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CAT 2017 Slot 1 — QA Question 7

Boats and StreamsEasy

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:

Answer & solution

Correct answer: √7 : 2

  • A

    √6 : √2

  • √7 : 2

  • C

    2√5 : 3

  • D

    3 : 2

Solution

Hard

Downstream speed is x+yx+y, upstream is x−yx-y (boat xx, river yy). Doubling the boat speed makes the new total time one quarter of the old (a 75%75\% cut). Write both round-trip times and equate.

1

Original and new round-trip times. Let one-way distance =d=d.

Told=dx+y+dx−yTnew=d2x+y+d2x−y(boat speed doubled)\begin{aligned} &T_{\text{old}} = \frac{d}{x+y}+\frac{d}{x-y}\\ &T_{\text{new}} = \frac{d}{2x+y}+\frac{d}{2x-y} \quad\text{(boat speed doubled)} \end{aligned}
2

Apply the 75%75\% reduction. New time is 14\tfrac14 of the old.

d2x+y+d2x−y=14 ⁣(dx+y+dx−y)⇒ 4dx4x2−y2=14⋅2dxx2−y2(combine each pair)\begin{aligned} &\frac{d}{2x+y}+\frac{d}{2x-y} = \frac14\!\left(\frac{d}{x+y}+\frac{d}{x-y}\right)\\ &\Rightarrow\ \frac{4dx}{4x^2-y^2} = \frac14\cdot\frac{2dx}{x^2-y^2} \quad\text{(combine each pair)} \end{aligned}
3

Simplify. Cancel dd and xx, cross-multiply.

44x2−y2=12(x2−y2)⇒ 8(x2−y2)=4x2−y2⇒ 4x2=7y2(collect terms)\begin{aligned} &\frac{4}{4x^2-y^2} = \frac{1}{2(x^2-y^2)}\\ &\Rightarrow\ 8(x^2-y^2) = 4x^2-y^2\\ &\Rightarrow\ 4x^2 = 7y^2 \quad\text{(collect terms)} \end{aligned}
4

The ratio. Take the square root.

x2y2=74⇒ xy=72\begin{aligned} &\frac{x^2}{y^2}=\frac{7}{4}\\ &\Rightarrow\ \frac{x}{y}=\frac{\sqrt7}{2} \end{aligned}
x:y=7:2x:y=\sqrt7:2

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CAT 2017 Slot 1 QA Q7: A man travels by a motor boat down a river to his office and back. With the speed of the river unchanged, if h — Solution | TheCATExam