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For numbers p, q, and r, (p × q × r) < 0 and Magoosh; Lele, [#permalink]
02 Dec 2018, 09:59

Expert's post

00:00

Question Stats:

66% (00:36) correct
33% (00:34) wrong based on 24 sessions

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

Re: For numbers p, q, and r, (p × q × r) < 0 and Magoosh; Lele, [#permalink]
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]
03 Dec 2018, 02:19

Expert's post

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, \(p*q>0.....A>B\)