**Question:**

Find all pairs of consecutive even positive integers both of

which are larger than 8 such that their sum is less than 25.

**Solution:**

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

$x+(x+2)<25$

$x+x+2<25$

$2 x+2<25$

Subtracting 2 from both the sides in above equation

$2 x+2-2<25-2$

$2 x<23$

Dividing both the sides by 2 in above equation

$\frac{2 x}{2}<\frac{23}{2}$

$x<11.5$

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).