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how do i solve this

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by grandh01 » Wed Aug 01, 2012 2:13 pm
11. If 6 gallons of gasoline are added to a tank that is already filled to 3/4 of its
capacity, the tank is then filled to 9/10 of its capacity. How many gallons does
the tank hold?
(A) 20
(B) 24
(C) 36
(D) 40
(E) 60

Thanks in advance
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by eagleeye » Wed Aug 01, 2012 2:18 pm
grandh01 wrote:11. If 6 gallons of gasoline are added to a tank that is already filled to 3/4 of its
capacity, the tank is then filled to 9/10 of its capacity. How many gallons does
the tank hold?
(A) 20
(B) 24
(C) 36
(D) 40
(E) 60

Thanks in advance
Let the capacity be x.

Then 6+(3/4)x = (9/10)x
=> (9/10-3/4)x = 6
=> (36-30)x/40 = 6
=> 6x/40 = 6
=> x = 40 gallons.

:)
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by GMATGuruNY » Wed Aug 01, 2012 5:08 pm
grandh01 wrote:11. If 6 gallons of gasoline are added to a tank that is already filled to 3/4 of its
capacity, the tank is then filled to 9/10 of its capacity. How many gallons does
the tank hold?
(A) 20
(B) 24
(C) 36
(D) 40
(E) 60

Thanks in advance
We can plug in the answers, which represent the capacity of the tank.
The capacity is almost certainly a multiple of the two denominators in the problem (4 and 10).
Thus, the correct answer is likely to be A, D or E.
We should start with the middle value.

Answer choice D: 40 gallons
Adding 6 gallons to 3/4 of this value, we get:
6 + (3/4)40 = 36.
Volume/capacity = 36/40 = 9/10.
Succcess!

The correct answer is D.
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by tisrar02 » Wed Aug 01, 2012 5:18 pm
A non algebraic way to think this problem through can be as follows:

The tank is already 3/4 or 75% full

Add 6 gallons and you get a total capacity of 9/10 or 90%.

90-75=15% is what 6 gallons represents in the tank

We know that 15X6=90% so 6x6=36.

We need 100% though, not 90% so it would be slightly larger than 36 hence 40 would be the correct answer.

NOTE: You only want to do this approach when the numbers are somewhat spread apart.
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