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Jackie has two solutions (OG20160

This topic has 8 expert replies and 1 member reply
boomgoesthegmat Senior | Next Rank: 100 Posts Default Avatar
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Jackie has two solutions (OG20160

Post Fri May 06, 2016 12:34 pm
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

OA: E

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Post Fri Jan 05, 2018 6:35 am
boomgoesthegmat 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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Post Fri Jan 05, 2018 7:17 am
boomgoesthegmat 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

OA: E
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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Post Sat May 07, 2016 6:54 am
boomgoesthegmat 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

OA: E
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

For more information on weighted averages, you can watch this video: http://www.gmatprepnow.com/module/gmat-statistics?id=805

Here are some additional practice questions related to weighted averages:
- http://www.beatthegmat.com/weighted-averages-t117237.html
- http://www.beatthegmat.com/weighted-average-t114506.html
- http://www.beatthegmat.com/average-weight-of-men-and-women-t57853.html
- http://www.beatthegmat.com/averages-question-t87118.html

Cheers,
Brent

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Post Thu May 12, 2016 12:38 am
x + y = 60

and

.02x + .12y = .05*(x + y)

The second equation simplifies to 7y = 3x, or y = (3/7)x. Plugging that into the first gives

x + (3/7)x = 60, or x = 42.

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Gurpreet singh Senior | Next Rank: 100 Posts Default Avatar
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Post Mon Jun 20, 2016 11:41 pm
Use allegation method for faster results

2-------------5--------------12
(Sol a) (Combined) (Sol b)

subtract 2 from 5 and 12 from 5 also ignore the signs and keep the result in opp direction

ie

2-------------5--------------12
(Sol a)=7 (Combined) (Sol b)=3



This gives you the ratio of the mixed sol ie 7:3.

Total sol is 60 also let the total sol be x

10(7+3)x=60 or x=6

multiply 7 by 6=42

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Post Tue Jun 21, 2016 4:34 am
boomgoesthegmat 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

OA: E
Or just use a little good old-fashioned logic. If the combined solution is 5%, you're going to have more than twice as much of the 2% solution as you will of the 12% solution. (5 is 3 units away from 2, and 7 units away from 12. 7/3 > 2. If you tested D, you'd have 36 liters of the 2% solution and 24 liters of the 12% solution, so 36 is too small, as it's not more than twice as much as 24. The answer must be larger than 36, so it's E.

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Post Tue Jun 21, 2016 5:22 am
boomgoesthegmat 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

OA: E
5% of 60 liters = (5/100)(60) = 3 liters of sulfuric acid.
An alternate approach is to PLUG IN THE ANSWERS, which represent the amount of 2-percent solution in the mixture.
When the correct answer choice is plugged in, the total amount of sulfuric acid = 3 liters.

D: 36 liters of 2-percent solution, implying 24 liters of 12-percent solution
Total sulfuric acid = (2/100)(36) + (12/100)(24) = (18/25) + (72/25) = 90/25 = more than 3 liters.
The total amount of sulfuric acid is TOO HIGH.
Eliminate D.

To reduce the amount of sulfuric acid in the mixture, we must use MORE 2-percent solution, which contains LESS sulfuric acid than does the 12-percent solution.

The correct answer is E.

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Post Thu Jun 23, 2016 4:08 pm
Gurpreet singh wrote:
Use allegation method for faster results

2-------------5--------------12
(Sol a) (Combined) (Sol b)

subtract 2 from 5 and 12 from 5 also ignore the signs and keep the result in opp direction

ie

2-------------5--------------12
(Sol a)=7 (Combined) (Sol b)=3



This gives you the ratio of the mixed sol ie 7:3.

Total sol is 60 also let the total sol be x

10(7+3)x=60 or x=6

multiply 7 by 6=42
You've got the right idea for sure, but that line is red would need to be revised.

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Post Fri Jan 05, 2018 7:18 am
boomgoesthegmat 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

OA: E
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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