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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

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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This post received KUDOS

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 (=300-40). 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

Re: A bullet train leaves Kyoto for Tokyo traveling 240 miles per hour at [#permalink]

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05 Jun 2006, 12:20

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.

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 _________________

"You have to find it. No one else can find it for you." - Bjorn Borg

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 300-x 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 300-x = 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: 300-160t=240t + 40 260 = 400t

t= 26/40 = 13/20 hours = (13/20)*60 = 39 mins.

12:10 + :39 = 12:49 _________________

"You have to find it. No one else can find it for you." - Bjorn Borg

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. _________________

The proof of understanding is the ability to explain it.

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.

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"

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

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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.

Re: A bullet train leaves Kyoto for Tokyo traveling 240 miles per hour at [#permalink]

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26 Feb 2016, 05:11

Expert's post

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= 300-40=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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