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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, 02:24
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Difficulty:

55% (hard)

Question Stats:

58% (01:38) correct 42% (01:21) wrong based on 120 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, 09:04.
Last edited by Bunuel on 15 Sep 2018, 02:24, edited 1 time in total.
Renamed the topic and edited the question.
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Joined: 26 Jul 2018
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, 09:32
2
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
Intern
Joined: 23 Oct 2017
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, 10:11
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, 18:59
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, 02:25
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

Similar questions to practice:
http://gmatclub.com/forum/m-is-the-sum- ... 43703.html
http://gmatclub.com/forum/if-k-is-the-s ... 45365.html
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, 10:05
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, 08:21
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  [#permalink]

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17 Nov 2018, 14:37
Hi All,

To start, this question is a 'lift' of an Official question (and it's based on the exact same concepts):
https://gmatclub.com/forum/if-s-is-the- ... 24847.html

Based on the wording of the prompt, you might think that you should add up the fractions 1/91 + 1/92 + .... 1/100, but the GMAT would NEVER require that you do that math. Instead, lets do some real basic estimation of what that sum would be LESS than and GREATER than.... Note that we're asked which of the three fractions is greater than 1/P...

There are 10 total fractions and 9 of them are GREATER than 1/100. So, at the 'lower end', let's just say that all 10 fractions are equal to 1/100 each....

(10)(1/100) = 10/100 = 1/10

Thus, we know that the sum of those 10 fractions will be GREATER than 1/10.

Similarly, we know that all 10 of those fractions are LESS than 1/90. So, at the 'higher end', let's just say that all 10 fractions are equal to 1/90 each...

(10)(1/90) = 10/90 = 1/9

Thus, we know that the sum of those 10 fractions will be LESS than 1/9.

Finally, we have to consider the value of 1/P. In simple terms, this requires us to 'invert' the fraction. Since 1/10 < P < 1/9, we know that 10 > 1/P > 9. Thus, only one of the three answers 'fits'....

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Re: If p is the sum of the reciprocals of the consecutive integers from 91 &nbs [#permalink] 17 Nov 2018, 14:37
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