mixed %

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

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

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.

Agree with fatalmilk

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

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