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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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Question Stats:

100% (00:51) correct
0% (00:00) wrong based on 11 sessions

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?

Re: A container holds 10 liters of a solution which is 20% acid. [#permalink]
05 Dec 2018, 13:25

1

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Expert's post

nadeem790 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

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 of ACID 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%

Answer: E

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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

nadeem790 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

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 = \(100*\frac{8}{16}=50\)%