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

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02 Mar 2011, 05:18

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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?

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

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.

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.

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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19 Sep 2013, 04:00

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

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03 Oct 2014, 12:20

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

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10 Oct 2015, 23:32

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

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02 Apr 2016, 17:29

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
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02 Apr 2016, 17:29

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