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Re: Mary passed a certain gas station on a highway while traveli
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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 : 6/5tt=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
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19 Mar 2017, 20:45
For me, there are two approaches to use.
1) Setting the time that Mary travels as t means that Paul travels for t15 mins. D=S*T and we want to find out when the distances from the station to the catchup point are equal for both Mary and Paul. 60*(t15)=50t 60t900=50t 10t=900 t=90 Paul travels for 15mins less than Mary and so Paul's time is 75mins.
2) Using relative speeds we know that Paul's speed in comparison to Mary's speed is 10mph. We need to know what distance Paul has to make up. Mary was travelling for 15mins before Paul arrived at the station and therefore traveled 50/4=12.5 miles. So, Paul will catch Mary up when he makes up the 12.5miles travelling at 10mph. T=D/S, 12.5/10=1.25h.
Kudos if you like the explanations. Also, please comment if you have alternative methods!



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Re: Mary passed a certain gas station on a highway while traveli
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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
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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 6050=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
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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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Re: Mary passed a certain gas station on a highway while traveli
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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
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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 T15. 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) * (T15)
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
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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"



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Re: Mary passed a certain gas station on a highway while traveli
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06 Mar 2019, 00:08
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 Mary's rate = 50 mph
Let her time = t hours
Distance = 50t miles
Paul's rate = 60 mph
Time when Paul over took Mary = t  1/4 (as Paul moved 15 minutes later)
Distance = 60(t  1/4) = 60t  15
As they travelled equal distance, we can equate their distances:
50t = 60t  15
t = 3/2
So Paul's time 3/2  1/4 = 5/4 * 60 = 75 minutes
Hence (D)
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Re: Mary passed a certain gas station on a highway while traveli
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