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30 May 2024, 12:28
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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.
(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.
31 May 2024, 08:40
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Bunuel
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
06 Sep 2024, 04:31
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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!
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Harsha
Passionate about all things GMAT | My website: gmatanchor.com
GMAT-Club-Forum-yzhe3950.png [ 215.53 KiB | Viewed 4847 times ]
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
Attachment:
GMAT-Club-Forum-yzhe3950.png [ 215.53 KiB | Viewed 4847 times ]
