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malkadhi
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 units

Assume 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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t=time of P
100t=120(t-1/4)
t=3/2 hour
(3/2)(100)=150 miles
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malkadhi
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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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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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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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
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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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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malkadhi
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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How do we know both the trains started at same time? This question requires some assumptions
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