Question

Find the pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

Answer 1

Let x be the smaller of the two consecutive odd positive integers, then the other is x + 2. 

According to the given conditions. 

x < 10, x + 2 < 10 and x + (x + 2) > 11

⇒ x < 10, x < 8

and 2x > 9 

x < 10, (∵ x < 8 automatically smallest of the lesser than) ... (1)

and x > (9/2) ... (2) 

From (1) and (2), we get 9

(9/2) < x < 8 

Also, x is an odd positive integer. x can take values 5 and 7.

So, the required possible pairs will be (x, x + 2) = (5, 7), (7, 9)