The average of 8 consecutive integers are 27/2 . What is the average of the largest four of these integers?
Show answer & explanation
Correct answer: D
Explanation
The correct answer is option 4 i.e 15.5
Let the first integer be x. Then, the eight consecutive integers will be x, x + 1, x + 2, x + 3, x + 4, x + 5 , x+ 6, and x + 7.
The average of these eight integers is:
(x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6) + (x + 7)) / 8
Simplifying, we get:
(8x + 28) / 8 = x + 7/2
We are given that the average of these eight integers is 27/2, so we can set up the equation:
x + 7/2 = 27/2
Solving for x, we get:
x = 10
Therefore, the eight consecutive integers are 10, 11, 12, 13, 14, 15, 16, and 17.
The largest four integers are 14, 15, 16, and 17, and their average is:
(14 + 15 + 16 + 17) / 4 = 15.5
Therefore, the average of the largest four of these integers is 15.5.