Aptitude - Boats and Streams - Discussion

But in question, there is still water mention not, the velocity of the boat in "still water" needs to be found not in downstream.

Why don't we take that 1 as upstream and 3/2 as downstream because they mention that boat covers a certain distance but not downstream are upstream please help me to understand.

@Rehman how you can do in short cut way?

Yes, right @Suraj Dev.

D1 = D2.
SO, U = X.

A = X+3, B = X-3.
A*T1 = B*T2.

HERE, T1 = 1, T2 = 3/2.

THEREFORE,
X=U=15KMPH.

I have one more way to solve it...

The boat goes downstream in 1 hour and returns back in 3/2 hours so we can conclude the boat covered the same distance.

Now, Sb = 1/2(Sd+Su)
Sb = 1/2(D/1 +D*2/3). As speed = Dist./Time
Sb = D(1/2+1/3)
Sb = D((5/6) -------> (1)

Also, Ss = 1/2(Sd-Su)
Ss = 1/2(D/1-D*2/3)
Ss = D(1/2-1/3)
Ss = D(1/6) -------> (2)

Now if we take ratio of (1) & (2) then,
Sb/Ss = [D(5/6)]/[D(1/6)].
Sb/3 = 5.
As speed of stream is 3kmph.
Sb = 15 kmph.

Here Sb is the speed of the boat.
Ss is the speed of the stream.
Sd is the speed of downstream.
Su is the speed of upstream.
D is distance.

Assume, Let the speed of the boat in still water be x kmph, Then
Speed downstream = (x + 3) kmph,
Speed upstream = (x - 3) kmph.
Given Time of downstream has 1 hour and while it comes back in
1(1/2) hours.i.e 1+(1/2)=3/2 hour ( Taking LCM)
we need to find out distance, Distance=velocity*Time
By Equating on both side,
Velocity of downstream * Time taken by downstream = Velocity of Upstream * Time taken by Upstream

(x + 3) x 1 = (x - 3)x 3/2
(x + 3) = (3x-9)/2
2(x+3) = (3x-9)
2x+6 = 3x-9
3x-2x = 6+9
x=15 km

Why we have not taken total time as 5/2? Please explain it.

SHORT-CUT

90-60=30/2=15KMH

Please explain in detail.