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Re: x > y > 4 [#permalink]
1
GreenlightTestPrep wrote:
Carcass wrote:
x > y > 4


Quantity A
Quantity B
four divided by the product of x and y
\(\frac{xy}{4}\)


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 take "four divided by the product of x and y" and express it as \(\frac{4}{xy}\)

So our two quantities are:
QUANTITY A: \(\frac{4}{xy}\)
QUANTITY B: \(\frac{xy}{4}\)

Since \(x\) and \(y\) are both positive, we can be certain that the product \(xy\) is also positive, which means we can safely multiply both quantities by \(xy\) to get:
QUANTITY A: \(4\)
QUANTITY B: \(\frac{x^2y^2}{4}\)

To eliminate the remaining fraction, we can multiply both quantities by \(4\) to get:
QUANTITY A: \(16\)
QUANTITY B: \(x^2y^2\)

Since \(x > 4\), we know that \(x^2 > 16\)
Similarly, since \(y > 4\), we know that \(y^2 > 16\)

Since \(x^2\) and \(y^2\) are each greater than \(16\), the product \(x^2y^2\) is definitely greater than 16

Answer: B[/quote]
Can you explain why the answer can’t be D?

I used the choose numbers strategy.
I chose x = 1 and y = 2.

For Option A, I got 2 and for Option B, I got 1/2.

In another scenario, I chose x = 2 and y = 3.
For Option A, I got 4/6 and for Option B, I got 6/4.

Thank you

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