Question
The speed of the boat in still water is 40% less than
the speed of the boat in downstream. The time taken by the boat to cover 468 km distance in upstream is (t+0.5) hours. If the speed of stream is 16 km/h, then find out the value of βtβ.Solution
Letβs assume the speed of boat in still water and the speed of stream are βBβ and βCβ respectively. The speed of the boat in still water is 40% less than the speed of the boat in downstream. B = (100-40)% of (B+C) B = 60% of (B+C) B = 0.6(B+C) If the speed of stream is 16 km/h. B = 0.6(B+16) B = 0.6B+96 B-0.6B = 9.6 0.4B = 9.6 B = 24 km/h The time taken by the boat to cover 468 km distance in upstream is (t+0.5) hours. 468/(t+0.5) = (B-C) Put the value of βBβ and βCβ in the above equation. 468/(t+0.5) = (24-16) 468/(t+0.5) = 8 468/8 = (t+0.5) (t+0.5) = 58.5 t = 58.5-0.5 Value of βtβ = 58
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