Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
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Solution:
Let the two consecutive odd positive integers be x and x + 2.
x < 10, x + 2 < 10, x + (x + 2) > 11
x + 2 < 10
x < 8
x + (x + 2) > 11
2x + 2 > 11
2x > 9
x > 9/2
i.e 8 > x > 9/2
Thus, the required pairs of consecutive odd positive integers are (5, 7) and (7, 5).
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Answered
Let x and x + 2 be the required pairs of consecutive odd positive integers.
Given: x + 2 < 10
⇒ x < 8 and
⇒ x + x + 2 > 11
⇒ 2x > 9
⇒ x > 4.5
∴ 4.5 < x < 8 x = 5, 7
∴ Required possible pairs are (5, 7), (7, 9).
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