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# Of the following sums, which is the greatest?

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Of the following sums, which is the greatest? [#permalink]  28 Sep 2013, 15:32
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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) $$\frac{1}{2}$$ + $$\frac{1}{3}$$ + $$\frac{1}{4}$$ + $$\frac{1}{5}$$
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Re: Of the following sums, which is the greatest? [#permalink]  29 Sep 2013, 09:09
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jlgdr wrote:
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) $$\frac{1}{2}$$ + $$\frac{1}{3}$$ + $$\frac{1}{4}$$ + $$\frac{1}{5}$$

$$\sqrt{2}<2<2^2$$, thus $$\frac{1}{\sqrt{2}}>\frac{1}{2}>\frac{1}{2^2}$$.

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Re: Of the following sums, which is the greatest? [#permalink]  12 May 2014, 17:12
Any other approach for this problem over here? I did not quite get your explanation Bunuel, how can we quickly eliminate the other answer choices?

Thanks!
Cheers
J
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Re: Of the following sums, which is the greatest? [#permalink]  12 May 2014, 19:24
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.
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Re: Of the following sums, which is the greatest? [#permalink]  12 May 2014, 19:47
Thanks missionphd.
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Re: Of the following sums, which is the greatest? [#permalink]  01 Aug 2014, 01:09
jlgdr wrote:
Any other approach for this problem over here? I did not quite get your explanation Bunuel, how can we quickly eliminate the other answer choices?

Thanks!
Cheers
J

One good thing of this question is numerator of all the expressions is 1

Now focus on denominators

More the value of denominator, less would be the sum

Least the denominator, more would be the sum value

$$\sqrt{2} = 2^{\frac{1}{2}}$$

$$2^{\frac{1}{2}} < 2^2$$

Reciprocal both sides

$$\frac{1}{2^{\frac{1}{2}}} > \frac{1}{2^2}$$

(Greater/smaller sign changes for reciprocals/negation)

Same for all other fractions

So highest addition would be of option A

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Re: Of the following sums, which is the greatest?   [#permalink] 01 Aug 2014, 01:09
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