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# If p is the sum of the reciprocals of the consecutive integers from 91

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Joined: 21 Aug 2018
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If p is the sum of the reciprocals of the consecutive integers from 91  [#permalink]

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Updated on: 15 Sep 2018, 03:24
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Difficulty:

65% (hard)

Question Stats:

56% (01:29) correct 44% (01:25) wrong based on 73 sessions

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If p is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of the following is more than p^(-1)?

I. 8
II. 9
III. 10

(A) Only I
(B) Only II
(C) Only III
(D) Only II and III
(E) I, II and III

Originally posted by patnaiksonal on 14 Sep 2018, 10:04.
Last edited by Bunuel on 15 Sep 2018, 03:24, edited 1 time in total.
Renamed the topic and edited the question.
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Posts: 4
Re: If p is the sum of the reciprocals of the consecutive integers from 91  [#permalink]

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14 Sep 2018, 10:32
1
patnaiksonal wrote:
If p is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of
the following is more than p^-1?

I. 8
II. 9
III. 10

(A) Only I
(B) Only II
(C) Only III
(D) Only II and III
(E) I, II and III

Here p= 1/91 + 1/92 + 1/93 + 1/94 + .. + 1/100

We can say that,
p < 1/90 + 1/90 +.. 1/90 (10 times)
p < 10/90
p < 1/9
reciprocal of p (p^-1) > 9

Also, p > 1/100 + 1/100 + 1/100 + ...+ 1/100

p> 10/100
p> 1/10
p^-1 < 10

Hence, 9 < p^-1 < 10

Hence, only 10 is greater than p^-1 as per option given. So the answer is C
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Posts: 7
Re: If p is the sum of the reciprocals of the consecutive integers from 91  [#permalink]

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14 Sep 2018, 11:11
quantumdweller wrote:
patnaiksonal wrote:
If p is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of
the following is more than p^-1?

I. 8
II. 9
III. 10

(A) Only I
(B) Only II
(C) Only III
(D) Only II and III
(E) I, II and III

Here p= 1/91 + 1/92 + 1/93 + 1/94 + .. + 1/100

We can say that,
p < 1/90 + 1/90 +.. 1/90 (10 times)
p < 10/90
p < 1/9
reciprocal of p (p^-1) > 9

Also, p > 1/100 + 1/100 + 1/100 + ...+ 1/100

p> 10/100
p> 1/10
p^-1 < 10

Hence, 9 < p^-1 < 10

Hence, only 10 is greater than p^-1 as per option given. So the answer is C

best way to solve the problem.
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Re: If p is the sum of the reciprocals of the consecutive integers from 91  [#permalink]

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14 Sep 2018, 19:59
patnaiksonal wrote:
If p is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of
the following is more than p^-1?

I. 8
II. 9
III. 10

(A) Only I
(B) Only II
(C) Only III
(D) Only II and III
(E) I, II and III

patnaiksonal Please go through the following post and provide an appropriate topic name. Refer #3

https://gmatclub.com/forum/rules-for-po ... 33935.html
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Re: If p is the sum of the reciprocals of the consecutive integers from 91  [#permalink]

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15 Sep 2018, 03:25
patnaiksonal wrote:
If p is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of the following is more than p^(-1)?

I. 8
II. 9
III. 10

(A) Only I
(B) Only II
(C) Only III
(D) Only II and III
(E) I, II and III

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https://gmatclub.com/forum/if-s-is-the- ... 24847.html
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Re: If p is the sum of the reciprocals of the consecutive integers from 91  [#permalink]

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20 Sep 2018, 11:05
patnaiksonal wrote:
If p is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of the following is more than p^(-1)?

I. 8
II. 9
III. 10

(A) Only I
(B) Only II
(C) Only III
(D) Only II and III
(E) I, II and III

OA:C

$$p=\frac{1}{91}+\frac{1}{92}+\frac{1}{93}+\frac{1}{94}+\frac{1}{95}+\frac{1}{96}+\frac{1}{97}+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}$$
$$p^{'}=\frac{1}{91}+\frac{1}{91}+\frac{1}{91}+\frac{1}{91}+\frac{1}{91}+\frac{1}{91}+\frac{1}{91}+\frac{1}{91}+\frac{1}{91}+\frac{1}{91}$$
$$p^{'}=\frac{10}{91}$$

$$p<p^{'}$$
As $$p$$ and $$p^{'}$$ are positive.

$$\frac{1}{p}>\frac{1}{p^{'}}$$

$$\frac{1}{p}>\frac{91}{10} \quad that \quad is \quad \frac{1}{p}>9.1$$

Only Option: C($$III. 10$$) satisfies the condition.
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Re: If p is the sum of the reciprocals of the consecutive integers from 91  [#permalink]

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21 Sep 2018, 09:21
If p is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, the we have p = 1/91 + 1/92 + 1/93 ... + 1/100 [Note that bigger the denominator (assuming the numerator is a constant, i.e. 1 in this case) then smaller the value fraction; therefore we have 1/91 > 1/92 > 1/93 ... > 1/100]

There are 10 terms in p. Therefore the smallest term is 1/100 and the biggest term is 1/91. The value of p will lie between 10 times both these values.

We have:-

10*(1/100) < p < 10*(1/91); or 1/10 < p < 1/9.1

If 1/10 < p : then 1/p is less than 10 [also 1/9.1 < p]

We can conclude that 10 is the only value which is more than 1/p.

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Re: If p is the sum of the reciprocals of the consecutive integers from 91 &nbs [#permalink] 21 Sep 2018, 09:21
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