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

Let's set up this problem step by step. A man travels downstream to his office and upstream back home. When he doubles his boat's speed, his total travel time reduces by 75%.

Key insight: When time reduces by 75%, the new time is only 25% of the original time (or 1/4 of the original time).

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Let's define:

$u$ = original speed of motor boat in still water

$v$ = speed of river current

$d$ = distance from home to office (one way)

Why these matter:

Downstream speed = $u + v$ (boat speed + current)

Upstream speed = $u - v$ (boat speed - current)

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Original journey time:

$T = \dfrac{d}{u+v} + \dfrac{d}{u-v}$

This represents: (time downstream) + (time upstream)

New journey time (with doubled boat speed):

$\dfrac{T}{4} = \dfrac{d}{2u+v} + \dfrac{d}{2u-v}$

Why T/4? A 75% reduction means new time = 100% - 75% = 25% of original = $\dfrac{T}{4}$

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Let's combine the fractions in the original equation:

$T = \dfrac{d}{u+v} + \dfrac{d}{u-v} = d \times \dfrac{(u-v) + (u+v)}{(u+v)(u-v)}$

When adding fractions, we need a common denominator $(u+v)(u-v)$

$T = d \times \dfrac{2u}{u^2 - v^2}$

Why $u^2 - v^2$? This comes from $(u+v)(u-v) = u^2 - v^2$ (difference of squares formula)

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Similarly, for the doubled speed:

$\dfrac{T}{4} = \dfrac{d}{2u+v} + \dfrac{d}{2u-v} = d \times \dfrac{(2u-v) + (2u+v)}{(2u+v)(2u-v)}$

$\dfrac{T}{4} = d \times \dfrac{4u}{4u^2 - v^2}$

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Now we substitute the expression for $T$ from the first equation into the second:

$d \times \dfrac{4u}{4u^2 - v^2} = \dfrac{1}{4} \times d \times \dfrac{2u}{u^2 - v^2}$

Since $d$ appears on both sides and $u > 0$, we can divide both sides by $d$ and $u$:

$\dfrac{4}{4u^2 - v^2} = \dfrac{2}{4(u^2 - v^2)}$

Simplifying the right side:

$\dfrac{4}{4u^2 - v^2} = \dfrac{1}{2(u^2 - v^2)}$

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Cross-multiplying:

$4 \times 2(u^2 - v^2) = 4u^2 - v^2$

$8(u^2 - v^2) = 4u^2 - v^2$

$8u^2 - 8v^2 = 4u^2 - v^2$

$8u^2 - 4u^2 = 8v^2 - v^2$

$4u^2 = 7v^2$

Therefore: $\dfrac{u^2}{v^2} = \dfrac{7}{4}$

Taking the square root: $\dfrac{u}{v} = \dfrac{\sqrt{7}}{2}$

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The ratio of the original speed of the motor boat to the speed of the river is $\sqrt{7}:2$.