Find all pairs of consecutive odd natural number, both of which are larger than 10, such that their sum is less than 40.

We need to assume two consecutive odd natural numbers.

So, let the smaller odd natural number be x.

Then, the other odd natural number will be (x + 2).

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

And their sum is less than 40. …(ii)

So,

From given statement (i),

x > 10 …(iii)

x + 2 > 10

⇒ x > 10 – 2

⇒ x > 8

Since, the number must be greater than 10, x > 8 can be ignored.

From given statement (ii),

Sum of these two consecutive odd natural numbers < 40

⇒ (x) + (x + 2) < 40

⇒ x + x + 2 < 40

⇒ 2x + 2 < 40

⇒ 2(x + 1) < 40

⇒ x + 1 < 20

⇒ x < 20 – 1

⇒ x < 19 …(iv)

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

x > 10 & x < 19

It can be merged and written as

10 < x < 19

From this inequality, we can say that x lies between 10 and 19.

So, the odd natural numbers lying between 10 and 19 are 11, 13, 15 and 17. (Excluding 19 as x < 19)

Now, let us find pairs of consecutive odd natural numbers.

Let x = 11, then (x + 2) = (11 + 2) = 13

Let x = 13, then (x + 2) = (13 + 2) = 15

Let x = 15, then (x + 2) = (15 + 2) = 17

Let x = 17, then (x + 2) = (17 + 2) = 19.

Hence, all such pairs of consecutive odd natural numbers required are (11, 13), (13, 15), (15, 17) and (17, 19).

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