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GMAT Quant — Quant (Inequalities)

Question

Let m, n, and p be nonzero real numbers such that \frac{m ^{2}}{n} < 0 and n p < 0 . Which of the following must be negative?

Answer Choices

A. $mn$
B. $n p$
C. $m p$
D. $m^{2} p$
E. $p^{2} m$

Steps

Explanation	Calculations	Help
$We begin by noting the given that m, n, and p are nonzero real numbers.$	$m is not equal to 0, n is not equal to 0, p is not equal to 0$
$Since m is nonzero, m^{2} is always positive, and we are given \frac{m ^{2}}{n} < 0, this implies that n must be negative.$	$m^{2} > 0$
$Continuing from the same inequality, \frac{m ^{2}}{n} < 0 can only hold if n is less than zero.$	$\frac{m ^{2}}{n} < 0, so n < 0$
$We use the fact that if the product of two numbers is negative then one of them is positive and the other is negative; here n p < 0 and n < 0, so p must be positive.$	$n p < 0$
$Restating the conclusion from the prior logic:$	$p > 0$
$Let's test option A: mn . Since n < 0, the sign of mn is the opposite of the sign of m, so it can be either positive or negative. It is not necessarily negative.$	$m > 0, n < 0, so mn < 0$
$This demonstrates that mn can be negative when m > 0 and n < 0 .$	$m > 0, n < 0, so mn < 0$	Theory & Tactics Method Card PN2-C Click to view full details
$Continuing the test for option A:$	$m < 0, n < 0, so mn > 0$
$This demonstrates that mn can be positive when both m < 0 and n < 0 .$	$m < 0, n < 0, so mn > 0$	Theory & Tactics Method Card PN2-B Click to view full details
$Let's test option B: n p . We are given directly that n p < 0, so this product must be negative.$	$n p < 0$
$Let's test option C: m p . Since p > 0, m p has the same sign as m and so can be either positive or negative. It is not necessarily negative.$	$m > 0, p > 0, so m p > 0$
$This demonstrates that m p can be positive when m > 0 and p > 0 .$	$m > 0, p > 0, so m p > 0$	Theory & Tactics Method Card PN2-A Click to view full details
$Continuing the test for option C:$	$m < 0, p > 0, so m p < 0$
$This demonstrates that m p can be negative when m < 0 and p > 0 .$	$m < 0, p > 0, so m p < 0$	Theory & Tactics Method Card PN2-C Click to view full details
$Let's test option D: m^{2} p . We know m^{2} > 0 and p > 0, so their product is positive and cannot be negative.$	$m^{2} > 0, p > 0, so m^{2} p > 0$
$This confirms that m^{2} p is always positive.$	$m^{2} > 0, p > 0, so m^{2} p > 0$	Theory & Tactics Method Card PN2-A Click to view full details
$Let's test option E: p^{2} m . Since p^{2} > 0, the sign of p^{2} m is the same as the sign of m, so it can be positive or negative and is not necessarily negative.$	$p^{2} > 0$
$Continuing the test for option E:$	$m > 0, so p^{2} m > 0$
$This demonstrates that p^{2} m can be positive when m > 0 .$	$m > 0, so p^{2} m > 0$	Theory & Tactics Method Card PN2-A Click to view full details
$And if m < 0,$	$m < 0, so p^{2} m < 0$
$This demonstrates that p^{2} m can be negative when m < 0 .$	$m < 0, so p^{2} m < 0$	Theory & Tactics Method Card PN2-C Click to view full details

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Final Answer

B

GMAT Quant — Quant (Inequalities)

Question

Let m, n, and p be nonzero real numbers such that \frac{m ^{2}}{n} < 0 and n p < 0 . Which of the following must be negative?

Answer Choices

A. $mn$
B. $n p$
C. $m p$
D. $m^{2} p$
E. $p^{2} m$

Steps

Explanation	Calculations	Help
$We begin by noting the given that m, n, and p are nonzero real numbers.$	$m is not equal to 0, n is not equal to 0, p is not equal to 0$
$Since m is nonzero, m^{2} is always positive, and we are given \frac{m ^{2}}{n} < 0, this implies that n must be negative.$	$m^{2} > 0$
$Continuing from the same inequality, \frac{m ^{2}}{n} < 0 can only hold if n is less than zero.$	$\frac{m ^{2}}{n} < 0, so n < 0$
$We use the fact that if the product of two numbers is negative then one of them is positive and the other is negative; here n p < 0 and n < 0, so p must be positive.$	$n p < 0$
$Restating the conclusion from the prior logic:$	$p > 0$
$Let's test option A: mn . Since n < 0, the sign of mn is the opposite of the sign of m, so it can be either positive or negative. It is not necessarily negative.$	$m > 0, n < 0, so mn < 0$
$This demonstrates that mn can be negative when m > 0 and n < 0 .$	$m > 0, n < 0, so mn < 0$	Theory & Tactics Method Card PN2-C Click to view full details
$Continuing the test for option A:$	$m < 0, n < 0, so mn > 0$
$This demonstrates that mn can be positive when both m < 0 and n < 0 .$	$m < 0, n < 0, so mn > 0$	Theory & Tactics Method Card PN2-B Click to view full details
$Let's test option B: n p . We are given directly that n p < 0, so this product must be negative.$	$n p < 0$
$Let's test option C: m p . Since p > 0, m p has the same sign as m and so can be either positive or negative. It is not necessarily negative.$	$m > 0, p > 0, so m p > 0$
$This demonstrates that m p can be positive when m > 0 and p > 0 .$	$m > 0, p > 0, so m p > 0$	Theory & Tactics Method Card PN2-A Click to view full details
$Continuing the test for option C:$	$m < 0, p > 0, so m p < 0$
$This demonstrates that m p can be negative when m < 0 and p > 0 .$	$m < 0, p > 0, so m p < 0$	Theory & Tactics Method Card PN2-C Click to view full details
$Let's test option D: m^{2} p . We know m^{2} > 0 and p > 0, so their product is positive and cannot be negative.$	$m^{2} > 0, p > 0, so m^{2} p > 0$
$This confirms that m^{2} p is always positive.$	$m^{2} > 0, p > 0, so m^{2} p > 0$	Theory & Tactics Method Card PN2-A Click to view full details
$Let's test option E: p^{2} m . Since p^{2} > 0, the sign of p^{2} m is the same as the sign of m, so it can be positive or negative and is not necessarily negative.$	$p^{2} > 0$
$Continuing the test for option E:$	$m > 0, so p^{2} m > 0$
$This demonstrates that p^{2} m can be positive when m > 0 .$	$m > 0, so p^{2} m > 0$	Theory & Tactics Method Card PN2-A Click to view full details
$And if m < 0,$	$m < 0, so p^{2} m < 0$
$This demonstrates that p^{2} m can be negative when m < 0 .$	$m < 0, so p^{2} m < 0$	Theory & Tactics Method Card PN2-C Click to view full details

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Final Answer

B