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easy method to slove this..

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by warlock » Sun Aug 03, 2008 8:48 am
A petroleum refinery lowered the oil level in one of its full storage tanks by draining out 7,326 gallons of oil, which was exactly 30 percent of the oil in the tank. Approximately what percent of the remaining oil would need to be drained in order to have drained a total of 40 percent of the oil that was originally in the tank?

a)10%
b)12%
c)14%
d)17%
e)20%

thank you.
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Source: — Problem Solving |

by eccentric » Sun Aug 03, 2008 10:07 am
Ans is C--- 14%
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by warlock » Sun Aug 03, 2008 10:18 am
ya...can you explain the method of solving..
thank you..
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14% is the answer

by Gaurav Tyagi » Mon Aug 04, 2008 2:28 pm
Suppose X is the total amount of oil.

.3X is drained and .7X is left.

Now, suppose Y is new percent need to be drained from the left oil.

so, Y*0.7X is to be drained.

The total amount to be drained has to be 40% of X i.e. 0.4X

We already drained 0.3X so we just need to drain

0.4X - 0.3X = 0.1X

so,

0.1 X = Y * 0.7X

Y = 0.1 / 0.7

Y = 100/7 (%)

Y = 14.2 ~ 14%
Thanks,

GT
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