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9x^2=y. x is an integer greater than 0.
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Updated on: 05 Oct 2020, 06:42
Updated on: 05 Oct 2020, 06:42
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\(9x^2=y\), and \(x\) is an integer greater than \(0\).
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.
Quantity A |
Quantity B |
\(x\) |
\(y\) |
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.
Re: 9x^2=y. x is an integer greater than 0.
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05 Oct 2020, 02:07
05 Oct 2020, 02:07
Expert Reply
\(x=\frac{\sqrt{y}}{3}\)
X is the result of a square root of Y divided by 3.
Therefore, Y MUST > x
B is the answer
Regards
X is the result of a square root of Y divided by 3.
Therefore, Y MUST > x
B is the answer
Regards
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9x^2=y. x is an integer greater than 0.
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05 Oct 2020, 05:58
05 Oct 2020, 05:58
1
Farina wrote:
\(9x^2=y\), and x is an integer greater than 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.
Quantity A |
Quantity B |
\(x\) |
\(y\) |
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)
First, replace \(y\) with \(9x^2\) to get:
Quantity A: \(x\)
Quantity B: \(9x^2\)
Since we're told \(x\) is POSITIVE, we can safely divide both quantities by \(x\) to get:
Quantity A: \(1\)
Quantity B: \(9x\)
Divide both quantities by \(9\) to get:
Quantity A: \(\frac{1}{9}\)
Quantity B: \(x\)
Since we're told x is an integer greater than 0, then x can equal 1, 2, 3, 4, ...etc
In every case, Quantity B is greater.
Answer: B
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