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# X grams of water were added to 80 grams of a strong solution

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X grams of water were added to 80 grams of a strong solution [#permalink]  01 Jun 2009, 09:12
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Xgrams of water were added to 80 grams of a strong solution of acid. If 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
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rampuria

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Re: DS-Mixture [#permalink]  01 Jun 2009, 09:37
E.

Statement 1)

When working with solutions, we need some way to figure out the amount of acid in the water in the original statement. if we know the total is 80 grams of solution, the possible % of acid of that solution is infinite. If you know that you add 80g water to 80g solution, this doesn't provide all the variables we need to know in order to determine the concentration of the original. We would need to know what % concentration the solution became after the 80g water is added. We have 3 variables, the 80g water added, the concentration of the original solution, and the % mixture [in the form of a fraction] after the water is added to the original solution. Knowing any 2 of the 3 can answer the question for you. Since after statement 1, we still only know 1 of the 3 variables, we do not have enough information, therefore INSUFFICIENT.

Statement 2)

The reason this is INSUFFICIENT is the same as why Statement 1 is insufficient. We need to know 2 of the 3 variables and only know 1 [Remember that we do not know how much water is added. We CANNOT keep information from Statement 1 when considering Statement 2].

Together.

INSUFFICIENT. If we have 80g to start and we add 80 g of water, then no matter what the original concentration is, we're going to be cutting the concentration in half.

Original Solution
40g acid, 40g water = 40g acid / 80 total = 50% original concentration. If we add 80g water to this, we have 40g acid / 160g total = 25%, which is 1/2 the original concentration.

If you have:

Original Solution
60g acid, 20g water = 60/80 = 75% solution. If we add 80g water, then we get 60g acid / 160g total = 37.5%, which is 1/2 of 75%, the original concentration.

When the end % concentration is relative to the original concentration, it does not give us enough information because there are too many variables that are not nailed down, and any combination of acid-water mix, when we double the volume by adding water, this will ALWAYS cut the original concentration in half.

rampuria wrote:
Xgrams of water were added to 80 grams of a strong solution of acid. If 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

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. Senior Manager Joined: 16 Jan 2009 Posts: 362 Concentration: Technology, Marketing GMAT 1: 700 Q50 V34 GPA: 3 WE: Sales (Telecommunications) Followers: 2 Kudos [?]: 60 [0], given: 15 Re: DS-Mixture [#permalink] 01 Jun 2009, 09:58 Statement 1. 80 gms of water added to 80 gms of acid solution. Here we do not know the concentration of acid in the NEW solution , so we cannot find y and so cannot reach to the original concentration. So insufficient Statement 2. X gms of water added to 80 gms of acid solution makes it ½ the concentration of the original solution. Since we do not know the weight of the new solution / acid concentration. So insufficient Taking Statement 1 and 2 together 80 gms of water added to 80 gms of acid solution is making it ½ the concentration of the original solution. However no information has been provided regarding the acid concentration of the resulting mixture. IMO E. I hope the question has been copied correctly. These type of questions usually end up in C. _________________ Lahoosaher Director Joined: 23 May 2008 Posts: 841 Followers: 3 Kudos [?]: 23 [0], given: 0 Re: DS-Mixture [#permalink] 01 Jun 2009, 19:24 rampuria wrote: Xgrams of water were added to 80 grams of a strong solution of acid. If 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 C because the concentration of water is always 1g/ml so let a = liters of water let b= liters of acid soln 80/(a+b)=1/y(80/b) from statements 1 and 2 80/(80+b)=80/2b b=800 0riginal concentration is 80g/800ml or .1g/liter SVP Joined: 30 Apr 2008 Posts: 1893 Location: Oklahoma City Schools: Hard Knocks Followers: 26 Kudos [?]: 391 [0], given: 32 Re: DS-Mixture [#permalink] 01 Jun 2009, 19:39 Where in the world did this come from? What if I have a solution of acid mixture that is 1g water and 9g acid? How does that fit in with "the concentration of water is always 1g/ml"? bigtreezl wrote: C because the concentration of water is always 1g/ml _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Re: DS-Mixture [#permalink]  01 Jun 2009, 21:56
jallenmorris wrote:
Where in the world did this come from? What if I have a solution of acid mixture that is 1g water and 9g acid? How does that fit in with "the concentration of water is always 1g/ml"?

bigtreezl wrote:
C because the concentration of water is always 1g/ml

then your concentration is 9g/ml of acid...its no longer water
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Re: DS-Mixture [#permalink]  01 Jun 2009, 22:05
Let A represent the acid.

ratio is A/80

Question:(\frac{A}{80})?

Given:

( NEW CONCENTRATION ) = \frac{1}{Y}*( OLD CONCENTRATION )

{frac({A}{80+X)} = {\frac{1}{Y}(\frac{A}{80})}

80AY = 80A + AX
80AY - 80A - AX = 0
A(80Y - 80 - X) = 0

Question:(\frac{A}{80})?
Question:({A})?

This implies that either:
A=0 or X+80=80Y

(1) X=80

Given implies that reither A=0 or ((80)+80=80Y ==> Y=2)

So we could have Y=2 and A is any value.

Insufficient.

(2) Y=2

Given implies that reither A=0 or (X+80=80(2) ==> X=80)

So we could have X=80 and A is any value.

Insufficient.

(1) 1&2 are really the same statements (stare hard).

Insufficient.

Final Answer, E.
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Last edited by Hades on 01 Jun 2009, 22:26, edited 1 time in total.
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Re: DS-Mixture [#permalink]  01 Jun 2009, 22:25
Hades wrote:
Let A represent the acid.

ratio is A/80

Question:(\frac{A}{80})?

Given:

( NEW CONCENTRATION ) = \frac{1}{Y}*( OLD CONCENTRATION )

{frac({A}{80+X)} = {\frac{1}{Y}(\frac{A}{80})}

80AY = 80A + AX
80AY - 80A - AX = 0
A(80Y - 80 - X) = 0

Question:(\frac{A}{80})?
Question:({A})?

This implies that either:
A=0 or X+80=80Y

(1) X=80

Given implies that reither A=0 or ((80)+80=80Y ==> Y=2)

So we could have Y=2 and A is any value.

Insufficient.

(2) Y=80

Given implies that reither A=0 or (X+80=80(2) ==> X=80)

So we could have X=80 and A is any value.

Insufficient.

(1) 1&2 are really the same statements (stare hard).

Insufficient.

Final Answer, E.

in statement 2..... y=2, not 80

ok..took a closrer look..you knew that ..just a typo

Last edited by bigtreezl on 01 Jun 2009, 22:28, edited 1 time in total.
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Re: DS-Mixture [#permalink]  01 Jun 2009, 22:25
You guys are overthinking it.

Convert it to equations and throw out all meaning, work purely with the equations. The harder questions will rarely make any 'logical' sense, but Mathematics is always logical
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Re: DS-Mixture [#permalink]  01 Jun 2009, 22:26
bigtreezl wrote:
in statement 2..... y=2, not 80

Thanks, typo. Fixed
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Re: DS-Mixture [#permalink]  02 Jun 2009, 06:59
I agree with overthinking it.

It's pretty easy to realize that no single statement will be sufficient.

When we look at them together, we have 80g water added to 80 grams solution. We're doubling the volume and keeping the volume of acid the same. This will always cut the strength the original solution in half, regardless of what it used to be.
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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

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Re: DS-Mixture [#permalink]  02 Jun 2009, 10:43
OA and source please.
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Re: DS-Mixture   [#permalink] 02 Jun 2009, 10:43
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# X grams of water were added to 80 grams of a strong solution

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