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Asked by Pehal Last Modified
Answer 
Vishaka Bhutia
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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 … (i)
Also, the sum of the two integers is more than 11.
∴x + (x + 2) > 11
⇒ 2x + 2 > 11
⇒ 2x > 11 – 2
⇒ 2x > 9
From (i) and (ii), we obtain 4.5<x<8.4.5<x<8.
Since x is an odd number, x can take the values, 5 and 7.
Thus, the required possible pairs are (5, 7) and (7, 9).
read lessLet 2x+1 and 2x +3 are two positive odd integers .
a/C to question ,
2x +1 < 10
=> 2x < 10-1 = 9
=> 2x < 9
x < 9/2
and 2x + 3 < 10
=> 2x < 7
x < 7/2
and sum of (2x+1) and (2x +3) >11
2x +1 + 2x +3 > 11
4x +4 > 11
4x > 7
x > 7/4
now, plotting all these values on number line .
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)
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