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A bullet train leaves Kyoto for Tokyo traveling 240 miles

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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 27 Sep 2008, 21:53
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


Nice method. I used the longer method (solving two simultaneous equations) to get (B)
+1 for you.
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Re: MGMAT - RATE PROBLEM [#permalink] New post 27 Sep 2008, 22:12
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 ?
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Re: MGMAT - RATE PROBLEM [#permalink] New post 28 Sep 2008, 02:30
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



R u sure that the calculation is right here? My answer is E.
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Re: MGMAT - RATE PROBLEM [#permalink] New post 28 Sep 2008, 02:40
scthakur 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



R u sure that the calculation is right here? My answer is E.



scthakur, the official answer is B
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Re: MGMAT - RATE PROBLEM [#permalink] New post 28 Sep 2008, 04:26
amitdgr wrote:
scthakur 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



R u sure that the calculation is right here? My answer is E.



scthakur, the official answer is B


I am still not sure how the calculation is giving 39 minutes? Can someone detail the steps? I think, I am missing even the basic arithmatic here.
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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, 09:48
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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
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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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Re: MGMAT - RATE PROBLEM [#permalink] New post 29 Sep 2008, 22:05
amitdgr wrote:
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



Thanks amitdgr....my silly mistake was that I was adding 240 and 160 to 300. :oops:
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Re: MGMAT - RATE PROBLEM [#permalink] New post 30 Sep 2008, 03:19
scthakur wrote:
Thanks amitdgr....my silly mistake was that I was adding 240 and 160 to 300. :oops:


I do that kinda mistake all the time :(
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Re: MGMAT - RATE PROBLEM   [#permalink] 30 Sep 2008, 03:19
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