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Time taken by train to cross electric post

CAT 2020 Slot 1 · Quantitative Ability · Easy · Speed, Time & Distance

This is an easy Quantitative Ability question from the CAT 2020 Slot 1 paper. It tests Speed, Time & Distance. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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Two persons are walking beside a railway track at respective speeds of 2 and 4 km per hour in the same direction. A train came from behind them and crossed them in 90 and 100 seconds, respectively. The time, in seconds, taken by the train to cross an electric post is nearest to

  1. A.

    87

  2. B.

    82

  3. C.

    78

  4. D.

    75

Answer

B

Explanation

Let LL be the length of the train (in km) and SS be its speed (in km/h).

Relative speed with respect to person 1 =(S−2)= (S - 2) km/h. Relative speed with respect to person 2 =(S−4)= (S - 4) km/h.

Since L=Relative Speed×TimeL = \text{Relative Speed} \times \text{Time}: L=(S−2)×903600L = (S - 2) \times \frac{90}{3600} L=(S−4)×1003600L = (S - 4) \times \frac{100}{3600}

Equating both expressions for LL: 90(S−2)=100(S−4)90(S - 2) = 100(S - 4) 9S−18=10S−40  ⟹  S=22 km/h9S - 18 = 10S - 40 \implies S = 22\text{ km/h}

Length of train L=(22−2)×903600=20×140=0.5 kmL = (22 - 2) \times \frac{90}{3600} = 20 \times \frac{1}{40} = 0.5\text{ km}.

Time taken to cross an electric post =LS=0.522 hours=0.5×360022=180022≈81.82 seconds≈82 seconds= \frac{L}{S} = \frac{0.5}{22}\text{ hours} = \frac{0.5 \times 3600}{22} = \frac{1800}{22} \approx 81.82\text{ seconds} \approx 82\text{ seconds}.

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