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
⇒ x > \(\frac{9}{2}\) …(v)
From inequalities (iv) & (v), we have
x < 8 & x > \(\frac{9}{2}\)
It can be merged and written as,
\(\frac{9}{2}\) < x < 8
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 & x > \(\frac{9}{2}\)
It can be merged and written as,
\(\frac{9}{2}\) < x < 10
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).
