Ram and Shyam Running Race

CAT 2005 Slot 1 · QA · Medium · Arithmetic

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Passage / data set

Ram and Shyam run a race between points A and B, 5 km apart. Ram starts at 9 a.m. from A at a speed of 5 km/hr, reaches B, and returns to A at the same speed. Shyam starts at 9:45 a.m. from A at a speed of 10 km/hr, reaches B and comes back to A at the same speed.

Question 1 of 2

At what time do Ram and Shyam first meet each other?

  1. A.

    10 a.m

  2. B.

    10:10 a.m

  3. C.

    10:20 a.m

  4. D.

    10:30 a.m.

Answer

B

Explanation

Ram starts at 9:00 AM @ 5 km/h. At 10:00 AM, Ram reaches B (5 km away) and turns back toward A. Shyam starts at 9:45 AM @ 10 km/h. At 10:00 AM, Shyam has traveled for 15 mins =2.5= 2.5 km towards B. At 10:00 AM, distance between Ram (at B) and Shyam (2.5 km from A) is 2.52.5 km, moving towards each other. Relative speed =5+10=15= 5 + 10 = 15 km/h. Time to meet =2.515 hours=10= \frac{2.5}{15} \text{ hours} = 10 minutes. So they meet at 10:10 AM.

Question 2 of 2

At what time does Shyam overtake Ram?

  1. A.

    10:20 a.m

  2. B.

    10:30 a.m

  3. C.

    10:40 a.m

  4. D.

    10:50 a.m

Answer

B

Explanation

Shyam reaches B (5 km from A) at 10:15 AM (takes 45 mins) and starts returning towards A. At 10:15 AM, Ram (moving back towards A) has been returning for 15 mins, covering 5×1560=1.255 \times \frac{15}{60} = 1.25 km from B. Shyam is at B, and Ram is 1.25 km ahead of Shyam heading towards A. Relative speed when both move towards A =10−5=5= 10 - 5 = 5 km/h. Time taken by Shyam to overtake Ram =1.255=0.25 hours=15= \frac{1.25}{5} = 0.25 \text{ hours} = 15 minutes. Overtaking time =10:15+15 mins=10:30= 10:15 + 15 \text{ mins} = 10:30 AM.

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