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10 Nov 2021, 10:00
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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:
55%
(hard)
Question Stats:
65% (02:49) correct
35%
(02:21)
wrong
based on 79
sessions
History
Date
Time
Result
Not Attempted Yet
If \(a ≠ 0\), \(\frac{x}{y}\) = \(\frac{-1}{a}\), and \(\frac{y}{z}\) = \(\frac{b}{3}\), is \(\frac{x}{z}\) > \(\frac{1}{2}\)?
(1) \(a^2 - 2a − 3 = 0\)
(2) \(b^2 - 4b + 4 = 1\)
(1) \(a^2 - 2a − 3 = 0\)
(2) \(b^2 - 4b + 4 = 1\)
10 Nov 2021, 23:58
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Bunuel
\(\frac{x}{y}\) = \(\frac{-1}{a}\) and \(\frac{y}{z}\) = \(\frac{b}{3}\)
But we are looking for \(\frac{x}{z}\), so multiply the above equations
\(\frac{x}{y}\)*\(\frac{y}{z}\) = \(\frac{-1}{a}\)* \(\frac{b}{3}.............\frac{x}{z}=\frac{-b}{3a}\)
The question becomes: Is \(\frac{-b}{3a}>\frac{1}{2}..........\frac{b}{a}<-\frac{3}{2}\)
When we look at statements individually, they give values of a and b separately, so surely none is sufficient individually.
If you are aware of change of signs in equations, statement I tells you that we will get two values of a with opposite signs. SO, you can mark E and move ahead.
But let us solve it.
(1) \(a^2 - 2a − 3 = 0........a^2-3a+a-3=(a-3)(a+1)=0\)
Two values of a: 3 and -1
Nothing about b?
But two values with opposite sign of a should give you E as the answer.
(2) \(b^2 - 4b + 4 = 1..........(b-3)(b-1)=0\)
Two values of b: 3 and 1
Nothing about a?
Combined
If b=3 and a=3.......\(\frac{b}{a}<-\frac{3}{2}\) ............ \(\frac{3}{3}<-\frac{3}{2}\)...No
If b=3 and a=-1.......\(\frac{b}{a}<-\frac{3}{2}\) ............. \(\frac{3}{-1}<-\frac{3}{2}\)...Yes
Insufficient
12 Nov 2021, 11:00
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Bunuel
Breaking Down the Info:
We are interested in \(\frac{x}{z}\), to find that we can multiply the two given equations to get \(\frac{x}{z} = -\frac{b}{3a}\).
Rephrase the question:
Is \(-\frac{b}{3a} > \frac{1}{2}\)?
Statement 1 Alone:
This gives us \(a = 3\) or \(a = -1\). Without knowing b, this is insufficient.
Statement 2 Alone:
This gives us \(b = 3\) or \(b = -1\). Without knowing a, this is insufficient.
Both Statements Combined:
Try to maximize \(-\frac{b}{3a}\) first. If \(a = -1\) and \(b = 3\) then \(-\frac{b}{3a} > \frac{1}{2}\).
We can also make the left-side-expression negative to make it less than one-half, so combined it is still insufficient.
Answer: E
