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# At noon, Paul stopped at a gas station and filled his car's

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Joined: 21 Jan 2007
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At noon, Paul stopped at a gas station and filled his car's [#permalink]

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25 Oct 2007, 05:23
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

At noon, Paul stopped at a gas station and filled his car's 50 litre tank to its capacity. After he drove 75 miles away from the station, the tank developed a leak and the car started to lose 1/5 litres of fuel per minute. If he is travelling at a constant speed of 50 miles per hour and his car consumes 10 litres every 100 miles, what time will his car stop moving?

4 pm
4:15 pm
4:30 pm
5:00 pm
5:20 pm
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Re: Challenge - Multi-step RT=D [#permalink]

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25 Oct 2007, 05:42
bmwhype2 wrote:
At noon, Paul stopped at a gas station and filled his car's 50 litre tank to its capacity. After he drove 75 miles away from the station, the tank developed a leak and the car started to lose 1/5 litres of fuel per minute. If he is travelling at a constant speed of 50 miles per hour and his car consumes 10 litres every 100 miles, what time will his car stop moving?

4 pm
4:15 pm
4:30 pm
5:00 pm
5:20 pm

12pm, 50 liter

After 75 miles, gas consumption only, 10/100 * 75 = 7.5 liter, so 42.5 liter left, time taken 1/50*75 = 1.5 hour, so in sum:
1.30pm, 42.5 liter

Beyond 75 mile, you have leak and gas consumption
leak = 1/5 liter/min
consume = 10/100 liter/mile
Set x = distance in miles
Set t = total time in hour

10/100 * x = total consume
x = 50t
So we have:
(1) 5t = total consume
(2) 1/5 * 60t = 12t = total leak

Looking for t:
42.5 - 12t - 5t = 0
42.5 = 17t
t = 2.5 hour

Therefore, the car will stop at 2.5 hour after 1.30pm
Ans = 4pm
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25 Oct 2007, 08:12
I get 4:00pm as well...

lets see

first he consumes 7.5 litres of fuel, i.e its already 1:30pm

he develops a leak

looses 1/5 * 60=12 litres/hr

he is burning 5litres/hr driving which is 17litres/hr

42.5litres/17=2.5 hr

totally time to burnout 1.5+2.5=4pm
25 Oct 2007, 08:12
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