C right?!!

[mehaksal]

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

[anujan007]

Let us consider the first statement.

(i). X=80 but no information given on Y hence insufficient.

(ii). Y=2 but no information given on X hence insufficient.

Considering both statements together --:

original acid solution's weight = 80
80 Grams of water were added. Hence new weight of the acid solution is = 160

Consider original conc of water as w. Hence new concentration is (w+x)

Consider original conc of acid as a. Hence new concentration is (a/2)

w + a = 80 ----> (I)

after the addition of 80 gms water,

(w+x) + (y/2) = 160 ---> (II)

There are 4 variables and 2 equations hence both together are not sufficient.

The answer I would choose is E.

[Bill@VeritasPrep]

Let us consider the first statement.

(i). X=80 but no information given on Y hence insufficient.

(ii). Y=2 but no information given on X hence insufficient.

Considering both statements together --:

original acid solution's weight = 80
80 Grams of water were added. Hence new weight of the acid solution is = 160

Consider original conc of water as w. Hence new concentration is (w+x)

Consider original conc of acid as a. Hence new concentration is (a/2)

w + a = 80 ----> (I)

after the addition of 80 gms water,

(w+x) + (y/2) = 160 ---> (II)

There are 4 variables and 2 equations hence both together are not sufficient.

The answer I would choose is E.

Agreed. Any original concentration will be cut in half by the addition of 80 grams of acid.

If it was originally 1/80, it's now 1/60. If it was originally 4/80, it's now 4/160. We can never find an exact value.

[mehaksal]

thanku