Question 71

The average of 8 consecutive integers is $$\frac{27}{2}$$. What is the average of the largest four of these integers?

Solution

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 = 27
2y = 27-7 = 20
y = 10

average of the largest four of these integers = $$\frac{(y+4)+(y+5)+(y+6)+(y+7)}{4}$$

= $$\frac{4y+22}{4}$$

= $$\frac{4\times10+22}{4}$$ [put the value of 'y'.]

= $$\frac{40+22}{4}$$

= $$\frac{62}{4}$$

= 15.5

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