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Mary passed a certain gas station on a highway while traveli

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

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New post 11 Feb 2017, 02:19
we can solve the above problem in just 15sec. if we look more closely to the problem as we can see that when they both meet they both have traveled the same distance so we do not need to calculate the distance , we know that when distance is constant then the speed is inversely proportional to the time, so form it we can calculate the speed is multiplied by 5/6 so no the time will multiplied by 6/5 and the difference between time between them equal to 15
such as : 6/5t-t=15
t=75 mins that equal to 1 hr.15 mins
hence the answer is D
thanks
if you liked my post please kudos :)
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Re: Mary passed a certain gas station on a highway while traveli  [#permalink]

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New post 20 Mar 2017, 01:15
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



after 15 minutes as paul is passing the gas station, Mast will be 50 *15/60 = 12.5 miles away
relative speed since they are moving in same direction will be 60 - 50 =10 mph

time taken to cover 12.5 miles = 12.5/10 = 1.25 hours = 1 hour 15 minute

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

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New post 11 Aug 2017, 01:28
So the speed of Mary is 50m/h and speed of Paul is 60m/hr . Also Paul travels for 15 minutes more than Mary. So the distance between the distance between the gas station and the meeting point is d and the relative speed is 60-50=10mh/hr
Can someone assist me after these steps. Thank you.
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Re: Mary passed a certain gas station on a highway while traveli  [#permalink]

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New post 11 Oct 2017, 12:03
Mary had started 15 mins before @ 50m/hr. Hence she had already covered 12.5m before Paul started. Hence the Paul has to cover 12.5m + the distance Mary is going to cover before he meets her. Let x be the time at which they both meet. Therefore the equation will be:
\(12.5 + 50x = 60x\)
-->\(10x = 12.5\)
-->\(x = 1.25\) i.e. 1 hr 15 mins.
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New post 23 Oct 2017, 02:16
A )At t=30
Mary: 25 miles Paul: 15 miles

B) At t= 45
Mary: 37,5 Paul: 30 miles

C) At t=1 hr
Mary: 50 Paul:45 miles

D) At t=1 hr 15
Mary: 62,5 Paul: 60

E) At t=1h30
Mary: 75 miles Paul: 75 miles

It seems like E is the answer. What is wrong with my line of reasoning? Would appreciate your help
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Re: Mary passed a certain gas station on a highway while traveli  [#permalink]

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New post 01 Feb 2018, 17:33
Another way of looking at it is to equate each driver's distance and solve for time. D=R*T, where Distance=D, Rate=R, and T=Time. We know that Paul traveled for 15 minutes less than Mary, so we can call Mary's time T and Paul's time T-15. We're told that Mary's rate was 50mph, and Paul's rate was 60mph. Because D=R*T, and Mary and Paul went the same distance, we can equate the products of their rates and times.

Mary's D = (50) * T
Paul's D = (60) * (T-15)

Mary's D = Paul's D

50T = 60T - 900

10T = 900

T = 90 (minutes)

So Mary drove for 90 minutes. Paul, however, only drove for 75 minutes, or 1.25 hours.
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Re: Mary passed a certain gas station on a highway while traveli  [#permalink]

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New post 02 Mar 2018, 18:03
Does the statement below not matter? We're not using "2hrs" to help calculate...

" If both drivers
maintained their speeds and both remained on the highway for at least 2 hours"
Re: Mary passed a certain gas station on a highway while traveli &nbs [#permalink] 02 Mar 2018, 18:03

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