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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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New post Updated on: 27 Jun 2013, 02:14
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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

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, 06:39
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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: Two trains started simultaneously from opposite ends of a  [#permalink]

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New post 19 Sep 2013, 05:00
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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: 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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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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New post 13 Apr 2012, 02:12
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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
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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
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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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New post 11 Apr 2018, 04:49
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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New post 11 Apr 2018, 05:19
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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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New post 12 Apr 2018, 10:24
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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...

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

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New post 30 Jul 2018, 23:25
Solved it this way:
train x: 100/5 = 20 mph
train y: 100/ 3 = 33.4 mph (approx)

relative speed = 100/53.4 = 25/13.1 = approx 2 hours

so time for X --> 20mph*2 = 40 --> but since its not actually 2 hours but close to it, it must closer to 40 not 40. Which is option A) 37.5

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

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New post 20 Oct 2018, 02:29
Train X travelling speed= 100/5 mph
Train Y travelling speed=100/3 mph
Relative speed as they are travelling in opposite direction= 100/5+100/3= 800/15 mph

Time after which X and Y would meet= distance/relative speed= 100/800/15= 100*15/800= 15/8 hr

X travelled distance when it met with train Y= speed*time they met= 100*15/5*8= 37.5 miles.

Ans (A)
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Re: Two trains started simultaneously from opposite ends of a  [#permalink]

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New post 09 Sep 2019, 10:58
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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


The combined distance traveled by the two trains was 100 miles. Each train traveled for t hours. We can create the distance equation:

100/5 * t + 100/3 * t = 100

Multiplying by 15 to clear the fractions from the equation, we have:

300t + 500t = 1500

800t = 1500

t = 15/8

Thus, train X had traveled 15/8 x 100/5 = 15/8 x 20 = 37.5 miles by the time it reached Y.

Answer: A
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Re: Two trains started simultaneously from opposite ends of a   [#permalink] 09 Sep 2019, 10:58
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