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# Mary passed a certain gas station on a highway while traveli

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

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01 Nov 2007, 10:28
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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
[Reveal] Spoiler: OA
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01 Nov 2007, 10:43
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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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01 Nov 2007, 14:48
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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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03 Mar 2010, 02:41
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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.
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11 Mar 2010, 02:43
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.

Nice explanation. But what if in the same question we have to calculate distance between the gas stattion and meeting point?
and what is the purpose of sayin they travelled for at least 2 hrs?
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11 Mar 2010, 04:45
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sidhu4u wrote:
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.

Nice explanation. But what if in the same question we have to calculate distance between the gas stattion and meeting point?
and what is the purpose of sayin they travelled for at least 2 hrs?

Since we know he travelled for 1.25 hours before catching up to her (after passing the gas station), its trivial to calculate the distance between the gas station and their meeting point. 1.25x60 = 75miles
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Mary passed a certain gas station on a highway while traveling west at [#permalink]

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01 May 2010, 08:31
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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
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Re: Mary passed a certain gas station on a highway while traveling west at [#permalink]

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01 May 2010, 09:54
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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01 May 2010, 10:28
Net distance traveled by Mary in 15 mins = 50*15/60
So relative distance between mary and Paul, when paul passed through station =50/4

Net time to meet mary is = relative distance/relative speed = (50/4)/(60-50) = 5/4 = 1+1/4 = 1 hr 15 min
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Re: Mary passed a certain gas station on a highway while traveling west at [#permalink]

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01 May 2010, 14:47
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 [#permalink]

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23 Feb 2014, 01:55
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Re: Mary passed a certain gas station on a highway while traveli [#permalink]

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20 Apr 2014, 22:04
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15 min Let After t hour they will meet each other.
A/Q, 50(t+1/4)=60t
T= 1.25 hour or 1 hour
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Re: Mary passed a certain gas station on a highway while traveli [#permalink]

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21 Apr 2014, 22:51
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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
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Re: Mary passed a certain gas station on a highway while traveli [#permalink]

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05 Oct 2014, 08:08
sidhu4u wrote:
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.

Nice explanation. But what if in the same question we have to calculate distance between the gas stattion and meeting point?
and what is the purpose of sayin they travelled for at least 2 hrs?

I also have the same query - what's the purpose of saying both remained on the highway for at least 2 hours?
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Re: Mary passed a certain gas station on a highway while traveli [#permalink]

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05 Oct 2014, 20:05
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windofchange wrote:
sidhu4u wrote:
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.

Nice explanation. But what if in the same question we have to calculate distance between the gas stattion and meeting point?
and what is the purpose of sayin they travelled for at least 2 hrs?

I also have the same query - what's the purpose of saying both remained on the highway for at least 2 hours?

That's the distraction. "2 Hours" is not required for computation purpose
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Re: Mary passed a certain gas station on a highway while traveli [#permalink]

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28 Oct 2014, 16:16
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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28 Oct 2014, 18:12
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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 23497 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 traveli [#permalink]

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19 Jan 2015, 12:37
Is my calculation correct?

You know she is 15 minutes further at 50 mph. This is 50/60 miles per minute = 5/6.
5/6 * 15 minutes is 75/6 = 12.5 miles.
He goes 10 mph faster(60-50) so the time het will pass her is 1 hour and 15 minutes later(12.5/10=1.25hour).
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Re: Mary passed a certain gas station on a highway while traveli [#permalink]

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03 Apr 2015, 06:03
I do not understand the logic of dividing 25/2 by 10 ? Can anybody help
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Re: Mary passed a certain gas station on a highway while traveli [#permalink]

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18 Apr 2015, 22:15
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mawus wrote:
I do not understand the logic of dividing 25/2 by 10 ? Can anybody help

Time=distance /speed

here 25/2 is the relative distance and 10 is the relative speed. Hence the division.
Re: Mary passed a certain gas station on a highway while traveli   [#permalink] 18 Apr 2015, 22:15

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