Aptitude - Boats and Streams - Discussion

Super @Sowmya. Your explanation is an easy way to get the answer.

We know,
Distance = Time x Speed.

Distance is same for both up and down streams => d = 36.
Speed of boat in still water is 10 => u = 10.

Let time for up stream be "t" and time for down stream becomes "t-(90/60)".

We need to find speed of the stream i.e; v = ? (say x).

From distance = Time x Speed => Time = Distance/Speed.

So, time for up stream t = 36/(10 - x) and,
Time for down stream (t - (90/60)) = 36/(10 + x).

For above to get 'x' value we need to subtract time for upstream and time for downstream.

=> t - (t - (90/60)) = [36/(10 - x)] - [36/(10 + x)].
=> 90/60 = 36{[1/(10 - x)] - [1/(10 + x)]}.
=> 90/60 = 36{ [(10 + x) - (10-x)]/[(10*10) - (x*x)]}.
=> 1/2 = 12 {[2x]/[100 - x2]}.
=> 100 - x2 = 12 * 2 * 2x.
=> 100 - x2 = 48x.
=> x2 + 48x - 100 = 0.

And here we got the complete equation with one variable "x" that is what we need to find so we can simply substitute options and check.

From the Option [A] will fit the above equation.

Yes. So option A is correct.

Well done @Sowmya.

Please give me any shortcuts to find the solution.

(90/60)min = (36/(10 - x)) - (36/(10 + x));

How the minus comes between this?

Given.

(up_stream time) - (down_stream time) = 90mins.

36(miles)/(10-x) - 36/(10+x) = 90/60.

We get equation form of
x^2 + 48x - 100 = 0.
Hence, x = 2 and x = -50.
x = 2 is the answer.

I too have the same doubt @Ram.

Please clear it.

@Venkat.

How
36=S/1.3
S= 2.

Explain this step.

D=S/T or D=S * T @ Venkat.