The average of 8 consecutive integers is $$\frac{27}{2}$$. What is the average of the largest four of these integers?
The average of 8 consecutive integers is $$\frac{27}{2}$$.
Let's assume the smallest integer is 'y'.
$$\frac{y+(y+1)+(y+2)+(y+3)+(y+4)+(y+5)+(y+6)+(y+7)}{8}=\frac{27}{2}$$
$$\frac{8y+28}{4}=27$$
2y+7 = 27average of the largest four of these integers =Â $$\frac{(y+4)+(y+5)+(y+6)+(y+7)}{4}$$
= $$\frac{4y+22}{4}$$
= $$\frac{4\times10+22}{4}$$ [put the value of 'y'.]
= $$\frac{40+22}{4}$$
= $$\frac{62}{4}$$
= 15.5
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