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# How many ounces of pure acid must be added to 20 ounces of a solution

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Math Expert
Joined: 02 Sep 2009
Posts: 49300
How many ounces of pure acid must be added to 20 ounces of a solution  [#permalink]

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05 Mar 2018, 21:30
00:00

Difficulty:

25% (medium)

Question Stats:

82% (01:42) correct 18% (01:45) wrong based on 124 sessions

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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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Math Expert
Joined: 02 Sep 2009
Posts: 49300
Re: How many ounces of pure acid must be added to 20 ounces of a solution  [#permalink]

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05 Mar 2018, 21:32
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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Hope it helps.
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How many ounces of pure acid must be added to 20 ounces of a solution  [#permalink]

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05 Mar 2018, 23:24
2
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]

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06 Mar 2018, 10:01
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.

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Re: How many ounces of pure acid must be added to 20 ounces of a solution  [#permalink]

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07 Mar 2018, 07:19
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]

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08 Mar 2018, 11:42
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

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Re: How many ounces of pure acid must be added to 20 ounces of a solution  [#permalink]

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11 Mar 2018, 02:46
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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Joined: 07 Dec 2014
Posts: 1087
Re: How many ounces of pure acid must be added to 20 ounces of a solution  [#permalink]

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11 Mar 2018, 13:28
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
Re: How many ounces of pure acid must be added to 20 ounces of a solution &nbs [#permalink] 11 Mar 2018, 13:28
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