A boat can cover ‘y’ km distance downstream in (t+2.5) hours. The same boat can cover ‘0.25y’ km distance upstream in ‘t’ hours. The time taken by the boat to cover 728 km in still water is (2t+3) hours. If the total time taken by the boat to cover ‘y’ km distance downstream and the same distance upstream is 65 hours, then find out the speed of the stream.

Let’s assume the speed of the boat in still water and the speed of the stream are ‘B’ and ‘C’ respectively.

A boat can cover ‘y’ km distance downstream in (t+2.5) hours.

(B+C) = y/(t+2.5)

y/(B+C) = (t+2.5)    Eq.(i)

The same boat can cover ‘0.25y’ km distance upstream in ‘t’ hours.

(B-C) = 0.25y/t

0.25y/(B-C) = t

y/[4(B-C)] = t

y/(B-C) = 4t    Eq.(ii)

If the total time taken by the boat to cover ‘y’ km distance downstream and the same distance upstream is 65 hours.

(y/(B+C))+(y/(B-C)) = 65

Put Eq.(i) and Eq.(ii) in the above equation.

(t+2.5)+4t = 65

5t+2.5 = 65

5t = 65-2.5

5t = 62.5

t = 12.5    Eq.(iii)

The time taken by the boat to cover 728 km in still water is (2t+3) hours.

728/B = (2t+3)

Put the value of ‘t’ from Eq.(iii) in the above equation.

728/B = (2x12.5+3)

728/B = (25+3)

728/B = 28

728/28 = B

B = 26    Eq.(iv)

Put the value of ‘B’ and ‘t’ in Eq.(i) and Eq.(ii).

y/(26+C) = (12.5+2.5)

y = 15(26+C)    Eq.(v)

y/(26-C) = 4x12.5

y/(26-C) = 50

y = 50(26-C)    Eq.(vi)

Equating Eq.(v) Eq.(vi).

15(26+C) = 50(26-C)

3(26+C) = 10(26-C)

78+3C = 260-10C

13C = 260-78

13C = 182

C = 14

So the speed of the stream = 14 km/h