Question
A boat can cover βyβ km distance downstream in
(t+2.5) hours. The same boat can cover β0.25yβ km distance upstream in βtβ 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 βyβ km distance downstream and the same distance upstream is 65 hours, then find out the speed of the stream.Solution
Letβs assume the speed of the boat in still water and the speed of the stream are βBβ and βCβ respectively.
A boat can cover βyβ km distance downstream in (t+2.5) hours.
(B+C) = y/(t+2.5)
y/(B+C) = (t+2.5)Β Β Eq.(i)
The same boat can cover β0.25yβ km distance upstream in βtβ hours.
(B-C) = 0.25y/t
0.25y/(B-C) = t
y/[4(B-C)] = t
y/(B-C) = 4tΒ Β Eq.(ii)
If the total time taken by the boat to cover βyβ 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 = 12.5Β Β 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 βtβ from Eq.(iii) in the above equation.
728/B = (2x12.5+3)
728/B = (25+3)
728/B = 28
728/28 = B
B = 26Β Β Eq.(iv)
Put the value of βBβ and βtβ in Eq.(i) and Eq.(ii).
y/(26+C) = (12.5+2.5)
y = 15(26+C)Β Β Eq.(v)
y/(26-C) = 4x12.5
y/(26-C) = 50
y = 50(26-C)Β Β 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
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