A motor boat goes from A to B and comes back to A, both places being on the bank of a river. If the speed of the boat in still water were doubled, the whole trip from A to B and back would take only 20% of the original time. The speed of the boat in still water is how many times the speed of the river?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\sqrt{\frac{3}{2}}\)
A round trip is one leg downstream at (boat + river) and one leg upstream at (boat − river). Writing the round-trip time for both boat speeds and using the 20% condition gives the ratio.
Let the boat speed be \(b\), the river speed \(r\) and AB \(=d\)
Original time \(\displaystyle =\frac{d}{b+r}+\frac{d}{b-r}=\frac{2bd}{b^{2}-r^{2}}\)
With speed \(2b\): time \(\displaystyle =\frac{d}{2b+r}+\frac{d}{2b-r}=\frac{4bd}{4b^{2}-r^{2}}\)
New time \(=\frac{1}{5}\) of the original: \(\displaystyle \frac{4bd}{4b^{2}-r^{2}}=\frac{1}{5}\times\frac{2bd}{b^{2}-r^{2}}\)
Cancelling \(2bd\): \(10(b^{2}-r^{2})=4b^{2}-r^{2}\)
\(6b^{2}=9r^{2}\), so \(\displaystyle \frac{b^{2}}{r^{2}}=\frac{3}{2}\)
\(\displaystyle \frac{b}{r}=\sqrt{\frac{3}{2}}\)
So the boat’s speed in still water is \(\sqrt{\frac{3}{2}}\) times the speed of the river.