Question -
Answer -
Let ‘x’ be the smaller of the two consecutive even positive integers. Then the other even integer is x + 2.
Given:
Both the even integers are greater than 5 and their sum is less than 23.
So,
x > 5 and x + x + 2 < 23
x > 5 and 2x < 21
x > 5 and x < 21/2
5 < x < 21/2
5 < x < 10.5
From this inequality, we can say that x lies between 5 and 10.5.
So, the even positive integers lying between 5 and 10.5 are 6, 8 and 10.
Now, let us find pairs of consecutive even positive integers.
Let x = 6, then (x + 2) = (6 + 2) = 8
Let x = 8, then (x + 2) = (8 + 2) = 10
Let x = 10, then (x + 2) = (10 + 2) = 12.
x = 6, 8, 10 [Since, x is even integer]
∴ The required pairs of even positive integer are (6, 8), (8, 10) and (10, 12)