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2+sqrt of 3 over (2-sqrt of 3)^3
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29 May 2024, 03:17
29 May 2024, 03:17
Expert Reply
2
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Question Stats:
85% (01:54) correct
15% (01:58) wrong based on 20 sessions
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Quantity A |
Quantity B |
\(\dfrac{2+\sqrt{3}}{(2-\sqrt{3})^3}\) |
\(\dfrac{10^6}{3^6}\) |
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Shorter GRE - Preparing for the Quantitative Reasoning Measure
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Free Materials for the GRE General Exam - Where to get it!!
Shorter GRE - Preparing for the Quantitative Reasoning Measure
Shorter GRE - Preparing for the Verbal Reasoning Measure
2+sqrt of 3 over (2-sqrt of 3)^3
[#permalink]
18 Nov 2024, 04:15
18 Nov 2024, 04:15
Expert Reply
We need to compare $\(\dfrac{2+\sqrt{3}}{(2-\sqrt{3})^3}\)$ with $\(\frac{10^6}{3^6}\)$
Column A has
$$
\( \frac{2+\sqrt{3}}{(2-\sqrt{3})^3}=\frac{2+\sqrt{3}}{(2-\sqrt{3})^3} \times \frac{(2+\sqrt{3})^3}{(2+\sqrt{3})^3}=\frac{(2+\sqrt{3})^2 \times(2+\sqrt{3})^2}{\left(2^2-\sqrt{3}^2\right)^3}=\frac{(7+4\sqrt{3})^2}{1^3}=(7+4 \sqrt{3})^2 \\
& =(7+4 \times 1.73)^2=(7+6.93)^2=(13.93)^2 \\
& \text { Column B } \frac{10^6}{3^6}=(3.33)^6=\left(3.33^3\right)^2=(36.93)^2
\)
$$
Clearly column B has greater quantity, so the answer is (B).
Column A has
$$
\( \frac{2+\sqrt{3}}{(2-\sqrt{3})^3}=\frac{2+\sqrt{3}}{(2-\sqrt{3})^3} \times \frac{(2+\sqrt{3})^3}{(2+\sqrt{3})^3}=\frac{(2+\sqrt{3})^2 \times(2+\sqrt{3})^2}{\left(2^2-\sqrt{3}^2\right)^3}=\frac{(7+4\sqrt{3})^2}{1^3}=(7+4 \sqrt{3})^2 \\
& =(7+4 \times 1.73)^2=(7+6.93)^2=(13.93)^2 \\
& \text { Column B } \frac{10^6}{3^6}=(3.33)^6=\left(3.33^3\right)^2=(36.93)^2
\)
$$
Clearly column B has greater quantity, so the answer is (B).
gmatclubot
2+sqrt of 3 over (2-sqrt of 3)^3 [#permalink]
18 Nov 2024, 04:15
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