# Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11

**Solution:**

Let x be the smaller of the two consecutive odd positive integers. Then, the other integer is (x + 2)

Since both the integers are smaller than 10,

x + 2 < 10

⇒ x < 10 - 2

⇒ x < 8 ....(1)

Also, the sum of the two integers is more than 11. Therefore,

x + ( x + 2) > 11

⇒ 2x + x > 11

⇒ 2x > 11 - 2

⇒ 2x > 9

⇒ x > 9/2

⇒ x > 4.5 ....(2)

From (1) and (2), we obtain

4.5 < x < 8

Since x is an odd positive integer, then values of x are 5 and 7

When x = 5 , the pair is (5, 7) and when x = 7 , the pair is (7, 9)

Thus, the required possible pairs are (5, 7) and (7, 9)

NCERT Solutions Class 11 Maths Chapter 6 Exercise 6.1 Question 23

## Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11

**Summary:**

A linear inequation x + ( x + 2) > 11 can be formed. We have found that the required possible pairs are (5, 7) and (7, 9)

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