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m09#04

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m09#04 [#permalink] New post 06 Apr 2011, 10:00
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At 12:00 in the afternoon Paul stopped at the gas station and filled his car's 50-liter tank to capacity. After Paul drove 75 miles away from the station, the tank developed a leak and the car started to lose 1/5 liters of fuel per minute. If Paul is traveling at a constant speed of 50 miles per hour and his car consumes 10 liters per every 100 miles, at what time of the day will Paul run out of gas?

A. 4:00 pm
B. 4:15 pm
C. 4:30 pm
D. 5:00 pm
E. 5:20 pm

[Reveal] Spoiler: OA
A

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Re: m09#04 [#permalink] New post 06 Apr 2011, 12:05
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Good question..
I would not try to solve this question by equations if it were an actual GMAT question.. I used the following approach

Rate of gas consumption = 1 ltrs for every 10 miles

Initial gas used before the leakage - 7.5 liters (75 miles would be covered in 90 mins at 50m/hr)
Time spent before leakage.

After this I picked up C as the choice and started to work with it. With C as an option the car would still travel for 3 more hours after the leakage, i.e.

Oil used dor driving 3 hours - 15 ltrs as Paul would cover 150 miles in 3 hrs
Oil lost dude to leakage - 0.2lit/min * 180mins =36

Total Oil used - 7.5 +36 +15 >50 (initial oil quantity)

Therefore we are sure that C,D,E can not be the answer. I then picked A (4:00 being better than 4:15 as an answer), followed the same approach and got the answer..

hope this helps
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Re: m09#04 [#permalink] New post 06 Apr 2011, 12:25
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Answer is (A)
At 12:00 p.m. Paul has 50 liters in his tank. He then starts driving at a speed of 50 mph and after 75 miles there is a leak in his car.

S=Distance/Time
50 = 75/Time
Time = 1.5 hours

So when there is a leak, Paul has already traveled for 1.5 hours. So the time when the tank started leaking, the time is 1:30 p.m.

75*10/100 = 7.5 liters get consumed till 1:30 p.m. as the car ran for 75 miles.
So at 1:30 p.m. there is 50 - 7.5 = 42.5 liters remaining in the fuel tank, since the distance traveled is 75 miles.

Now once the tank started leaking, the car is losing fuel at the rate of 1/5 th liter per minute which is 1 liter in 5 minutes, but at the same time Paul is driving the car as well and the car consumes 10 liters for every 100 miles run.

Speed = 50 miles per hour. So in 2 hours Paul will drive for 100 miles.
So now say it is 3:30 p.m. and Paul has driven for 2 more hours and the car has lost 10 liters + 24 liters (1/5th liter per minute) get consumed till 3:30 p.m.

So remaining fuel in the car is 50 - 7.5 - 10 - 24 = 8.5 liters

Now say, Paul continues driving so till 4:00 p.m. the car consumes 2.5 liters (10 liters for 100 miles so for 25 miles 2.5 liters get consumed) + 6 liters (1/5th liter in 1 minutes, so its 6 liters in 30 minutes) = 8.5 liters

So the car's tank is empty at 4:00 p.m.

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Re: m09#04 [#permalink] New post 06 Apr 2011, 12:32
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fluke wrote:
At 12:00 in the afternoon Paul stopped at the gas station and filled his car's 50-liter tank to capacity. After Paul drove 75 miles away from the station, the tank developed a leak and the car started to lose 1/5 liters of fuel per minute. If Paul is traveling at a constant speed of 50 miles per hour and his car consumes 10 liters per every 100 miles, at what time of the day will Paul run out of gas?

A. 4:00 pm
B. 4:15 pm
C. 4:30 pm
D. 5:00 pm
E. 5:20 pm

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - Hardest GMAT questions


For initial 75 miles, fuel consumed is 7.5 liters and time taken is 1.5 hrs.

We need to find the time by which remaining 42.5 liters fuel would be used up.

Normal usage is 5 liters per hour as it travels 50 miles per hour and it takes 1 liter for every 10 miles.

Leakage rate is 12 liters per hour (1/5 liters per minute), so total consumption is 17 liters per hour

Hence 42.5 liters will run out in 42.5/17 = 2.5 hours.

So, total time after 12:00 is 1.5+2.5 = 4.0 hrs and hence answer is A.
Re: m09#04   [#permalink] 06 Apr 2011, 12:32
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