Question

Find all 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 integer is x + 2.

Since both the integers are smaller than 10,

x + 2 < 10

⇒ x < 10 – 2

⇒ x < 8  ....(i)

Also, the sum of the two integers is more than 11.

∴ x + (x + 2) > 11

⇒ 2x + 2 > 11

⇒ 2x > 11 – 2

⇒ 2x > 9

= `x > 9/2`

= x > 4.5     ....(ii)

From (i) and (ii), we obtain.

Since x is an odd number, x can take the values 5 and 7.

Thus, the required possible pairs are (5, 7) and (7, 9).