A certain fraction of a 15% salt solution by weight was replaced by a 25% salt solution by weight. The resulting solution has 21 % salt by weight. What fraction of the original solution was replaced?

A. 1/2
B. 3/5
C. 2/3
D. 7/10
E. 3/4

Hi Turkish,

This question is essentially a "weighted average" question, which means that it can be solved in a variety of ways. Here's an approach that uses the Weighted Average Formula:

We're mixing a certain amount of a 15% salt solution with a certain amount of a 25% salt solution and ending up with a total solution that is 21% salt.

X = # of ounces of 15% solution
Y = # of ounces of 25% solution

(.15X + .25Y)/(X+Y) = .21

.15X + .25Y = .21X + .21Y
.04Y = .06X
4Y = 6X
2Y = 3X

This ratio tells us that for every 2 ounces of 15% solution, we have 3 ounces of 25% solution. So, we can have ANY totals that maintain this ratio. For the sake of simplicity, I'm going to choose a TOTAL of 5 ounces (2 ounces of 15% and 3 ounces of 25%)

The original question is phrased in a way that implies that ALL the 5 ounces were 15% salt, but some of those ounces were removed/replaced with 25% salt. If we start with 5 total ounces and replace 3 of them with a 25% salt solution, then we get the 21% solution above. Thus 3/5 of the original solution was replaced.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________

Hi Celestial09,

In the first sentence of the question, we're told that a fraction of a solution was REPLACED, so the original number of ounces stays the same. We're NOT adding or subtracting ounces - we're replacing them.

In my example, the original mixture is 5 ounces total, so some fraction "out of 5" was replaced.... To get the 21% solution, 3 of those original ounces are replaced by the 25% solution, thus 3 of the 5 (or 3/5) of the original solution is replaced.

GMAT assassins aren't born, they're made,
Rich
_________________

This "replacement" problem is just like a regular mixture problem and should be treated as such.

15% salt solution was mixed with 25% salt solution to give 21% salt solution. So we can find the ratio of the weights in which the two were mixed.

w1/w2 = (25 - 21)/(21 - 15) = 4/6 = 2/3

So for every 2 parts of 15% solution, we have 3 parts of 25% solution. So 3/5th of the original solution was replaced.

Answer (B)

Check this post for an explanation of the formula: http://www.veritasprep.com/blog/2011/03 ... -averages/
and this one for another similar question: http://www.veritasprep.com/blog/2012/01 ... -mixtures/
_________________

hi pacifist...
the reason is that the average will be closer to the value which has more number....
that is .. an example..
x class has 20 students with 3 books each and y class has 50 students with 10 books each...
so avg books per student=(20*3+10*50)/(20+50)=560 / 70 = 8 books...
so look at the spread from average of 8.... (8-3):(10-8)=5:2...
so students x:Y=20:50=2:5 and books =5:2... both are inverse..
simmilarly for this question
_________________

Answer = B. 3/5

Using allegation method, refer diagram below:



Answer \(= \frac{6}{6+4} =\frac{6}{10} = \frac{3}{5}\)
_________________

ok shrey,
say you have 20 red balls .. you take out few of them and replace with blue balls..
now blue balls are 12 ie 3/5 of 20 ... what portion of initial mixture was changed..
blue balls have gone from 0 to 12... that is 12 of initial red balls have been changed... or in other words 12/20 (3/5) of initial mixture has been changed..
it is the same with this question.. red balls are 15% and blue balls are 25%
_________________

We can let the original amount of salt in the solution = x, and the amount taken out and replaced = y. Thus we have:

0.15x - 0.15y + 0.25y = 0.21(x - y + y)

0.15x + 0.1y = 0.21x

15x + 10y = 21x

10y = 6x

y/x = 6/10 = 3/5

Answer: B
_________________

Alligation:

15_______21________25

4__________________6

The ratio of the 15 percent solution to the 25 percent solution is 2:3

When you replace the same total amount remains. At first there were 5x grams of 15 percent solution. Then there were 2x grams of the 15 percent solution and 3x of the 25 percent solution. So 3/5 of the solution was replaced.

Alternative solution:
Assume there are 100 grams of the 15 percent solution. After a certain amount is removed there will be 100x grams left and 15x grams of the 15 percent solution. There needs to be 100-100x grams replaced. That is the 25 percent solution so 25-25x grams of alcohol get added.
Since the final mixture is 100 grams and has 21 percent alcohol there aren 21 grams of alcohol.

15x+25-25x=21

-10x=-4

X=.4

So 10x= 4

That implies 6 grams were removed before it was replaced. So 3/5 was replaced.

Alternate solution:

.15x+.25y=.21(x+y)

Posted from my mobile device

Thank you. This kept bugging me