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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