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Bunuel
An elephant in a zoo is fed a diet that consists solely of a mixture of 2 different foods, Food A and Food B, which contains 30 percent fat and 10 percent fat, respectively. The elephant consumes a total of 50 pounds of food daily. How many pounds of Food A are in the elephant's daily food mixture if the mixture contains exactly 11 pounds of fat?

(A) 50
(B) 30
(C) 20
(D) 9
(E) 2

I found a neat trick that helped me solve this question quickly.

The final mixture contains 11 pounds of fat.

\(\frac{11}{50}*100 = 22%\)

From this we can infer that more of food A is present in the mixture. That is A is more than 25 pounds.
If the foods were mixed in the ratio of 1:1 then the resultant would be 20% of fat (Avg of 30 and 10).
If the entire mixture was A, then the fat would be 30 percent. Which is not the case.

So the answer should be B.
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This is a problem of aggregate and the answer can be given as

Take A be the weight of food A given to the elephant and B be the weight of food B given to the elephant

A+B=50
0.3A+0.1B=11

Solving A= 30

Hence A
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Bunuel
An elephant in a zoo is fed a diet that consists solely of a mixture of 2 different foods, Food A and Food B, which contains 30 percent fat and 10 percent fat, respectively. The elephant consumes a total of 50 pounds of food daily. How many pounds of Food A are in the elephant's daily food mixture if the mixture contains exactly 11 pounds of fat?

(A) 50
(B) 30
(C) 20
(D) 9
(E) 2

We are told that the total weight of the mixture is 11 pounds of fat out of the total 50 pounds of food. The final mixture will be 11/50 = 22%

Setting up a mixture matrix:

Percent of Fat Number of pounds Total

Food A 30% x 0.3x
Food B 10% 50-x 5-0.1x
Total 22% 50 11

0.3x + 5-0.1x = 11 —> multiply by 10 to get rid of decimals

3x + 50 - x = 110
2x = 60
x = 30


IMO Option B

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