A candy shop owner must mix three types of sweetenersA, B, and Cin

Question

A candy shop owner must mix three types of sweeteners—A, B, and C—in a certain ratio to get the desired mixture in his candies. What is the ratio of the weights of Sweetener A to Sweetener B to Sweetener C in the final mixture?

(1) Sweeteners A, B, and C cost $40, $50, and $60 per pound. The average (arithmetic mean) cost of the final mixture is $50 per pound.

(2) In the final mixture, the average (arithmetic mean) price of Sweeteners A and B together is $45, and the average (arithmetic mean) price of Sweeteners B and C together is $55.

chetan2u · Accepted Answer

If the quantities are A, B and C respectively, then we are looking for A:B:C.

(1) Sweeteners A, B, and C cost $40, $50, and $60 per pound. The average (arithmetic mean) cost of the final mixture is $50 per pound. 
We can say that A and C are in the same ratio, but B could be anything.
Possible ratios => 1:1:1 ; 1:2:1 ; 2:1:2 ; 100:1:100 and so on

(2) In the final mixture, the average (arithmetic mean) price of Sweeteners A and B together is $45, and the average (arithmetic mean) price of Sweeteners B and C together is $55. 
We do not know anything about individual costs.

Combined
If cost of 40 and 50 leads to average of 45, then quantities of A and B is same.
If cost of 50 and 60 leads to average of 55, then quantities of B and C is same.
Thus, A:B = 1:1 and B:C = 1:1 or A:B:C = 1:1:1
Sufficient

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=82869 S1) arithmetic mean = middle term -> only means that the two extreme terms should have equal weightage (thus ensuring that the mean does not get dragged to the left or to the right from the middle term). This can mean multiple ratios such as 1:1:1, 4:5:4, etc. So, S1 is insufficient. S2) We only know the average price of (A and B together) and (B and C together). Without knowing the actual price of A, B, and C, this can result in several weightage or ratio combinations. For instance, between A and B, if both A and B have a price of 45, the weightage of A to B is 1:1. If the weight of A is 50 and that of B is 35, the ratio is 2:1. We should already be able to see that S2 is also insufficient. S1 and S2 together) This should work, 1:1:1 is the only possibility here. Hence, choice C. Hope this helps! --- Harsha Passionate about all things GMAT | My website: gmatanchor.com GMAT-Club-Forum-yzhe3950.png

PReciSioN · Answer

B should mention average price  per pound.  Is there a typo in the question posted?

guddo · Answer

https://gmatclub.com/forum/download/file.php?id=82297 A candy shop owner must mix three types of sweeteners—A, B, and C—in a certain ratio to get the desired mixture in his candies. What is the ratio of the weights of Sweetener A to Sweetener B to Sweetener C in the final mixture? Sweeteners A, B, and C cost $40, $50, and $60 per pound. The average (arithmetic mean) cost of the final mixture is $50 per pound. In the final mixture, the average (arithmetic mean) price of Sweeteners A and B together is $45, and the average (arithmetic mean) price of Sweeteners B and C together is $55. Applied Problems Three sweeteners A, B, and C are mixed as ingredients for candy. We are asked to determine the ratio among the weights of each based on information about cost per pound for each of the three ingredients and information about average (arithmetic mean) cost per pound for mixtures of two or three of the ingredients. This information will not allow us to calculate weights of individual ingredients in the mixture, or even a ratio between individual weights. Therefore, (1) is not sufficient by itself; NOT sufficient. This information will not allow us to calculate weights of individual ingredients in the mixture or even a ratio between the weights of individual ingredients. Therefore, (2) is not sufficient by itself; NOT sufficient. If we combine the information provided in (1) and (2), we will be able to calculate the ratio between the weights of each of the three ingredients in the mixture. Let a be the weight in pounds of Sweetener A, b the weight in pounds of Sweetener B, and c the weight in pounds of Sweetener C. From (1) and (2), we get the following equations: (40a + 50b) = 45(a + b) (50b + 60c) = 55(b + c). From the first, we can infer 5b – 5a = 0, and therefore that a = b. Similarly, from the second, we can infer that b = c. So a:b:c = 1:1:1. So (1) and (2) are sufficient together. The correct answer is C; both statements together are sufficient. GMAT-Club-Forum-9dz9jz9v.png

chetan2u · Answer

Average price itself means average price per pound here.

It is similar to average speed.

Posted from my mobile device

Elite097 · Answer

chetan2u  whats the significance of arithmetic mean here, what if we used weighted mean instead

OmerKor · Answer

what about statement 2 with this case? (im trying to refute it)

(2)
a) 1 : 1 : 1 - clearly
b) 1 : 3 : 1
 30$ : 50$ : 70$ so Average: A-B=45$ BC=55$

(1)+(2) the two statments also could be insufficient?

Bunuel · Answer

Check highlighted parts. Your prices for the sweeteners and the prices given in the first statement are not the same.

avohra223 · Answer

Hi, this is my exact doubt! what do u mean by this? How do we know that the 45 and 55 values are average price per pound, and not something else? besides the fact that it is a reasonable assumption to make, what's the mathematically valid answer here?

sofiwerner1 · Answer

Why is statemnent 2 alone not enough? I know that a:b = 1:1 and b:c also 1:1 --> a:b:c = 1:1:1. In what scenarios is this refuted fro S2 not to be sufficient

Bunuel · Answer

How do you know that "a:b = 1:1 and b:c = 1:1" just from Statement 2 alone?

From (2), we know (Pa * a + Pb * b)/(a + b) = 45 and (Pb * b + Pc * c)/(b + c) = 55, where Pa, Pb, and Pc are the prices per pound of A, B, and C, respectively.

How does this imply that a:b:c = 1:1:1?

findingmyself · Answer

How I solved it:

Let ratio of weights be X:Y:Z

Statement 1 says:
(40X+50Y+60Z)/3=50
4X+5Y+6Z=15
X Y Z could be 1:1:1 or 2:1:2 or any ration for that matter since X Y and Z can have various combination other than 1 which will satisfy this equation, Hence not sufficient

Statement 2 says
Price(A)*X(weight A)+Price(B)*Y(Weight B)=90, Price(B)*Y(Weight B)+Price(C)*C(Weight C)=110
Which in itself not giving any answers

Combining both statements
40X+50Y=90
50Y+60Z=110
4X+5Y=9
5Y+6Z=11
Summing this 4X+10Y+6Z=20
We already have 4X+5Y+6Z=15
Subtracting first from second
5Y=5
Y=1, which also gives 4X=9-5Y=4
X=1
Y=1, which also gives 6Z=11-5Y
Z=1
X:Y:Z=1:1:1

A candy shop owner must mix three types of sweetenersA, B, and Cin

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