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Which is greater 2/^3sqrt6 + 3/^3sqrt6

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Which is greater 2/^3sqrt6 + 3/^3sqrt6 [#permalink]  29 Nov 2019, 02:47
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Question Stats:

63% (01:28) correct 36% (01:32) wrong based on 22 sessions
 Quantity A Quantity B $$\frac{2}{\sqrt[3]{6}} + \frac{3}{\sqrt[3]{6}}$$ $$\frac{\sqrt[3]{6}}{2} + \frac{\sqrt[3]{6}}{3}$$

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

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[Reveal] Spoiler: OA

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Intern
Joined: 05 Oct 2019
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Re: Which is greater 2/^3sqrt6 + 3/^3sqrt6 [#permalink]  29 Nov 2019, 18:44
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Since there is no negative sign involved multiple both sides by the cube root of 6.
QA resolves to 5
QB becomes 5/cuberoot(6), since cuberoot(6)>1. QB <QA (thus option A)
Intern
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Re: Which is greater 2/^3sqrt6 + 3/^3sqrt6 [#permalink]  29 Nov 2019, 19:18
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I like the answer given by user Luffy, just an alternative solution here.

A: sum it and its results to 5/cuberoot(6). hold it there, we'll check quantity B.
B: cross multiply and solving gives 5/[cuberoot(6)*cueroot(6)].

Now, as cuberoot(6) >1, cuberoot(6)<[cuberoot(6)*cuberoot(6)] and these values go in the denominator, so 1/cuberoot(6) > 1/[cuberoot(6)*cuberoot(6)].
From this we can conclude that A > B, so the correct answer is option A.
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Re: Which is greater 2/^3sqrt6 + 3/^3sqrt6 [#permalink]  30 Nov 2019, 16:56
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Yet another answer: cuberoot (6) <2, so in A, the numberators are bigger than the denominator. In B, the denominators are bigger.

1/big = small, and 1/small = big
Re: Which is greater 2/^3sqrt6 + 3/^3sqrt6   [#permalink] 30 Nov 2019, 16:56
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