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Bird A starts flying from P to Q at 9.00 A.M. and bird B starts flying from Q to P at 10.00 A.M. B is 50% faster than A. What is the time at which they meet if P and Q are 300kms apart and A’s speed is 50kmph.
a) 12 noon b) 12.30pm c) 11.30am d) 11.00am
Read Solution (Total 6)
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- it should be 12 noon
- 9 years agoHelpfull: Yes(6) No(0)
- Distance traveled by A in one hour+Distance traveled by A after B starts flying+Distance traveled by B=300
50*1+50*t+75*t=300
t=2 hrs
So, 12 noon - 9 years agoHelpfull: Yes(3) No(0)
- 11:00am
50[y+1]+3/2*50y=300
y=2. - 11 years agoHelpfull: Yes(1) No(6)
- they meet at 12 noon
- 9 years agoHelpfull: Yes(1) No(0)
- d) 11:00 am
- 9 years agoHelpfull: Yes(0) No(0)
- a) 12 noon
- 8 years agoHelpfull: Yes(0) No(0)
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