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Ship A departs dock X and heads towards dock Y at a constant speed of
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Ship A departs dock X and heads towards dock Y at a constant speed of [#permalink] New post  20 Sep 2019, 05:03
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Ship A departs dock X and heads towards dock Y at a constant speed of 10 miles per hour. 2 later, ship B departs dock X and travels along the same route at a constant speed of 14 miles per hour. Ship A does not reach dock Y before ship B. How many hours will it take for ship B to catch up with ship A?

A. 3 hrs
B. 3.5 hrs
C. 4 hrs
D. 4.5 hrs
E. 5 hrs
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Re: Ship A departs dock X and heads towards dock Y at a constant speed of [#permalink] New post  29 Sep 2019, 21:32
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Relative distance between both ships when ship B departs the dock= 2*10=20 miles
Relative speed of ship B with respect to ship A= 14-10= 4mph

Time taken by ship B to catch up ship A= 20/4=5 hours



whollymoses wrote:
Ship A departs dock X and heads towards dock Y at a constant speed of 10 miles per hour. 2 later, ship B departs dock X and travels along the same route at a constant speed of 14 miles per hour. Ship A does not reach dock Y before ship B. How many hours will it take for ship B to catch up with ship A?

A. 3 hrs
B. 3.5 hrs
C. 4 hrs
D. 4.5 hrs
E. 5 hrs
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Re: Ship A departs dock X and heads towards dock Y at a constant speed of [#permalink] New post  30 Sep 2019, 15:31
whollymoses wrote:
Ship A departs dock X and heads towards dock Y at a constant speed of 10 miles per hour. 2 later, ship B departs dock X and travels along the same route at a constant speed of 14 miles per hour. Ship A does not reach dock Y before ship B. How many hours will it take for ship B to catch up with ship A?

A. 3 hrs
B. 3.5 hrs
C. 4 hrs
D. 4.5 hrs
E. 5 hrs


from given info
10t=(t+2)14
solve for t = -7
so time for B ; -7+2 ; -5 time cannot be -ve
5 hrs
IMO E
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Re: Ship A departs dock X and heads towards dock Y at a constant speed of [#permalink] New post  03 Mar 2022, 13:06
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Re: Ship A departs dock X and heads towards dock Y at a constant speed of [#permalink]
03 Mar 2022, 13:06
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