Aptitude - Boats and Streams - Discussion

Very well Explained Seetha ...Thank you.....

By solving the fraction 4*2=8, n 8+1=9, so the ans is 9/2

How we get 9/2 there please tell me.

Here only one direction distance is given. Time which requires to go and back position is given. How is possible to put total time to only one direction distance.

How can you multiply in place of addition ?

Total time=(distance of downstream/speed of downstream)+(distance of downstream/speed of downstream)

Therefore, (30/(15+x))+(30/(15-x)) = 9/2 hrs

Then find the value of x by simplifying the above equation.

What dou mean by still water?

Here we have been given the speed of the boat is 15 kmph that too in still water and the boat goes and comes back in 4 and half hour and that distance is 30 kms so we have to find the speed of the stream in kmph.
Let us think that the speed of the stream be X then
We know that Speed of the downstream is (u+v) here u is speed of the boat and v is speed of the stream so
Speed of downstream is (15+x)Kmph
Speed of upstream is (15-x)Kmph
We know that
speed in still water is 1/2(a+b) kmph
so
30/(15+x) + 30/(15-x) = 4 1/2

30*30/(15+x)(15-x) = 4 1/2 [Since we know (a+b)(a-b) = (a2 - b2)

900 / 225 - x2 = 9/2

cross multiply the two fractions

then 900 * 2 = 9(225 -x2)

1800 = 2025 - 9x2

9x2 = 2025-1800

9x2 = 225
x2 = 225/9
x2 = 25
x = 5

This is the solution

Simply follow formula when only distance the, time t and speed of boat x given.. and asked for speed of stream.... use in formula,

d^2/(x^2-y^2)=t ,

9/2=(225-y^2)/15, ==> y^2=25 , y=5

I don't really understand this question. Some one should please help me out.