hsampath wrote:

Hmm, I get C.

If "a" grams are in 80 grams originally, given:

a/80-(a/(80+X)) = Y/100

a(80+X-80)/(80*(80+X)) = Y/100

So, knowing X and Y we can solve for a.

Although, solving, I do get 80 ie: 100% solution initially

y is the % change from original so would equal:

100*(orig conc A - New conc A)/Orig conc A

You've equated Y to 100 times the absolute change in conc, not relative change.

We could pick any concentrations for A: If A = 50% acid then in 80g of solution there would be 40g of acid.

If 80g of water were added we'd have 40g of acid in 160g of solution or 40/160 = 25% concentration or 50% DECREASE from original concentration.

If A = 20% acid in 80g of solution then we'd have 16g of acid.

If 80g of water were added we'd have 16g of acid in 160g of solutin or 16/160 = 10% concentration or 50% DECREASE from orginal concentration.