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can anyone solve this problem and explain me [#permalink]
24 Apr 2011, 10:44

A train X starts from Meerut at 4 P.M. and reaches Ghaziabad at 5 P.M. while another train Y starts from Ghaziabad at 4 P.M. and reaches Meerut, at 5.30 PM. At what time the two trains meet each other. .

Re: can anyone solve this problem and explain me [#permalink]
24 Apr 2011, 11:16

We know that S = D/t , also D is fixed Therefore, for train running from meerut to gazi… Sa = D / ta but ta = 1 hr , therefore Sa = D Now, for train running from Gazi… to meerut … Sb = D / tb But tb = 1 ½ hr = 3/2 hr Therefore, Sb = 2D/3 However, D = Sa therefore Sb = 2Sa/3

Now, to calculate the time at which the trains will meet we need to calculate relative speed Since, the trains are moving towards each other , Relative speed = Sa + Sb = Sa + 2Sa/3 = 5Sa /3 Distance will be “D” because it is fixed And therefore Time at which the trains will meet i.e t = D / (5Sa /3) but D = Sa therefore t = Sa / (5Sa /3) = 3 / 5 Hr = 3/5 * 60 = 36min

Re: can anyone solve this problem and explain me [#permalink]
24 Apr 2011, 11:35

tparanidharan wrote:

A train X starts from Meerut at 4 P.M. and reaches Ghaziabad at 5 P.M. while another train Y starts from Ghaziabad at 4 P.M. and reaches Meerut, at 5.30 PM. At what time the two trains meet each other. .

Sol:

Let the distance between two points be "D miles"

Speed of X = D/1= D miles/h Speed of Y = D/1.5= 2D/3 miles/h

Suppose X meets Y after completing "x miles" traveling for \(t_x\) hours and y traveled \(t_y\) hours

\(t_x=t_y\)

\(time=\frac{Distance}{Speed}\)

\(\frac{x}{D}=\frac{3(D-x)}{2D}\)

\(2x=3(D-x)\)

\(5x=3D\)

\(x=\frac{3}{5}D\)

Now, x travels D miles in 1 hour To travel (3/5)D miles, it would need 3/5 hours = 36minutes.

Thus; the trains meet at 4:36PM. _________________

Re: can anyone solve this problem and explain me [#permalink]
24 Apr 2011, 15:32

1

This post received KUDOS

Expert's post

tparanidharan wrote:

A train X starts from Meerut at 4 P.M. and reaches Ghaziabad at 5 P.M. while another train Y starts from Ghaziabad at 4 P.M. and reaches Meerut, at 5.30 PM. At what time the two trains meet each other. .

Great solutions above. Let us look at a solution without equations too. X takes 1 hr to cover the distance that Y covers in 1.5 hrs. That is a ratio of 2:3. So the ratio of their speeds is 3:2 since the distance they cover is the same. Now imagine the situation when they meet. They start at 4 and meet after some time. In this time, they have covered distance in the ratio 3:2 since their speed is in the ratio 3:2.

Attachment:

Ques2.jpg [ 3.53 KiB | Viewed 2502 times ]

Think of X now. It has covered 3/5th of the distance that it is supposed to cover to G. It needs to cover another 2/5th. So out of its 1 hr journey, it has covered 3/5th ie. 36 mins journey (because 3/5 * 60 min = 36 min) So they met at 4:36.

Re: can anyone solve this problem and explain me [#permalink]
25 Apr 2011, 03:09

VeritasPrepKarishma wrote:

tparanidharan wrote:

A train X starts from Meerut at 4 P.M. and reaches Ghaziabad at 5 P.M. while another train Y starts from Ghaziabad at 4 P.M. and reaches Meerut, at 5.30 PM. At what time the two trains meet each other. .

Great solutions above. Let us look at a solution without equations too. X takes 1 hr to cover the distance that Y covers in 1.5 hrs. That is a ratio of 2:3. So the ratio of their speeds is 3:2 since the distance they cover is the same. Now imagine the situation when they meet. They start at 4 and meet after some time. In this time, they have covered distance in the ratio 3:2 since their speed is in the ratio 3:2.

Attachment:

Ques2.jpg

Think of X now. It has covered 3/5th of the distance that it is supposed to cover to G. It needs to cover another 2/5th. So out of its 1 hr journey, it has covered 3/5th ie. 36 mins journey (because 3/5 * 60 min = 36 min) So they met at 4:36.

Re: can anyone solve this problem and explain me [#permalink]
25 Apr 2011, 05:02

We can use the same principle to answer something more complex... give it a go!

A train leaves Manchester at 10:00 and arrives in London at 12:15, a second train leaves London at 9:00 and arrives in Manchester at 11:30. If both trains travel a constant speed throughout the journey, at what time do the trains pass each other?

The first train takes 2hrs 15 minutes (135 minutes) to complete its journey. The second train takes 2hrs 30 minutes (150 minutes) to complete its journey.

Working out the ratio of speeds: Time A / Time B = 135/150 = 27/30 = 9/10. The distances travelled by the trains will be in the ratio 10:9. So Train A travels 10/19 in the same time Train B travels 9/19 of the distance.

If they started at the same time, they'd pass after (10/19)*135 = 71 minutes... but B set off an hour earlier! ....

If train B travels for an hour, it will cover (60/150) = 2/5 of the distance in one hour, leaving 3/5 to be travelled.

Calculating the time A will have to travel before it meets B, therefore, we use: (3/5) * (10/19)*135 = 42 + 12/19

So A will meet B after A travels just over 42 and a half minutes, or, rounding to the nearest minute, at around 10:43.

Re: can anyone solve this problem and explain me [#permalink]
25 Apr 2011, 05:27

bleemgame wrote:

We can use the same principle to answer something more complex... give it a go!

A train leaves Manchester at 10:00 and arrives in London at 12:15, a second train leaves London at 9:00 and arrives in Manchester at 11:30. If both trains travel a constant speed throughout the journey, at what time do the trains pass each other?

Karishma: Thanks for the easier way to solve question. Bleemgame: I was able to do half question but got stuck in the different timing concept. Thanks for the explanation. _________________

Re: can anyone solve this problem and explain me [#permalink]
25 Apr 2011, 07:14

That's ok if you got halfway you're doing great!! (sorry if it confused!!)

Different timings are thrown in the mix sometimes - in my example, all we actually did was find out how much closer the early train was before the other train started to move, and then applied the same workings as in your question

(as the early train covered 2/5 already, we only need to work with the 3/5 of the distance. 3/5 of the distance would take 3/5 of the time)

Re: can anyone solve this problem and explain me [#permalink]
26 Apr 2011, 00:30

bleemgame wrote:

That's ok if you got halfway you're doing great!! (sorry if it confused!!)

Different timings are thrown in the mix sometimes - in my example, all we actually did was find out how much closer the early train was before the other train started to move, and then applied the same workings as in your question

(as the early train covered 2/5 already, we only need to work with the 3/5 of the distance. 3/5 of the distance would take 3/5 of the time)

Posted from my mobile device

Yeah.. I have done such questions but it has been quite a long time.. Guess need to Refresh the concepts, which I'm currently in progress. _________________

Re: can anyone solve this problem and explain me [#permalink]
08 May 2011, 03:59

Going to explain how I calculated it.

The first train travels the entire distance in 1 hour. The other train travels only 2/3 of the distance in one hour. When the trains meet they will together have covered the entire distance, from each others ends. The speeds of the trains are 1 times the distance per hours and 2/3 of the distance per hour. After traveling at their respective speeds for time X both trains will together have covered the entire distance.

Therefore:

1X + (2/3)X = 1.

(5/3)X = 1

X = 1/(5/3)

X = 3/5

Both trains will meet after 3/5 hours, as they will have covered the entire distance together at this point in time. 1/5 of an hour is 12 minutes. 12 x 3 is 36. The trains will meet at 4:36. _________________

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