The average of 10 consecutive integers is $$\frac{11}{2}$$. What is the average of the "smallest four" of these integers?

The average of 10 consecutive integers is $$\frac{11}{2}$$.

Let's assume the smallest integer among them is 'y'.

$$\frac{y+\left(y+1\right)+\left(y+2\right)+\left(y+3\right)+\left(y+4\right)+\left(y+5\right)+\left(y+6\right)+\left(y+7\right)+\left(y+8\right)+\left(y+9\right)}{10}\ =\ \frac{11}{2}$$

$$\frac{y+\left(y+1\right)+\left(y+2\right)+\left(y+3\right)+\left(y+4\right)+\left(y+5\right)+\left(y+6\right)+\left(y+7\right)+\left(y+8\right)+\left(y+9\right)}{10}\ =\ 5.5$$

10y = 55-45

10y = 10

y = 1

So the "smallest four" numbers will be 1, 2, 3 and 4

Average of the "smallest four" of these integers = $$\frac{1+2+3+4}{4}$$

= $$\frac{10}{4}$$

= 2.5