Answer :

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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