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Time Comparison for Equal Time vs Equal Distance

CAT 2019 Slot 1 · Quantitative Ability · Medium · Speed, time and distance

This is a medium Quantitative Ability question from the CAT 2019 Slot 1 paper. It tests Speed, time and 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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One can use three different transports which move at 10, 20, and 30 kmph, respectively. To reach from A to B, Amal took each mode of transport 1/3 of his total journey time, while Bimal took each mode of transport 1/3 of the total distance. The percentage by which Bimal's travel time exceeds Amal's travel time is nearest to

  1. A.

    22

  2. B.

    19

  3. C.

    21

  4. D.

    20

Answer

A

Explanation

Amal's average speed =AM(10,20,30)=10+20+303=20 km/h= \text{AM}(10, 20, 30) = \frac{10 + 20 + 30}{3} = 20\text{ km/h}. Bimal's average speed =HM(10,20,30)=3110+120+130=31160=18011≈16.36 km/h= \text{HM}(10, 20, 30) = \frac{3}{\frac{1}{10} + \frac{1}{20} + \frac{1}{30}} = \frac{3}{\frac{11}{60}} = \frac{180}{11} \approx 16.36\text{ km/h}. Ratio of times TB/TA=SA/SB=20180/11=220180=119≈1.2222T_B / T_A = S_A / S_B = \frac{20}{180/11} = \frac{220}{180} = \frac{11}{9} \approx 1.2222. Percentage excess =22.22%≈22%= 22.22\% \approx 22\%.

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