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

Question

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

quantumdweller · Accepted Answer

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

samartharya · Answer

best way to solve the problem.

Nikhil · Answer

patnaiksonal Please go through the following post and provide an appropriate topic name. Refer #3 https://gmatclub.com/forum/rules-for-po ... 33935.html

Bunuel · Answer

Similar questions to practice: https://gmatclub.com/forum/m-is-the-sum- ... 43703.html https://gmatclub.com/forum/if-k-is-the-s ... 45365.html https://gmatclub.com/forum/if-s-is-the- ... 24847.html

Princ · Answer

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.

CAMANISHPARMAR · Answer

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

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

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

Therefore correct answer C

EMPOWERgmatRichC · Answer

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'....

Final Answer:

Show Spoiler

C

GMAT assassins aren't born, they're made,
Rich

MagnusSamIBA · Answer

The goal is to find the reciprocal of the sum of the reciprocals from 91 to 100, then determine which value is greater than its reciprocal.

Estimate the sum of reciprocals, p:

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

Since each term is roughly between 1/91 and 1/100, we can make a quick estimate by approximating each reciprocal as close to 1/95, the middle of the range.
So, we can approximate:

p ≈ 10 * (1/95) = 10/95 = 2/19.

Find the reciprocal of p, p^(-1):

Now,  p ≈ 2/19, so p^(-1) ≈ 19/2 = 9.5.

This means p^(-1) is somewhere between 9 and 10.

Looking at the options, we need to identify which value is greater than p^(-1). Since 9.5 is between 9 and 10, only 10 is greater than p^(-1).

Answer: C (Only III)

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

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

Prep Toolkit

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