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Question 7 : 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:

- √6 : √2
- √7 : 2
- 2√5 : 3
- 3 : 2

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Let us take the speed of the boat and river to be ‘b’ and ‘x’ respectively and let the distance be ‘d’.

As per the condition in the question,

⟹ \\frac{d}{x + b}) + \\frac{d}{x - b}) is the normal time taken.

⟹ \\frac{d}{2x + b}) + \\frac{d}{2x - b}) is the special time taken i.e if the speed gets doubled.

Since the time gets reduced by 75%

⟹ \\frac{1}{4}) [\\frac{d}{x + b}) + \\frac{d}{x - b})] = [\\frac{d}{2x + b}) + \\frac{d}{2x - b})]

We have to find the ratio of the speed of the motor boat to the speed of the river = \\frac{x}{b})

Let us divide throughout by b and assume \\frac{x}{b}) = k

⟹ [\\frac{d}{x + b}) + \\frac{d}{x - b})] = 4[\\frac{d}{2x + b}) + \\frac{d}{2x - b})]

⟹ [\\frac{1}{k + 1}) + \\frac{1}{k - 1})] = 4[\\frac{1}{2k + 1}) + \\frac{1}{2k - 1})]

⟹ \\frac{(k - 1)(k + 1)}{k^2 - 1}) = \\frac{2k + 1 + 2k - 1}{4k^2 - 1})

⟹ \\frac{2k}{k^2 - 1}) = \\frac{16k}{4k^2 - 1})

⟹ 8k^{2} – 8 = 4k^{2} – 1

⟹ 4k^{2} = – 1 +8

⟹ k^{2} = \\frac{7}{4})

⟹ k = \\frac{√7}{√4})

⟹ k = \\frac{√7}{2}) where \\frac{x}{b}) = k

Hence the ratio of the original speed of the motor boat to the speed of the river is equal to √7 : 2

The question is **"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:" **

Choice B is the correct answer.

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