Sites: Global Q&A | Wits | MathsGee Club | Joburg Libraries | StartUps | Zimbabwe | OER

MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB

0 like 0 dislike
129 views
If the sum of the reciprocals of two consecutive odd integers is $\frac{12}{35}$, then the greater of the two integers is:
| 129 views

1 like 0 dislike
Let the smaller of the two odd integers be n, such that the larger one is n + 2

(1/n) + (1/(n + 2) = (12/35)

((n + 2 + n)/(n(n+2)) = (12/35)

((2n + 2)/(n(n+2)) = (12/35)

Equating the numerators both sides give: 2n + 2 = 12

2n = 12 - 2 = 10

n = 5

Therefore the greater of the two integers is n+2 = 5 + 2 = 7.
by Diamond (40,854 points)
selected by

0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
1 like 0 dislike
1 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike