Let the pair of consecutive even positive integers be x and x + 2.
So, it is given that both the integers are greater than 8
Therefore,
x > 8 and x + 2 > 8
When,
x + 2 > 8
Subtracting 2 from both the sides in above equation
x + 2 – 2 > 8 – 2
x > 6
Since x > 8 and x > 6
Therefore,
x > 8
It is also given that sum of both the integers is less than 25
Therefore,
x + (x + 2) < 25
x + x + 2 < 25
2x + 2 < 25
Subtracting 2 from both the sides in above equation
2x + 2 – 2 < 25 – 2
2x < 23
Dividing both the sides by 2 in above equation
\(\frac{2{\text{x}}}{2}< \frac{23}{2}\)
x < 11.5
Since x > 8 and x < 11.5
So, the only possible value of x can be 10
Therefore, x + 2 = 10 + 2 = 12
Thus, the required possible pair is (10, 12).