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Two locations A and B are at diametrically opposite ends of a circular track. Rekha starts running along the track from location A in the clockwise direction. Sajal starts running simultaneously along the track in the anticlockwise direction from location B. If the length of the circular track is 14 km, and the speeds of Rekha and Sajal are in the ratio 5: 2, then the distance, in km, travelled by Rekha, when they meet at location B for the first time, is

Solution

Correct Option: 2

AA and BB are diametrically opposite, so the arc from AA to BB (either direction) is 142=7\dfrac{14}{2} = 7 km.

Let speeds be 5v5v (Rekha) and 2v2v (Sajal). They run in opposite directions.


For them to meet at BB, both must be at BB at the same time tt.

Sajal starts at BB and moves anticlockwise, so she returns to BB after each full lap. Her distances at BB are 0,14,28,0, 14, 28, \ldots, that is multiples of 1414.

Rekha starts at AA and moves clockwise, reaching BB first after 77 km, then every 1414 km after that. Her distances at BB are 7,21,35,7, 21, 35, \ldots, i.e. 7+14k7 + 14k.


Since speeds are in ratio 5:25 : 2, distances covered in the same time are in ratio 5:25 : 2.

If Sajal covers 14m14m km, Rekha covers 5214m=35m\dfrac{5}{2} \cdot 14m = 35m km.

Rekha must be at BB, so 35m35m must be of the form 7+14k7 + 14k:

35m=7+14k35m = 7 + 14k, i.e. 5m2k=15m - 2k = 1


Smallest positive integer solution: m=1m = 1, k=2k = 2.

Rekha's distance =35×1=35= 35 \times 1 = 35 km.

Answer =35= 35

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