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# Mary passed a certain gas station on a highway while traveling west at

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Re: Mary passed a certain gas station on a highway while traveling west at [#permalink]
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rags wrote:
Mary passed a certain gas station on a highway while traveling west at a constant
speed of 50 miles per hour. Then, 15 minutes later, Paul passed the same gas station
while traveling west at a constant speed of 60 miles per hour. If both drivers
maintained their speeds and both remained on the highway for at least 2 hours, how
long after he passed the gas station did Paul catch up with Mary?
A. 30 min
B. 45 min
C. 1 hr
D. 1 hr 15 min
E. 1 hr 30 min

i get D

d = rt

M: r = 50mph, t = t + 1/4hr
d = 50 (t + 1/4)

P: r = 60, t = t
d = 60t

since they went the same distance:
50t + 50/4 = 60t
10t = 50/4
t = 1.25 or 1 hr, 15min
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Re: Mary passed a certain gas station on a highway while traveling west at [#permalink]
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Quote:
Mary passed a certain gas station on a highway while traveling west at a constant speed of 50 mph. Then, 15 minutes later, Paul passed the same gas station while traveling west at a constant speed of 60 mph. If both travelers maintained their speed and both remained on the highway for at least 2 hours, how long after he passed the gas station did Paul catch up with Mary?

A 35 mins
B 45 mins
C 1 hour
D 1 hour 15 mins
E 1 hour 30 mins

Mary passes gas station and travels at 50 mph for 15 minutes
So, Mary drove for 1/4 hours
Distance traveled = (time)(speed)
= (1/4)(50)
= 50/4
= 12.5 miles

Paul passes the same gas station and travels at 60 mph
IMPORTANT: At this point, Paul is 12.5 miles behind Mary.
However, since Paul is traveling 10 mph FASTER than Mary, the gap between them SHRINKS at a rate of 10 mph.
So, at this point, we need to determine the time it takes for the 12.5 mile gap to shrink to zero.
Time = (distance)/(speed)
= 12.5/10
= 1.25 hours
= 1 hour 15 mins
= D

Cheers,
Brent
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Re: Mary passed a certain gas station on a highway while traveling west at [#permalink]
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rags wrote:
Mary passed a certain gas station on a highway while traveling west at a constant
speed of 50 miles per hour. Then, 15 minutes later, Paul passed the same gas station
while traveling west at a constant speed of 60 miles per hour. If both drivers
maintained their speeds and both remained on the highway for at least 2 hours, how
long after he passed the gas station did Paul catch up with Mary?
A. 30 min
B. 45 min
C. 1 hr
D. 1 hr 15 min
E. 1 hr 30 min

First you must realize that Mary went 50mph for 15minutes after she passed the gas station. @ exactly 15min Paul passes the gas station.

15min = 1/4hr So mary went 50/4 miles.

To find out how long it will take Paul to reach Mary, just make Mary's speed 0 and Paul's speed 10.

So t=50/4/10 ---> t=1.25 or 1 1/4 hrs which is 1 hr 15min
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Re: Mary passed a certain gas station on a highway while traveling west at [#permalink]
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aimeehittinger wrote:
Mary passed a certain gas station on a highway while traveling west at a constant speed of 50 mph. Then, 15 minutes later, Paul passed the same gas station while traveling west at a constant speed of 60 mph. If both drivers maintained their speeds and both remained on the highway for at least 2 hours, how long after he passed the gas station did Paul catch up with Mary?

A. 30 min
B. 45 min
C. 1 hr
D. 1 hr 15 min
E. 1 hr 30 min

We can classify this problem as a “catch-up” rate problem, for which we use the formula:

distance of Mary = distance of Paul

We are given that Mary passed a certain gas station traveling at a rate of 50 mph and that 15 minutes later Paul passed the same gas station traveling at a rate of 60 mph.

Since Mary passed the gas station 15 minutes before Paul, we can let Paul’s time = t hours and Mary’s time = t + 1/4 hours.

Since distance = rate x time, we can calculate each distance in terms of t.

Mary’s distance = 50(t + 1/4) = 50t + 50/4 = 50t + 25/2

Paul’s distance = 60t

We can equate the two distances and determine t.

50t + 25/2 = 60t

25/2 = 10t

t = (25/2)/10 = 25/20 = 5/4 = 1¼ hours = 1 hour and 15 minutes

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Re: Mary passed a certain gas station on a highway while traveling west at [#permalink]
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angelfire213 wrote:
I am still confused on this. How do we know they caught up after 1 hour?

I tried with the relative speed approach and realized he was gaining on her at 10mph. but after that I don't get the logic/leap saying total time is 1 hr+ 15 mins? how do we know that is when they caught up?

Refer diagram below:
Attachment:

ma.jpg [ 28.41 KiB | Viewed 141530 times ]

Let the distance between gas station & meeting point = d

Mary travels for 15 Minutes at speed of 50 Miles/hr before Paul crosses the gas station means

Mary has travelled distance of 50/4 = 12.5 when Paul starts from the gas station

Now, Mary is at point P which is 12.5 Miles away from gas station, so now the remaining distance = (d - 12.5) for which Paul will chase

Time required for Paul to reach the catch point = Time required by Mary to reach the catch point

Equation setup will be as follows

$$\frac{d}{60} = \frac{d-12.5}{50}$$

d = 75 (Distance between Gas station & Catch point)

Speed of Paul = 60 Miles/Hr = 1 Miles/ Minute

Time required by Paul = 75 * 1 = 75 Minutes = 1 Hour 15 Minutes

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Re: Mary passed a certain gas station on a highway while traveling west at [#permalink]
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Hi aimeehittinger,

This is an example of a 'chase down' question. The 'key' to solving it is to realize that the faster person will 'catch up' to the slower one at a constant rate. Here, since Paul is going 60mph and Mary is going 50mph, Paul will catch up 10 miles each hour that they both travel.

Using the gas station as a starting point, Mary has a 15-minute 'lead' on Paul. That 15 minutes translates into...

D = (R)(T)
D = (50mph)(1/4 hour)
D = 50/4 miles = 12.5 miles

So Mary is 12.5 miles ahead of Paul when he passes the gas station. Now we can figure out how much time it will take Paul to catch up...

D = (R)(T)
12.5 miles = (10mph)(T)
12.5/10 = T
1.25 hours = T
T = 1 hour 15 minutes

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Re: Mary passed a certain gas station on a highway while traveling west at [#permalink]
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When Paul is at the station, Mary is 50/4 miles ahead on the highway. (the distance she drove in 15 min)

every hour, paul drives 10miles more than Mary. how many hours will it takes him to drive 50/4 miles more? the answer is (50/4)/10 = 25/20=1h15min.

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Re: Mary passed a certain gas station on a highway while traveling west at [#permalink]
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Relative speed = 60-50 = 10miles/hour

relative distance between them = distance traveled by Mary in 15 min = 50 * 15/60

50 * 15/60 = 10 t

=> t = 1 hour 15 mins
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Re: Mary passed a certain gas station on a highway while traveling west at [#permalink]
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$$\frac{d-12.5}{50} = \frac{d}{60}$$

d = 75

$$t = \frac{75}{60} * 60$$ Minutes

= 75 Minutes

= 01 Hr 15 Minutes
Attachments

ma.jpg [ 15.84 KiB | Viewed 144857 times ]

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Re: Mary passed a certain gas station on a highway while traveling west at [#permalink]
I am still confused on this. How do we know they caught up after 1 hour?

I tried with the relative speed approach and realized he was gaining on her at 10mph. but after that I don't get the logic/leap saying total time is 1 hr+ 15 mins? how do we know that is when they caught up?
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Re: Mary passed a certain gas station on a highway while traveling west at [#permalink]
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Hi,
anyone know a similar problem ?
maybe Bunuel

thx so much
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Re: Mary passed a certain gas station on a highway while traveling west at [#permalink]
ScottTargetTestPrep wrote:
aimeehittinger wrote:
Mary passed a certain gas station on a highway while traveling west at a constant speed of 50 mph. Then, 15 minutes later, Paul passed the same gas station while traveling west at a constant speed of 60 mph. If both drivers maintained their speeds and both remained on the highway for at least 2 hours, how long after he passed the gas station did Paul catch up with Mary?

A. 30 min
B. 45 min
C. 1 hr
D. 1 hr 15 min
E. 1 hr 30 min

We can classify this problem as a “catch-up” rate problem, for which we use the formula:

distance of Mary = distance of Paul

We are given that Mary passed a certain gas station traveling at a rate of 50 mph and that 15 minutes later Paul passed the same gas station traveling at a rate of 60 mph.

Since Mary passed the gas station 15 minutes before Paul, we can let Paul’s time = t hours and Mary’s time = t + 1/4 hours.

Since distance = rate x time, we can calculate each distance in terms of t.

Mary’s distance = 50(t + 1/4) = 50t + 50/4 = 50t + 25/2

Paul’s distance = 60t

We can equate the two distances and determine t.

50t + 25/2 = 60t

25/2 = 10t

t = (25/2)/10 = 25/20 = 5/4 = 1¼ hours = 1 hour and 15 minutes

ScottTargetTestPrep

I tried your approach on this problem: https://gmatclub.com/forum/a-gang-of-cr ... 39444.html
but was unfortunately very confused how to approach this given that times were not given.

Using your approach for the criminal problem that you used on this problem, would it be

Police Rate=80 mi/1 hr
Let time=t

Train rate=50 mi / 1 hr
Let t=t+1 --> adding one hour because the police is 50 miles behind the train and the train does 50 miles in 1 hour

Solving: 80t=50t+50 --> t=5/3 hours

Thank you for all of your time and help
80t=50t+50
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Re: Mary passed a certain gas station on a highway while traveling west at [#permalink]
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woohoo921 wrote:

I tried your approach on this problem: https://gmatclub.com/forum/a-gang-of-cr ... 39444.html
but was unfortunately very confused how to approach this given that times were not given.

Using your approach for the criminal problem that you used on this problem, would it be

Police Rate=80 mi/1 hr
Let time=t

Train rate=50 mi / 1 hr
Let t=t+1 --> adding one hour because the police is 50 miles behind the train and the train does 50 miles in 1 hour

Solving: 80t=50t+50 --> t=5/3 hours

Thank you for all of your time and help
80t=50t+50

As a matter of fact, your answer is correct, because 5/3 hours = 100 minutes = 1 hour 40 minutes, which is the correct answer. However, I wouldn't solve the question this way. You are saying that times were not given, but actually we are told that the police car started moving at the exact same time the hijacking happened, which means that both the police and the hijackers travel for the same amount of time. Thus, we can use the variable t to represent the time of both parties. On the other hand, since the police started the chase from 50 miles behind, the police will have traveled 50 miles more than the hijackers when the two parties meet. Thus, after determining the distance of police to be 80t miles and the distance of gangsters to be 50t miles, we should add 50 miles to the distance of gangsters so that it equals the distance of the police. We will end up getting 80t = 50t + 50, which is exactly the same equation you get. My only objection to your mehtod is that it involves an unintuitive conversion (50 miles gets converted to one extra hour of travel for the gangsters), which could lead to mistakes if you tried the same method in other question. Your method is otherwise correct. Actually, your method is exactly how one would solve the question if the police and the gangsters started from the same point, but the gangsters left one hour earlier than the police.
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Mary passed a certain gas station on a highway while traveling west at [#permalink]
BrentGMATPrepNow wrote:
Quote:
Mary passed a certain gas station on a highway while traveling west at a constant speed of 50 mph. Then, 15 minutes later, Paul passed the same gas station while traveling west at a constant speed of 60 mph. If both travelers maintained their speed and both remained on the highway for at least 2 hours, how long after he passed the gas station did Paul catch up with Mary?

A 35 mins
B 45 mins
C 1 hour
D 1 hour 15 mins
E 1 hour 30 mins

Mary passes gas station and travels at 50 mph for 15 minutes
So, Mary drove for 1/4 hours
Distance traveled = (time)(speed)
= (1/4)(50)
= 50/4
= 12.5 miles

Paul passes the same gas station and travels at 60 mph
IMPORTANT: At this point, Paul is 12.5 miles behind Mary.
However, since Paul is traveling 10 mph FASTER than Mary, the gap between them SHRINKS at a rate of 10 mph.
So, at this point, we need to determine the time it takes for the 12.5 mile gap to shrink to zero.
Time = (distance)/(speed)
= 12.5/10
= 1.25 hours
= 1 hour 15 mins
= D

Cheers,
Brent

Great explanation Brent BrentGMATPrepNow
To clarify so sentence like this "If both travelers maintained their speed and both remained on the highway for at least 2 hours" can be ignored in speed question then as it's not being used or useless here? Thanks Brent.

Originally posted by Kimberly77 on 09 Nov 2022, 12:53.
Last edited by Kimberly77 on 22 Jan 2023, 03:11, edited 2 times in total.
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Re: Mary passed a certain gas station on a highway while traveling west at [#permalink]
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Kimberly77 wrote:

Great explanation Brent BrentGMATPrepNow
To clarify so sentence like this "If both travelers maintained their speed and both remained on the highway for at least 2 hours" can be ignored in speed question then as it's not being used or useless here? Thanks Brent.

That's correct.
Since the question asks "how long after he passed the gas station did Paul catch up with Mary?", Paul would never catch up to Mary if he just stopped once he reached the gas station.
That's why we need to have that extra part ("If both travelers maintained their speed and both remained on the highway for at least 2 hours") in the question
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Re: Mary passed a certain gas station on a highway while traveling west at [#permalink]
Hi! I wonder does this piece of infomation hint anything " both remained on the highway for at least 2 hours" ?
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