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Mary passed a certain gas station on a highway while traveli

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New post 11 Jun 2015, 05:19
So, I am not sure if what I did is correct:

Using the R*T=D, what we get for Mary is:
50*(T+0.25) = 50T + 12.5 = D. That was the Distance she had covered when she passed the gas station.

What we get for Paul is:
60T. Again the Distance he had covered when he passed the gas station.

We know however that they both travelled for at least 2 hours after having passed the gas station. So, what this means for each is this:
For Mary: 50*2 = 100. She travelled 100 miles more.
For Paul: 60*2 = 120. He travelled 120 miles more.

Now we want to find the distance when Pail cought up with Mary, so we add the initial Distance each one covered with the distance each one covered after that gas station and equate the two:
50T + 12.5 + 100 = 60T +12O, which leads to 0.75. This is 60 minutes plus 15 minutes, which is 1h 15'.

Does it make sense..?
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New post 14 Jun 2015, 00:52
Ergenekon wrote:
Bunuel, what is the purpose of 2 hours in this sentence?


hi Ergenekon,
mention of two hours is to stress on the point that Paul and mary kept travelling on the road with constant speed atleast till the time paul crossed mary..
it has no significance on the answer...
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New post 14 Jun 2015, 01:00
pacifist85 wrote:
So, I am not sure if what I did is correct:

Using the R*T=D, what we get for Mary is:
50*(T+0.25) = 50T + 12.5 = D. That was the Distance she had covered when she passed the gas station.

What we get for Paul is:
60T. Again the Distance he had covered when he passed the gas station.

We know however that they both travelled for at least 2 hours after having passed the gas station. So, what this means for each is this:
For Mary: 50*2 = 100. She travelled 100 miles more.
For Paul: 60*2 = 120. He travelled 120 miles more.


Now we want to find the distance when Pail cought up with Mary, so we add the initial Distance each one covered with the distance each one covered after that gas station and equate the two:
50T + 12.5 + 100 = 60T +12O, which leads to 0.75. This is 60 minutes plus 15 minutes, which is 1h 15'.

Does it make sense..?


Hi pacifist85,
you are correct in your approach but gone wrong in the coloured portion..
It is just by luck that you have come to right answer since 0.75 hour will be equal to 0.75*60=45 min and not 1hr 15 minutes, which will be equal to 1.25 hour...
dont get distracted by mention of 2 hrs, it has nothing to do with tha answer..
We haave to find the time paul takes to cross mary and as you have found it is 1hr 15 min, so how does it make a difference if they travelled for 2 hours or 15 hours...
50T + 12.5 = 60T ... 10T=12.5
T=1.25hour =1.25*60min=75min or 1 hour 15min..
hope it is clear..
Regards .........chetan
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New post 21 Nov 2015, 19:38
My way is slightly different than most others and a lot easier.

Mary went .25 of an hour
Paul is catching up at 10 miles an hour.

So it would take 1 plus .25 hours = 1hr 15min
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New post 21 Nov 2015, 22:06
DJ1986 wrote:
My way is slightly different than most others and a lot easier.

Mary went .25 of an hour
Paul is catching up at 10 miles an hour.

So it would take 1 plus .25 hours = 1hr 15min



Hi,
how did you get "So it would take 1 plus .25 hours "..
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New post 22 Nov 2015, 07:43
chetan2u wrote:
DJ1986 wrote:
My way is slightly different than most others and a lot easier.

Mary went .25 of an hour
Paul is catching up at 10 miles an hour.

So it would take 1 plus .25 hours = 1hr 15min



Hi,
how did you get "So it would take 1 plus .25 hours "..



Sorry I skipped a step! Let me re-write it:

Mary went .25 of an hour @ 50mph = 12.5 miles
Paul is catching up at 10 miles an hour

It takes 1.25 hours at 10 miles per hour to make up that distance.

I think that while it's important to know the equations for how to solve these problems especially for the more difficult problems, the easier ones can be solved more quickly if you take a second to breath and think about it before just chugging away and plugging in numbers into the equations.

I'm struggling to get a 650 on the GMAT and part of my issue is always running out of time so right now I'm working on solving easier problems more quickly. I'm always searching for short cuts and guesstimations. More often than not, the GMAT rewards you for it!
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New post 22 Nov 2015, 08:11
DJ1986 wrote:
chetan2u wrote:
DJ1986 wrote:
My way is slightly different than most others and a lot easier.

Mary went .25 of an hour
Paul is catching up at 10 miles an hour.

So it would take 1 plus .25 hours = 1hr 15min



Hi,
how did you get "So it would take 1 plus .25 hours "..



Sorry I skipped a step! Let me re-write it:

Mary went .25 of an hour @ 50mph = 12.5 miles
Paul is catching up at 10 miles an hour

It takes 1.25 hours at 10 miles per hour to make up that distance.

I think that while it's important to know the equations for how to solve these problems especially for the more difficult problems, the easier ones can be solved more quickly if you take a second to breath and think about it before just chugging away and plugging in numbers into the equations.

I'm struggling to get a 650 on the GMAT and part of my issue is always running out of time so right now I'm working on solving easier problems more quickly. I'm always searching for short cuts and guesstimations. More often than not, the GMAT rewards you for it!



Hi,
this is what is done earlier. so how is it a shortcut? this is a Q which requires these 2-3 steps to get the answer..
and i dont think it can have any other shortcut which is lesser than these 2-3 steps..
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New post 22 Nov 2015, 09:47
Hi,
this is what is done earlier. so how is it a shortcut? this is a Q which requires these 2-3 steps to get the answer..
and i dont think it can have any other shortcut which is lesser than these 2-3 steps..[/quote]

You're right. It is basically the same thing you did I just didn't bother multiplying by minutes or any of that. I just saw 12.5 and 10 so I just moved the decimal to get 1.25 hours.
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New post 22 Nov 2015, 14:13
mary is 50/4=12.5 miles ahead of paul
paul gains 10 mph on mary
12.5 miles/10 mph=1.25 hr=1 hr 15 min
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New post 31 Jan 2016, 15:17
Another way to solve is to say he drove 15 miles in those 15 minutes (60mph/15min). Take the distance he drove and divide by the rate he's catching up with her, so 15miles/10mph = 1.5 hours. He already drove 15 mins, so 1 hr and 15 mins is left
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New post 31 Jan 2016, 15:23
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

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Mary passed a certain gas station on a highway while traveli  [#permalink]

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New post 30 Mar 2016, 20:43
Attached is a visual that should help.
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Screen Shot 2016-03-30 at 8.43.22 PM.png [ 174.03 KiB | Viewed 1706 times ]


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New post 04 Apr 2016, 15:47
why can't this problem be solved by the equation 60(t-15)=50t ?
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New post 04 Apr 2016, 16:23
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bullfromist wrote:
why can't this problem be solved by the equation 60(t-15)=50t ?


You could use that equation, but if you did, then you would be solving for Mary's time, not Paul's time (remember that Mary had a 15-minute head start). Mary's time = 90 min (1.5 hours), subtract 15 to get Paul's time of 75 (1 hour, 15 minutes).

In general, it's best to solve for the variable asked for by the question, and to avoid unnecessary extra steps such as these.
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New post 10 Jul 2016, 11:31
rohityes wrote:
Problems like these are best solved using relative speeds. (as done in the above solution).

Paul's speed relative to Mary = 60 - 50 = 10mph
Distance to cover = 15 mins x 50 mph = 25/2 miles

Time = (25/2) / 10 = 25/20 = 5/4 = 1hr 15 mins.

Ans D.



Hi,
This is an efficient approach but how are we getting 25/2 miles? Does anybody know?
Thanks
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New post 10 Sep 2016, 11:54
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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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New post 10 Sep 2016, 13:13
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


In 15 minutes Mary covered 50/4 => 12.5 miles

Attachment:
Untitled.png
Untitled.png [ 5 KiB | Viewed 1110 times ]


Distance Paul needs to travel is ( d + 12.5 ) and speed is 60
Distance Mary needs to travel is d and speed is 50

Since time to travel is same for both we can frame an equation as under -

\(\frac{( d + 12.5 )}{60}\) \(=\) \(\frac{d}{50}\)

60d = 50d + 625

Or, 10d = 625

Or, d = 62.50

Total Distance Paul needs to cover is ( 62.50 + 12.50) 75 miles

Time taken to reach meeting point is 75/60 => 1 Hour 15 Minutes...

Hence answer will be (D)
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Re: Mary passed a certain gas station on a highway while traveli  [#permalink]

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New post 27 Sep 2016, 15:44
Mary has been driving for 15 min (0,25 hours) before Paul starts driving. As Distance=Speed*Time, she has driven 12,5 miles before Paul starts.
So we need to find lapse of time in which the difference of their distances is 12,5 miles.
Let's start with 1 hour. Mary covers 50 miles in an hour and Paul 60 miles in an hour. The difference is still not 12,5 miles, so it has to be more than an hour.
Trying 1h15m (or 5/4 hours) you get that Mary has covered 62,5 miles and Paul 75. 75-62,5=12,5 so that's when they catch up.
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Re: Mary passed a certain gas station on a highway while traveli  [#permalink]

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New post 02 Feb 2017, 12:15
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

Answer: D
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