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

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A container holds 10 liters of a solution which is 20% acid. [#permalink]  05 Dec 2018, 03:23
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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
[Reveal] Spoiler: OA

Last edited by chetan2u on 05 Dec 2018, 19:56, edited 2 times in total.
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Re: A container holds 10 liters of a solution which is 20% acid. [#permalink]  05 Dec 2018, 13:25
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Expert's post
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

Let's keep track of the acid

A container holds 10 liters of a solution which is 20% acid.
20% of 10 liters = 2 liters
So, there are 2 liters to start

6 liters of pure acid are added to the container
2 + 6 = 8
So, there are now 8 liters of acid in the RESULTING solution.

What percent of the resulting mixture is acid?
We added 6 liters to 10 liters, to get a RESULTING solution with volume = 16 liters
8/16 = 1/2 = 50%

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Re: A container holds 10 liters of a solution which is 20% acid. [#permalink]  05 Dec 2018, 19:59
Expert's post
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

20% acid in 10 liter of solution means $$10*\frac{20}{100}=2$$..

When we add 6 liters of acid..
Acid becomes 2+6=8
Total solution becomes : 10+6=16

so acid as % of resulting mixture = $$100*\frac{8}{16}=50$$%

E
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Re: A container holds 10 liters of a solution which is 20% acid.   [#permalink] 05 Dec 2018, 19:59
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