Question

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.

Class X1 - Maths -Linear Inequalities Page 122

Answer 1

Let (2x +1) and (2x +3) are two positive integers .
a/C to question,
(2x +1) < 10
=> 2x < 10 - 1
=> 2x < 9
=> x < 9/2 ------------(1)

and (2x +3) < 10
2x < 10 - 3
2x < 7
x < 7/2 --------------(2)

and sum of (2x +1) and (2x +3) > 11
(2x + 1) + (2x +3) > 11
4x + 4 > 11
4x > 11 - 4 = 7
x > 7/4 -----------------(3)

from eqns (1), (2) and (3)
7/4 < x < 7/2
hence, possible value of x = 2, 3
when x = 2 , then , (2×2+1,2×2+3) = (5,7)
when x = 3 then , (2×3+1 , 2×3+3) = (7,9)

Answer 2

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Let , x be the first odd natural number , so that the other one is x + 2

The positive odd integers are smaller than 10 , so,

x < 10 ----- (I)

x + 2 < 10 i.e x < 8 ----- (II)

And

The sum of both numbers is more than 11 , so ,

x + (x + 2) > 11

2x > 9

x > 4.5 -------- (III)

From (II) and (III), we get

4.5 < x < 8

Since , x is an odd numbers , x can take the values 5 and 7

Hence , the required possible pairs will be (5,7) and (7,9)