1. (a) Determine the average of the six integers $22,23,23,25,26,31$.
(b) The average of the three numbers $y+7,2 y-9,8 y+6$ is 27 . What is the value of $y$ ?
(c) Four positive integers, not necessarily different and each less than 100, have an average of 94 . Determine, with explanation, the minimum possible value for one of these integers.
1. (a) The average of the 6 integers given is $\frac{22+23+23+25+26+31}{6}=\frac{150}{6}=25$.
(b) Since the average of the three numbers $y+7,2 y-9$ and $8 y+6$ is 27 , then the sum of the three numbers is $3(27)=81$.
Therefore, $(y+7)+(2 y-9)+(8 y+6)=81$ or $11 y+4=81$, and so $11 y=77$ or $y=7$.
(c) Since the average of four integers is 94 , then their sum is $4(94)=376$.
Since the sum of the integers is constant, then for one of the integers to be as small as possible, the other three integers must be as large as possible.
To see this algebraically, we can call the four integers $a, b, c, d$ with $a$ the smallest.
Since $a+b+c+d=376$, then $a=376-b-c-d$.
To make $a$ as small as possible, we want to subtract as much as possible from 376 .
Since each of the four integers is smaller than 100 , then the largest that each can be is 99 . Therefore, to make one of the four integers as small as possible, we set the other three integers equal to 99 .
This means that the fourth integer equals $376-3(99)=376-297=79$.