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17 Jan 2010, 10:43
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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:
45%
(medium)
Question Stats:
67% (01:46) correct
33%
(02:04)
wrong
based on 1562
sessions
History
Date
Time
Result
Not Attempted Yet
If x(x - 5)(x + 2) = 0, is x negative?
(1) x^2 – 7x ≠ 0
(2) x^2 –2x –15 ≠ 0
(1) x^2 – 7x ≠ 0
(2) x^2 –2x –15 ≠ 0
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17 Jan 2010, 13:24
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If x(x - 5)(x + 2) = 0, is x negative?
\(x(x - 5)(x + 2) = 0\) --> \(x=0\) or \(x=5\) or \(x=-2\). Question is \(x<0\) or is \(x=-2\)?
(1) x^2 – 7x ≠ 0 --> \(x(x-7)\neq{0}\)--> \(x\neq{0}\) and \(x\neq{7}\), so \(x\) can be 5 or -2 (from the stem as \(x=0\) is out). Not sufficient.
(2) x^2 –2x –15 ≠ 0 --> \((x+3)(x-5)\neq{0}\) --> \(x\neq{-3}\) and \(x\neq{5}\), so \(x\) can be 0 or -2 (from the stem as \(x=5\) is out). Not sufficient.
(1)+(2) As \(x=0\) and \(x=5\) are out, only value left is \(x=-2\), so \(x\) is negative. Sufficient.
Answer: C.
\(x(x - 5)(x + 2) = 0\) --> \(x=0\) or \(x=5\) or \(x=-2\). Question is \(x<0\) or is \(x=-2\)?
(1) x^2 – 7x ≠ 0 --> \(x(x-7)\neq{0}\)--> \(x\neq{0}\) and \(x\neq{7}\), so \(x\) can be 5 or -2 (from the stem as \(x=0\) is out). Not sufficient.
(2) x^2 –2x –15 ≠ 0 --> \((x+3)(x-5)\neq{0}\) --> \(x\neq{-3}\) and \(x\neq{5}\), so \(x\) can be 0 or -2 (from the stem as \(x=5\) is out). Not sufficient.
(1)+(2) As \(x=0\) and \(x=5\) are out, only value left is \(x=-2\), so \(x\) is negative. Sufficient.
Answer: C.
04 Jul 2013, 08:23
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For the given equation to be equal to 0 x must either equal 0, 5 or -2.
Statement 1: Tells us that x cannot equal 0 or 7. But it could equal 5 or -2. Not sufficient
Statement 2: Tells us that x cannot equal 5 or -3. But it could equal -2 or 0.
Taken together x cannot equal 0 or 5. Therefore, x must equal -2. Answer: C
Statement 1: Tells us that x cannot equal 0 or 7. But it could equal 5 or -2. Not sufficient
Statement 2: Tells us that x cannot equal 5 or -3. But it could equal -2 or 0.
Taken together x cannot equal 0 or 5. Therefore, x must equal -2. Answer: C
