mixed %

gmat2010 · Posted: Fri Sep 05, 2008 11:27 am Post subject: mixed %

X grams of water were added to the 80 grams of a strong solution of acid. As a result, the concentration of acid in the solution became 1/Y times of the initial concentration. What was the concentration of acid in the original solution?

1. X = 80
2. Y = 2

schumi_gmat · Posted: Fri Sep 05, 2008 12:29 pm Post subject:

Ans is E

based on given condition we have -
let c1 be concentration

wt. of solute in the mixture = 80c1/100 gm

hence the new concentration c2 after adding X gm of water

((80c1/100) * 100) / (80+x) = c2 = c1/Y

hence we get,

80y = 80+x

From 1, we get x = 80 and y =2

but we still cant find the concentration

From 2, same as above

Hence E

lali_mew · Posted: Fri Sep 05, 2008 5:01 pm Post subject:

I think the OA is C. Can somebody please verify.
This is how -

Let the solution contain Z gms of Acid.
Thus concentration is Z/80

X gms water added, thus total solution = 80 + X
Thus concentration is Z/(80+X)

Z/(80+X) = (1/Y)(Z/80)

1) X = 80. Insufficient
2) Y = 2. Insufficient

Replacing both X and Y gives us the vlaue of Z. And thus can find Z/80

Answer is C

fatalmilk · Posted: Sat Sep 06, 2008 12:14 am Post subject:

lali_mew,

Z/160=Z/160 after replacing x and y.

Z can be any value, so you can't find the concentration.

IMO E

Please correct me if I'm wrong.

mals24 · Posted: Sat Sep 06, 2008 2:19 am Post subject:

Agree with fatalmilk

The answer should be E as both the statements gives us the same information.

lali_mew · Posted: Sat Sep 06, 2008 1:38 pm Post subject:

You are right fatalmilk!
I overlooked that point. Both sides amount to the same.
The answer is E.
Thanks for pointing it out.