsto\tt{\dfrac{32}{24-R} -\dfrac{32}{24+R} =1}\\\\\\\\\mapsto\tt{\dfrac{32(24+R)-32(24-R)}{(24-R)(24+R)} =1}\\\\\\\\\mapsto\tt{\dfrac{\cancel{768}+32R\cancel{-768}+32R}{(24-R)(24+R)} =1}\\\\\\\\\mapsto\tt{64R=(24-R)(24+R)}\\\\\\\\\mapsto\tt{64R=576\cancel{+24R} \cancel{-24R}-R^{2} }\\\\\\\\\mapsto\tt{64R=576-R^{2} }\\\\\\\\\mapsto\tt{R^{2} +64R-576=0}\\\\\\\\\mapsto\tt{R^{2} +72R-8R-576=0}$

$\mapsto\tt{R(R+72)-8(R+72)=0}\\\\\\\\\mapsto\tt{(R+72)(R-8)=0}\\\\\\\\\mapsto\tt{R+72=0\:\:\:\:\:Or\:\:\:\:\:R-8=0}\\\\\\\\\mapsto\tt{\red{R=-72\:\:\:\:\:Or\:\:\:\:\:\:R=8}}$

We know that negative value isn't acceptable.

Then,

$\bf{\large{\orange{\underline{\sf{The\:speed\:of\:the\:stream\:is\:8\:km/hrs.}}}}}$