Suppose ‘x’ be the smaller of the two consecutive odd positive integers. Now the other odd integer is x + 2.
Given as
Both the integers are smaller than 10 and their sum is more than 11.
Therefore,
x + 2 < 10 and x + (x + 2) > 11
x < 10 – 2 and 2x + 2 > 11
x < 8 and 2x > 11 – 2
x < 8 and 2x > 9
x < 8 and x > 9/2
9/2 < x < 8
Note the odd positive integers lying between 4.5 and 8.
x = 5, 7 [Since, x is an odd integer]
x < 10 [From the given statement]
9/2 < x < 10
Note that, the upper limit here has shifted from 8 to 10. Then, x is odd integer from 4.5 to 10.
Therefore, the odd integers from 4.5 to 10 are 5, 7 and 9.
Then, let us find pairs of consecutive odd integers.
Let x = 5, then (x + 2) = (5 + 2) = 7.
Let x = 7, then (x + 2) = (7 + 2) = 9.
Suppose x = 9, then (x + 2) = (9 + 2) = 11. But, 11 is greater than 10.
Hence, the required pairs of odd integers are (5, 7) and (7, 9)
