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Two trains started simultaneously from opposite ends of a

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Two trains started simultaneously from opposite ends of a [#permalink]

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Two trains started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; train Y, travelling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had X traveled when it met train Y?

A. 37.5
B. 40.0
C. 60.0
D. 62.5
E. 77.5
[Reveal] Spoiler: OA

Originally posted by Madelaine88 on 02 Mar 2011, 06:18.
Last edited by Bunuel on 27 Jun 2013, 02:14, edited 2 times in total.
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Re: 2 trains [#permalink]

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New post 02 Mar 2011, 06:30
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Train X : 100 m in 5 hrs
speed = 100/5 = 20mph
Train Y : 100 m in 3 hrs
speed = 100/3 mph

The distance shrinking at effective speed (20 + 100/3 ) mph
Time of intersect = 100 / (20+100/3) = 15/8 hrs
Distance travelled by X < Distance travelled by Y

Distance travelled by X = time * speed = 15/8 * 20 = 75/2
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Re: 2 trains [#permalink]

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Madelaine88 wrote:
Two trains started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; train Y, travelling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had X traveled when it met train Y?

A/ 37.5
B/ 40.0
C/ 60.0
D/ 62.5
E/ 77.5


As the ratio of the rates of X and Y is 3 to 5 then the distance covered at the time of the meeting (so after traveling the same time interval) would also be in that ratio, which means that X would cover 3/(3+5)=3/8 of 100 miles: 100*3/8=37.5 miles.

Answer: A.
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Re: 2 trains [#permalink]

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New post 13 Apr 2012, 00:42
Bunuel, could you please explain this. It is not clear to me. thanks!
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Re: 2 trains [#permalink]

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New post 13 Apr 2012, 02:02
vel x/ vel y=3/5

let they met after t time, distance travelled by X=x
then,

x/vel x=(100-x)/ vel y
=> x/(100-x)= 3/5
=> x= 300/8= 37.5
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Re: 2 trains [#permalink]

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mymbadreamz wrote:
Bunuel, could you please explain this. It is not clear to me. thanks!


Since, the ratio of times of X and Y to cover the same distance of 100 miles is is 5:3, then the ratio of their rates is 3:5. Consider this, say the rates of trains X and Y are X and Y respectively, then:

Distance=Rate*Time --> X*5=Y*3 --> ratio of the rates is X:Y=3:5. At the time they meet, so after they travel the same time interval the ratio of distances covered by X and Y would also be in that ratio (for example if X=3 mph and Y=5 mph then they would meet in 100/(3+5)=100/8 hours, hence train X would cover 300/8 miles and train Y would cover 500/8 miles --> ratio of distances covered (300/8):(500/8)=3:5).

Now, since the the ratio of distances covered by X and Y is 3:5 then X covered 3/(3+5)=3/8 of the total distance.

Hope it's clear.
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Re: Two trains started simultaneously from opposite ends of a [#permalink]

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New post 19 Sep 2013, 02:37
Why is that when the meet they would have covered 100 miles? This bit is little confusing.
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Re: Two trains started simultaneously from opposite ends of a [#permalink]

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New post 19 Sep 2013, 04:55
theGame001 wrote:
Why is that when the meet they would have covered 100 miles? This bit is little confusing.


When they meet one train covers some part of 100-mile distance and another covers the remaining part of 100-mile distance, so combined they cover 100 miles.
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Re: Two trains started simultaneously from opposite ends of a [#permalink]

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Speed of first train: 100/5 = 20 Mph
Speed of second train: 100/3 = 33.33 Mph

Now in 1 hr distance covered by X is 20 & that by Y is 33.33.In the next hr it will be 40 & 66.66 resp.
40+66.66=106.6> the distance between them...So they would have met by now & the answer would be something less than 40..which leaves us with A
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Re: Two trains started simultaneously from opposite ends of a [#permalink]

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New post 30 Dec 2015, 10:48
Speed of train X= Distance/Time
= 100/5
= 20 mph

Speed of train Y= Distance/Time
= 100/3
= 33.33 mph

Relative Speed (opposite direction) = 20+33.33
= 53.33 mph

Now,
Time when X & Y meet:

Distance = Relative Speed x Time

Time= D/RS
= 100/53.33
= 1.87 hrs

After 1.87 hrs X & Y meet at some distance

In 1.87 hrs, X traveled

Distance= 20 x 1.87
=37.50 miles (Answer)

And Y traveled,

Distance= 33.33 x 1.87
= 62.3 (approx.)

Answer is A.

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Re: Two trains started simultaneously from opposite ends of a [#permalink]

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New post 02 Apr 2016, 18:29
RxT = D. D/R = T, SO WHEN YOU SEE TIME = 100 AND THE CHALLENGE IS THAT BOTH TRAINS MUST MEET SO YOU NEED TO ADD (+) THE INDIVIDUAL RATES. ALSO, TO GET THE INDIVIDUAL RATES USE FOR X = 100/5 = 20 M/H, RATE FOR Y = 100/3 M/H. SO IN YOUR FORMULA YOU HAVE: \(\frac{100}{(20+(100/3))} = T\) => T = 15/8, MEANING BOTH TRAINS MET AT 15/8 TIME. TO FIND OUT THE DISTANCE AT WHICH X MET Y, YOU DO R x T = D => 20 x (15/8) => D = 37.5 MILES FOR X WHET IT MET Y.
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Re: Two trains started simultaneously from opposite ends of a [#permalink]

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Bunuel wrote:
Madelaine88 wrote:
Two trains started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; train Y, travelling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had X traveled when it met train Y?

A/ 37.5
B/ 40.0
C/ 60.0
D/ 62.5
E/ 77.5


As the ratio of the rates of X and Y is 3 to 5 then the distance covered at the time of the meeting (so after traveling the same time interval) would also be in that ratio, which means that X would cover 3/(3+5)=3/8 of 100 miles: 100*3/8=37.5 miles.

Answer: A.


Hello Bunuel can you please advise if my approach is correct :)

Speed X = 100/5= 20
Speed y = 100/3 = 33.33

relative speed 20+33 =53

time when they meet = 100/53 =1.8

Distance done by X when meeting Y --> 20*1.8 = 36 (I got approximate answer , is it correct ? )

thank you! :)
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Two trains started simultaneously from opposite ends of a [#permalink]

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dave13 wrote:
Hello Bunuel can you please advise if my approach is correct :)

Speed X = 100/5= 20
Speed y = 100/3 = 33.33

relative speed 20+33 =53

time when they meet = 100/53 =1.8

Distance done by X when meeting Y --> 20*1.8 = 36 (I got approximate answer , is it correct ? )

thank you! :)


Hey dave13 ,

Yes, your approach is 100% correct. If you get the time to two decimal places, you will get more closer answer to 37.5.

Thanks!
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Re: Two trains started simultaneously from opposite ends of a [#permalink]

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dave13 wrote:
Bunuel wrote:
Madelaine88 wrote:
Two trains started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; train Y, travelling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had X traveled when it met train Y?

A/ 37.5
B/ 40.0
C/ 60.0
D/ 62.5
E/ 77.5


As the ratio of the rates of X and Y is 3 to 5 then the distance covered at the time of the meeting (so after traveling the same time interval) would also be in that ratio, which means that X would cover 3/(3+5)=3/8 of 100 miles: 100*3/8=37.5 miles.

Answer: A.


Hello Bunuel can you please advise if my approach is correct :)

Speed X = 100/5= 20
Speed y = 100/3 = 33.33

relative speed 20+33 =53

time when they meet = 100/53 =1.8

Distance done by X when meeting Y --> 20*1.8 = 36 (I got approximate answer , is it correct ? )

thank you! :)


dave13 Perfect approach :-D .Relative speed concept is used nicely with proper simplification of numbers.

However if you keep the relative speed as is: 100 ( 1/5 + 1/3 ) = 100*8/15 (Speed)

time = 100/(100*8/15) = 15/8 hr

distance traveled by 20*15/8 = 75/2 = 37.5

It would help if the options were very close. Supposed we have 36 & 37.5 both in the options...

Best,
Gladi
Re: Two trains started simultaneously from opposite ends of a   [#permalink] 12 Apr 2018, 10:24
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