prashanths wrote:
A man travels by a motor boat down a river to his office and back. With the speed of the river unchanged, if he doubles the speed of his motor boat, then his total travel time gets reduced by 75%. The ratio of the original speed of the motor boat to the speed of the river is:
A) 6^0.5 : 2^0.5
B) 7^0.5 : 2^0.5
C) 2 x 5^0.5 : 3^0.5
D) 3:2
E) 7^0.5 : 2
Let \(Sb\) be the speed of boat and \(Sr\) be the speed of river.
When the man is travelling downstream, his speed is = \(Sb + Sr\)
and when he is travelling upstream, his speed is = \(Sb - Sr\)
The average speed of both upstream and downstream is:
\(\frac{(2*(Sb+Sr)(Sb-Sr))}{(Sb+Sr+Sb-Sr)}\)
\(=\frac{(Sb^2-Sr^2)}{Sb}\)
Let D be the distance and t be the time
\(\frac{(Sb^2-Sr^2)}{Sb} = \frac{D}{t}\)
When the speed of boat is doubled -
Downstream speed = \(2Sb+Sr\)
Upstream speed = \(2Sb-Sr\)
Average speed = \(\frac{2*(4Sb^2-Sr^2)}{4Sb}\)
\(t1=0.25t\)
Distance will remain same.
\(\frac{(4Sb^2-Sr^2)}{2Sb} = \frac{D}{0.25t}\)
\(\frac{(4Sb^2-Sr^2)}{8Sb} = \frac{D}{t}\)
Equating the two equations -
\(\frac{(Sb^2-Sr^2)}{Sb} = \frac{(4Sb^2-Sr^2)}{8Sb}\)
\(8*(Sb^2-Sr^2) = 4Sb^2-Sr^2\)
\(4Sb^2 = 7Sr^2\)
\(Sb^2/Sr^2 = 7/4\)
\(Sb/Sr = 7^\frac{1}{2}/2\)
Answer- E
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