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Which is greater 2/^3sqrt6 + 3/^3sqrt6 : Quantitative Comparison Questions

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Re: Which is greater 2/^3sqrt6 + 3/^3sqrt6 [#permalink]
2
Yet another answer: cuberoot (6) <2, so in A, the numerators are bigger than the denominator. In B, the denominators are bigger.

1/big = small, and 1/small = big

Originally posted by Chakolate on 30 Nov 2019, 16:56.
Last edited by Chakolate on 17 Dec 2019, 08:22, edited 1 time in total.
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Re: Which is greater 2/^3sqrt6 + 3/^3sqrt6 [#permalink]
Carcass wrote:
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.

Kudos for the right answer and explanation




\(\frac{2}{\sqrt[3]{6}} + \frac{3}{\sqrt[3]{6}}\) = \(\frac{\sqrt[3]{6}}{2} + \frac{\sqrt[3]{6}}{3}\)


=> \(\frac{5}{\sqrt[3]{6}}\) = \(\frac{\sqrt[3]{6} * 5 }{6}\) (as \(\frac{\sqrt[3]{6} * 3 + \sqrt[3]{6} * 2 } {6}\) = \(\frac{\sqrt[3]{6} * (3 + 2 = 5) } {6}\))

=> \(\frac{1}{\sqrt[3]{6}}\) = \(\frac{\sqrt[3]{6} * 1 }{6}\) (remove 5 from both sides)

So, A > B
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Re: Which is greater 2/^3sqrt6 + 3/^3sqrt6 [#permalink]
3
Carcass wrote:
Quantity A
Quantity B
\(\frac{2}{\sqrt[3]{6}} + \frac{3}{\sqrt[3]{6}}\)
\(\frac{\sqrt[3]{6}}{2} + \frac{\sqrt[3]{6}}{3}\)



Let's start by estimating the value of \(\sqrt[3]6\)

We know that:
\(\sqrt[3]{1} = 1\)
\(\sqrt[3]{8} = 2\)

Since 6 is BETWEEN 1 and 8, we know that \(\sqrt[3]6\) is BETWEEN 1 and 2.
We can say \(\sqrt[3]6\) = 1.something

We get:
QUANTITY A: (2/1.something + (3/1.something)
QUANTITY B: (1.something/2) + (1.something/3)

We can now use the fact that:
2/1.something = some number greater than 1
3/1.something = some number greater than 1
1.something/2 = some number less than 1
1.something/3 = some number less than 1
NOTE: We don't need to be any more precise than this

We now have:
QUANTITY A: (some number greater than 1) + (some number greater than 1)
QUANTITY B: (some number less than 1) + (some number less than 1)

Answer: A

Cheers,
Brent
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Re: Which is greater 2/^3sqrt6 + 3/^3sqrt6 [#permalink]
^3sqrt6 will be approximately 1.4 if we plug that in we can easily find the answer
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Re: Which is greater 2/^3sqrt6 + 3/^3sqrt6 [#permalink]
vyascd wrote:
^3sqrt6 will be approximately 1.4 if we plug that in we can easily find the answer


How did you get 1.4?
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Re: Which is greater 2/^3sqrt6 + 3/^3sqrt6 [#permalink]
1
Let's break it down logically

2^3 = 8
cube root of 8 = 2
cube root of 6 will be less than 2

In A --> Denominator is always less than numerator --> each fraction would be greater than 1

In B --> Denominator is always more than numerator --> each fraction would be less than 1

Hence, A is the Answer.
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Re: Which is greater 2/^3sqrt6 + 3/^3sqrt6 [#permalink]
1
The use of estimating values is helpful in this.
We know that the square root of 4 is 2.
Cube root of 8 is 2. Now we know that the cube root of 6 is less than 2.

Quantity A : 5/ (value less than 2) = value greater than 2

Quantity B
: (less than 2) / 2 + (less than 2) / 3 = value less than 1

Hence the answer is Quantity A
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Re: Which is greater 2/^3sqrt6 + 3/^3sqrt6 [#permalink]
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Re: Which is greater 2/^3sqrt6 + 3/^3sqrt6 [#permalink]
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