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The question is as follows: Train A left Centerville Station, heading toward Dale City Station at 3:00pm. Train B left Dale City Station heading toward Centerville Station at 3:20pm. on the same day. The trains rode on straight tracks that were parallel to each other. If train A traveled at a constant speed of 30 miles per hour, Train B traveled at a constant rate of 10 miles per hour, and the distance between Centerville Station and Dale City station is 90 miles, when did the trains pass each other? A.)4:45pm B.)5:00pm C.)5:20pm D.)5:35pm E.)6:00pm Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums! ### GMAT/MBA Expert Brent@GMATPrepNow GMAT Instructor Joined 08 Dec 2008 Posted: 10886 messages Followed by: 1215 members Thanked: 5211 times GMAT Score: 770 Wed Mar 11, 2015 1:22 pm datonman wrote: Train A left Centerville Station, heading toward Dale City Station at 3:00pm. Train B left Dale City Station heading toward Centerville Station at 3:20pm. on the same day. The trains rode on straight tracks that were parallel to each other. If train A traveled at a constant speed of 30 miles per hour, Train B traveled at a constant rate of 10 miles per hour, and the distance between Centerville Station and Dale City station is 90 miles, when did the trains pass each other? A.)4:45pm B.)5:00pm C.)5:20pm D.)5:35pm E.)6:00pm 3:00pm - Train A (traveling 30 miles per hour) left Centerville Station, heading toward Dale City Station If the train travels 30 miles every HOUR, then it travels 10 miles every 20 minutes. In other words, at 3:20pm, Train A has already traveled 10 miles. So, at 3:20, the trains are now 80 miles apart. 3:20pm - Train B (traveling 10 miles per hour) left Dale City Station heading toward Centerville Station At this point (at 3:20pm), the trains are 80 miles apart. Also, every hour, Train A moves 30 miles toward train B, and Train B moves 10 miles towards train A. This means the gap between them is decreasing at a rate of 40 miles per hour. At this rate, the 80-mile gap will be reduced to zero miles (i.e., the trains meet) in 2 hours. So, the trains will meet at 3:20pm + 2 hours = 5:20pm Answer: C Cheers, Brent _________________ Brent Hanneson â€“ Founder of GMATPrepNow.com Use our video course along with Check out the online reviews of our course Come see all of our free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! ### GMAT/MBA Expert Rich.C@EMPOWERgmat.com Elite Legendary Member Joined 23 Jun 2013 Posted: 8832 messages Followed by: 463 members Thanked: 2823 times GMAT Score: 800 Wed Mar 11, 2015 3:56 pm Hi datonman, Brent's solution is spot-on, so I won't rehash any of that here. Instead, I want to point out a few aspects of these types of questions: 1) The Distance Formula is Distance = (Rate)(Time). However you choose to 'manipulate' it, THAT is the formula. 2) This prompt is an example of a Combined Rate question. These types of questions are relatively rare on Test Day (you'll likely see just one) and they're almost always 'story problems.' As such, they take longer than average to solve and are worth less than most other categories to your overall score. 3) The key to dealing with these types of prompts involves figuring out what happens when both 'entities' (trains, cars, people, etc.) are moving at the same time. If one entity moves for a certain amount of time before the other gets moving, then you have to deal with that part of the 'math' first. GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at Rich.C@empowergmat.com Thanked by: umasarath52, asherkunal ### GMAT/MBA Expert DavidG@VeritasPrep Legendary Member Joined 14 Jan 2015 Posted: 2380 messages Followed by: 115 members Thanked: 1121 times GMAT Score: 770 Tue Mar 17, 2015 3:15 am Brent's method is probably the fastest way to approach this question. Alternatively, you can plug and chug in a R*T = D chart. If we know that A left 20 minutes earlier, then it traveled for 1/3 of an hour longer than B did. Call their times 't + 1/3' and 't' respectively. Our chart will look like this: ttp://postimg.org/image/64q9xmejd/" target="_blank"> Now simply solve the distance column to get the following equation: 30t + 10 + 10t = 90 40t = 80 t = 2 If they pass once train B has been traveling for 2 hours, it will be 5:20 _________________ Veritas Prep | GMAT Instructor Veritas Prep Reviews Save$100 off any live Veritas Prep GMAT Course

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Sun Mar 22, 2015 10:58 pm
DavidG@VeritasPrep wrote:
Brent's method is probably the fastest way to approach this question. Alternatively, you can plug and chug in a R*T = D chart.
This is a great method to remember. If you forget the formula, you can almost always still solve this problem anywhere if you just plot the trains - and you can often approximate with the answers to save time.

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YTarhouni Newbie | Next Rank: 10 Posts
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Tue Sep 05, 2017 7:12 pm
at 3:20. Train A would have traveled 10 miles (Speed*time=30*1/3=10). the distance now is only 80 miles.
now the two trains are traveling toward each others at a combined speed of 30+10=40. the 80 miles will be traveled in a 2 hours (distance/speed=80/40=2).
3:20+2:00=5:20

Imperiex Junior | Next Rank: 30 Posts
Joined
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Posted:
26 messages
Tue Sep 05, 2017 10:32 pm
datonman wrote:
The question had me stumped because i forgot which formula i had to use, either d*r/t or r*d/t.

The question is as follows:

Train A left Centerville Station, heading toward Dale City Station at 3:00pm. Train B left Dale City Station heading toward Centerville Station at 3:20pm. on the same day. The trains rode on straight tracks that were parallel to each other. If train A traveled at a constant speed of 30 miles per hour, Train B traveled at a constant rate of 10 miles per hour, and the distance between Centerville Station and Dale City station is 90 miles, when did the trains pass each other?

A.)4:45pm
B.)5:00pm
C.)5:20pm
D.)5:35pm
E.)6:00pm
Solution:
Train A is moving at a speed of 30 miles per hours. It means in 20 minutes it will cover 10 miles. Therefore at 3:20 pm, When Train B starts moving, Train A have already covered a distance of 10 miles and remaining distance between both the trains will be 80 miles at 3:20 pm as shown in the image below.
ttps://postimg.org/image/f72icdgyt/" target="_blank">
Let D be the point where both the trains meet each other.
Since Train A is moving with a speed 30 m/h and it covers a distance of x miles in t hours, then we can write,
Distance = (Speed)(Time)
x = 30 t --- (i)

Similarly, Train B is moving with a speed of 10 m/h and covers a distance of (80 - x) miles to meet train A, therefore,
Distance = (Speed)(Time)
80 - x = 10 t
80 = x + 10t
x = 80 - 10t --- (ii)

Using Eq (i) and (ii), we can write
30t = 80 - 10t
30t + 10t = 80
40t = 80
t = 80/40
t = 2 hours.

Therefore, Train A and B meet each other after 2 hours.
Hence, the time will be
3:20 pm + 2 hours = 5:20 pm

Option C is the correct answer.

Imperiex Junior | Next Rank: 30 Posts
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Tue Sep 05, 2017 11:07 pm
Here's a more technical math solution as well:
Based on the given values, let the coordinates at Centerville Station be 0 and that at Dale City Station be 90 miles, and

Train A:
x0A = 0
t0A = 3 hr
v0A = vA = +30 miles/hour

Train B:
x0B = 90 miles
t0B = (3 + 1/3) hr = (10 / 3) hr
v0B = vB = -10 miles/hour

(Notice that we converted 20 minutes into (20/60) = 1/3 hr.)
(Notice also that since Train B is traveling in the opposite direction with respect to Train A, we therefore put negative sign in the velocity.)
Both would pass each other at x and at time t.

Using the following kinematic equation (or also known as equation of motion):

x - x0 = v0 (t - t0) + (1/2)a(t - t0)^2,

where acceleration a = 0 (due to constant velocity) here in this case, we have, respectively for Train A and Train B,

x - x0A = v0A (t - t0A),

x - x0B = v0B (t - t0B).

Simultaneously solving this system of two equations and solving for time t (say, we eliminate x by subtracting sides of one equation from that of the other), we obtain,

t = ( -x0A + x0B - ( v0B t0B - v0A t0A ) ) / ( v0A - v0B )

Hence,

t = ( -0 + 90 - ( (-10) (10/3) - (30)(3) ) ) / ( 30 - (-10) )

= (90 + (10)(10/3) + (30)(3)) / 40

= (9 + (10/3) + 9) / 4

= 16/3

= 5 and 1/3

Therefore, the trains pass each other at 5:20 pm.

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Mon Sep 11, 2017 10:15 am
datonman wrote:
The question had me stumped because i forgot which formula i had to use, either d*r/t or r*d/t.

The question is as follows:

Train A left Centerville Station, heading toward Dale City Station at 3:00pm. Train B left Dale City Station heading toward Centerville Station at 3:20pm. on the same day. The trains rode on straight tracks that were parallel to each other. If train A traveled at a constant speed of 30 miles per hour, Train B traveled at a constant rate of 10 miles per hour, and the distance between Centerville Station and Dale City station is 90 miles, when did the trains pass each other?

A.)4:45pm
B.)5:00pm
C.)5:20pm
D.)5:35pm
E.)6:00pm
Since 20 minutes = 1/3 hour, we can let the time of train B = t hours, and thus the time of train A = t + 1/3 hours.

Using the formula distance = rate x time, the distance for train A is 30(t + 1/3) = 30t + 10 and the distance for train B is 10t. The distances of the two trains sum to 90 miles; thus:

30t + 10 + 10t = 90

40t = 80

t = 2 hours

So, the trains passed each other at 3:20 p.m. + 2 hours = 5:20 p.m.

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