find all pairs of consecutive odd natural number both of which are larger then 10 such that their sum is less then 40
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Answered by
45
heya
10< x + x+2 <40
10<2x+2<40
10<2x+2 and 2x+2<40
8<2x. and. 2x<38
4<x. and x<19
so x equal to 11,13,15,17
10< x + x+2 <40
10<2x+2<40
10<2x+2 and 2x+2<40
8<2x. and. 2x<38
4<x. and x<19
so x equal to 11,13,15,17
Answered by
2
Answer:
(11, 13), (13, 15), (15, 17), (17,19)
Step-by-step explanation:
Let x be the smaller of the two consecutive odd natural number, so that the other one is x + 2. Given x > 10 and x + (x + 2) < 40 ⇒ 2x < 38 ⇒ x < 19 ∴10 < x < 19 Since x is an odd number, x can take the values 11,13, 15 and 17. ∴ Required possible pairs are (11, 13), (13, 15), (15, 17), (17,19).
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