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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]

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18 Nov 2008, 11:16
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95% (hard)

Question Stats:

38% (02:16) correct 62% (02:02) wrong based on 528 sessions

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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) $$Y = 2$$
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Re: X grams of water were added to the 80 grams of a strong  [#permalink]

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09 Dec 2014, 23:01
15
16
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]

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18 Nov 2008, 12:54
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]

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19 Nov 2008, 10:15
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]

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19 Nov 2008, 22:34
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]

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20 Nov 2008, 10:48
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]

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30 Oct 2014, 06:31
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]

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30 Oct 2014, 08:31
1
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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]

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31 Oct 2014, 02:25
1
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]

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31 Oct 2014, 03:27
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]

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31 Oct 2014, 06:27
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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]

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01 Nov 2014, 08:05
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]

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26 Jun 2015, 23:56
1
1
Can someone please tell me if my logic is correct?

According to the statements:
if a is the weight of acid and w is the weight of water then total weight is a + w = 80

Initial concentration of acid is a/(a+w) = a/80 >> this is what we need to find.

If x gms of water is added to the mixture then,

a/(80+x) = (1/y) a/80; which gives
1/80+x = 1/y80

Substituting value of x from statement 1 we get 1/160 = 1/160 >> Not Sufficient
Substituting vale of y from statement 2 again gives the same fraction >> Not sufficient

Combining 1 & 2 does not give us any new info; hence E

This approach is correct?
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Re: X grams of water were added to the 80 grams of a strong  [#permalink]

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27 Jun 2015, 00:09
AjChakravarthy wrote:
Can someone please tell me if my logic is correct?

According to the statements:
if a is the weight of acid and w is the weight of water then total weight is a + w = 80

Initial concentration of acid is a/(a+w) = a/80 >> this is what we need to find.

If x gms of water is added to the mixture then,

a/(80+x) = (1/y) a/80; which gives
1/80+x = 1/y80

Substituting value of x from statement 1 we get 1/160 = 1/160 >> Not Sufficient
Substituting vale of y from statement 2 again gives the same fraction >> Not sufficient

Combining 1 & 2 does not give us any new info; hence E

This approach is correct?

This is absolutely fine approach.

Only the reasoning of highlighted part is not very clear.

We must be clear that, We only need to know the value of either a or W to answer the question

Statement 1: Gives us the value of y but no information about values of a or W hence Not Sufficient
Statement 2: Gives us the value of y but no information about values of a or W hence Not Sufficient

Combining the two gives us the value of y only so NOT Sufficient
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Re: X grams of water were added to the 80 grams of a strong  [#permalink]

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27 Jun 2015, 01:15
GMATinsight wrote:
AjChakravarthy wrote:
Can someone please tell me if my logic is correct?

According to the statements:
if a is the weight of acid and w is the weight of water then total weight is a + w = 80

Initial concentration of acid is a/(a+w) = a/80 >> this is what we need to find.

If x gms of water is added to the mixture then,

a/(80+x) = (1/y) a/80; which gives
1/80+x = 1/y80

Substituting value of x from statement 1 we get 1/160 = 1/160 >> Not Sufficient
Substituting vale of y from statement 2 again gives the same fraction >> Not sufficient

Combining 1 & 2 does not give us any new info; hence E

This approach is correct?

This is absolutely fine approach.

Only the reasoning of highlighted part is not very clear.

We must be clear that, We only need to know the value of either a or W to answer the question

Statement 1: Gives us the value of y but no information about values of a or W hence Not Sufficient
Statement 2: Gives us the value of y but no information about values of a or W hence Not Sufficient

Combining the two gives us the value of y only so NOT Sufficient

Thanks very much for the clarification

Regarding the highlighted part : just saying that wen we substitute the values and simplify, in each case we get 1/160 = 1/160....
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Re: X grams of water were added to the 80 grams of a strong  [#permalink]

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20 Aug 2017, 07:17
1
Hello Karishma,

I recently went through your article on solving weighted average and mixture problems.
Please tell me if I am applying the approach correctly on this problem.

Let A1=a (conc. of acid in original solution)
A2=100 (conc. of water added to the solution)
A(avg.) = a/Y
w1=80
w2=X
Using Allegation :
(80/X) = (100-(a/Y)) / ((a/Y)-a)
We need to find "a".

(1) X=80
Cannot get the value of "a" from the above equation with only the value of X.
Insufficient

(2) Y=2
Same as (1)
Insufficient

(1) + (2)
(80/80)=(100-(a/2)) / ((a/2)-a)
=> (a/2)-a = 100-(a/2)
"a" gets cancelled out hence not solvable

Ans. (E)
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Re: X grams of water were added to the 80 grams of a strong  [#permalink]

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11 Nov 2018, 06:54
A+W=80

Stmt 1 => x=80
A+w+80=160

Acid quantity remains the same in both cases

A/160=1/Y*A/80

Y=2 which is same as stmt 2

Stmt 1 or 2 no way to reach concetration of acid of original solution
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Re: X grams of water were added to the 80 grams of a strong  [#permalink]

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10 Jan 2019, 01:05
Consider Variables A1 and A2 for the acid
Initial concentration is A1/80 and final concentration is nothing but A2/(80+x)
It is given that A2/(80+x) = 1/Y*(A1/80)
We have to find A1/80
We need the values of A2, X, Y
Combining both the statements we have only X and Y
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Re: X grams of water were added to the 80 grams of a strong  [#permalink]

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15 Mar 2019, 01:09
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) $$Y = 2$$

Can't we conclude the concentration to be 100% using 1 and 2 combined?

when 80grms of water was added to the acid its concentration reduced to half?
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Re: X grams of water were added to the 80 grams of a strong  [#permalink]

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11 May 2019, 18:52
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/

HI Karishma,

Can you clarify how we know from the question that the amount of acid remains the same after we add water?

Thanks.
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Re: X grams of water were added to the 80 grams of a strong   [#permalink] 11 May 2019, 18:52

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