Question

32. A motor boat whose speed in
still water is 18 km/h, takes 1 hour
more to go 24 km upstream than
to return downstream to the same
spot. Find the speed of the stream.
​

Answer 1

Step-by-step explanation:

Given:-

Speed of boat =18km/hr

Distance =24km

Let x be the speed of stream.

Let t

1

and t

2

be the time for upstream and downstream.

As we know that,

speed=

time

distance

⇒time=

speed

distance

For upstream,

Speed =(18−x)km/hr

Distance =24km

Time =t

1

Therefore,

t

1

=

18−x

24

For downstream,

Speed =(18+x)km/hr

Distance =24km

Time =t

2

Therefore,

t

2

=

18+x

24

Now according to the question-

t

1

=t

2

+1

18−x

24

=

18+x

24

+1

⇒

18−x

1

−

18+x

1

=

24

1

⇒

(18−x)(18+x)

(18+x)−(18−x)

=

24

1

⇒48x=(18−x)(18+x)

⇒48x=324+18x−18x−x

2

⇒x

2

+48x−324=0

⇒x

2

+54x−6x−324=0

⇒x(x+54)−6(x+54)=0

⇒(x+54)(x−6)=0

⇒x=−54 or x=6

Since speed cannot be negative.

⇒x



=−54

∴x=6

Thus the speed of stream is 6km/hr

Hence the correct answer is 6km/hr.

Answer 2

Given: Speed of Motorboat is 18km/hr.

❏ Let the speed of the stream be x km/hr.

Therefore,

Speed of Motorboat in downstream = (18 + x) km/hr.

And,

Speed of Motorboat in upstream = (18 - x) km/hr.

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⠀⠀⠀

$\underline{\boldsymbol{According\: to \:the\: Question :}}$ ⠀

⠀⠀⠀

$:\implies\sf \dfrac{24}{18 - x} - \dfrac{24}{18+x} = 1 \\\\\\:\implies\sf \dfrac{24(18 + x) - 24(18 - x)}{(18 - x) (18 +x)} = 1 \\\\\\:\implies\sf \dfrac{24( \:\cancel{18} + x - \:\cancel{18} + x}{(18 - x) (18 +x)} = 1 \\\\\\:\implies\sf \dfrac{24(2x)}{324 - x^2} = 1\\\\\\:\implies\sf 324 - x^2 = 48x\\\\\\:\implies\sf -x^2 - 48x + 324 = 0\\\\\\:\implies\sf x^2 + 48x - 324 = 0\\\\\\:\implies\sf x^2 - 6x + 54x - 324 = 0\\\\\\:\implies\sf x(x - 6) +54(x - 6) = 0\\\\\\:\implies\sf (x -6) (x + 54) = 0\\\\\\:\implies{\underline{\boxed{\frak{\purple{ x = 6 \: and \: -54}}}}}\:\bigstar$

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Ignoring negative value, because speed can't be negative.

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$\therefore\:{\underline{\sf{Hence, \: speed \: of \ the \: stream \: is\: \bf{6 km/hr}.}}}$