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A container holds 10 liters of a solution which is 20% acid. If 6 lite

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Bunuel
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A container holds 10 liters of a solution which is 20% acid. If 6 lite [#permalink] New post   05 Dec 2018, 21:56
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A container holds 10 liters of a solution which is 20% acid. If 6 liters of pure acid are added to the container, what percent of the resulting mixture is acid?

A. 5

B. 10

C. 20

D. \(33 \frac{1}{3}\)

E. 50
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A container holds 10 liters of a solution which is 20% acid. If 6 lite [#permalink] New post   05 Dec 2018, 22:00
10 liters of a solution of which 20% acid, i.e, 2 liters of acid.
6 liters added, then acid content becomes 8 liters.
Now the mixture contains 8 liters of acid and 8 liters of non-acid.
Percentage of acid = 50%.

E is the answer.
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Re: A container holds 10 liters of a solution which is 20% acid. If 6 lite [#permalink] New post   05 Dec 2018, 22:03
(100%-x%)/(x%-20%)= 10/6
X%= 50

E is correct.
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Re: A container holds 10 liters of a solution which is 20% acid. If 6 lite [#permalink] New post   06 Dec 2018, 04:11
Bunuel wrote:
A container holds 10 liters of a solution which is 20% acid. If 6 liters of pure acid are added to the container, what percent of the resulting mixture is acid?

A. 5

B. 10

C. 20

D. \(33 \frac{1}{3}\)

E. 50



in present solution the acid : 2 ltrs
and 6 ltrs is added , then acid is 8 ltrs ; total solution is 16 ltrs which is 50% acid.. IMO E
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Re: A container holds 10 liters of a solution which is 20% acid. If 6 lite [#permalink] New post   05 Jun 2019, 06:56
Bunuel wrote:
A container holds 10 liters of a solution which is 20% acid. If 6 liters of pure acid are added to the container, what percent of the resulting mixture is acid?

A. 5

B. 10

C. 20

D. \(33 \frac{1}{3}\)

E. 50

Total Soln.= 10
Initial Acid= 2
Final ACID= 2+6

% of acid= 8/16 so 50%
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Re: A container holds 10 liters of a solution which is 20% acid. If 6 lite [#permalink] New post   06 Jun 2019, 18:06
Expert Reply
Bunuel wrote:
A container holds 10 liters of a solution which is 20% acid. If 6 liters of pure acid are added to the container, what percent of the resulting mixture is acid?

A. 5

B. 10

C. 20

D. \(33 \frac{1}{3}\)

E. 50


We can create the equation:

(10 x 0.2 + 6)/16 = 8/16 = 1/2 = 0.5 = 50%

Alternate Solution:

We have a container with 10 liters of 20% acid. We add 6 liters of 100% acid, and the result is 16 liters of an unknown percentage (x) of acid. Let’s summarize this information into an equation:

10(0.20) + 6(1.00) = 16x

2 + 6 = 16x

8 = 16x

0.5 = x

Therefore, the resulting mixture is 50% acid.

Answer: E
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Re: A container holds 10 liters of a solution which is 20% acid. If 6 lite [#permalink] New post   07 Jun 2019, 21:11
Expert Reply
Given:

A container holds 10 liters of a solution which is 20% acid.

The volume of Acid = 20% of 10 Liters = \(\frac{20}{100}\) * 10 = \(\frac{1}{5}\) * 10 = 2 Liters

The volume of water = Total volume - Volume of Acid = 10 - 2 = 8 Liters

Now, 6 liters of pure acid is added to the container:

New volume of Acid = 2 + 6 = 8 Liters

The new volume of water (Same as old as no water is added) = 8 Liters

Total new Volume = 16 Liters

The percent of the acid in the resulting mixture = ( New Volume of Acid/Total New Volume ) * 100 = \(\frac{8}{16}\) * 100 = \(\frac{1}{2}\) * 100 = 50%

The correct answer is E
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Re: A container holds 10 liters of a solution which is 20% acid. If 6 lite [#permalink]
07 Jun 2019, 21:11
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