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29 Jun 2022, 01:30
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
71% (01:04) correct
29%
(00:54)
wrong
based on 21
sessions
History
Date
Time
Result
Not Attempted Yet
Is \(xz = 3\)?
(1) \(\frac{z}{4} = \frac{3}{x}\)
(2) \(x^2yz = 3xy\)
Are You Up For the Challenge: 700 Level Questions
(1) \(\frac{z}{4} = \frac{3}{x}\)
(2) \(x^2yz = 3xy\)
Are You Up For the Challenge: 700 Level Questions
29 Jun 2022, 02:02
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1) Since x and z are part of opposite side of fractions and both fractions are positive so either x and z are both positive or both negative so product will be positive
xz = 12 which is a definite answer that its not 3
SUFFICIENT
2) x^2yz = 3xy
We cannot divide both sides by x and y since we are not given that they are not equal to 0 and dividing by 0 is undefined.
x^2yz - 3xy = 0
xy (xz - 3) = 0
xy = 0 or xz = 3
So if y is 0 then xz can be 3
But if x is 0 then xz is also 0
NOT SUFFICIENT
Answer : A
Posted from my mobile device
xz = 12 which is a definite answer that its not 3
SUFFICIENT
2) x^2yz = 3xy
We cannot divide both sides by x and y since we are not given that they are not equal to 0 and dividing by 0 is undefined.
x^2yz - 3xy = 0
xy (xz - 3) = 0
xy = 0 or xz = 3
So if y is 0 then xz can be 3
But if x is 0 then xz is also 0
NOT SUFFICIENT
Answer : A
Posted from my mobile device
29 Jun 2022, 02:29
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(1)Only
xz=12
We can definitely answer "No" to the question posed.
(2)Only
Tip: While we should not include (1) in evaluating (2)Only, we know (1) and (2) should not contradict with each other. So, definitely xz=12 is true.
Now, with x^2*y*z=3xy
If we assume x, y, z are non-zero numbers, we actually got xz=3
So xz can be 3, but xz can (actually is) be 12.
Not sufficient.
The answer is thus (A).
xz=12
We can definitely answer "No" to the question posed.
(2)Only
Tip: While we should not include (1) in evaluating (2)Only, we know (1) and (2) should not contradict with each other. So, definitely xz=12 is true.
Now, with x^2*y*z=3xy
If we assume x, y, z are non-zero numbers, we actually got xz=3
So xz can be 3, but xz can (actually is) be 12.
Not sufficient.
The answer is thus (A).
