Aptitude - Boats and Streams - Discussion

Time taken by upstream is 90 minutes more then the time taken by downstream.

I solved like this;
36/t =10+x.
36/t+1.3=10-x.I am not getting the answer. Please anyone help me.

Please tell me the shortcut way to get the answer.

If we put x=2 in 36/(10-x)-36/(10+x) = 90/60,
We get 1=3/2.

Which is invalid, then how can we say that x= 2?
Anyone can tell me?

x = 10.
36/(10-x) - 36/(10+x) = 90/60,
36/(10-x) - 36/(10-x) = 3/2.
then 12*4 = 36.
divide all no. * 12.

12/(10-x) - 12/(10-x) = 1/2.
(12/(x-y)- 12/(x+y)) / ((10-x)(10+x)) = 1/2.

Multiply by * 2 all equation.
24(10-x) - 24(10+x) = (10-x)(10+x),
240 + 24x - 240+ 24x = (10-x)(10+x),
24x + 24x = (10-x)(10+x),
48x = 100+10x-10x-x^2,
48x = 100- x^2,
48 = 100-x,
x= (100/48),
x = 2.

Can anyone say please, how can we equate both the upstream and downstream to 90, if downstream takes less than 90 minutes? Please tell me.

@Siddesh.

Boat takes 90 minutes less to travel 36 miles downstream than to travel the same distance upstream.

It means while going downstream the time will be less and will while going upstream (i.e against) it will take 90 min more;
so t=d/s.

In question time taken is 90 min less. How you can add or substrate time data with speed? Please tell me.

Let stream is considered as "S"

Speed Downstream = (10+S).
Speed Upstream =(10-S).

Upstream Speed - Downstream Speed = 90/60 (minutes converted to hour).

36/(10-S) - 36(10+S) = 90/60.

36(10+S) - 36(10-S) / (10+S) (10-S) = 3/2 ( Divided those) {(10+S) and (10-S) as LCM}.

360+36S-360+36S / 100-10S+10S-S^2= 3/2 (360 and 10s Subtracted).

72S / (100-S^2) = 3/2.

144S = 300 - 3S^2 (cross Multiplication).

3S^2 + 144S - 300 = 0.
S^2 + 48S - 100 = 0 (ALL are divided by 3).

S^2 + 50S - 2S -100 =0.
S( S+50) - 2(S+50) = 0.
(S+50) (S-2) = 0.

S = - 50 (Is not granted).
S = 2.

So the speed of the stream is 2.

Hi, I am unable to understand this step that is 72X x 60 =90(100 - X^2).
X^2 +48X - 100 = 0.
(X +50) (X - 2).
X=2mph.