Sarai purchased $13 \frac{1}{12}$ pounds of apples, bananas, and kiwis. Apples cost $3.50 per pound, bananas cost $0.75 per pound, and kiwis cost $2.00 per pound. If the number of pounds of apples purchased was $\frac{4}{6}$ more than the number of pounds of bananas purchased and she spent $33 in total, how many pounds of each type of fruit did she purchase?

A. Apples: $7 \frac{1}{4}$ pounds, Bananas: $3 \frac{1}{3}$ pounds, Kiwis: $2 \frac{3}{12}$ pounds
B. Apples: $8 \frac{1}{2}$ pounds, Bananas: $4 \frac{1}{6}$ pounds, Kiwis: 0
C. Apples: 6 pounds, Bananas: 2 pounds, Kiwis: $5 \frac{1}{2}$ pounds
D. Apples: 5 pounds, Bananas: 3 pounds, Kiwis: $4 \frac{1}{12}$ pounds

Using the given data, we can establish two equations that represent the total weight and cost of Sarai's fruit purchase. Solving these equations simultaneously leads to the conclusion that Sarai bought 5 1/2 pounds of apples, 1 1/3 pounds of bananas, and 6 5/12 pounds of kiwis.

Explanation:

Given the details of Sarai's spending on fruits, the most efficient way to solve this problem is by establishing two equations with two variables, a for the amount of apples and b for bananas, and then solving them simultaneously. We know from the problem that the total weight of the fruits purchased equals 13 1/12 pounds (converted which is 157/12 pounds), and we presented this in our first equation as: a + b + k (where k represents kiwis).

Simultaneously, we learned from the problem that Sarai spent $33 in total, and the cost of each fruit per pound is provided (apples $3.50, bananas $0.75, and kiwis $2). Therefore, we formed our second equation: 3.5a + 0.75b + 2k = 33. After turning the equations into a solvable system, we concluded with the following amounts of each fruit: Apples: 5 1/2 pounds, Bananas: 1 1/3 pounds, Kiwis: 6 5/12 pounds. So, none of the provided options in the question are correct.

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