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Simultaneous Motion

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Senior Manager
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Simultaneous Motion [#permalink] New post 13 Mar 2011, 08:37
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Question Stats:

60% (02:28) correct 40% (01:01) wrong based on 15 sessions
14) A bullet train leaves kyoto for Tokyo traveling 240 miles per hour at 12 noon. Ten minutes later, a train leaves Tokyo for Kyoto traveling 160 miles per hour.If Tokyo and Kyoto are 300 miles apart, at what time ill the trains pass each other?

a) 12.40 pm
b) 12.49 pm
c) 12.55 pm
d) 1.00 pm
e) 1.05 pm

Catch up time= Distance between the 2/Speed of Train from K->T + Speed of Train from T->K

300/240+160 = 45 mins
as Train from T->k started 10 mins late catch up time will be 12.55
If i calculate with other method

240(t+1/6)+ 160t = 300

i got t as 39 mins

as Train from T->k started 10 mins late catch up time will be 12.49

WHEN CAN WE APPLY THE FORMULA TO CALCULATE Catch up time because in this case its not giving correct answer.
[Reveal] Spoiler: OA

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Re: Simultaneous Motion [#permalink] New post 13 Mar 2011, 09:28
300/240+160 = 45 mins ---> This is wrong
You are assuming that train K->T started the same instant that the train T->K started. So train T->K traveled 1/6 * 160 miles more and hence the "kiss" time went down by 4 mins. i.e. 45 mins. However the actual time to intersect is 49 mins.

as Train from T->k started 10 mins late catch up time will be 12.55 --->

When using the relative speed you assumed if one train were still, the other train will catch up at effectively 400 mph. So 400/6 miles is covered in 1/6 hr. This cannot be undone by simple arithmetic as adding 10 mins to the time of intersect. This doesn't work.

In fact the correct assumption is that for the first 1/6 hr (10 mins), the slower train was still and only the faster train was moving.

GMATD11 wrote:
14) A bullet train leaves kyoto for Tokyo traveling 240 miles per hour at 12 noon. Ten minutes later, a train leaves Tokyo for Kyoto traveling 160 miles per hour.If Tokyo and Kyoto are 300 miles apart, at what time ill the trains pass each other?

a) 12.40 pm
b) 12.49 pm
c) 12.55 pm
d) 1.00 pm
e) 1.05 pm

Catch up time= Distance between the 2/Speed of Train from K->T + Speed of Train from T->K

300/240+160 = 45 mins
as Train from T->k started 10 mins late catch up time will be 12.55
If i calculate with other method

240(t+1/6)+ 160t = 300

i got t as 39 mins

as Train from T->k started 10 mins late catch up time will be 12.49

WHEN CAN WE APPLY THE FORMULA TO CALCULATE Catch up time because in this case its not giving correct answer.
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Re: Simultaneous Motion [#permalink] New post 13 Mar 2011, 11:49
Last part that is after 45 minutes is not clear. Please help.
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Last edited by Baten80 on 15 Mar 2011, 19:19, edited 1 time in total.
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Re: Simultaneous Motion [#permalink] New post 13 Mar 2011, 20:01
Let the trains meet after x hours.

240x + (x - 1/6)160 = 300

24x + 16x - 16/6 = 30

40x = 16/6 + 30

=> 40x = 196/6

=> 10x = 49/6

So it wil be after 49/60 hr or 49 min, so 12:49, hence the answer is 12:49 PM
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Re: Simultaneous Motion [#permalink] New post 14 Mar 2011, 09:18
train A-------------------B
r: 240 ------------------160
t: t+10/60=t+1/6 --------t (cause A moved 10 mins ealier)
S: 300-[240*(t+1/6)-----160 t
300-240t-40=160t so t=1 hour and 5 mins

from what i learned from MGMAT this is " the kiss/ crash"

train ----rate-----time-----distance
A---------a--------t--------at= A's distance
B---------b---------t-------bt=B's distance
total-----(a+b)-----t-------total distance covered
(add) ----the same----add
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Re: Simultaneous Motion [#permalink] New post 28 Mar 2011, 08:31
Time taken to meet = Distance/relative speed
Second train started 10 minutes later, the first train travels in 10 minutes = 240/60*10=40 miles
so they have to travel 300-40=260 miles
now, time= [260/(240+160)]*60= 39 minutes
as 10 minutes later 39+10=49 minutes

Ans. B
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Re: Simultaneous Motion [#permalink] New post 18 Apr 2011, 07:40
240(t +1/6) +160t = 300
t = 39 mins
t+10 = 49 mins
Answer is B.

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Re: Simultaneous Motion   [#permalink] 18 Apr 2011, 07:40
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