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Some part of a 50% solution of acid was replaced with an equal amount

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Joined: 30 Jan 2019
Posts: 1
Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

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New post 11 Feb 2019, 23:07
my thought process was to use the process of elimination strategy, so I started with the middle answer of 1/2.

if half of the original solution was replaced that means I am left with:

50% of 50% acid solution + 50% of 30% solution = new acid solution (which happens to be 40% acid)

.5x.5 = .25
.5x.3 = .15

.25+.15 = .4. So, my first attempt was the answer.


Let's say the new acid solution was 35% instead of 40%, then after trying the 1/2 answer, I would realize that I need more of the 30% solution to bring down the amount of acid. From this point, I can eliminate A,B, and C.

I would then try 3/4 of the 30% solution.

.25*.5 + .75*.3 = new solution
.125+.225 = .35, which is the answer I was looking for.

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Re: Some part of a 50% solution of acid was replaced with an equal amount  [#permalink]

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New post 11 Feb 2019, 23:28
chrissy28 wrote:
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?


A. \(\frac{1}{5}\)

B. \(\frac{1}{4}\)

C. \(\frac{1}{2}\)

D. \(\frac{3}{4}\)

E. \(\frac{4}{5}\)


M25Q30


so we have a mixture in which
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid.

40% solution of acid was obtained

By allegation method, we get the ratio
M1--------M2
50----------30
-------40-----
10-----------10

Mixture will be in the ratio of 1:1

what part of the original solution was replaced?

Total mixture = 2

1/2 was replaced

C
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Quote which i can relate to.
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Re: Some part of a 50% solution of acid was replaced with an equal amount   [#permalink] 11 Feb 2019, 23:28

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Some part of a 50% solution of acid was replaced with an equal amount

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