yellowho wrote:A chemist combined a milliliters of a solution that contained 20
percent substance S, by volume, with b milliliters of a solution
that contained 8 percent substance S, by volume, to product c
milliliters of a solution that was 10 percent substance S, by
volume. What is the value of a?
(1) b = 20
(2) c = 24

How is D(OA) possible?

The following method is called alligation. It's an easy way to handle weighted average problems.

To combine a 20% solution with an 8% solution to yield a 10% solution:

The proportion needed of each starting percentage is the positive difference between the other 2 percentages.

Proportion needed of a (20% solution) = 10-8 = 2.
Proportion needed of b (8% solution) = 20-10 = 10.

Thus, a:b = 2:10 = 1:5.
We also know that a+b = c.

Statement 1: b = 20
Since a:b = 1:5 and b = 20, we know that a = 4, yielding a ratio of 4:20 = 1:5.
Sufficient.

Statement 2: c = 24
a+b = 24.
Since a:b = 1:5, a = 4 and b = 20.
Sufficient.

The correct answer is D.

An important take-away:

Since all the percentages are given -- the two original percentages as well as the percentage in the final mixture -- knowing any one of the values (a, b, or c) will allow us to determine the other 2 values. Thus, without doing any math, we can quickly see that each statement is sufficient.

The proportion needed of each starting percentage is the positive difference between the other 2 percentages.


@Mitch: This holds good for all the alligation problems? can u plz elaborate on this?

hi Rohu, as a follower of Mitch introducing many useful methods here I'm asking your attention for this post (study, do take-aways and have a two-second click with such kind of questions) https://www.beatthegmat.com/those-of-you ... 71770.html

as a matter of fact we need no data to solve this DS if we are already given %one material, %another material, %combined material /material~substance~concentration here/ other than one element a,b or c
st(1) b=20 --> assign 2% for a [10%-20%], assign 10% for b [10%-8%] //cross assignment+high-low method//; 10%+2%=12%; (c/12%)*10%=b <-- eliminate % symbol and keep only numbers; 10c/12=b --> c=12b/10 OR 240/10=20. Now find a from c --> (2c/12)=c/6 OR 24/6= 4 Sufficient
st(2) you know c=24, much easier here --> 2c/12=a And 10c/12=b <> 2*24/12=a=4 And 10*24/12=b=20; Sufficient

answer is D

____________
yesterday the lovely CAT GMAT gave no mixture problems to me...

The proportion needed of each starting percentage is the positive difference between the other 2 percentages.


@Mitch: This holds good for all the alligation problems? can u plz elaborate on this?[/quote]

A chemist combined a milliliters of a solution that contained 20 percent substance S, by volume, with b milliliters of a solution that contained 8 percent substance S, by volume, to product c milliliters of a solution that was 10 percent substance S, by volume. What is the value of a?
(1) b = 20
(2) c = 24

Mitch and Night Reader,

Great tips! Alligation turns a 2 minute problem into a 10 second one on DS (and probably only 30 seconds on PS!).

thanks Night Reader

I guess everyone read this problem as mixture a plus mixture b = mixture c. I guess you read "product" as "produce." I cut and paste the problem exact. The problem said product.

yea, seems mistyping contained in the orignal source - to product was automatically understood as to produce (otherwise one word is missing, i.e. to make product)

yellowho wrote:I guess everyone read this problem as mixture a plus mixture b = mixture c. I guess you read "product" as "produce." I cut and paste the problem exact. The problem said product.

lmaooooooooo....................long live alligation technique ......