Question

Three consecutive even numbers have a sum where one half of that sum is between 90 and 105.
a. Write an inequality to find the three numbers. Let n represent the smallest even number.
b. Solve the inequality.

Answer 1

a. The three consecutive even numbers can be represented as n, n+2, and n+4.
To find the sum of these numbers, we can add them together: n + (n+2) + (n+4).
To find one half of the sum, we can divide the sum by 2: (n + (n+2) + (n+4))/2.

We're given that one half of the sum is between 90 and 105, so we can write the inequality as:
90 ≤ (n + (n+2) + (n+4))/2 ≤ 105.

b. To solve the inequality, we'll simplify the expression step by step:
90 ≤ (n + (n+2) + (n+4))/2 ≤ 105
Multiply both sides of the inequality by 2 to get rid of the denominator:
180 ≤ n + (n+2) + (n+4) ≤ 210
Combine like terms on both sides:
180 ≤ 3n + 6 ≤ 210
Subtract 6 from all sides of the inequality:
174 ≤ 3n ≤ 204
Divide all sides of the inequality by 3:
58 ≤ n ≤ 68

So the smallest even number, n, lies between 58 and 68. Therefore, the three consecutive even numbers are 58, 60, and 62.