Proportion: Train Speed and Compartments

CAT 1999 Slot 1 · QA · Medium · Ratios and Proportions

All CAT 1999 Slot 1 questionsMore Ratios and Proportions questionsAll CAT QA

The speed of a railway engine is 42 kmph42 \text{ kmph} when no compartment is attached, and the reduction in speed is directly proportional to the square root of the number of compartments attached. If the speed of the train carried by this engine is 24 kmph24 \text{ kmph} when 99 compartments are attached, the maximum number of compartments that can be carried by the engine is

  1. A.

    49

  2. B.

    48

  3. C.

    46

  4. D.

    47

Answer

B

Explanation

Reduction in speed Δv=kN\Delta v = k \sqrt{N}. When N=9N = 9, Δv=42−24=18 kmph\Delta v = 42 - 24 = 18 \text{ kmph}. So 18=k9=3k  ⟹  k=618 = k \sqrt{9} = 3k \implies k = 6. To stop the engine completely (v=0v = 0), Δv=42  ⟹  6N=42  ⟹  N=7  ⟹  N=49\Delta v = 42 \implies 6 \sqrt{N} = 42 \implies \sqrt{N} = 7 \implies N = 49. Since at N=49N=49 the train stops, the maximum number of compartments it can carry (move with) is 49−1=4849 - 1 = 48.

Practise this under exam conditions

Sign in to solve it with a live timer, the on-screen CAT calculator, and streak and accuracy tracking across every question you attempt.

Solve in the workspace