How Many Ounces of Pure Acid Must Be Added to 20 Ounces of a Solution to Make
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How many ounces of pure acid must be added to 20 ounces of a solution [#permalink] 
05 Mar 2018, 20:30
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How many ounces of pure acid must be added to 20 ounces of a solution that is 5% acid to strengthen it to a solution that is 24% acid?
(A) 5/2
(B) 5
(C) 6
(D) 15/2
(E) 10
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Re: How many ounces of pure acid must be added to 20 ounces of a solution [#permalink] 05 Mar 2018, 20:32
Bunuel wrote:
How many ounces of pure acid must be added to 20 ounces of a solution that is 5% acid to strengthen it to a solution that is 24% acid?
(A) 5/2
(B) 5
(C) 6
(D) 15/2
(E) 10
18. Mixture Problems
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Weighted Average and Mixture Problems on the GMAT
Tips and Tricks: Mixtures
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For more check Ultimate GMAT Quantitative Megathread
Hope it helps.
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How many ounces of pure acid must be added to 20 ounces of a solution [#permalink] 05 Mar 2018, 22:24
Bunuel wrote:
How many ounces of pure acid must be added to 20 ounces of a solution that is 5% acid to strengthen it to a solution that is 24% acid?
(A) 5/2
(B) 5
(C) 6
(D) 15/2
(E) 10
ANS: (B)
5% of 20 ounces=1
so now the question says what quantity of acid must be added to this 1 ounce of acid so that it becomes 24% of the new solution.
let the new quantity of acid to be added be x
total new quantity of acid=1+x
new quantity of solution = 20+x
Therefore 1+x= 24/100* (20+x) = solve and the answer is 5 ounces.
hope this is of some help! 
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How many ounces of pure acid must be added to 20 ounces of a solution [#permalink] 06 Mar 2018, 09:01
Bunuel wrote:
How many ounces of pure acid must be added to 20 ounces of a solution that is 5% acid to strengthen it to a solution that is 24% acid?
(A) 5/2
(B) 5
(C) 6
(D) 15/2
(E) 10
To find the volume of one solution in a resultant mixture, weighted average works well. I use this formula:
\((Concen_{A})(Vol_{A}) + (Concen_{B})(Vol_{B} = (Concen_{A+B})(Vol_{A+B})\)
Per the formula, in setting up the equation:
A = the solution that is 5% acid
B = the solution that is pure acid (100% = 1)
Let \(x\)= # of ounces of pure acid (B) to be added
Desired concentration of resultant mix: 24%=.24
A + B (volume) = 20 + x
\(.05(20) + 1(x) = .24 (20 + x)\)
\(1 + x = .24(20) +.24(x)\)
\(1 + x = 4.8 + .24x\)
\(.76x = 3.8\)
\(x =\frac{3.8}{.76}=\frac{380}{76}=5\)
We need 5 ounces of pure acid.
Answer B
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Re: How many ounces of pure acid must be added to 20 ounces of a solution [#permalink] 07 Mar 2018, 06:19
The Equation:
100x + 20*5 = 24(x+20)
Simplifying, we get
76x = (24-5)*20 = 19*20
Therefore, x=5.
Ans B
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Re: How many ounces of pure acid must be added to 20 ounces of a solution [#permalink] 08 Mar 2018, 10:42
Bunuel wrote:
How many ounces of pure acid must be added to 20 ounces of a solution that is 5% acid to strengthen it to a solution that is 24% acid?
(A) 5/2
(B) 5
(C) 6
(D) 15/2
(E) 10
We can let n = the number of ounces of pure (100%) acid added to the solution.
(0.05)(20) + n = 0.24(20 + n)
1 + n = 4.8 + 0.24n
0.76n = 3.8
n = 5
Answer: B
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Re: How many ounces of pure acid must be added to 20 ounces of a solution [#permalink] 11 Mar 2018, 01:46
Bunuel wrote:
How many ounces of pure acid must be added to 20 ounces of a solution that is 5% acid to strengthen it to a solution that is 24% acid?
(A) 5/2
(B) 5
(C) 6
(D) 15/2
(E) 10
Keep it Simple!
=> Non Acid before = Non Acid After
=> 95% of 20 = 76% of (20+x)
=> 19 = 76/100*(20+x)
=> 25 = 20+x
=> x = 5
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Re: How many ounces of pure acid must be added to 20 ounces of a solution [#permalink] 11 Mar 2018, 12:28
Bunuel wrote:
How many ounces of pure acid must be added to 20 ounces of a solution that is 5% acid to strengthen it to a solution that is 24% acid?
(A) 5/2
(B) 5
(C) 6
(D) 15/2
(E) 10
let a=ounces of acid to be added
.05*20+a=.24*(20+a)
a=5 ounces
B
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Re: How many ounces of pure acid must be added to 20 ounces of a solution [#permalink] 06 Oct 2019, 04:29
let soln = x
x+.05*20 = .24*(20+x)
1+x=4.8+.24x
x=5
IMO B
Bunuel wrote:
How many ounces of pure acid must be added to 20 ounces of a solution that is 5% acid to strengthen it to a solution that is 24% acid?
(A) 5/2
(B) 5
(C) 6
(D) 15/2
(E) 10
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Re: How many ounces of pure acid must be added to 20 ounces of a solution [#permalink] 06 Oct 2019, 06:08
Bunuel wrote:
How many ounces of pure acid must be added to 20 ounces of a solution that is 5% acid to strengthen it to a solution that is 24% acid?
(A) 5/2
(B) 5
(C) 6
(D) 15/2
(E) 10
5% of 20 ounces= 1 ounces
Final quantity=1+x
1+x=0.24(20+x)
x=5 ounces
B:)
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Re: How many ounces of pure acid must be added to 20 ounces of a solution [#permalink] 02 Jan 2021, 01:38
Bunuel wrote:
How many ounces of pure acid must be added to 20 ounces of a solution that is 5% acid to strengthen it to a solution that is 24% acid?
(A) 5/2
(B) 5
(C) 6
(D) 15/2
(E) 10
How can the allegation approach be applied here?
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Re: How many ounces of pure acid must be added to 20 ounces of a solution [#permalink]
02 Jan 2021, 01:38
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How Many Ounces of Pure Acid Must Be Added to 20 Ounces of a Solution to Make
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