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27 Oct 2021, 05:15
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
68% (02:26) correct
32%
(02:18)
wrong
based on 755
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If \(x^3y<0\), and \(\frac{y}{z}>0\), then which of the following must be less than 1?
A. \(\sqrt[3]{x}\)
B. \(\frac{y}{x^2}\)
C. \(x^3z^4\)
D. \(x^2yz\)
E. \(xy^2z^3\)
A. \(\sqrt[3]{x}\)
B. \(\frac{y}{x^2}\)
C. \(x^3z^4\)
D. \(x^2yz\)
E. \(xy^2z^3\)
27 Oct 2021, 05:47
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We need x^3y<0 and y/z>0.
For x^3y<0 we need either x<0 and y>0 or x>0 and y<0.
For y/z>0 we need either y and z positive or y and z negative.
We need both x^3y<0 and y/z>0 so we need to combine our requirements above, like so:
x^3y<0 and y/z>0
if x<0, y>0 and z>0 (let's call it scenario 1, or s1)
or x>0, y<0 and z<0 (scenario 2, s2)
Let's take a look at the possible answers. We are first confused with the "less than 1" part of the question, but in our requirements we have only found whether x, y and z are positive or negative, so we are possibly looking at "positive or negative" results.
A. In s1, this is negative, but is positive in s2. Inconclusive.
B. Denominator is always positive, so it depends on the sign of y, which depends on the scenario. Inconclusive.
C. z^4 is always positive so it depends on the sign of x, which depends on the scenario. Inconclusive.
D. x^2 is always positive so it depends on the signs of y and z. y and z must both be positive (s1) or both negative (s2), which means that yz is always positive. Not our answer.
E. y^2 is positive. It will depend on the signs of x and z. In s1, x>0 and z<0. In s2, x<0 and z>0. This means that in both scenarios, xz is negative. xy^2z^3 is always negative (and also less than 1) so this is our answer.
ANSWER E
For x^3y<0 we need either x<0 and y>0 or x>0 and y<0.
For y/z>0 we need either y and z positive or y and z negative.
We need both x^3y<0 and y/z>0 so we need to combine our requirements above, like so:
x^3y<0 and y/z>0
if x<0, y>0 and z>0 (let's call it scenario 1, or s1)
or x>0, y<0 and z<0 (scenario 2, s2)
Let's take a look at the possible answers. We are first confused with the "less than 1" part of the question, but in our requirements we have only found whether x, y and z are positive or negative, so we are possibly looking at "positive or negative" results.
A. In s1, this is negative, but is positive in s2. Inconclusive.
B. Denominator is always positive, so it depends on the sign of y, which depends on the scenario. Inconclusive.
C. z^4 is always positive so it depends on the sign of x, which depends on the scenario. Inconclusive.
D. x^2 is always positive so it depends on the signs of y and z. y and z must both be positive (s1) or both negative (s2), which means that yz is always positive. Not our answer.
E. y^2 is positive. It will depend on the signs of x and z. In s1, x>0 and z<0. In s2, x<0 and z>0. This means that in both scenarios, xz is negative. xy^2z^3 is always negative (and also less than 1) so this is our answer.
ANSWER E
27 Oct 2021, 06:25
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Inequalities like x^3 y < 0 and y/z > 0 only tell us about the signs of x, y and z. The inequality x^3 y < 0 just means x and y have opposite signs, and y/z > 0 means y and z have the same sign (and since x and y have opposite signs, x and z must also have opposite signs). Since we know nothing at all about how big or small x, y and z are, we have no chance of determining how big, in absolute value, any of the answer choices are. So if we can determine that one of the answers is less than 1, it will only be because we can determine that one of the answer choices is negative; we'd never be able, with the information in the question, to determine that one of the answers was between 0 and 1. And since x and z have opposite signs, xz must be negative, and if we multiply xz by y^2 z^2, which is positive, we'll still get a negative quantity, so x y^2 z^3 must be negative and the answer is E.
