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Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively. [#permalink]
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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.

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Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively. [#permalink]
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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

-------|--D----|---R---|---T------|
P--------x-------100---- t-5----
-------|--------|--------|---------|
P--------x-------120---- t+10----
-------|--------|--------|---------|

100t-500 = 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
\(\frac{5}{6} = \frac{(t-15)}{t}\)

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. [#permalink]
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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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Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively. [#permalink]
t=time of P
100t=120(t-1/4)
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. [#permalink]
Answer would be 150 Km ..Using options we can reach the answer . P will travel for 15 more mins than Q


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Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively. [#permalink]
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 (d-25)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/(d-25)
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. [#permalink]
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 - 1/6-1/12)

Distance = Speed * Time
100t = 120 *(t-1/6-1/12)
100t = 120t -20-10
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

-------|--D----|---R---|---T------|
P--------x-------100---- t-5----
-------|--------|--------|---------|
P--------x-------120---- t+10----
-------|--------|--------|---------|

100t-500 = 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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Train P and Q travel from A to B at 100 mph, and 120 mph respectively. [#permalink]
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 , \(D = 120 (t-15)\) ---------- (ii)

Distance is same for both trains. Therefore equating (i) and (ii) we get;

\(100*t = 120(t-15)\)

\(100t = 120t - 1800\)

\(20t = 1800\)

\(t = 90\) mins or \(1.5\) hours

Substituting value of '\(t\)' in distance formula, we get;

\(D = 100t = 100 * 1.5\) hours \(= 150\) km

Answer (D)...


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Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively. [#permalink]
How do we know both the trains started at same time? This question requires some assumptions
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Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively. [#permalink]
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Re: Train P and Q travel from A to B at 100 mph, and 120 mph respectively. [#permalink]
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