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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, 07:46
Argh this is really frustrating me. I don't know If its just because i'm going wrong with the math or what but i get:

0.02x + 0.12y
---------------- = 3 (5% of 60 is 3)
x+y


This becomes 0.02x +0.12y = 3x + 3y which becomes 2.98x = 2.88y

What am I doing wrong here?

Thanks!
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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  [#permalink]

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New post 06 Sep 2018, 03:36
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Hi Bunuel math expert! how can I practice more questions on this topic i.e. mixtures ? Thanks
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Re: Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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

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New post 03 Nov 2018, 12:37
2x + 12y = 300
x + y = 60

Solving both these equations, you get what's x (i.e.., 42). Please let me know if you feel my approach is wrong.
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Re: Jackie has two solutions that are 2 percent sulfuric acid  [#permalink]

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New post 15 Jan 2019, 20:36
VeritasKarishma wrote:
Akgmat85 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


Use weighted average:

2% and 12% solutions mix to give 5% solution.

w1/w2 = (A2 - Aavg)/(Aavg - A1) = (12 - 5)/(5 - 2) = 7/3

You need 7 parts of 2% solution and 3 parts of 12% solution to get 10 parts of 5% solution.
If total 5% solution is actually 60 litres, you need 7*6 = 42 litres of 2% solution and 3*6 = 18 litres of 12% solution.

Answer (E)

The formula and its application in mixtures are discussed in the following two posts:
http://www.veritasprep.com/blog/2011/03 ... -averages/
http://www.veritasprep.com/blog/2011/04 ... -mixtures/


Hello everyone!

I've got a question related to mixtures,

w1/w2 = (A2 - Aavg)/(Aavg - A1) = (12 - 5)/(5 - 2) = 7/3

is A2 = W2 and A1 =W1?, Is it always true?

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

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New post 27 Jan 2019, 19:21
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


Using the Scale Method (Easiest for Me)


2%------------------------------------------12%
^
7

If there were equal parts of 2% and 12% solution, the average would be 7% [(2+12)/2]. With a split of 30 liters for each solution to create 60 liters
The question stem stated that when the solutions are mixed, 60 liters will have a concentration of 5%
This means we need more than 30 liters of the 2% solution. So we can cross of A, B, C.

2%------------------------------------------12%
(5-2)=3 ^ (12-5)=7
5%

Ratio is 3x:7x = 10x
If we know there's 60 Liters, then x in 10x will equal to 6
3*6 =18 liters for the 12% solution
7*6 = 42 liters for the 2% solution

Answer is E
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Re: Jackie has two solutions that are 2 percent sulfuric acid   [#permalink] 27 Jan 2019, 19:21

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