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


Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

Let x be the number of liters of 2% solution in the mixture
Since there are 60 liters in total, 60 - x will equal number of liters of 12% solution in the mixture

Now apply the formula:
5 = (x/60)(2) + [(60-x)/60](12)
Multiply both sides by 60 to get: 300 = 2x + (60-x)(12)
Expand: 300 = 2x + 720 - 12x
Rearrange: -420 = -10x
Solve: x = 42

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


Another approach is to keep track of the acid
Let x = number of liters of 2% solution needed
So, 60 - x = number of liters of 12% solution needed

2% of x = 0.02x
So, 0.02x = the number of liters of PURE acid in the 2% solution

12% of 60 - x = 0.12(60 - x) = 7.2 - 0.12x
So, 7.2 - 0.12x = the number of liters of PURE acid in the 12% solution

Now let's COMBINE the two solutions.
Total volume of PURE acid = 0.02x + 7.2 - 0.12x
= 7.2 - 0.1x
So, our NEW solution contains 7.2 - 0.1x liters of PURE acid
Also, the NEW solution has a total volume of 60 liters

Since the NEW solution is 5% PURE acid, we can write: (7.2 - 0.1x)/60 = 5/100
Cross multiply to get: 100(7.2 - 0.1x) = 5(60)
Expand: 720 - 10x = 300
Add 10 x to both sides: 720 = 300 + 10x
Subtract 300 from both sides: 420 = 10x
Solve: x = 42

Answer: E
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Re: Jackie has two solutions that are 2 percent sulfuric acid an [#permalink]
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Equation with litres:
x+y=60
Equation with percents:
0.02x + 0.12y = .05(x+y)
0.02x+0.12y = 0.05 (60)
remove decimal
2x/100 + 12y/100 = 3
2x + 12y = 300
divide by 2
x + 6y = 150
compare with litres equation
x+y=60 and
x+6y=150
by solving we get y = 18 and x = 42
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Re: Jackie has two solutions that are 2 percent sulfuric acid an [#permalink]
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Another approach:

x----------------------y
2%-------5%--------12%

5-2=3------------12-5 = 7

write in cross direction:
x/y = 7/3
(we can take x = 7 and y = 3)

Now,
part/whole * total
7/7+3 * 60
=42 option E
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Re: Jackie has two solutions that are 2 percent sulfuric acid an [#permalink]
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Another approach.


Let x= number of liters of 2% solution needed
so, 60-x = number of 12% solution needed

Now in the final 5% solution,the volume of pure acid = 5*60/100 = 3 liters

we can write,

2*x/100 + 12*(60-x)/100 = 3

Multiply both side by 100:

2*x +720 -12*x = 300
Rearrange: 10*x = 420
Solve: x=42

Answer: E


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