Question : The average of 8 consecutive integers is $\frac{27}{2}$. What is the average of the largest four of these integers?
Option 1: 16.5
Option 2: 15
Option 3: 14.5
Option 4: 15.5
Correct Answer: 15.5
Solution :
Let the 8 consecutive integers be $x, (x + 1), (x+2), (x+3), (x+4), (x+5), (x+6), (x+7)$
Average $=\frac{x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)}{8} = \frac{27}{2}$
⇒ $\frac{8x+28}{8} = \frac{27}{2}$
⇒ $8x = 108 - 28$
⇒ $x = \frac{80}{8} = 10$
$\therefore$ The sum of the largest four of these integers = $(x+4) + (x+5) + (x+6) + (x+7)$
= 14 + 15 + 16 + 17
So, the average of the largest four of these integers
= $\frac{14 + 15 + 16 + 17}{4}$
= $\frac{62}{4}$
= 15.5
Hence, the correct answer is 15.5.
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