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Re: A 5-liter jug contains 4 liters of a saltwater solution that [#permalink]
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Without performing any tedious calculation
We know the jug is five litre and the solution is 4ltr then 1.5 litre is removed making the remain solution 2.5litre
To fill the jug full we need 2.5more litre

Adding the same quantity of water will reduce the solution concentration by half
Hence we have 7.5 as the new concentration

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Re: A 5-liter jug contains 4 liters of a saltwater solution that [#permalink]
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Carcass wrote:
A 5-liter jug contains 4 liters of a saltwater solution that is 15 percent salt. If 1.5 liters of the solution spills out of the jug, and the jug is then filled to capacity with water, approximately what percent of the resulting solution in the jug is salt?

(A) \(7\frac{1}{2}\)%
(B) \(9\frac{3}{8}\)%
(C) \(10\frac{1}{2}\)%
(D) \(12\)%
(E) \(15\)%

-----------ASIDE------------
IMPORTANT CONCEPT: If EQUAL VOLUMES of 2 solutions are combined, the concentration of the resulting mixture will be the AVERAGE of the 2 solutions.
Example: If 3 liters of a 10% alcohol solution are combined with 3 liters of a 30% alcohol solution, the resulting solution will be 20% alcohol.
Likewise, if 11 liters of 6% alcohol solution are combined with 11 liters of a 7% alcohol solution, the resulting solution will be 6.5% alcohol.
-------------------------------

Once we pour out 1.5 liters, we're left with 2.5 liters of 15% salt solution
Since the jug holds 5 liters, we then add 2.5 liters of water (aka a 0% salt solution)

Since we're combining EQUAL VOLUMES of 15% and 0% solutions, the salt content of the resulting solution will equal the average of 15% and 0%
(15 + 0)/2 = 7.5

So, the resulting solution will be 7.5% salt.

Answer: A
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Re: A 5-liter jug contains 4 liters of a saltwater solution that [#permalink]
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