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In a stream that is running at 2 kmph, a man goes 10 km upstream and

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Bunuel
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In a stream that is running at 2 kmph, a man goes 10 km upstream and [#permalink] New post   03 Jan 2020, 02:31
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In a stream that is running at 2 kmph, a man goes 10 km upstream and comes back to the starting point in 55 minutes. Find the speed of the boat in still water.

A. 16 kmph
B. 18 kmph
C. 20 kmph
D. 22 kmph
E. 24 kmph
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D
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Re: In a stream that is running at 2 kmph, a man goes 10 km upstream and [#permalink] New post   03 Jan 2020, 03:23
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Speed of boat = X
Speed when boat is going upstream = X-2 kmph
Speed when the boat is going Downstream = X+2 kmph
Total Time taken by boat= 55min= 11/12 hr
Equating both sides
10/(x-2) +10/(x+2)=11/12
Using hit and trial or solving equation we get X= 22kmph

Answer D.
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Re: In a stream that is running at 2 kmph, a man goes 10 km upstream and [#permalink] New post   03 Jan 2020, 03:32
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Let X be the speed of the boat in still water. Now, we know that - speed in upstream: X - speed of the stream (2 km/h here)
Then the man comes back the same distance on downstream. Speed in Downstream: X + speed of the stream (2 km/h as well)
Again, we know d = r*t . So we can set up an equation like this:
1. For upstream: 10/(X-2)
2. For downstream: 10/(X+2)

As per the question: 10/(X-2) + 10/(X+2) = 55/60 (as the total time taken in 55 mints)

Solving this, we get: (X-22) (11X+2) = 0

So X = 22 (Ans: D)
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Re: In a stream that is running at 2 kmph, a man goes 10 km upstream and [#permalink] New post   03 Jan 2020, 03:54
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Distance to cover = 11 km
Let speed of boat = S
Speed in upstream = \(S - 2\)
Speed in downstream = \(S + 2\)
Now,
\(\frac{10}{S-2}+\frac{10}{S+2} = \frac{55}{60}\)
\(10 \times (\frac{2S}{(S-2)(S+2)}) = \frac{11}{12}\)
Cross multiplying, we get

\(10 \times 2S \times 12 = 11 \times (S-2) (S+2)\)
Now, we can see there is 1 power of 11 present in the RHS but no power of 11 in the LHS.
So, to bring one power of 11 in the LHS, the only option is D

OA,D

If we wish to check, put S = 22 and we will get

\(10 \times 2 \times 22 \times 12 = 11 \times 20 \times 24\)
And we can see both the sides are equal, hence our answer was correct.


Bunuel wrote:
In a stream that is running at 2 kmph, a man goes 10 km upstream and comes back to the starting point in 55 minutes. Find the speed of the boat in still water.

A. 16 kmph
B. 18 kmph
C. 20 kmph
D. 22 kmph
E. 24 kmph
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Re: In a stream that is running at 2 kmph, a man goes 10 km upstream and [#permalink] New post   13 Oct 2023, 00:12
xxxxniaz wrote:
Let X be the speed of the boat in still water. Now, we know that - speed in upstream: X - speed of the stream (2 km/h here)
Then the man comes back the same distance on downstream. Speed in Downstream: X + speed of the stream (2 km/h as well)
Again, we know d = r*t . So we can set up an equation like this:
1. For upstream: 10/(X-2)
2. For downstream: 10/(X+2)

As per the question: 10/(X-2) + 10/(X+2) = 55/60 (as the total time taken in 55 mints)

Solving this, we get: (X-22) (11X+2) = 0

So X = 22 (Ans: D)



can somebody tell me how to easily get (x-22)(11x+2) from 11x^2-240x-44???
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Bunuel
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Re: In a stream that is running at 2 kmph, a man goes 10 km upstream and [#permalink] New post   13 Oct 2023, 00:23
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tabishkazi wrote:
xxxxniaz wrote:
Let X be the speed of the boat in still water. Now, we know that - speed in upstream: X - speed of the stream (2 km/h here)
Then the man comes back the same distance on downstream. Speed in Downstream: X + speed of the stream (2 km/h as well)
Again, we know d = r*t . So we can set up an equation like this:
1. For upstream: 10/(X-2)
2. For downstream: 10/(X+2)

As per the question: 10/(X-2) + 10/(X+2) = 55/60 (as the total time taken in 55 mints)

Solving this, we get: (X-22) (11X+2) = 0

So X = 22 (Ans: D)



can somebody tell me how to easily get (x-22)(11x+2) from 11x^2-240x-44???


Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm
Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm

Hope it helps.
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Re: In a stream that is running at 2 kmph, a man goes 10 km upstream and [#permalink] New post   13 Oct 2023, 00:47
Given: In a stream that is running at 2 kmph, a man goes 10 km upstream and comes back to the starting point in 55 minutes.
Asked: Find the speed of the boat in still water.

Let the speed of the boat in still water be x kmph.

10/(x-2) + 10/(x+2) = 55/60 = 11/12
10 *2x/(xˆ2-4) = 11/12
11xˆ2 - 240x - 44 = 0
x = 22 kmph

IMO D
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Re: In a stream that is running at 2 kmph, a man goes 10 km upstream and [#permalink]
13 Oct 2023, 00:47
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