Question

A motorboat whose speed 20km/h in still water takes 1 hr more to go 48 km upstream than to return downstream to the same spot.Find the speed of the stream​

Answer 1

Answer:

Let, the speed of the stream be x km/hr

Speed of boat in still water =20 km/hr

∴Speed of boat with downstream 20+x km/hr

∴ Speed of boat with upstream 20−x km/hr

As per given condition

20−x48−20+x48=1

⟹48[20−x1−20+x1]=1

⟹[(20−x)(20+x)20+x−20+x]=481

⟹400−x22x=481

⟹96x=400−x2

⟹x2+96x−400=0

⟹x2+100x−4x−400

⟹x(x+100)−4(x+100)=0

⟹(x−4)(x+100)=0

Either, x=4 or x=−100

∵ Speed cannot be negative ∴x=4 km/hr is considered.

∴ the speed of the stream =4 km/hr

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Answer 2

Answer:-

Let the speed of the stream be x km/h.

Speed of boat in still water is 20km/h

• Speed of boat in upstream

$:\implies$ ( 20-X ) km/h

• Speed of boat in downstream

$:\implies$ ( 20 + X) km/h

According to the Given Condition

Motorboat whose speed 20km/h in still water takes 1 hr more to go 48 km upstream than to return downstream to the same spot.

$:\implies$ 48/(20-X) - 48/(20+X) = 1

Taking common from numerator

$:\implies$ 48 [ 1/20-X - 1/(20+X] = 1

cross multiplication

$:\implies$ 48 [ 20+x/(20-x) - (20-x)/(20+x) ] = 1

$:\implies$ 48 [ (20 + x -20 + x)/400-x² ] = 1

$:\implies$ 48 [ 2x/400-x² ] = 1

$:\implies$ 2x/400-x² = 1/48

Cross Multiplication

$:\implies$ 2x × 48 = 400-x²

$:\implies$ 96x = 400 - x²

$:\implies$ x² + 96 -400 = 0

Splitting the middle term

$:\implies$ x² +100x -4x - 400 = 0

Taking Common

$:\implies$ x ( x +100) -4 (x+100) = 0

$:\implies$ (x+100) (x-4) = 0

$:\implies$ x+100 = 0 or x-4 = 0

$:\implies$ x = -100 or x = 4

Since, the speed can't be negative

So, the value of x = 4

• Hence, the speed of the stream is 4km/h.