Q. 13.8( 6 Votes )

# Find all pairs of

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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