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If D is the sum of the reciprocals of the consecutive

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If D is the sum of the reciprocals of the consecutive [#permalink] New post 28 Jul 2007, 08:45
17. If D is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of the following is less than D?
I. 1/8
II. 1/9
III. 1/10
A. None
B. I only
C. III only
D. II and III only
E. I, II, and III
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 [#permalink] New post 28 Jul 2007, 09:23
C.

The average of those recipricals is ((1/91)+(1/100))/2=(1/95.5). There are 10 integers between 1/91 and 1/100 inclusive. So the sum of the integers is 10/95.5 or 100/955>100/1000=1/10. So III is less than D.

For II, 1/9=100/900>100/955. This means that I and II are greater than D.
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Re: Integers [#permalink] New post 28 Jul 2007, 10:56
jet1445 wrote:
17. If D is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of the following is less than D?
I. 1/8
II. 1/9
III. 1/10
A. None
B. I only
C. III only
D. II and III only
E. I, II, and III


D = 1/91 + 1/92 + ... + 1/100

hence, 10/91 > D > 10/100
or, 10/91 > D > 1/10

Since 1/9 (or 10/90) > 10/91, so D is less than both 1/9 and 1/8.

So, only 1/10 is less than D.

Answer is C.
Re: Integers   [#permalink] 28 Jul 2007, 10:56
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