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For numbers p, q, and r, (p × q × r) < 0

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Carcass
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For numbers p, q, and r, (p × q × r) < 0 [#permalink] New post   02 Dec 2018, 09:59
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For numbers \(p\), \(q\), and \(r\), \((p * q * r) < 0\) and \(\frac{(p * q)^2}{r} < 0\)

Quantity A
Quantity B
\(p * q\)
\(0\)



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
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A
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Fatemeh
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Re: For numbers p, q, and r, (p × q × r) < 0 and Magoosh; Lele, [#permalink] New post   02 Dec 2018, 20:36
Answer: A
We have (p*q)^2 / r < 0. Since anything in power of 2 or any even is positive then as this expression is negative we can construe that r must be negative.

We have p*q*r <0 so either one or 3 of them are negative.
We know that r is negative. Two approaches are possible either r is negative and p&q are positive, then (p*q) is positive and The answer is A. Or the two of p&q are negative then (p*q) is again positive and bigger than 0, So, the answer is A.
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Re: For numbers p, q, and r, (p × q × r) < 0 and Magoosh; Lele, [#permalink] New post   03 Dec 2018, 02:19
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Carcass wrote:
For numbers \(p\), \(q\), and \(r\), \((p * q * r) < 0\) and \(\frac{(p * q)^2}{r} < 0\)

Quantity A
Quantity B
\(p * q\)
\(0\)



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


\(\frac{(p * q)^2}{r} < 0\)..
since \((p*q)^2\) is positive, r<0...

Now \(p*q*r<0.....(p*q)*r<0\)..
As we know r<0, \(p*q>0.....A>B\)

A
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Re: For numbers p, q, and r, (p × q × r) < 0 [#permalink] New post   26 Sep 2023, 18:38
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: For numbers p, q, and r, (p × q × r) < 0 [#permalink]
26 Sep 2023, 18:38
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