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Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
If x is the sum of the reciprocals of the consecutive integers from 51
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23 Sep 2021, 00:25
23 Sep 2021, 00:25
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If \(x\) is the sum of the reciprocals of the consecutive integers from 51 to 60, inclusive and \(y\) is the sum of the reciprocals of the consecutive integers from 61 to 70, inclusive, which of the following is correct?
I. \(\frac{1}{x} > 6\)
II. \(\frac{1}{y} > 7\)
III. \(\frac{1}{y} > \frac{1}{x}\)
(A) Only I
(B) Only II
(C) Only III
(D) Only II and III
(E) I, II and III
Source: Manhattan Review, GMAT Question Bank
I. \(\frac{1}{x} > 6\)
II. \(\frac{1}{y} > 7\)
III. \(\frac{1}{y} > \frac{1}{x}\)
(A) Only I
(B) Only II
(C) Only III
(D) Only II and III
(E) I, II and III
Source: Manhattan Review, GMAT Question Bank
Re: If x is the sum of the reciprocals of the consecutive integers from 51
[#permalink]
23 Sep 2021, 11:20
23 Sep 2021, 11:20
2
motion2020 wrote:
If \(x\) is the sum of the reciprocals of the consecutive integers from 51 to 60, inclusive and \(y\) is the sum of the reciprocals of the consecutive integers from 61 to 70, inclusive, which of the following is correct?
I. \(\frac{1}{x} > 6\)
II. \(\frac{1}{y} > 7\)
III. \(\frac{1}{y} > \frac{1}{x}\)
I. \(\frac{1}{x} > 6\)
II. \(\frac{1}{y} > 7\)
III. \(\frac{1}{y} > \frac{1}{x}\)
Given \(x = \frac{1}{51} + \frac{1}{52} + .... + \frac{1}{60}\)
Since there are 10 integers, minimum is \(\frac{1}{60}\) and maximum is \(\frac{1}{51}\), so the value of x lies between \(\frac{10}{51} > x > \frac{10}{60}\) and taking inverse \(\frac{51}{10} < \frac{1}{x} < \frac{60}{10}\)
which translates to \(5.1 < \frac{1}{x} < 6\)
Similarly for y, \( 6.1 < \frac{1}{y} < 7\)
So from the choices, only III is true.
gmatclubot
Re: If x is the sum of the reciprocals of the consecutive integers from 51 [#permalink]
23 Sep 2021, 11:20
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