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X grams of water were added to the 80 grams of a strong

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Senior Manager
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X grams of water were added to the 80 grams of a strong [#permalink] New post 02 Dec 2006, 07:43
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A
B
C
D
E

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X grams of water were added to the 80 grams of a strong solution acid. As a result, the concentration of acid in the solution dropped by Y percent. What was the original concentration of acid in the original solution?

(1) X = 80

(2) Y = 50
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 [#permalink] New post 02 Dec 2006, 07:49
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I'm getting E)

From 1) 80g of water + 80g of solution of unknown concentration
We've doubled the volume so halved the concentration (y must equal 50) but without knowing any figure for resulting concentrating then we cant calculate original

From 2) y = 50% This just tells us that the volume of solution has doubled so x = 80 Therefore it doesn't tell us any more than 1)

Taking both together tells us no more than either individually
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Re: DS: Acid Solution [#permalink] New post 02 Dec 2006, 08:03
Hermione wrote:
X grams of water were added to the 80 grams of a strong solution acid. As a result, the concentration of acid in the solution dropped by Y percent. What was the original concentration of acid in the original solution?

(1) X = 80

(2) Y = 50


If we take "80 grams of a strong solution acid" as pure acid, then we dont need any information to solve the question.. :wink:

Original concentration is 100%.

However, that can't be the right way to solve this question. Hence E seems to be appropriate.
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 [#permalink] New post 02 Dec 2006, 08:30
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
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 [#permalink] New post 02 Dec 2006, 08:48
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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


Thats cool..u seem to be right.
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 [#permalink] New post 02 Dec 2006, 08:56
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.
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 [#permalink] New post 02 Dec 2006, 09:17
MBALad, I think I understand now, thank you.

dropped by Y % does transalte to [(orig-new)/orig]*100 = Y
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 [#permalink] New post 02 Dec 2006, 13:48
Basically, this is how we can go about it using a equation:

if "a" gms is the original acid, then

(Orig. - New)/Orig. *100 = Y

or (1 - New/Orig.) *100 = Y

New = a/(80 + X), Orig = a/80

Substituting, 1 - 80/(80 + X) = Y/100,

Since a gets cancelled out, answer is E. :lol:
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 [#permalink] New post 02 Dec 2006, 22:59
it is E

you can get an equation which cancels out the concentration, so you'll never be able to work it out.
  [#permalink] 02 Dec 2006, 22:59
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