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29 Oct 2021, 05:45
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
78% (01:10) correct
22%
(01:31)
wrong
based on 367
sessions
History
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If \(S = \frac{1}{10} + \frac{1}{20} + \frac{1}{30} + \frac{1}{40} + \frac{1}{50} + \frac{1}{60} + \frac{1}{70}\), which of the following must be true?
I. \(S < \frac{1}{10}\)
II. \(S > \frac{1}{70}\)
III. \(S > \frac{1}{40}\)
(A) I only
(B) II only
(C) I and II
(D) II and III
(E) I, II, and III
I. \(S < \frac{1}{10}\)
II. \(S > \frac{1}{70}\)
III. \(S > \frac{1}{40}\)
(A) I only
(B) II only
(C) I and II
(D) II and III
(E) I, II, and III
PyjamaScientist
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29 Oct 2021, 06:13
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Bunuel
Given, \(S = \frac{1}{10} + \frac{1}{20} + \frac{1}{30} + \frac{1}{40} + \frac{1}{50} + \frac{1}{60} + \frac{1}{70}\)
Take \(\frac{1}{10 }\) common out from the RHS. Cross Multiply by 10.
It gives,
\(10S = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}\)
This gives => \(10S = 1 + constant\) => \(10S >1\) -----------(1)
Compare answer choices with (1),
I. \(S < \frac{1}{10}\) => \(10S < 1\) FALSE
II. \(S > \frac{1}{70}\) => \(10S > \frac{1}{7}\) TRUE
III. \(S > \frac{1}{40}\) => \(10S > \frac{1}{4}\) TRUE
Option (D) is correct.
29 Oct 2021, 07:14
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Without solving, we observe that S =sum of 7 positive numbers.
Hence, S will always be greater than any of those individual number.
Thus 1 is out and 2 & 3are correct.
Option d is correct.
Q. E. D.
Posted from my mobile device
Hence, S will always be greater than any of those individual number.
Thus 1 is out and 2 & 3are correct.
Option d is correct.
Q. E. D.
Posted from my mobile device
