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Re: rs = \sqrt 3 [#permalink]
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I am not quite sure where is the division indicated by you Sir

Could you point out ??
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rs = \sqrt 3 [#permalink]
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condition: \(rs=\sqrt{3}\)

Quantity A:\(\frac{2r \sqrt{12}}{r^2s \sqrt{72}}\) = \(\frac{2r \sqrt{12}}{rrs \sqrt{72}}\) = \(\frac{2r \sqrt{4 \times 3}}{r \sqrt{3} \sqrt{9 \times 8}}\) = \(\frac{2.2 \sqrt{3}}{\sqrt{3}.3.2\sqrt{2}}\) = \(\frac{2}{3\sqrt{2}}\) = \(\frac{\sqrt{2}}{3}\)


Quantity B: \(\frac{14rs^2}{42s}\) = \(\frac{14rss}{42s}\) = \(\frac{14rs}{42}\) = \(\frac{14rs}{7 \times 6}\) = \(\frac{2rs}{6}\) = \(\frac{2\sqrt{3}}{2 \times 3}\) = \(\frac{\sqrt{3}}{3}\)


Clearly Quantity B is greater.
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rs = \sqrt 3 [#permalink]
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Aprazors wrote:
GreenlightTestPrep wrote:
Carcass wrote:
\(rs = \sqrt{3}\)

Quantity A
Quantity B
\(\frac{2r \sqrt{12}}{r^2s \sqrt{72}}\)
\(\frac{14rs^2}{42s}\)


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.


\(\sqrt{12}=(\sqrt{4})(\sqrt{3})=2\sqrt{3}\)
\(\sqrt{72}=(\sqrt{36})(\sqrt{2})=6\sqrt{2}\)

We get:
QUANTITY A: \(\frac{2r \sqrt{12}}{r^2s \sqrt{72}}=\frac{(2r)(2 \sqrt{3})}{(r)(rs)(6\sqrt{2})}=\frac{(4r)(\sqrt{3})}{(r)(\sqrt{3})(6\sqrt{2})}=\frac{4}{6\sqrt{2}}=\frac{2}{3\sqrt{2}}\)

QUANTITY B: \(\frac{14rs^2}{42s}=\frac{14(rs)(s)}{42s}=\frac{rs}{3}=\frac{\sqrt{3}}{3}\)

At this point, we can use matching operations

Multiply both quantities by 3 to get:
QUANTITY A: \(\frac{2}{\sqrt{2}}\)

QUANTITY B: \(\sqrt{3}\)



Multiply both quantities by \(\sqrt{2}\) to get:
QUANTITY A: \(2\)

QUANTITY B: \(\sqrt{6}\)

Since \(\sqrt{6}>2\), the correct answer is B


RELATED VIDEO FROM MY COURSE




for quantity A why not just divide by 12 to 72 so it will become root(1/6)



Yes. You can follow that route too. Then the fraction becomes \(\frac{2r\sqrt{1}}{r^2s.\sqrt{6}} = \frac{2}{rs.\sqrt{6}} = \frac{2}{\sqrt{3}.\sqrt{6}} = \frac{2}{\sqrt{18}} = \frac{2}{\sqrt{9 \times 2}} = \frac{2}{3.\sqrt{2}} = \frac{\sqrt{2}}{3}\)

You will get the same answer.
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