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A bullet train leaves Kyoto for Tokyo traveling 240 miles per hour at [#permalink]
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05 Jun 2006, 11:54
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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 will the trains pass each other? A. 12:40pm B. 12:49pm C. 12:55pm D. 1:00pm E. 1:05pm
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Last edited by Bunuel on 11 May 2015, 02:51, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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Re: A bullet train leaves Kyoto for Tokyo traveling 240 miles per hour at [#permalink]
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05 Jun 2006, 12:13
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buckkitty wrote: 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 wil the trains pass each other?
A. 12:40pm B. 12:49pm C. 12:55pm D. 1:00pm E. 1:05pm
I would love to know the approach you guys take to solving this. I approached it a little differently from the book. The source is MGMAT word problem translations book.
B 
The bullet train will have travelled 240 x (10/60) miles after the first 10 minutes = 40 miles.
At 12:10 , the bullet train and the train are 260 miles apart (=30040). To calculate when they will pass:
260 = 240*(60)*x + 160*(60)*x where x = the time it takes for them to reach one another in minutes
x = 39 minutes for the trains to pass each other AFTER 12:10.
Therefore the trains must have passed each other at 12:49



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Re: A bullet train leaves Kyoto for Tokyo traveling 240 miles per hour at [#permalink]
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05 Jun 2006, 12:20
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Let distance travelled by A be x miles
Speed A = 240/60 = 4 miles per minue
Speed B = 160/60 = 8/3 miles/min
time = distance/speed
Distance travelled by A in 10 minutes = 40 miles.
Distance remaining = 260 miles.
3x/8 = (260x)/4
12x = 260.8  8x
x = 26.4 = 104
time = 39mins
Hence B
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Re: A bullet train leaves Kyoto for Tokyo traveling 240 miles per hour at [#permalink]
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05 Jun 2006, 15:30
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Train A: going from K to T
Train B: going from T to K
First I calculated how far Train A will have travelled by the time Train B starts at 12:10:
10 minutes at 240 mph
1/6*240=40miles
They have 30040=260 miles to travel to meet each other.
I found the combined rate of the two trains
Rate(A) + Rate(B) = 400mph
Divide Distance/Rate to find total time each will travel:
260/400=39/60 >>> 39 Minutes to meet
12:10+39 minutes = 12:49 or Answer B



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A bullet train leaves Kyoto for Tokyo traveling 240 miles [#permalink]
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27 Sep 2008, 01:04
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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 wil the trains pass each other? A. 12:40pm B. 12:49pm C. 12:55pm D. 1:00pm E. 1:05pm
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Re: MGMAT  RATE PROBLEM [#permalink]
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27 Sep 2008, 01:30
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B
in 10 minutes tha train travels 40 miles the rest is 260.
x=v*t
=> 260= (160+240)*t
=> t=39 minutes
39+10=49
thus the answer is b



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Re: MGMAT  RATE PROBLEM [#permalink]
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27 Sep 2008, 18:20
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rino wrote: B
in 10 minutes tha train travels 40 miles the rest is 260.
x=v*t
=> 260= (160+240)*t
=> t=39 minutes
39+10=49
thus the answer is b thats an interesting , shorter approach, I got the same answer but im sure it took me longer R=T*D 240t=d 160(t1/6)=300d 160t160=300240t t=49/60



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Re: MGMAT  RATE PROBLEM [#permalink]
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28 Sep 2008, 06:18
amitdgr wrote: 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 will the trains pass each other?
A. 12:40pm B. 12:49pm C. 12:55pm D. 1:00pm E. 1:05pm !)Train K starts from Kyoto at 12:00 noon. It moves at the speed of 240 mph. Say it covers 300x miles 2)Train T starts from Tokyo at 12:10 PM at a speed of 160 mph. Say it covers the rest of x miles From (1) we have 300x = 240(t+1/6) [ we have "+1/6" here since Train K travels 10 minutes or 1/6 hours longer than train T] and From (2) we have x=160t combining above two equations: 300160t=240t + 40 260 = 400t t= 26/40 = 13/20 hours = (13/20)*60 = 39 mins. 12:10 + :39 = 12:49
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Re: MGMAT  RATE PROBLEM [#permalink]
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28 Sep 2008, 12:04
rino wrote: amitdgr wrote: rino wrote: B
in 10 minutes tha train travels 40 miles the rest is 260.
x=v*t
=> 260= (160+240)*t
=> t=39 minutes
39+10=49
thus the answer is b That is a pretty neat method rino. is 160+240 the relative speed between two trains ? Quote: yes it's the relative speed Good method. I did the long method too (t+1/6) 240 + 160 t = 300



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Simultaneous Motion [#permalink]
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13 Mar 2011, 09:37
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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]
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13 Mar 2011, 10:28
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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]
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13 Mar 2011, 21:01
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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]
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14 Mar 2011, 10:18
train AB 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 300240t40=160t so t=1 hour and 5 mins
from what i learned from MGMAT this is " the kiss/ crash"
train ratetimedistance Aatat= A's distance Bbtbt=B's distance total(a+b)ttotal distance covered (add) the sameadd



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Re: Simultaneous Motion [#permalink]
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28 Mar 2011, 09:31
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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 30040=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]
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18 Apr 2011, 08:40
240(t +1/6) +160t = 300 t = 39 mins t+10 = 49 mins Answer is B.
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Re: Simultaneous Motion [#permalink]
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Re: A bullet train leaves Kyoto for Tokyo traveling 240 miles per hour at [#permalink]
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26 Feb 2016, 04:51
buckkitty wrote: 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 will the trains pass each other?
A. 12:40pm B. 12:49pm C. 12:55pm D. 1:00pm E. 1:05pm This is a very beautiful question for any student of clear thinking. You could apply the right strategy/formular and blow the solution by an error in the not very short solution process. Rate x Time = Distance Either \(240t + 160(t  1/6) = 300\) OR \(240(t + 1/6) + 160t = 300\) will give you the solution; the first gives you t = 49 and the second gives you t = 39. The first is straight forward without a trap cos it came straight out from the wording of the question without manipulation. You might say the second is great since you were aware that you must add \(\frac{(1}{6)}\)hours to it. Well,in a CAT, if you had got an answer choice that says 12:39pm then you might have fallen into the trap and FAILED THE QUESTION pitifully with your head held high. I would always want to follow the question prompt like I'm not too smart.



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Re: A bullet train leaves Kyoto for Tokyo traveling 240 miles per hour at [#permalink]
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26 Feb 2016, 05:11
Nez wrote: buckkitty wrote: 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 will the trains pass each other?
A. 12:40pm B. 12:49pm C. 12:55pm D. 1:00pm E. 1:05pm This is a very beautiful question for any student of clear thinking. You could apply the right strategy/formular and blow the solution by an error in the not very short solution process. Rate x Time = Distance Either \(240t + 160(t  1/6) = 300\) OR \(240(t + 1/6) + 160t = 300\) will give you the solution; the first gives you t = 49 and the second gives you t = 39. The first is straight forward without a trap cos it came straight out from the wording of the question without manipulation. You might say the second is great since you were aware that you must add \(\frac{(1}{6)}\)hours to it. Well,in a CAT, if you had got an answer choice that says 12:39pm then you might have fallen into the trap and FAILED THE QUESTION pitifully with your head held high. I would always want to follow the question prompt like I'm not too smart. Hi Nez, You are absolutely correct that a Q like this can be blown to pieces if you take one wrong step.. If I were to do this, I will not get into making equation... i will see how much distance is left between two when both are in motion.. 1) in 10 min, train from K to T would have travelled= 240*10/60=40.. 2) so distance between the two when Train starts from T to K= 30040=260.. 3) combined speed as both travelling towards each other= 240+160=400. 4) they will meet after= 260/400 * 60=39 minutes.. time after 39 minutes= 12:10+39 min=12:49.. This method too finally does what the equation does but slightly less error prone
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Re: A bullet train leaves Kyoto for Tokyo traveling 240 miles per hour at [#permalink]
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A bullet train leaves Kyoto for Tokyo traveling 240 miles per hour at [#permalink]
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13 May 2017, 08:35
buckkitty wrote: 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 will the trains pass each other?
A. 12:40pm B. 12:49pm C. 12:55pm D. 1:00pm E. 1:05pm [b]AS, later train start ten minutes later , so the first train(from Kyoto to Tokyo) cover distance in ten minutes equals (240/60*10 ) 40 miles. Rest of the distance 30040 = 260.therefore, two train together will cover up this distance Relative speed will be (240 + 160) 400 Miles per hour. required time = (260/400*60) 39 minutes. Total time would be 39+10 = 49 minutes . they will pass each other at 12: 49 pm. Kudos if it helps.Thanks




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