Question : The average of 10 consecutive integers is $\frac{33}{2}$. What is the average of the first three integers?
Option 1: 12
Option 2: 13
Option 3: 15
Option 4: 11
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Correct Answer: 13
Solution : Denote the first integer, so the next consecutive integers would be $(x + 1), (x + 2), (x + 3),$, and so on up to $(x + 9)$. Average = $\frac{(x + (x + 1) + (x + 2) + ... + (x + 9))}{10}$ ⇒ $\frac{33}{2} = \frac{(x + (x + 1) + (x + 2) + ... + (x + 9))}{10}$ ⇒ $\frac{33}{2} \times 10 = x + (x + 1) + (x + 2) + ... + (x + 9)$ ⇒ $165 = 10x + (1 + 2 + ... + 9)$ ⇒ $165 = 10x + 45$ ⇒ $10x = 120$ $\therefore x = 12$ To find the average of the first three integers, we can calculate: Average = $\frac{(12 + 13 + 14)}{3}=\frac{39}{3}= 13$ Hence, the correct answer is 13.
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