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# M25-30

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Math Expert
Joined: 02 Sep 2009
Posts: 46264

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16 Sep 2014, 01:24
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Difficulty:

25% (medium)

Question Stats:

75% (01:12) correct 25% (01:52) wrong based on 249 sessions

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Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as the result, a 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. $$\frac{4}{5}$$

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Math Expert
Joined: 02 Sep 2009
Posts: 46264

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16 Sep 2014, 01:24
3
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Official Solution:

Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as the result, a 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. $$\frac{4}{5}$$

Since the concentration of acid in resulting solution (40%) is the average of 50% and 30% then exactly half of the original solution was replaced.

Alternatively, you can do the following, say $$x$$ part of the original solution was replaced then: $$(1-x)*0.5+x*0.3=1*0.4$$, which gives $$x=\frac{1}{2}$$.

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Joined: 29 Mar 2014
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28 Feb 2015, 13:41
Bunuel 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. $$\frac{4}{5}$$

Hi,

Can I got the answer right but I hope my approach is correct in tackling this question. Let x equal to the original solution, now if we remove 50% from the original we get:

100%x - 50%x = 50%x remaining solution

from the remaining solution, 30% was added. Therefore we get 50%x+30%x = 80%x now we know the remaining is 40% solution. So we can create an equation of 80%x=40% and now if we isolate x we get 1/2 -> Answer choice C
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Joined: 05 Jul 2015
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GMAT 1: 600 Q33 V40
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20 Feb 2016, 21:45
At first I thought that this was too easy so maybe I fell for a trick. But after triple checking, yeah its just half.
Current Student
Joined: 12 Nov 2015
Posts: 60
Location: Uruguay
Concentration: General Management
Schools: Goizueta '19 (A)
GMAT 1: 610 Q41 V32
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07 Mar 2016, 19:49
Another way to solve this: When I add 30% acid I end up with 40% acid so the "x" amount i subtracted at the beginning leaves a 10% acid of the original solution and a 90% water of the original solution.
let's say I had 10 acid and 10 water>I extract> I now have 1 acid and and 9 water. I have 10 units vs the 20 I had to start off with, so I extracted 50%.
This would also answer the question, what % of water and acid did I extract? 40% acid and 10% water,
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Joined: 19 Jul 2016
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24 Jan 2017, 11:04
why are we doing (1−x)∗0.5? why it shouldn't be x*0.5?
pls explain
thnx
Math Expert
Joined: 02 Sep 2009
Posts: 46264

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25 Jan 2017, 06:23
gupta87 wrote:
why are we doing (1−x)∗0.5? why it shouldn't be x*0.5?
pls explain
thnx

If x part of the original solution was replaced then (1 - x) part of the original solution is left in which 50% is acid.
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Joined: 08 Oct 2014
Posts: 4

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27 Oct 2017, 19:21
I have a slightly extended version for a question like this:
1/2L of 30% acid solution was replaced with 1/4L of 50% solution. If the original solution had 40% solution of acid, what was is the amount of acid in the new solution?

Intern
Joined: 05 Jun 2017
Posts: 7
Location: India
GPA: 3.67
WE: Science (Non-Profit and Government)

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22 Nov 2017, 01:38
I agree with explanation. other method can be alligation:

since 50% and 30% solutions giving 40%. therefore, new ratio is 1:1 (meaning both original solutions formed equal part). therefore, 50% solution's only half part was replaced (1/2. as other half formed by 30% solution)
Re M25-30   [#permalink] 22 Nov 2017, 01:38
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# M25-30

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