HCL
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Numerical Ability
Time Distance and Speed
One young and one old man reach a same distance in 20min and 30min respectively. If old starts at 10.00am and youngman starts at 10.05am. When they meet. Pls provide solution
Read Solution (Total 7)
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- The real ans is 10:15am
Explanation
Let the dis be 30 m
Old man travel 1m/min
Young travel 1.5m/min
At 10:05 old man = 5m
young = 0m
At 10:10 old man = 10m
young = 7.5m
At 10:15 old man= 15m
young = 15m - 9 years agoHelpfull: Yes(22) No(1)
- 30 mins of old man=20 min of young man=same dist
15 min of old man=10 min of young man=same dist
Old man starts at 10.00 whereas young man starts at 10.05.
Thus at 10.15 young will catch up the old
- 8 years agoHelpfull: Yes(4) No(0)
- can anyone explain the answer properly please.............?
- 9 years agoHelpfull: Yes(3) No(1)
- at 10.06 am.
- 9 years agoHelpfull: Yes(0) No(3)
- At 10.12 am
- 9 years agoHelpfull: Yes(0) No(2)
- let distance in meter, for 20 min 10 m and 30 min 10 m , when younger start, old man reached at 1.65 m distance, after that both walking and meet at 10:14 .
- 9 years agoHelpfull: Yes(0) No(1)
- the correct answer is 10.14
take the lcm of 20.30 and consider it as distance it is 60
for applying relative speed make the two stations time equal which is 10.05
at 10.05 the older one gets the distance 15m front then the remaining distance is 45m for 10.10 the younger one and older moves the distance 25 and the remaining distance is 20 then for 4 miutes the older one moves for 12m and the youger moves to another 8m so at 10.14 they will meet - 8 years agoHelpfull: Yes(0) No(1)
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