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10 Jan 2008, 00:13
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B
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:
76% (01:48) correct
24%
(01:53)
wrong
based on 181
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Is \(abc = 0\)?
(1) \(a^2 = 2a\)
(2) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\)
M03-19
(1) \(a^2 = 2a\)
(2) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\)
M03-19
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Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
26 Oct 2023, 09:00
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Official Solution:
Is \(abc = 0\)?
For \(abc = 0\) to be true, at least one of the variables must be zero.
(1) \(a^2 = 2a\).
Rearrange: \(a^2-2a=0\). Factor out \(a\): \(a(a-2)=0\). Either \(a=0\) or \(a=2\). If \(a=0\), then the answer is YES, but if \(a=2\), then \(abc\) may not be equal to zero (for example, consider: \(a=2\), \(b=3\), and \(c=4\)). This statement is not sufficient.
(2) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\)
\(b= \frac{(\sqrt{c}*(a+b))^2}{(a+b)^2} - c\)
\(b= \frac{(\sqrt{c})^2*(a+b)^2}{(a+b)^2} - c\)
\(b= \frac{c*(a+b)^2}{(a+b)^2} - c\);
\(b=c-c\);
\(b=0\).
Sufficient.
Answer: B
Is \(abc = 0\)?
For \(abc = 0\) to be true, at least one of the variables must be zero.
(1) \(a^2 = 2a\).
Rearrange: \(a^2-2a=0\). Factor out \(a\): \(a(a-2)=0\). Either \(a=0\) or \(a=2\). If \(a=0\), then the answer is YES, but if \(a=2\), then \(abc\) may not be equal to zero (for example, consider: \(a=2\), \(b=3\), and \(c=4\)). This statement is not sufficient.
(2) \(b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c\)
\(b= \frac{(\sqrt{c}*(a+b))^2}{(a+b)^2} - c\)
\(b= \frac{(\sqrt{c})^2*(a+b)^2}{(a+b)^2} - c\)
\(b= \frac{c*(a+b)^2}{(a+b)^2} - c\);
\(b=c-c\);
\(b=0\).
Sufficient.
Answer: B
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