Answer :

We need to assume two consecutive even positive integers.


So, let the smaller even positive integer be x.


Then, the other even positive integer will be (x + 2).


Given: Both these numbers are larger than 5. …(i)


And their sum is less than 23. …(ii)


So,


From given statement (i),


x > 5 …(iii)


x + 2 > 5


x > 5 – 2


x > 3


Since, the number must be larger than 5, x > 3 can be ignored.


From given statement (ii),


Sum of these two consecutive even positive integers < 23


(x) + (x + 2) < 23


x + x + 2 < 23


2x + 2 < 23


2(x + 1) < 23



x + 1 < 11.5


x < 11.5 – 1


x < 10.5 …(iv)


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


x > 5 & x < 10.5


It can be merged and written as


5 < x < 10.5


From this inequality, we can say that x lies between 5 and 10.5.


So, the even positive integers lying between 5 and 10.5 are 6, 8 and 10.


Now, let us find pairs of consecutive even positive integers.


Let x = 6, then (x + 2) = (6 + 2) = 8


Let x = 8, then (x + 2) = (8 + 2) = 10


Let x = 10, then (x + 2) = (10 + 2) = 12.


Hence, all such pairs of consecutive even positive integers required are (6, 8), (8, 10) and (10, 12).


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