A person leaves a place A to place B at 6 a m and reaches place B at 10 a m Another person leaves B ...........

Quantitative Aptitude ( CCS) Time & distance

A person leaves a place A to place B at 6 a.m. and reaches place B at 10 a.m. Another person leaves B at 7.30 a.m. and reaches A at 11 a.m. They will meet each other at

7.40 a.m.

8.20 a.m

8.40 a.m.

9.20 a.m.

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Explanation:

If AB = x km, then
Speed of person starting from A = x/4 kmph
Speed of person starting from B = x/(7/2) = 2x/7 kmph
Relative speed = x/4 + 2x/7 = (7x + 8x)/28 = 15x/28 kmph
Distance covered by first person till 7.30 a.m. = x/4 × 3/2 = 3x/8 km
Remaining distance = x – 3x/8 = 5x/8 km.
Time of meeting = Remaining distance/ Relative speed
(5x/8)/(15x/28) = 5x/8 × 28/15x = 7/6 hours
=7/6 × 60 = 70 minutes
8.40 am