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Joined: 15 Sep 2015
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Train P and Q travel from A to B at 100 mph, and 120 mph respectively.
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01 Dec 2015, 09:42
Question Stats:
65% (02:39) correct 35% (02:56) wrong based on 206 sessions
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Train P and Q travel from A to B at 100 mph, and 120 mph respectively. Train Q stops for 10 minutes at station C, but reaches Station B, 5 Minutes before Train P. what is the distance between A and B. A) 50 km B) 100 km C) 120 km D) 150 km This is not GMAT Material, so only 4 choices.
So I was going through a couple of varied DRT problems, and this one took longer than it should. My question is, how can I approach it differently? Or more efficiently?
So, what I initially tried do was the following
DRT Px100 t5  Px120 t+10 
100t500 = 120t + 1200
Which is obviously wrong. Then I thought about what I'm actually doing
P's time is five minutes behind T's time after T stops for 10 minutes
P's Time  5 = Q's Time + 10
P's Time = \(\frac{x}{100}\) and Q's Time = \(\frac{x}{120}\)
So we have:
\(\frac{x}{100}\) = \(\frac{x}{120}\) + \(\frac{15}{60}\)
x=150
So, I suppose I was wondering if anyone has any suggestions on tackling these kind of problems? Or maybe a different approach.
Thanks
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Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively.
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01 Dec 2015, 09:55
malkadhi wrote: So I was going through a couple of varied DRT problems, and this one took longer than it should. My question is, how can I approach it differently? Or more efficiently?
Train P and Q travel from A to B at 100 mph, and 120 mph respectively. Train Q stops for 10 minutes at station C, but reaches Station B, 5 Minutes before Train P.
A) 50 km B) 100 km C) 120 km D) 150 km
This is not GMAT Material, so only 4 choices.
So, what I initially tried do was the following
DRT Px100 t5  Px120 t+10 
100t500 = 120t + 1200
Which is obviously wrong. Then I thought about what I'm actually doing
P's time is five minutes behind T's time after T stops for 10 minutes
P's Time  5 = Q's Time + 10
P's Time = \(\frac{x}{100}\) and Q's Time = \(\frac{x}{120}\)
So we have:
\(\frac{x}{100}\) = \(\frac{x}{120}\) + \(\frac{15}{60}\)
x=150
So, I suppose I was wondering if anyone has any suggestions on tackling these kind of problems? Or maybe a different approach.
Thanks Make sure to follow posting guidelines (link in my signatures). All your analyses should either go under "spoilers" or in the next post. This will not dilute the discussion. We do not usually encourage posting of non GMAT questions as this will only develop bad habits. But for a one time discussion, your method is absolutely fine. Realize that the for PS questions in GMAT, you must use the options to your advantage. This is not only provide you the correct option but will also help you in spending lesser time than usual. Time management is of utmost importance GMAT. That being said, once you set up the equation: P's time  5 min = Q's time + 10 minutes > x/100  5/60 = x/120 + 10/60 > x/100x/120 = 15/60 . Now use the values given. Only D will satisfy this.



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Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively.
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02 Dec 2015, 22:00
Rich malkadhi wrote: Train P and Q travel from A to B at 100 mph, and 120 mph respectively. Train Q stops for 10 minutes at station C, but reaches Station B, 5 Minutes before Train P. what is the distance between A and B.
A) 50 km B) 100 km C) 120 km D) 150 km
Hi malkadhi, What is the source of this question (and does the original source have the typos in it or is it just your transcription)? I ask because the prompt only has 4 answer choices and the answers are in KILOMETERS, while the speeds are in MILES per hour. There are plenty of quality GMAT materials that you can use during your studies, so you might want to stop using this resource. GMAT assassins aren't born, they're made,
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Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively.
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07 Dec 2015, 05:46
malkadhi wrote: Train P and Q travel from A to B at 100 mph, and 120 mph respectively. Train Q stops for 10 minutes at station C, but reaches Station B, 5 Minutes before Train P. what is the distance between A and B.
A) 50 km B) 100 km C) 120 km D) 150 km
Since every thing is given in the form of time, let us form an equation in time. Make sure to keep everything in the same unitsAssume the distance between A and B to be x miles. Time taken by Train P = \(\frac{x}{100}\) Time taken by Train Q = \(\frac{x}{120}\) + 10/60 We are given that Train P reaches 5 minutes after Train Q Hence \(\frac{x}{100}\)  5/60 = \(\frac{x}{120}\) + 10/60 \(\frac{x}{100}\)  \(\frac{x}{120}\) = 15/60 \(\frac{6x  5x}{600}\) = 15/60 x = 150 Option D



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Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively.
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07 Dec 2015, 08:20
t=time of P 100t=120(t1/4) t=3/2 hour (3/2)(100)=150 miles



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Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively.
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28 May 2017, 01:40
malkadhi wrote: Train P and Q travel from A to B at 100 mph, and 120 mph respectively. Train Q stops for 10 minutes at station C, but reaches Station B, 5 Minutes before Train P. what is the distance between A and B. A) 50 km B) 100 km C) 120 km D) 150 km This is not GMAT Material, so only 4 choices.
So I was going through a couple of varied DRT problems, and this one took longer than it should. My question is, how can I approach it differently? Or more efficiently?
So, what I initially tried do was the following
DRT Px100 t5  Px120 t+10 
100t500 = 120t + 1200
Which is obviously wrong. Then I thought about what I'm actually doing
P's time is five minutes behind T's time after T stops for 10 minutes
P's Time  5 = Q's Time + 10
P's Time = \(\frac{x}{100}\) and Q's Time = \(\frac{x}{120}\)
So we have:
\(\frac{x}{100}\) = \(\frac{x}{120}\) + \(\frac{15}{60}\)
x=150
So, I suppose I was wondering if anyone has any suggestions on tackling these kind of problems? Or maybe a different approach.
Thanks Quick way of doing this is by using ratios. Let the number of minutes that it takes P = t minutes. So, for Q it is = t  15 minutes. Ratio of their speeds = 5:6 So, ratio of their times = 6:5 \(\frac{5}{6} = \frac{(t15)}{t}\) t = 90 minutes..or 1.5 hrs.. Thus, distance is \(1.5*100 = 150 km\) (D)
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Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively.
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28 May 2017, 01:45
Answer would be 150 Km ..Using options we can reach the answer . P will travel for 15 more mins than Q Sent from my iPhone using GMAT Club Forum
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Joined: 09 Nov 2015
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Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively.
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28 May 2017, 07:28
If Q had not stopped for 10 minutes, it would have reached Station B 15 minutes earlier than P. That is, when Q reaches B, P is still 100/4=25 miles behind. Which means that in the time that Q covers the distance AB (lets assume this to be d miles) P covers (d25)miles. Since the ratios of the speeds of the two trains is equal to ratio of the distances covered by them in the same time, we can write: 120/100=d/(d25) Thus, d=150. Answer: D



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Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively.
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28 Jul 2017, 13:49
For Train P S = 100mph D = x miles T = t hrs For Train Q. S= 120mph D = xmiles T = t 1/6 1/12 (Since Train Q stopped for 10 min, i.e. 1/6hrs and reached 5 min earlier, i.e, 1/12 hrs; Therefore total time the Train Q took was t  1/61/12) Distance = Speed * Time 100t = 120 *(t1/61/12) 100t = 120t 2010 20t =30 t= 1.5 hrs Distance = X miles = 100*t = 150 miles malkadhi wrote: Train P and Q travel from A to B at 100 mph, and 120 mph respectively. Train Q stops for 10 minutes at station C, but reaches Station B, 5 Minutes before Train P. what is the distance between A and B. A) 50 km B) 100 km C) 120 km D) 150 km This is not GMAT Material, so only 4 choices.
So I was going through a couple of varied DRT problems, and this one took longer than it should. My question is, how can I approach it differently? Or more efficiently?
So, what I initially tried do was the following
DRT Px100 t5  Px120 t+10 
100t500 = 120t + 1200
Which is obviously wrong. Then I thought about what I'm actually doing
P's time is five minutes behind T's time after T stops for 10 minutes
P's Time  5 = Q's Time + 10
P's Time = \(\frac{x}{100}\) and Q's Time = \(\frac{x}{120}\)
So we have:
\(\frac{x}{100}\) = \(\frac{x}{120}\) + \(\frac{15}{60}\)
x=150
So, I suppose I was wondering if anyone has any suggestions on tackling these kind of problems? Or maybe a different approach.
Thanks



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Joined: 23 Jul 2015
Posts: 157

Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively.
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06 Aug 2017, 03:40
Total time taken by Q =\(\frac{D}{120} + \frac{10}{60}\) Total time taken by P =\(\frac{D}{100}\) = Total time take by Q + \(\frac{5}{60}\) \(\frac{D}{120} + \frac{10}{60}\) + \(\frac{1}{12}\) = \(\frac{D}{100}\)
\(\frac{D+30}{120} = \frac{D}{100}\)
Solving for D > D = 150



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Train P and Q travel from A to B at 100 mph, and 120 mph respectively.
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06 Aug 2017, 21:05
malkadhi wrote: Train P and Q travel from A to B at 100 mph, and 120 mph respectively. Train Q stops for 10 minutes at station C, but reaches Station B, 5 Minutes before Train P. what is the distance between A and B.
A) 50 km B) 100 km C) 120 km D) 150 km Let the distance from \(A\) to \(B = D\)
Let the time for Train \(P\) to travel from \(A\) to \(B = t\)
Speed of train \(P = 100\) mph
Distance , \(D = 100*t\)  (i)
Time for Train \(Q\) to travel from \(A\) to \(B = t  (10+5) = t  15\)
Speed of train \(Q = 120\) mph
Distance , \(D = 120 (t15)\)  (ii)
Distance is same for both trains. Therefore equating (i) and (ii) we get;
\(100*t = 120(t15)\)
\(100t = 120t  1800\)
\(20t = 1800\)
\(t = 90\) mins or \(1.5\) hours
Substituting value of '\(t\)' in distance formula, we get;
\(D = 100t = 100 * 1.5\) hours \(= 150\) km
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Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively.
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14 Sep 2018, 21:34
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Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively. &nbs
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