Q. 2

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

Answer :

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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