Question

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. \(\sqrt{6}:\sqrt{2}\)
• B. \(\sqrt{7}:2\)
• C. \(2\sqrt{5}:3\)
• D. \(3:2\)

Answer 1

The Correct Option is B

Let the original speed of the motorboat be \(b\) (in km/h) and the speed of the river be \(r\) (in km/h). When traveling downstream, the effective speed is \(b+r\), and while traveling upstream, it is \(b-r\). Let the distance to the office be \(d\) km.

Total travel time for the original speed is given by: 

\(T = \frac{d}{b+r} + \frac{d}{b-r}\)

If the speed of the motorboat is doubled, the speed becomes \(2b\). Hence, the new travel time becomes:

\(T' = \frac{d}{2b+r} + \frac{d}{2b-r}\)

We know that the new travel time \(T'\) is 75% less than the original travel time \(T\), which gives:

\(T' = \frac{1}{4}T\)

Substituting the times, we have:

\(\frac{d}{2b+r} + \frac{d}{2b-r} = \frac{1}{4}\left(\frac{d}{b+r} + \frac{d}{b-r}\right)\)

Canceling \(d\) from both sides:

\(\frac{1}{2b+r} + \frac{1}{2b-r} = \frac{1}{4}\left(\frac{1}{b+r} + \frac{1}{b-r}\right)\)

Cross-multiplying, we get:

\(4[(2b-r)(2b+r)] = [(b-r)(b+r)]\)

\(4(4b^2-r^2) = b^2 - r^2\)

\(16b^2 - 4r^2 = b^2 - r^2\)

\(15b^2 = 3r^2\)

\(\frac{b^2}{r^2} = \frac{1}{2}\)

\(\frac{b}{r} = \sqrt{\frac{7}{2}}\)

Thus, the ratio of the original speed of the motorboat to the speed of the river is \(\sqrt{7}:2\).