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

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X grams of water were added to the 80 grams of a strong [#permalink]  18 Nov 2008, 10:16
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$$X$$ grams 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 $$\frac{1}{Y}$$ times of the initial concentration, what was the concentration of acid in the original solution?

(1) $$X = 80$$

(2) $$Y = 2$$
[Reveal] Spoiler: OA
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Re: X grams of water were added to the 80 grams of a strong [#permalink]  18 Nov 2008, 11:54
let us assume that originally 80 gms of solution had a gms of acid

Hence from statement 1

we get a/80 - a/160 = y/100 -------equation can t be solved hence insufficient

from 2 we get

a/80 - a/(80+x) = 50/100 = 1/2 --------------again can t be solved insuff

combining can be solved

Hence C
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Re: X grams of water were added to the 80 grams of a strong [#permalink]  19 Nov 2008, 09:15
I would go with E..we dont know the original concentration..
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Re: X grams of water were added to the 80 grams of a strong [#permalink]  19 Nov 2008, 21:34
taking stmts 1 and 2 in consideration and the given statements the solution is

a 100% solution of srong acid (all 80 gms is acid) reduces in concetration to 50 % ( 80 gm acid + 80 gms water)

Hence C
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Re: X grams of water were added to the 80 grams of a strong [#permalink]  20 Nov 2008, 09:48
Even i would go with E.
We do not know the original concentration of the acid.Strong does not mean concentration is 100%.
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Re: X grams of water were added to the 80 grams of a strong [#permalink]  30 Oct 2014, 05:31
Bunuel

Can you please give a detailed explanation to this question ?
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Re: X grams of water were added to the 80 grams of a strong [#permalink]  30 Oct 2014, 07:31
Expert's post
prasun84 wrote:
$$X$$ grams 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 $$\frac{1}{Y}$$ times of the initial concentration, what was the concentration of acid in the original solution?

(1) $$X = 80$$

(2) $$Y = 2$$

Statements (1) and (2) combined are insufficient. Denote the original concentration as $$C$$. Construct an equation using S1 and S2:
$$80C + 80*0 = \frac{160C}{2}$$
$$80C = 80C$$

$$C$$ cancels out, so we cannot determine the answer.

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Re: X grams of water were added to the 80 grams of a strong [#permalink]  31 Oct 2014, 01:25
Bunuel

But, I must admit my weakness on "Mixture" type of questions. I didn't find the MGMAT mathbook sufficient to face different types of Mixture problems.

Can you please explain any CORE FORMULA regarding this type of problem, or how to set the equations in such cases ?

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Re: X grams of water were added to the 80 grams of a strong [#permalink]  31 Oct 2014, 02:27
Another question:

I set the equation using st1 and st2 as...

80C = C/2 (80+80) [ old weight.old concentration = new weight.new concentration ]
80C= 80C

Is there any flaw/mistake above ?
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Re: X grams of water were added to the 80 grams of a strong [#permalink]  31 Oct 2014, 05:27
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Expert's post
Tanvr wrote:
Bunuel

But, I must admit my weakness on "Mixture" type of questions. I didn't find the MGMAT mathbook sufficient to face different types of Mixture problems.

Can you please explain any CORE FORMULA regarding this type of problem, or how to set the equations in such cases ?

All DS Mixture Problems to practice: search.php?search_id=tag&tag_id=43
All PS Mixture Problems to practice: search.php?search_id=tag&tag_id=114

Hope this helps.
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Re: X grams of water were added to the 80 grams of a strong [#permalink]  01 Nov 2014, 07:05
Bunuel wrote:
prasun84 wrote:
$$X$$ grams 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 $$\frac{1}{Y}$$ times of the initial concentration, what was the concentration of acid in the original solution?

(1) $$X = 80$$

(2) $$Y = 2$$

Statements (1) and (2) combined are insufficient. Denote the original concentration as $$C$$. Construct an equation using S1 and S2:
$$80C + 80*0 = \frac{160C}{2}$$
$$80C = 80C$$

$$C$$ cancels out, so we cannot determine the answer.

Hi Bunuel

Didn't quite get what you did up there.

This is what I did.

let in the original solution there be U unknown liquid and 80 gms of acid.

combining (1) & (2) --> {80/(80+u)}*(1/y) = {80/(80+80+u)}
=> u = 0
That's how i got C.

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Re: X grams of water were added to the 80 grams of a strong [#permalink]  09 Dec 2014, 22:01
Expert's post
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BOOKMARKED
prasun84 wrote:
$$X$$ grams 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 $$\frac{1}{Y}$$ times of the initial concentration, what was the concentration of acid in the original solution?

(1) $$X = 80$$

(2) $$Y = 2$$

Responding to a pm:
Quote:
I made this equation: A1/(80+x) = 1/Y *(A2/80).
Is this equation right? Where A1 is the new concentration of acid, and A2 is the original one. I realize my initial wrror , which was to put both A1 and A2 as A (lack of concentration I suppose).

I don't know how you get this equation.

Amount of acid remains same after you add water so

Amount of acid = C1*V1 (initial) = C2*V2 (final)
C1*80 = C1/Y * (80+x)
80 = (80+x)/Y

The equation is independent of C1.

Look at the statements now:
(1) $$X = 80$$
If X = 80, Y = 2
But you don't get the value of C1.

(2) $$Y = 2$$
If Y = 2, you get x = 80
But you don't get the value of C1.

For more details, check here: http://www.veritasprep.com/blog/2012/01 ... -mixtures/
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Re: X grams of water were added to the 80 grams of a strong   [#permalink] 09 Dec 2014, 22:01
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# X grams of water were added to the 80 grams of a strong

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