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A bullet train leaves Kyoto for Tokyo traveling 240 miles per hour at [#permalink]
05 Jun 2006, 10:54

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

56% (03:04) correct
44% (02:41) wrong based on 16 sessions

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]
05 Jun 2006, 11:13

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]
05 Jun 2006, 11:20

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]
27 Sep 2008, 00:04

1

This post was BOOKMARKED

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

Re: MGMAT - RATE PROBLEM [#permalink]
28 Sep 2008, 05:18

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 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

Simultaneous Motion [#permalink]
13 Mar 2011, 08:37

2

This post was BOOKMARKED

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.

Re: Simultaneous Motion [#permalink]
13 Mar 2011, 09:28

1

This post was BOOKMARKED

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.

Re: Simultaneous Motion [#permalink]
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"

Re: Simultaneous Motion [#permalink]
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

Re: Simultaneous Motion [#permalink]
10 May 2015, 00:30

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