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Jackie has two solutions that are 2 percent sulfuric acid

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Re: Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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New post 29 Sep 2017, 10:16
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blockman wrote:
Jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume, respectively. If these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid, approximately how many liters of the 2 percent solution will be required?

A) 18
B) 20
C) 24
D) 36
E) 42


We can let the amount of 2% sulfuric acid solution = x and the amount of 12% sulfuric acid solution = y. Thus:

x + y = 60

y = 60 - x

and

0.02x + 0.12y = 0.05(x + y)

2x + 12y = 5x + 5y

7y = 3x

Thus:

7(60 - x) = 3x

420 - 7x = 3x

420 = 10x

42 = x

Alternate Solution:

We will mix x liters of 2% sulfuric acid with (60 – x) liters of 12% sulfuric acid to produce 60 liters of 5% sulfuric acid. We can create an equation from this information and solve for x:

0.02x + 0.12(60 – x) = (0.05)(60)

0.02x + 7.2 – 0.12x = 3

-0.10x = -4.2

x = 42

Answer: E
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Re: Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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New post 07 Dec 2017, 15:28
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Hi calappa1234,

You made a small math mistake (trying to combine 2 steps into 1 step). Here's what the initial equation should look like:

(.02X + .12Y)/(X + Y) = .05

The prompt tells us that (X + Y) = 60, so you could 'substitute' that into the denominator if you like. However - if you're going to approach this algebraically - based on the wording of the prompt, you would likely find it easier to combine 'like' terms and simplify:

.02X + .12Y = .05X + .05Y
.07Y = .03X
7Y = 3X
7/3 = X/Y

Thus, the amount of X is a bit more than DOUBLE the amount of Y. Since the total of X+Y is 60, that means X = a little more than 40 and Y = a little less than 20.

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Re: Jackie has two solutions that are 2 percent sulfuric acid &nbs [#permalink] 07 Dec 2017, 15:28

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