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x>0, y>0, z>0
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Updated on: 07 Oct 2020, 03:34
Updated on: 07 Oct 2020, 03:34
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37% (00:52) wrong based on 107 sessions
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\(x>0 \)
\(y>0\)
\(z>0\)
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
\(y>0\)
\(z>0\)
Quantity A |
Quantity B |
\(\frac{\sqrt{yx} \sqrt{x} }{\sqrt{y}}\) |
\(\frac{\sqrt{z}}{\sqrt{x^{-2}z}}\) |
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
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x>0, y>0, z>0
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07 Oct 2020, 05:39
07 Oct 2020, 05:39
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Farina wrote:
\(x>0 \)
\(y>0\)
\(z>0\)
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
\(y>0\)
\(z>0\)
Quantity A |
Quantity B |
\(\frac{\sqrt{yx} \sqrt{x} }{\sqrt{y}}\) |
\(\frac{\sqrt{z}}{\sqrt{x^{-2}z}}\) |
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
We can solve this question using matching operations (my favorite Quantitative Comparison strategy)
Key property: \((\sqrt{x})(\sqrt{y}) = \sqrt{xy}\)
Given:
Quantity A: \(\frac{\sqrt{yx} \sqrt{x} }{\sqrt{y}}\)
Quantity B: \(\frac{\sqrt{z}}{\sqrt{x^{-2}z}}\)
Multiply both quantities by \(\sqrt{y}\) to get:
Quantity A: \((\sqrt{yx})(\sqrt{x})\)
Quantity B: \(\frac{(\sqrt{z})(\sqrt{y})}{\sqrt{x^{-2}z}}\)
Multiply both quantities by \(\sqrt{x^{-2}z}\) to get:
Quantity A: \((\sqrt{yx})(\sqrt{x})(\sqrt{x^{-2}z})\)
Quantity B: \((\sqrt{z})(\sqrt{y})\)
Apply the above property to get:
Quantity A: \(\sqrt{(yx)(x)(x^{-2}z)}\)
Quantity B: \(\sqrt{yz}\)
Simplify:
Quantity A: \(\sqrt{yz}\)
Quantity B: \(\sqrt{yz}\)
Answer: C
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Re: x>0, y>0, z>0
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07 Oct 2020, 04:06
07 Oct 2020, 04:06
Thanks @Carcass for editing. I was confused in this question but while typing the confusion I got my answer 
