Liqueur A contains 24% of alcohol. What is the alcohol concentration

Concentration of Alcohol in A = 24%
Concentration of Alcohol in B = B%
Concentration of Alcohol in the mixed solution = M%

St1: Mixing Ratio of A and B = 1:4 = x : 4x
Not Sufficient to calculate the concentration of Alcohol in the mixed solution as we can have a number of different values for x.

St2: B = 2.5*M --> Clearly insufficient as we have two unknowns

Combine St1 and St2: Using the allegation technique we have,
24 - 4x = M
M - B = x --> M = x + B
24 - 4x = x + B
24 - 5x = 2.5(24 - 4x)
We can find the value of x and determine the value of M.
Sufficient

Answer: C

Bunuel can you please explain this?

Bunuel

Can you pls explain this.

TIA

weighted average DS

wa/wb= Ab-weighted average of mix / weighted average -0.24

what is WEIGHTED AVERAGE?? we need wa,wb,Ab,

from 1

wa:wb = 1:4 ... insuff

from 2

Weighted average = 2/3 wb.. insuff

both we have wa:wb and we have wb in terms of weighted average ( equation has one unknown now) .....C

didn't get it... pls explain..?

I am using a weighted average formula , Since the question is asking for a final concentration of a mix that is the same as asking for the weighted average of the 2 solutions concentration of alcohol.

Weight A concentration of alcohol in B - Weighted average of the final mix
--------- = ------------------------------------------------------
weight B Weighted average of the final mix - Concentration of alcohol in A (given as 24%)

Question asks for the weighted average of alcohol concentration in the final mix? thus we need Weight A, Weight B , Concentration of alcohol in B

FROM 1

Weight A 1
--------- = --- We still need concentration of alcohol in B......... INSUFF
weight of B 4

FROM 2

Concentration in B = 1.5 *0.24............. INSUFF since we need Weight A and Weight B

Both together

we have all 3 missing unknowns to solve the equation.........thus suff

C

Bunuel

could you please, if you find time, show us how you would come up with the correct equation here.
I know it is a DS problem but I still would like to solve it for the actual solution, no matter how I write the equation I don't get a result on this one as it is very confusing with the % and actual values.

Liqueur A contains 24% of alcohol. What is the alcohol concentration of the mixed cocktail of liqueur A and B?

(1) The mixing ratio of liqueur A and B is 1:4
(2) The alcohol concentration of liqueur B is 1.5 times greater than the alcohol concentration of the mixed cocktail.

1)
M=Mixed Cocktail
Mathematical Equation:
0.2*M = A
0.8*M = B

No information of B => INSUFFICIENT

2)
x = % alcohol concentration of B
Mathematical Equation:
1.5(0.24*0.2M+x*0.8M)/M=x
<=> 1.5*0.24*0.2+x*0.8*1.5=x
<=> 36/500=-2x/10
<=> -36/100=x = -36% which makes no sense...

I think there is a typo mistake in the question.

Even I am getting negative concentration can some expert provide the final concentration values along with solution. In the question the concentration of B is told to be 1.5 times greater so it means B is 2.5 of the concentration of cocktail right.

Combining the statements

A has 24% alcohol
Mixture has x% alcohol
B has 1.5x% alcohol

i.e. 24% of A + 1.5x% of B = x% pf (A+B)

But A:B = 1:4
i.e. If A = 1 then B = 4

Hence, i.e. (24/100)*1 + (1.5x/100)*4 = (x/100)*(1+4)

i.e. 24 + 6x = 5x

i.e. x = -24

Bunuel I think there is some error here as pointed out by other people too. x is -ve