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

Look at it this way.

Total part is 1. Out of this, x part was replaced.

Hence, (1-x)*50 (percent) + x*30 = 1*40
or, 50 - 20x = 40
or, 10 = 20x
or, x = 1/2....or, 50% of original solution was replaced.

C. 1/2

In the table below, this refers to the original solution, not the acid concentration. Sorry about the formatting...it should be a 3X4 table with one expression in each box.

Original Removed Added Result
Concentration 0.50 0.50 0.70 0.60
Amount 1 x x 1
Multiply 0.50 0.50x 0.70x 0.60

0.50 (Original amount) - 0.50x (Removed) + 0.70x (Added) = 0.60 (End Result)
0.20x = 0.10
x = 1/2

Explanation:
Because the question asks for the original solution, you can change the numbers to reflect that. If 50% of the solution is acid, then 50% of the solution is the original solution. Then you're removing 50% concentration of acid (which is also 50% original solution) and then adding the same amount back but with 30% concentration of acid (also 70% concentration of original solution). What I mean by that is if you remove 50% of the acid, then you are also removing 50% concentration of the original solution. If you are adding an amount with 30% solution of acid, then you are also adding 70% of the original solution. The end result should be 40% acid---60% original solution, which is the focus since the question asks for the original amount of the solution.

Looking at the table, the original amount is 50% of let's say, 1 unit. The original solution is 0.50. You remove 50% concentration of original solution of some amount x and then you add back the same amount x but with 70% of original solution concentration. The end result is 60% original solution. Multiply down, then remove and add terms and solve for x. The amount you replaced, or x, is 1/2.

Try this problem: mixture-problem-78345.html and see how you do.

I had trouble with mixture problems, too, but I came across the user Economist's link, https://www.onlinemathlearning.com/mixture-problems.html, and this is where I got the above strategy from. You should check it out.

Another way to look at this problem could be this.

Let the total solution be T and the part removed be P.

Now T has 50% acid
And P is removed from T.

Therefore left over acid (by quantity) would be:
\(\frac{50}{100} * (T-P) = \frac{T-P}{2}\)

to this P was added which had 30% acid in it. Hence added acid quanity is:
\(\frac{30}{100} * P\)

The resultant mixture had 40% acid in it. Hence the resultant acid quantity is:
\(\frac{40}{100} * T = \frac{T-P}{2} + \frac{30}{100}* P\)


Solving for P, we get \(P = \frac{T}{2}\) which gives the part as \(\frac{1}{2}\) of Original

Therefore answer is C

Hope this helps!

Cheers!
JT

since the replaced amount is equal, let x be the amount replaced.

so, A - xA + xB

A - xA + xB = 0.4

so,

0.5 - 0.5x + 0.3x = 0.4
-0.5x + 0.3x = -0.1
-0.2x = -0.1
\(x = \frac{1}{2}\)

IMO, it's C

Alligative Method is such a time-saving method.

Here we go:

30%-----40%-----50%
Distances 1:1

This means that both contribute an equal amount of the final solution.

Straight C: 1/2

Cheers,
Der alte Fritz.

If 50% solution becomes 40% solution, then 40% solution must be equally weighted between 50% solution and 30% solution - i.e. \(\frac{1}{2}\) of 50% solution and \(\frac{1}{2}\) of 30% solution. If the solution must be equally weighted, then 30% solution must replace \(\frac{1}{2}\) of 50% solution (assuming original solution was 50% solution of acid).

Just another way of saying the same thing. Cheers

Did using allegation method

Refer diagram below

Answer = 1/2

Bunuel, kindly update the OA.

Thanks in advance

Responding to a pm:

It is worded to look like a replacement question but essentially, it is a simple mixtures problem only.

You have some 50% solution of acid and you remove some of it. Now to that 50% solution, you are adding 30% solution to get 40% solution. The ratio in which you mix the 50% solution and the 30% solution is given by w1/w2 = (50 - 40)/(40 - 30) = 1/1
Basically, you mix equal parts of 50% solution and 30% solution.
So initially you must have had 2 parts of 50% solution out of which you removed 1 part and replaced with 1 part of 30% solution. So you replaced half of the original solution.

The first question of this post discusses exactly this concept: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/01 ... -mixtures/

Hello chetan2u .

I was unable to interpret the question and the above solutions .
could you please explain
Thank you.

Hi..

The question means that initially a 50% solution of acid is there. 50%solution of acid means the percentage of acid is 50%, so of the solution is 2 litre , the acid is 50% or 1 litre.
Later some of it is changed with a 30% solution that is a solution which has 30% as acid, so if you took 2 litre, acid was 2000*30/100=600ml
And finally it became 40% acid...

These questions are best done through weighted average method..
So two solutions 50% and 30% are mixed together to get 40%..
Since 40% is half way between 30 and 50, we can say that we took half of each..
Otherwise the weight of 30% = (50-40)/(50-30)=1/2 ....

Let the original amount of the 50% solution of acid = 10 liters (so 5 liters are acid). Let x = the amount of solution that was replaced. So 0.5x acid was removed and replaced with 0.3x acid. We can create the equation:

(5 - 0.5x + 0.3x)/(10 - x + x) = 0.4

(5 - 0.2x)/10 = 4/10

5 - 0.2x = 4

1 = 0.2x

5 = x

Since 5/10 = 1/2, we see that 1/2 of the original solution was replaced.

Answer: C

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.

bing bing bing

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

We know this,

w1/w2=A2-Avg/Avg-A1

ie

w1/w2=50-40/40-30

w1/w2=1

originally 50% alcohol was 1
now it is 1/2

so reduction is 1/2

(50-40)/ (40 -30) = 1/1

meaning they are equal in contribution so C. - 1/2

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