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A and B start from Opladen and Cologne respectively at the

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A and B start from Opladen and Cologne respectively at the  [#permalink]

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New post Updated on: 05 Oct 2013, 07:02
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Question Stats:

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A and B start from Opladen and Cologne respectively at the same time and travel towards each other at constant speeds along the same route. After meeting at a point between Opladen and Cologne, A and B proceed to their destinations of Cologne and Opladen respectively. A reaches Cologne 40 minutes after the two meet and B reaches Opladen 90 minutes after their meeting. How long did A take to cover the distance between Opladen and Cologne?

(A) 1 hour
(B) 1 hour 10 minutes
(C) 2 hours 30 minutes
(D) 1 hour 40 minutes
(E) 2 hours 10 minutes

Found this question on veritas gmat blog. Could not undertand their solution. I was trying to solve the question using algebra and assigning variables but could not get the answer.

Originally posted by sammy04 on 05 Oct 2013, 06:59.
Last edited by Bunuel on 05 Oct 2013, 07:02, edited 1 time in total.
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Re: A and B start from Opladen and Cologne respectively at the  [#permalink]

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New post 12 Oct 2013, 03:43
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sammy04 wrote:
A and B start from Opladen and Cologne respectively at the same time and travel towards each other at constant speeds along the same route. After meeting at a point between Opladen and Cologne, A and B proceed to their destinations of Cologne and Opladen respectively. A reaches Cologne 40 minutes after the two meet and B reaches Opladen 90 minutes after their meeting. How long did A take to cover the distance between Opladen and Cologne?

(A) 1 hour
(B) 1 hour 10 minutes
(C) 2 hours 30 minutes
(D) 1 hour 40 minutes
(E) 2 hours 10 minutes

Found this question on veritas gmat blog. Could not undertand their solution. I was trying to solve the question using algebra and assigning variables but could not get the answer.


OK. So, Let Speed of A=Va and Speed of B= Vb. Also, they meet after travelling for t minutes.
Hence, Va*t = distance travelled by A before meeting B ( which happens to be the same as distance travelled by B after meeting and reaching Cologne in 90 mins )
Vb*t = distance travelled by B before meeting A ( which happens to be the same as distance travelled by A after meeting and reaching Opladen in 40 mins )

Hence, 90 mins = Va*t/ Vb = > Va/Vb= 90/t
and 40 mins = Vb*t /Va => Va/Vb = t/40

Hence, 90/t=t/40
t^2 = 3600
t=60 mins
Hence, total time for which A travels = t+40 = 60 + 40 = 1 hours and 40 mins.
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Re: A and B start from Opladen and Cologne respectively at the  [#permalink]

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New post 05 Oct 2013, 07:24
1
v1 and V2 are speeds.
v1.t /90 = v2

v2.t/40 = v1
v1/v2 = 3/2
which train A would 90. 2/3 mins to cover the same distance

40 + 60 = 100 mins
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Re: A and B start from Opladen and Cologne respectively at the  [#permalink]

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New post 06 Oct 2013, 00:43
1. If before meeting they had traveled equal distance they would have taken equal time before and after meeting. If A had traveled more distance before meeting, then it means it had traveled faster and then would take less time after meeting to reach its destination.
2. If A traveled let us say 200km before meeting, and B had traveled 50 km before the meeting i.,e A had traveled x times more , in this case 4 times more, then after the meeting A by itself would travel 1/x of 200 km i.e, 50 km after the meeting and B would travel x times more x * 50 = 4*50=200km after the meeting.
3. The time taken by A decreases after the meeting by 1/x and the time taken by B increases after the meeting by 1/x.
4. In our case, we know the ratio of the time taken by the A and B after the meeting are 1:2.25 min resp. So we have from (3), the ratio of the time taken by A and B before meeting as 1*x : 2.25/x . Because the time taken before meeting is equal 1*x =2.25/x . So x=1.5. That is the time taken before the meeting for A is 40*1.5=60 min
5. So totally 60 +40= 100 min for A to travel the distance
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Re: A and B start from Opladen and Cologne respectively at the  [#permalink]

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New post 17 Oct 2014, 21:49
1
other approach
t-time taken for A&B to meet
(Va+Vb)t=Va*40+Vb*90;
Va(t-40)=Vb(90-t)
This means 40<t<90
so ans will be 40+t
Only D fits this eqn
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Re: A and B start from Opladen and Cologne respectively at the  [#permalink]

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New post 14 Nov 2015, 11:36
1
let t=time to meeting
t/40=90/t
t=60 minutes
60+40=1 hour 40 minutes
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Re: A and B start from Opladen and Cologne respectively at the  [#permalink]

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New post 15 Nov 2015, 09:56
gracie wrote:
let t=time to meeting
t/40=90/t
t=60 minutes
60+40=1 hour 40 minutes


Hi gracie

Can u please explain the method used to solve?

Thanks!
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A and B start from Opladen and Cologne respectively at the  [#permalink]

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New post 23 Nov 2015, 15:24
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sytabish wrote:
gracie wrote:
let t=time to meeting
t/40=90/t
t=60 minutes
60+40=1 hour 40 minutes


Hi gracie

Can u please explain the method used to solve?

Thanks!


Hi sytabish,
Distance varies directly with time. The ratio between the distance from O to meeting and the distance from meeting to C equals
the ratio between the time from O to meeting and the time from meeting to C, or
distance from O to m/distance from m to C=t/40=90/t
You can confirm these relationships by plugging in a speed of 60 mph for A, 40 mph for B, and a total distance of 100 miles.
I hope this helps.
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Re: A and B start from Opladen and Cologne respectively at the  [#permalink]

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New post 23 Jul 2016, 02:47
let Sa = speed of A.
Sb = speed of B.
Da = distance travelled by A till time t.
Db = distance travelled by B till time t.
t = time the two object collide.

Sa = Db / (2/3) **2/3 = 40 mins
Sb = Da / (3/2) **3/2 = 90 mins

since at time t = Da / Sa = Db / Sb

Da^2 = Db^2 (9/4)
Da = (3/2) Db

t = (Da + Db) / Sa
= (3/2) Db + Db / Sa
= 5/3 hrs.
= 100 mins.
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A and B start from Opladen and Cologne respectively at the  [#permalink]

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New post 07 Apr 2017, 20:29
gracie wrote:
sytabish wrote:
gracie wrote:
let t=time to meeting
t/40=90/t
t=60 minutes
60+40=1 hour 40 minutes


Hi gracie

Can u please explain the method used to solve?

Thanks!


Hi sytabish,
Distance varies directly with time. The ratio between the distance from O to meeting and the distance from meeting to C equals
the ratio between the time from O to meeting and the time from meeting to C, or
distance from O to m/distance from m to C=t/40=90/t
You can confirm these relationships by plugging in a speed of 60 mph for A, 40 mph for B, and a total distance of 100 miles.
I hope this helps.
gracie


Would you please explain this t/40=90/t? Why t is in both denominator and numerator?

OK got it from https://www.veritasprep.com/blog/2013/0 ... -formulas/
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Re: A and B start from Opladen and Cologne respectively at the  [#permalink]

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New post 08 Apr 2017, 03:27
Hi guys, I'm facing a lot of problem with these questions. Where can get proper material to understand the basics of these problems and practice?
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Re: A and B start from Opladen and Cologne respectively at the  [#permalink]

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New post 08 Apr 2017, 03:32
avinsethi wrote:
Hi guys, I'm facing a lot of problem with these questions. Where can get proper material to understand the basics of these problems and practice?


Theory on Distance/Rate Problems: http://gmatclub.com/forum/distance-spee ... 87481.html

All DS Distance/Rate Problems to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=44
All PS Distance/Rate Problems to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=64


For more on relative speed, check: http://www.veritasprep.com/blog/2012/07 ... elatively/
http://www.veritasprep.com/blog/2012/08 ... -speeding/

Hope it helps.
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Re: A and B start from Opladen and Cologne respectively at the  [#permalink]

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