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

Question

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

Loser94 · Accepted Answer

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.

xxxxniaz · Answer

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)

uchihaitachi · Answer

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 	imes (\frac{2S}{(S-2)(S+2)}) = \frac{11}{12} 
Cross multiplying, we get

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

tabishkazi · Answer

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

Bunuel · Answer

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

Kinshook · Answer

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

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