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23. Find all pairs of consecutive odd positive integers both of which are smaller than10 such that their sum is more than 11. |
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Answer» ANSWER: 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 ⇒x > 9/2 ⇒x > 4.5 ...... (ii) From (i) and (ii), we obtain Since x is an odd NUMBER, x can take VALUES, 5 and 7. Thus, the required possible pairs are (5,7) and (7,9).
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