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A bullet train leaves Kyoto for Tokyo traveling 240 miles per hour at

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

Last edited by Bunuel on 11 May 2015, 01:51, 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] New post 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
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Re: A bullet train leaves Kyoto for Tokyo traveling 240 miles per hour at [#permalink] New post 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.

Distance remaining = 260 miles.

3x/8 = (260-x)/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] New post 05 Jun 2006, 14:30
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 300-40=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] New post 27 Sep 2008, 00:04
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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] New post 27 Sep 2008, 00:30
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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] New post 27 Sep 2008, 17:20
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(t-1/6)=300-d

160t-160=300-240t

t=49/60
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Re: MGMAT - RATE PROBLEM [#permalink] New post 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
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Re: MGMAT - RATE PROBLEM [#permalink] New post 28 Sep 2008, 11:04
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] New post 13 Mar 2011, 08:37
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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] New post 13 Mar 2011, 09:28
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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] 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] New post 10 May 2015, 00:30
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Re: Simultaneous Motion   [#permalink] 10 May 2015, 00:30
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