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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