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A and B start from Opladen and Cologne respectively at the
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Updated on: 05 Oct 2013, 08:02
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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.
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Originally posted by sammy04 on 05 Oct 2013, 07:59.
Last edited by Bunuel on 05 Oct 2013, 08:02, edited 1 time in total.
RENAMED THE TOPIC.




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Re: A and B start from Opladen and Cologne respectively at the
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12 Oct 2013, 04:43
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
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05 Oct 2013, 08:24
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
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06 Oct 2013, 01:43
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
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17 Oct 2014, 22:49
other approach ttime taken for A&B to meet (Va+Vb)t=Va*40+Vb*90; Va(t40)=Vb(90t) 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
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14 Nov 2015, 12:36
let t=time to meeting t/40=90/t t=60 minutes 60+40=1 hour 40 minutes



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15 Nov 2015, 10:56
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
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23 Nov 2015, 16:24
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



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Re: A and B start from Opladen and Cologne respectively at the
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23 Jul 2016, 03:47
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
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07 Apr 2017, 21:29
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
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08 Apr 2017, 04:27
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
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08 Apr 2017, 04:32
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? 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
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17 Feb 2019, 05:53
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 When two elements travel at different speeds, their TIME RATIO to travel the same distance will always be the same. If A takes 1/2 as long as B to travel 10 miles, then A will take 1/2 as long as B to travel 1000 miles. If A takes 3 times as long as B to travel 500 miles, then A will take three times as long as B to travel 2 miles. Let M = the meeting point. Let t = the time for A and B each to travel to M. Train A: O  t > M > 40 > C Train B: O < 90 M < t  C Since A takes t minutes to travel the blue portion, while B takes 90 minutes, the time ratio for A and B to travel the blue portion = t/90. Since A takes 40 minutes to travel the red portion, while B takes t minutes, the time ratio for A and B to travel the red portion = 40/t. Since the time ratio in each case must be the same, we get: t/90 = 40/t t² = 3600 t = 60. Thus: A's total time = t+40 = 60+40 = 100 minutes = 1 hour 40 minutes. .
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Re: A and B start from Opladen and Cologne respectively at the
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18 Feb 2019, 03:55
GMATGuruNY wrote: 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 When two elements travel at different speeds, their TIME RATIO to travel the same distance will always be the same. If A takes 1/2 as long as B to travel 10 miles, then A will take 1/2 as long as B to travel 1000 miles. If A takes 3 times as long as B to travel 500 miles, then A will take three times as long as B to travel 2 miles. Let M = the meeting point. Let t = the time for A and B each to travel to M. Train A: O  t > M > 40 > C Train B: O < 90 M < t  C Since A takes t minutes to travel the blue portion, while B takes 90 minutes, the time ratio for A and B to travel the blue portion = t/90. Since A takes 40 minutes to travel the red portion, while B takes t minutes, the time ratio for A and B to travel the red portion = 40/t. Since the time ratio in each case must be the same, we get: t/90 = 40/t t² = 3600 t = 60. Thus: A's total time = t+40 = 60+40 = 100 minutes = 1 hour 40 minutes. . Dear GMATGuruNYI had the same result but based on the time is reciprocal with speed. Let T= time to meeting point, VA=speed of A, VB = speed of B VA/VB = 90/40 = 3x/2xTotal distance = VA*T + VB *T = 5x * T...........1 But Total distance = VA (T+40) = 3x * T + 120x.......2 By equating 1& 2 5x * T = 3x * T + 120x............cancel x 5 * T = 3 * T + 120 2T = 120......T =60 minutes Then, A will take 1 hour & 40 minutes to reach its distance. Is my assumption highlighted valid? or is coincidence in this question? Thanks in advance



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Re: A and B start from Opladen and Cologne respectively at the
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19 Feb 2019, 08:01
Mo2men wrote: VA/VB = 90/40 = 3x/2x
Is my assumption highlighted valid?
The red portion is incorrect: \(\frac{90}{40} = \frac{9}{4}\) The ratio above does not accurately reflect the information given in the prompt. My earlier solution yields t=60. Thus: A's total time = t+40 = 60+40 = 100 minutes B's total time = 90+t = 90+60 = 150 minutes Since the time ratio for A and B \(= \frac{100}{150} = \frac{2}{3}\), the rate ratio for A and B = \(\frac{3}{2}\) \(\frac{9}{4}\) does not accurately reflect the rate ratio for A and B.
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Re: A and B start from Opladen and Cologne respectively at the
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19 Feb 2019, 08:16
GMATGuruNY wrote: Mo2men wrote: VA/VB = 90/40 = 3x/2x
Is my assumption highlighted valid?
The red portion is incorrect: \(\frac{90}{40} = \frac{9}{4}\) The ratio above does not accurately reflect the information given in the prompt. My earlier solution yields t=60. Thus: A's total time = t+40 = 60+40 = 100 minutes B's total time = 90+t = 90+60 = 150 minutes Since the time ratio for A and B \(= \frac{100}{150} = \frac{2}{3}\), the rate ratio for A and B = \(\frac{3}{2}\) \(\frac{9}{4}\) does not accurately reflect the rate ratio for A and B. Thanks a lot for explanation.




Re: A and B start from Opladen and Cologne respectively at the
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