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jlgdr
Joined: 06 Sep 2013
Last visit: 24 Jul 2015
Posts: 1,311
Given Kudos: 355
Concentration: Finance
Schools: Wharton '17 (A) HBS '17 (WA$)
28 Sep 2013, 16:32
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Of the following sums, which is the greatest?
(A) \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{3}}\) + \(\frac{1}{\sqrt{4}}\) + \(\frac{1}{\sqrt{5}}\)
(B) \(\frac{1}{2^2}\) + \(\frac{1}{3^2}\) + \(\frac{1}{4^2}\) + \(\frac{1}{5^2}\)
(C) \(\frac{1}{2^2}\) + \(\frac{1}{2^3}\) + \(\frac{1}{2^2}\) + \(\frac{1}{2^5}\)
(D) 1- \(\frac{1}{2}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\)
(E) \(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) + \(\frac{1}{5}\)
(A) \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{3}}\) + \(\frac{1}{\sqrt{4}}\) + \(\frac{1}{\sqrt{5}}\)
(B) \(\frac{1}{2^2}\) + \(\frac{1}{3^2}\) + \(\frac{1}{4^2}\) + \(\frac{1}{5^2}\)
(C) \(\frac{1}{2^2}\) + \(\frac{1}{2^3}\) + \(\frac{1}{2^2}\) + \(\frac{1}{2^5}\)
(D) 1- \(\frac{1}{2}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\)
(E) \(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) + \(\frac{1}{5}\)
29 Sep 2013, 10:09
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jlgdr
\(\sqrt{2}<2<2^2\), thus \(\frac{1}{\sqrt{2}}>\frac{1}{2}>\frac{1}{2^2}\).
Answer: A.
missionphd
GMAT 1: 720 Q50 V36
Posts: 52
12 May 2014, 20:24
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Lets compare A & E first,
sqrt(2)= 1.4 which is <2
As we know,
1/smaller > 1/larger
=>1/sqrt(2) > 1/2
so will be the case for 3,4,5
So clearly A is greater than E,
D<E, clearly because there are negative terms
B & C increase the denominators by increasing the powers, so again B & C are even lesser than E
Hence, A is the greatest
Hope this helps. Hit kudos it it does.
sqrt(2)= 1.4 which is <2
As we know,
1/smaller > 1/larger
=>1/sqrt(2) > 1/2
so will be the case for 3,4,5
So clearly A is greater than E,
D<E, clearly because there are negative terms
B & C increase the denominators by increasing the powers, so again B & C are even lesser than E
Hence, A is the greatest
Hope this helps. Hit kudos it it does.
