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# A scientist has 400 units of a 6% phosphoric acid solution, and an unl

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Math Expert
Joined: 02 Sep 2009
Posts: 49430
A scientist has 400 units of a 6% phosphoric acid solution, and an unl  [#permalink]

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03 Mar 2015, 07:03
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A scientist has 400 units of a 6% phosphoric acid solution, and an unlimited supply of 12% phosphoric acid solution. How many units of the latter must she add to the former to produce a 10% phosphoric acid solution?

A. 200
B. 400
C. 500
D. 600
E. 800

Kudos for a correct solution.

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Re: A scientist has 400 units of a 6% phosphoric acid solution, and an unl  [#permalink]

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03 Mar 2015, 22:35
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Bunuel wrote:
A scientist has 400 units of a 6% phosphoric acid solution, and an unlimited supply of 12% phosphoric acid solution. How many units of the latter must she add to the former to produce a 10% phosphoric acid solution?

A. 200
B. 400
C. 500
D. 600
E. 800

Kudos for a correct solution.

Using the weighted average formula,

w1/w2 = (12 - 10)/(10 - 6) = 1/2

So every unit of 6% solution, he should add 2 units of 12% solution. Since he has 400 units of 6% solution, he should add 800 units of 12% solution.

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Joined: 26 Mar 2010
Posts: 11
Re: A scientist has 400 units of a 6% phosphoric acid solution, and an unl  [#permalink]

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03 Mar 2015, 13:21
2
2
With mixture problems I usually take following approach

10%=Total mix of acid/Total amount of mixture.

With X= amount of 12% acidic solution we have:

10%=(400*0.06+.12*X)/400+X)

40+0.1X=24+0.12X
16=0.002X
X=800

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Joined: 22 Aug 2014
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Re: A scientist has 400 units of a 6% phosphoric acid solution, and an unl  [#permalink]

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03 Mar 2015, 23:46
1
Bunuel wrote:
A scientist has 400 units of a 6% phosphoric acid solution, and an unlimited supply of 12% phosphoric acid solution. How many units of the latter must she add to the former to produce a 10% phosphoric acid solution?

A. 200
B. 400
C. 500
D. 600
E. 800

Kudos for a correct solution.

The acid solution level in the first is half of the second one. So the ratio of the acid solution in S1:S2 = 1:2

Now the simple way to go about this apart from weighted average would be assuming that you have 1 unit of 6% acid solution.

A General rule when the mixtures are in the ratio 1:2

When you add 1 part of second solution, the mixed solution will contain 50% more than the first solution. i.e, 1.5*6 = 9
When you add 2 parts of second solution, the mixed solution will contain 66% more than the first solution. i.e, 1.6*6 = 10

Thus the required solution = 400*2 = 800

Option E
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Re: A scientist has 400 units of a 6% phosphoric acid solution, and an unl  [#permalink]

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03 Mar 2015, 23:46
2
Hi All,

While an Algebra approach (using the "weighted average" formula) would work nicely on this prompt, you can also answer it rather quickly by TESTing THE ANSWERS and using a bit of logic.

Here, we're going to mix 400 ounces of a 6% acid solution with X ounces of a 12% acid solution to form a 10% acid solution. We're asked for the value of X. The answers are all nice, round numbers, so we can take advantage of them....

IF.....
X = 400 ounces
Then we'd have the same amount of each solution: 400 ounces of 6% and 400 ounces of 12% --> this would produce a (6%+12%)/2 = 9% mixture, which is TOO SMALL. X must be BIGGER. Eliminate Answers A and B.

IF....
X = 600 ounces
Then with 400 ounces of 6% and 600 ounces of 12%, we'd have...

[(400)(.06) + (600)(.12)]/(400 + 600) =
(24 + 72)/1000 = 96/1000 = 9.6%, which is TOO SMALL. X must be BIGGER. Eliminate C and D.

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A scientist has 400 units of a 6% phosphoric acid solution, and an unl  [#permalink]

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04 Mar 2015, 04:13
1
Or using the scaling method:

6%.........10%...........12%
|________|__________|
.......4................2

We draw the number line, from lowest to greatest as this is the expected, and find the difference between the first solution and the wanted and the second solution and the wanted.

$$\frac{6%}{12%}$$ $$=$$$$\frac{2}{4}$$ $$=$$$$\frac{1}{2}$$

When we write the ratios we make sure to flip, as seen above. So, it is 1 unit for the 6% solution and 2 units for the 12% solution, instead of the opposite. This is derived from the weighted averages formula.

Then we have:

$$\frac{6%}{12%}$$ =$$\frac{1}{2}$$ = $$\frac{400}{x}$$ $$=$$$$2*400$$ $$=$$ $$800$$
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Re: A scientist has 400 units of a 6% phosphoric acid solution, and an unl  [#permalink]

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04 Mar 2015, 22:04
1

Say "x" units of 12% solution is added.

Equation setup would be as follows:

$$\frac{6}{100} * 400 + \frac{12}{100}* x = \frac{10}{100}(400+x)$$

x = 800
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Re: A scientist has 400 units of a 6% phosphoric acid solution, and an unl  [#permalink]

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08 Mar 2015, 14:33
Bunuel wrote:
A scientist has 400 units of a 6% phosphoric acid solution, and an unlimited supply of 12% phosphoric acid solution. How many units of the latter must she add to the former to produce a 10% phosphoric acid solution?

A. 200
B. 400
C. 500
D. 600
E. 800

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

We could backsolve from the numerical answer choices, but let’s use a straight algebra approach. Let X equal the units of 12% phosphoric acid solution we use, and let Y be the units of 10% sulfuric acid solution that result.

The volume equation is:
400 + X = Y

In the first solution, we have 6% of 400, or 24 units of phosphoric acid.

In the second solution, we have 12% of X = 0.12*X of phosphoric acid.

In the resultant solution, we have 10% of Y = 0.10*Y of phosphoric acid.

The concentration equation is:
24 + 0.12*X = 0.10*Y

Multiply this by 100, to clear the decimals:
2400 + 12X = 10Y

Everything is even, so divide by 2 to simplify:
1200 + 6X = 5Y

We want X, so let’s multiply the volume equation by -5 and add that to this equation we just got:

1200 + 6x = 5y
-2000 - 5x = -5y
_____________
800 - x = 0

x = 800.

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Re: A scientist has 400 units of a 6% phosphoric acid solution, and an unl  [#permalink]

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24 May 2018, 09:35
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