Question

Find the all pairs of consecutive even positive
integers, both of which are less than 11 and
their sum is more than 12.

Answer 1

Answer:

Let x be the smaller of the two consecutive even positive integers .

Then the other integer is x+2.

Since both the integers are larger than 5,x>5 ....(1)

Also the sum of the two integers is less than 23.

x+(x+2)<23

⇒2x+2<23

⇒2x<23−2

⇒2x<21

⇒x<221

⇒x<10.5....(2)

From (1) and (2) we obtain 5<x<10.5.

Since x is an even number, x can take the values 6,8 and 10.

Thus the required possible pairs are (6,8),(8,10) and (10,12).