Question : The average of 10 consecutive integers is $\frac{11}{2}$. What is the average of the "smallest four" of these integers?
Option 1: 2.5
Option 2: 3.5
Option 3: 3
Option 4: 2
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Correct Answer: 2.5
Solution : Let the 10 consecutive integers be $x, (x + 1), (x+2), (x+3), (x+4), (x+5), (x+6), (x+7), (x+8),(x+9)$ Average $=\frac{x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)(x+8)(x+9)}{10} = \frac{11}{2}$ ⇒ $10x + 45 = \frac{11}{2} × 10$ ⇒ $10x = 55 - 45$ ⇒ $x = \frac{10}{10} = 1$ The Smallest four integers = $x + (x + 1) + (x+2) + (x+3)$ = 1 + 2 + 3 + 4 = 10 Average of the smallest four integers = $\frac{10}{4}$ = 2.5 Hence, the correct answer is 2.5.
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