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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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

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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