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).

Rate this question :

Solve each of theRD Sharma - Mathematics

A solution is to RD Sharma - Mathematics

A company manufacRD Sharma - Mathematics

<span lang="EN-USRS Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal - Mathematics

Solve each of theRS Aggarwal - Mathematics

Solve x + 5 > RS Aggarwal - Mathematics

Find the solutionRS Aggarwal - Mathematics

If |x – 1| > 5Mathematics - Exemplar