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A chemist is mixing a solution of acetone and water. She currently has

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Bunuel
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A chemist is mixing a solution of acetone and water. She currently has [#permalink] New post   03 Jun 2018, 12:39
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A chemist is mixing a solution of acetone and water. She currently has 30 ounces mixed, 10 of which are acetone. How many ounces of acetone should she add to her current mixture to attain a 50/50 mixture of acetone and water if no additional water is added?

(A) 2.5
(B) 5
(C) 10
(D) 15
(E) 20
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C
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Re: A chemist is mixing a solution of acetone and water. She currently has [#permalink] New post   05 Jun 2018, 10:32
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Bunuel wrote:
A chemist is mixing a solution of acetone and water. She currently has 30 ounces mixed, 10 of which are acetone. How many ounces of acetone should she add to her current mixture to attain a 50/50 mixture of acetone and water if no additional water is added?

(A) 2.5
(B) 5
(C) 10
(D) 15
(E) 20


We can let n = the additional amount of acetone added and create the equation:

(10 + n)/(30 + n) = 1/2

2(10 + n) = 1(30 + n)

20 + 2n = 30 + n

n = 10

Answer: C
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Re: A chemist is mixing a solution of acetone and water. She currently has [#permalink] New post   03 Jun 2018, 13:57
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Bunuel wrote:
A chemist is mixing a solution of acetone and water. She currently has 30 ounces mixed, 10 of which are acetone. How many ounces of acetone should she add to her current mixture to attain a 50/50 mixture of acetone and water if no additional water is added?

(A) 2.5
(B) 5
(C) 10
(D) 15
(E) 20



Old acetone = 10
Acetone added = x
Old total = 30


---> New acetone/New total = .5

---> New acetone = Old acetone + acetone added = 10 + x

---> New total = old total + acetone added = 30 + x

(10 + x)/(30 + x) = .5
-------> 15 + .5x = 10 + x -------> 5 = .5x ---------> x = 10

Answer: D
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A chemist is mixing a solution of acetone and water. She currently has [#permalink] New post   03 Jun 2018, 16:55
Bunuel wrote:
A chemist is mixing a solution of acetone and water. She currently has 30 ounces mixed, 10 of which are acetone. How many ounces of acetone should she add to her current mixture to attain a 50/50 mixture of acetone and water if no additional water is added?

(A) 2.5
(B) 5
(C) 10
(D) 15
(E) 20


let x=ounces of acetone to be added
10+x=.5(30+x)
x=10
C
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Re: A chemist is mixing a solution of acetone and water. She currently has [#permalink] New post   04 Jun 2018, 02:19
How does 0.5 come? Please explain this point.

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A chemist is mixing a solution of acetone and water. She currently has [#permalink] New post   04 Jun 2018, 02:26
viveknegi wrote:
How does 0.5 come? Please explain this point.

Posted from my mobile device


Hi viveknegi

The total concentration of the acetone must be 50% of the total solution.

Currently, we have 10 ounces of acetone and 30 ounces of the solution.
If we were to add x ounces of pure acetone to the solution, acetone will
be \(\frac{1}{2}\) or \(0.5\) of the total solution.

That's the reason gracie has formed the equation: 10+x = 0.5(30 + x)

Hope this helps you!
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Re: A chemist is mixing a solution of acetone and water. She currently has [#permalink] New post   04 Jun 2018, 16:01
Bunuel wrote:
A chemist is mixing a solution of acetone and water. She currently has 30 ounces mixed, 10 of which are acetone. How many ounces of acetone should she add to her current mixture to attain a 50/50 mixture of acetone and water if no additional water is added?

(A) 2.5
(B) 5
(C) 10
(D) 15
(E) 20



A= 10 oz.
W= 20 oz.
Total = 30 oz.

If A is to equal 50% then....
A + x = W
10 + x = 20
x = 10

Ans = C
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A chemist is mixing a solution of acetone and water. She currently has [#permalink] New post   04 Jun 2018, 16:35
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Bunuel wrote:
A chemist is mixing a solution of acetone and water. She currently has 30 ounces mixed, 10 of which are acetone. How many ounces of acetone should she add to her current mixture to attain a 50/50 mixture of acetone and water if no additional water is added?

(A) 2.5
(B) 5
(C) 10
(D) 15
(E) 20

For mixture problems, this weighted average formula works well:

(% A)(Vol A)+ (% B)(Vol B) = (% A+B)(Vol A+B)

% = concentration of a substance or that substance as a fraction of the whole

A = current solution, acetone = \(\frac{1}{3}\)
A's volume = 30
B = solution to be added, acetone = 100% = 1
B's volume = x
Resultant solution (A+B), acetone: \(\frac{1}{2}\)
Vol of (A+B) = (30 + x)

\((\frac{1}{3})*(30) + 1*x = \frac{1}{2}*(30+x)\)
\(10 + x = 15 + \frac{1}{2}x\)
\(\frac{1}{2}x=5\)
\(x = 10\)

She needs 10 ounces of pure acetone

Answer C
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Re: A chemist is mixing a solution of acetone and water. She currently has [#permalink] New post   04 Jun 2018, 17:34
Concentration 1 * Volume 1 = Conentration 2 * Volume 2 => C1*V1 = C2*V2

1/3*30 = 1/2*V2

V2= 20

Hence 10 ounces were added

Answer C
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A chemist is mixing a solution of acetone and water. She currently has [#permalink] New post   04 Jun 2018, 21:24
Given:
Total mix(water+acceton)=30

Acceton=10.


So water content:30-10=20

After adding x acceton Ration become 50:50,no additional water.

So 10+x/20=1
Hence x=10



Sent from my iPad using GMAT Club Forum mobile app
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Re: A chemist is mixing a solution of acetone and water. She currently has [#permalink] New post   07 Jun 2018, 17:42
currently,

10 ounce acetone so remaining 20 is water.
In order to make the mixture 50-50 additional acetone needed is 10
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Re: A chemist is mixing a solution of acetone and water. She currently has [#permalink] New post   06 Jul 2019, 14:02
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Re: A chemist is mixing a solution of acetone and water. She currently has [#permalink]
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