Question:medium

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

Updated On: Jan 23, 2026
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Solution and Explanation

Let:

The smaller odd positive integer be x.

Since the numbers are consecutive odd integers,
the next integer is x + 2.

Given that both integers are smaller than 10,
x + 2 < 10


Condition: Sum is more than 11

x + (x + 2) > 11

2x + 2 > 11

2x > 9

x > 4.5


Possible values of x:

x is an odd positive integer such that

5 < x ≤ 7

So,
x = 5, 7


Corresponding pairs:

(5, 7) → sum = 12
(7, 9) → sum = 16


Final Answer:

The required pairs of consecutive odd positive integers are:
(5, 7) and (7, 9)

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