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Re: A motorboat whose speed is 20 Kmph in still water goes 30 km downstrea [#permalink]
tapas3016 wrote:
Time = \(\frac{Distance}{Speed}\)

5 = \(\frac{Total Distance travelled}{Speed in still water + Speed in stream}\) + \(\frac{Total Distance travelled}{Speed in still water - Speed in stream}\)

Let the speed in stream be X kmph

5 = \(\frac{30}{20+x}\) +\(\frac{30}{20-x }\)

On Solving, We get

400 -\(x^2\)= 240
\(x^2\) = 160

Taking Square root, it will be approx 4*3.16 which is greater than 10.

Hence A

Sir, I did not understand this formula. Thanks in Advance :heart
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A motorboat whose speed is 20 Kmph in still water goes 30 km downstrea [#permalink]
Let's take a speed of boat in a still water = \(x\) & speed of current = \(y\)

Now, when the boat travels downstream, the relative speed = \(|x + y|\)
& when the boat travels upstream, the relative speed = \(|x - y|\)


kumarneupane4344 wrote:
tapas3016 wrote:
Time = \(\frac{Distance}{Speed}\)

5 = \(\frac{Total Distance travelled}{Speed in still water + Speed in stream}\) + \(\frac{Total Distance travelled}{Speed in still water - Speed in stream}\)

Let the speed in stream be X kmph

5 = \(\frac{30}{20+x}\) +\(\frac{30}{20-x }\)

On Solving, We get

400 -\(x^2\)= 240
\(x^2\) = 160

Taking Square root, it will be approx 4*3.16 which is greater than 10.

Hence A

Sir, I did not understand this formula. Thanks in Advance :heart
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Re: A motorboat whose speed is 20 Kmph in still water goes 30 km downstrea [#permalink]
rx10 wrote:
Let's take a speed of boat in a still water = \(x\) & speed of current = \(y\)

Now, when the boat travels downstream, the relative speed = \(|x + y|\)
& when the boat travels upstream, the relative speed = \(|x - y|\)


kumarneupane4344 wrote:
tapas3016 wrote:
Time = \(\frac{Distance}{Speed}\)

5 = \(\frac{Total Distance travelled}{Speed in still water + Speed in stream}\) + \(\frac{Total Distance travelled}{Speed in still water - Speed in stream}\)

Let the speed in stream be X kmph

Thank you :blushing: :blushing:

5 = \(\frac{30}{20+x}\) +\(\frac{30}{20-x }\)

On Solving, We get

400 -\(x^2\)= 240
\(x^2\) = 160

Taking Square root, it will be approx 4*3.16 which is greater than 10.

Hence A

Sir, I did not understand this formula. Thanks in Advance :heart
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Re: A motorboat whose speed is 20 Kmph in still water goes 30 km downstrea [#permalink]
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kumarneupane4344

Let me reply on his behalf!

Let the speed of boat be \(b\) and that of the stream be \(x\)

Total Time taken for the trip = Time taken to travel downstream + Time taken to travel upstream

Time taken to travel downstream = \(\frac{d}{b+x} = \frac{30}{20+x}\)

Time taken to travel upstream = \(\frac{d}{b-x} = \frac{30}{20-x}\)

We have been given total time as five hours. So,

\(5 = \frac{30}{20+x} + \frac{30}{20-x}\)
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Re: A motorboat whose speed is 20 Kmph in still water goes 30 km downstrea [#permalink]
KarunMendiratta wrote:
kumarneupane4344

Let me reply on his behalf!

Let the speed of boat be \(b\) and that of the stream be \(x\)

Total Time taken for the trip = Time taken to travel downstream + Time taken to travel upstream

Time taken to travel downstream = \(\frac{d}{b+x} = \frac{30}{20+x}\)

Time taken to travel upstream = \(\frac{d}{b-x} = \frac{30}{20-x}\)

We have been given total time as five hours. So,

\(5 = \frac{30}{20+x} + \frac{30}{20-x}\)



Thank you sir. Kudos!
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Re: A motorboat whose speed is 20 Kmph in still water goes 30 km downstrea [#permalink]
BrushMyQuant wrote:
Theory

Downstream: If a boat with speed v in still water is going down on a stream of water which is flowing at speed u then the speed of the boat downstream will become v + u

Upstream: If a boat with speed v in still water is going up on a stream of water which is flowing at speed u then the speed of the boat Upstream will become v - u

Let Speed of the stream is u kmph

Downstream
Distance = 30 km
Speed = 20 + u kmph
Time = \(\frac{Distance}{Speed}\) = \(\frac{30 km}{20+u kmph}\)
= \(\frac{30 }{ 20+u}\) hours

Upstream
Distance = 30 km
Speed = 20 - u kmph
Time = \(\frac{Distance}{Speed}\) = \(\frac{30 km}{20-u kmph}\)
= \(\frac{30 }{ 20-u}\) hours

Total Time Downstream + Upstream = \(\frac{30 }{ 20+u}\) hours + \(\frac{30 }{ 20-u}\) hours = 5 hours [given]

=> 30*(20-u) + 30*(20+u) = 5*(20-u)*(20+u) [ Divide both sides by 5 we get]
=> 6*(20-u) + 6*(20+u) = \(20^2\) - \(u^2\) [ Using (a-b)*(a+b) = \(a^2\) - \(b^2\) ]
=> 120 - 6u + 120 + 6u = 400 - \(u^2\)
=> 240 = 400 - \(u^2\)
=> \(u^2\) = 400 - 240 = 160
=> u = 12.64

=> Quantity A (12.64) > Quantity B (10)

So, answer will be A
Hope it helps!



Well explanation, Thank you ankit sir!
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Re: A motorboat whose speed is 20 Kmph in still water goes 30 km downstrea [#permalink]
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