Motor Boat and River Speed Ratio

CAT 2017 Slot 1 · QA · Hard · Speed, Time and Distance

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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%75\%. The ratio of the original speed of the motor boat to the speed of the river is:

  1. A.

    6:2\sqrt{6} : \sqrt{2}

  2. B.

    7:2\sqrt{7} : 2

  3. C.

    25:32\sqrt{5} : 3

  4. D.

    3:23 : 2

Answer

B

Explanation

Let speed of boat = bb, speed of river = rr. Original time T1=db+r+db−r=2dbb2−r2T_1 = \frac{d}{b+r} + \frac{d}{b-r} = \frac{2db}{b^2 - r^2}. New boat speed = 2b2b. New time T2=d2b+r+d2b−r=4db4b2−r2T_2 = \frac{d}{2b+r} + \frac{d}{2b-r} = \frac{4db}{4b^2 - r^2}. Given T2=0.25T1=T14T_2 = 0.25 T_1 = \frac{T_1}{4}. 4db4b2−r2=14×2dbb2−r2  ⟹  44b2−r2=12(b2−r2)\frac{4db}{4b^2 - r^2} = \frac{1}{4} \times \frac{2db}{b^2 - r^2} \implies \frac{4}{4b^2 - r^2} = \frac{1}{2(b^2 - r^2)} 8b2−8r2=4b2−r2  ⟹  4b2=7r2  ⟹  br=728b^2 - 8r^2 = 4b^2 - r^2 \implies 4b^2 = 7r^2 \implies \frac{b}{r} = \frac{\sqrt{7}}{2}.

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