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16 Sep 2014, 00:39
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At 12:00 in the afternoon, Paul left his home with a full 50-liter tank in his car. He had driven 75 miles when the tank developed a leak and the car started to lose \(\frac{1}{5}\) liters of fuel per minute. If Paul is traveling at a constant speed of 50 miles per hour and his car consumes 10 liters for 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
A. 4:00 PM
B. 4:15 PM
C. 4:30 PM
D. 5:00 PM
E. 5:20 PM
16 Sep 2014, 00:39
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Official Solution:
At 12:00 in the afternoon, Paul left his home with a full 50-liter tank in his car. He had driven 75 miles when the tank developed a leak and the car started to lose \(\frac{1}{5}\) liters of fuel per minute. If Paul is traveling at a constant speed of 50 miles per hour and his car consumes 10 liters for 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
Given that Paul's speed was 50 miles per hour, the tank began to leak \(\frac{75}{50}=1.5\) hours after he left home, which was at 1:30 PM. At this time, the tank contained \(50 - 10*0.75 = 42.5\) liters of fuel. From that point forward, the car was losing a total of 17 liters per hour: \(\frac{60}{5} = 12\) liters due to the leak and \(\frac{50}{100}*10 = 5\) liters for engine operation. Consequently, after 1:30 PM the fuel would only last for \(\frac{42.5}{17} = 2.5\) hours. Therefore, Paul will run out of gas at \(1:30 + 2:30 = 4:00\) PM.
Answer: A
At 12:00 in the afternoon, Paul left his home with a full 50-liter tank in his car. He had driven 75 miles when the tank developed a leak and the car started to lose \(\frac{1}{5}\) liters of fuel per minute. If Paul is traveling at a constant speed of 50 miles per hour and his car consumes 10 liters for 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
Given that Paul's speed was 50 miles per hour, the tank began to leak \(\frac{75}{50}=1.5\) hours after he left home, which was at 1:30 PM. At this time, the tank contained \(50 - 10*0.75 = 42.5\) liters of fuel. From that point forward, the car was losing a total of 17 liters per hour: \(\frac{60}{5} = 12\) liters due to the leak and \(\frac{50}{100}*10 = 5\) liters for engine operation. Consequently, after 1:30 PM the fuel would only last for \(\frac{42.5}{17} = 2.5\) hours. Therefore, Paul will run out of gas at \(1:30 + 2:30 = 4:00\) PM.
Answer: A
General Discussion
16 Jul 2016, 23:00
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I think this is a high-quality question and I agree with explanation. great perfect type of GMAT Style questions that was helpful!
