Responding to a pm:


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.

Answer (E)

For more details, check here: http://www.veritasprep.com/blog/2012/01 ... -mixtures/
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I would go with E..we dont know the original concentration..

Even i would go with E.
We do not know the original concentration of the acid.Strong does not mean concentration is 100%.

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.


Answer: E.
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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 ?

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.

please help!
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Illegitimi non carborundum.

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
Answer: OptionE
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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)

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
Answer is E