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can anyone solve this problem and explain me

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can anyone solve this problem and explain me  [#permalink]

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New post 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.
.

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New post Updated on: 24 Apr 2011, 12:34
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

Thus, trains will meet at 4:36

Originally posted by vyassaptarashi on 24 Apr 2011, 11:16.
Last edited by vyassaptarashi on 24 Apr 2011, 12:34, edited 1 time in total.
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Re: can anyone solve this problem and explain me  [#permalink]

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New post 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.
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Re: can anyone solve this problem and explain me  [#permalink]

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New post 24 Apr 2011, 15:32
2
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
Ques2.jpg [ 3.53 KiB | Viewed 23403 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.

Check out this detailed concept at
http://www.veritasprep.com/blog/2011/03 ... of-ratios/
http://www.veritasprep.com/blog/2011/03 ... os-in-tsd/
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Re: can anyone solve this problem and explain me  [#permalink]

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New post 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.

...


A clear and quick solution! :)
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Re: can anyone solve this problem and explain me  [#permalink]

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New post 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.

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New post 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.
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New post 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) :)

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New post 25 Apr 2011, 22:43
Hi, is 42 min and 38 sec the answer for 2nd Ques.?
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New post 26 Apr 2011, 00:28
tparanidharan: Yes. It is the answer.
U can see the explanation under [Reveal] Spoiler Area.
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New post 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. :-D
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New post 27 Apr 2011, 12:01
Great explanation Karishma. Thanks.
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New post 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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