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

We need to assume two consecutive odd positive integers.

So, let the smaller odd positive integer be x.

Then, the other odd positive integer will be (x + 2).

Given: Both these numbers are smaller than 10. …(i)

And their sum is more than 11. …(ii)

So,

From given statement (i),

x < 10 …(iii)

x + 2 < 10

⇒ x < 10 – 2

⇒ x < 8 …(iv)

From given statement (ii),

Sum of these two consecutive odd positive integers > 11

⇒ (x) + (x + 2) > 11

⇒ x + x + 2 > 11

⇒ 2x + 2 > 11

⇒ 2(x + 1) > 11

…(v)

From inequalities (iv) & (v), we have

x < 8 &

It can be merged and written as

This means that x lies between 9/2 (or 4.5) and 8.

Note the odd positive integers lying between 4.5 and 8.

They are 5 and 7.

Also, consider inequality (iii), that is, x < 10.

So, from inequalities (iii) & (v), we have

x < 10 &

It can be merged and written as

Note that, the upper limit here has shifted from 8 to 10. Now, x is odd integer from 4.5 to 10.

So, the odd integers from 4.5 to 10 are 5, 7 and 9.

Now, 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.

Let x = 9, then (x + 2) = (9 + 2) = 11. But, 11 is greater than 10.

Hence, all such pairs of odd consecutive positive integers required are (5, 7) and (7, 9).

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