Three containers A, B and C have wine concentrations a, b and c, respe

I came to the conclusion of E.

Neither statement compares A,B,C to P,Q,R. P,Q,R could be greater or less.

Thanks

It is C.

Statement 1: Insufficient

a - b = b -c tells us the difference in concentrations is same but we do not know the ratio in which these are mixed and hence cannot tell if resulting concentration is greater than b.

Statement 2: Insufficient

p > q > r tells us solution with concentration a has the highest proportion in the resulting solution but without knowing the concentrations of each, we do not know whether the eventual concentration is greater than b or not.

Combining 1 and 2, we know concentrations are equidistant and b lies exactly in the middle. Now resulting solution will depend on the proportion that is mixed and if p is greatest, we know the resulting concentration is tilted towards a and not towards c. Hence, resulting concentration is less than b.

Shouldn't the answer depend on values of a,b,c & p,q,r ?

It does. Say we have concentrations 60%, 40% and 20% and we mix them in equal quantity, resulting concentration would be 40%.
But if we have these concentrations where we add more of 60% solution (statement 2), concentration would sway from the mean towards the 60% concentration.

The solution is:
Amt of wine after taking p,q,r of each wine is ap+bq+cr. Let this be A
A=Concentration(C)*Volume(V)
C=A/V= (ap+bq+cr)/p+q+r
Question is C>b --> is (ap+bq+cr)/p+q+r>b --> ap+bq+cr>bp+bq+br -->ap+cr>b(p+r) .....(1)

Stmt 1 a-b=b-c..no info on p,r so NOT suff
Stmt 2 p>q>r, no info on a,b,c...NOT suff
Using both
from eqn 1
IS ap-bp>br-cr
IS p(a-b)>r(b-c)
using both stmt a-b=b-c, and p>r, STMT is true. so BOTH are sufficient. Answer is C

It took me a while to get to C!

Let's simply the question, we have below table:

Cont.-------%-----Vol
A------------a-------p
B------------b-------q
C------------c--------r
with a>b>c

Analysis:
- To simplify the question, we can assume the total volume is 1: p+q+r =1. The assumption to be 1 (or you can say 100%) does not affect the final out put % of resulting concentration.

The concentration of the mixture can be written as follow:

% = (qp+bq+cr)/(p+q+r) = qp+bq+cr. (with a>b>c).

We need to prove % > b

Statement #1: a-b = b-c
>> INSUFFICIENT >> this is obvious since we know the final % must be depended to the volume p, q, r (above equation)

Statement #2: p>q>r
>> INSUFFICIENT >> same thing. Missing a,b,c info to calculate final %.

Combining both Statement:

From 1), we have c=2b-a. Replace a from % by c=2b-a, the % will be:

% = ap + bq + (2b-a)r = a(p-r)+b(q+2r)

now, with 2), p>r hence p-r>0. Thus, a(p-r)>b(p-r) (a>b>0)

Applying above, % will be:

% > b(p-r) + b(q+2r)
~ % > b(p-r+q+2r)
~ % > b(p+r+q)
and since p+q+r = 1 ; % > b >> Choose C

VK.

Statement 1: a − b = b − c
\(a+c = 2b\)
\(b = \frac{a+c}{2}\)
Here, b is equal to the average of a and c and thus is HALFWAY between a and c, as in the following example:
a=30%..........b=20%..........c=10%
Let's apply this example to Cases 1 and 2 below.

Case 1: p=r, while a=30%, b=20% and c=10%
In this case, the mixture contains equal amounts of A and C, yielding a percentage for the mixture that is HALWAY BETWEEN A AND C and thus is EQUAL TO B.
For example:
If p=r=1 liter and b=2 liters, the average percentage for the 4-liter mixture \(= \frac{(1*30) + (2*20) + (1*10)}{4} = \frac{80}{4} = 20\).
The percentage for the mixture is equal to b.

Case 2: p > r, while a=30%, b=20% and c=10%
In this case, the mixture contains more A than C, yielding a percentage for the mixture that is CLOSER TO A THAN TO C and thus is GREATER THAN B.
For example:
If p=7 liters, q=2 liters and r=1 liter, the average percentage for the 10-liter mixture \(= \frac{(7*30) + (2*20) + (1*10)}{10} = \frac{260}{10} = 26\).
The percentage of the mixture is greater than b.

Since the answer to the question stem is NO in Case 1 but YES in Case 2, INSUFFICIENT.

Statement 2: p > q > r
The example in Case 2 satisfies Statement 2 as well as Statement 1.
In Case 2, the answer to the question stem is YES.

Case 3: p=3, q=2 and r=1, while a=21%, b=20%, and c=2%
Here, the average percentage for the 6-liter mixture \(= \frac{(3*21)+(2*20) + (1*2)}{6} = \frac{105}{6} = 17.5\).
The percentage for the mixture is less than b.

Since the answer to the question stem is YES in Case 2 but NO in Case 3, INSUFFICIENT.

Statements combined:
As illustrated by the example in Case 2, the statements combined yield an average percentage that is closer to a than to c and thus is greater than b.
The answer to the question stem is YES.
SUFFICIENT

B. a has the highest concentration and the highest volume, thus the final concentration will always be above the concentration of b!

Bunuel what is the correct solution?

I have seen all the algebraic solutions in the thread. But if we apply general logic of weighted avg (B) should be sufficient by itself, at least that's what I think.

As, a>b>c and we are mixing in the ratio p>q>r

So this shifts the concentration towards "a", as it's weight is highest.

Bunuel can you please help me understand where am I getting this wrong ?

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