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26 Dec 2010, 14:28
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
39% (01:33) correct
61%
(01:14)
wrong
based on 203
sessions
History
Date
Time
Result
Not Attempted Yet
26 Dec 2010, 15:07
Kudos
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xy = z
1) y = 2z
=> x*2z = z
=> x*2z -z = 0
=> z(x*2 -1)= 0
z=0 or (x*2-1)=0 not sufficient
2)
y2=4
=> y=2
=> x*2=z
=> not sufficient
with 1+2)
y=2z & y=2
=>2z=2
=>z=1
z=1 for z(x*2 -1)= 0
=> 1(x*2-1) = 0
=> 2x-1=0
=> 2x = 1
=> x = 1/2
Sufficient
1) y = 2z
=> x*2z = z
=> x*2z -z = 0
=> z(x*2 -1)= 0
z=0 or (x*2-1)=0 not sufficient
2)
y2=4
=> y=2
=> x*2=z
=> not sufficient
with 1+2)
y=2z & y=2
=>2z=2
=>z=1
z=1 for z(x*2 -1)= 0
=> 1(x*2-1) = 0
=> 2x-1=0
=> 2x = 1
=> x = 1/2
Sufficient
26 Dec 2010, 15:46
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rxs0005
I guess it's y^2=4 in statement (2).
If xy = z, what is the value of x ?
(1) y =2z --> \(2zx=z\) --> \(z(2x-1)=0\) --> either \(z=0\) (and \(x\) can take any value) or \(x=\frac{1}{2}\) (and \(z\) can take any value). Not sufficient.
(2) y^2 = 4 --> \(y=2\) or \(y=-2\). Clearly insufficient.
(1)+(2) As \(y^2=4\) and \(y=2z\) then it's clear that \(z\neq{0}\), so from (1) must be true that \(x=\frac{1}{2}\). Sufficient.
Answer: C.
