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20 Mar 2019, 01:52
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MathRevolution
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Since we have \(1\) variable (\(n\)) and \(0\) equations, D is most likely to be the answer. So, we should consider each condition on its own first.
Condition 1)
If \(n = 15\), then \(\sqrt{15n} = \sqrt{15^2} = 15\) is an integer and the answer is “yes”.
If \(n = 3\), then \(\sqrt{15n} = \sqrt{45} = 3\sqrt{5}\) is not an integer and the answer is “no”.
Thus, condition 1) is not sufficient, since it does not yield a unique solution.
Condition 2)
We have \(n = k^2\) for some integer \(k\).
So, \(\sqrt{15n} = \sqrt{15k^2}=k\sqrt{15}\) is not an integer.
Since ‘no’ is also a unique answer by CMT (Common Mistake Type) 1, condition 2) is sufficient.
Therefore, B is the answer.
Answer: B
If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
20 Mar 2019, 01:53
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[GMAT math practice question]
(number properties) Is \(n\) an odd number?
1) \(n\) and \(n+2\) are prime numbers
2) \(n+5\) is a multiple of \(5\)
(number properties) Is \(n\) an odd number?
1) \(n\) and \(n+2\) are prime numbers
2) \(n+5\) is a multiple of \(5\)
20 Mar 2019, 08:58
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MathRevolution
\(? = xy\)
\(\left( 1 \right)\,\,\,{\left( {x - 2} \right)^2} = - \left| {y - 3} \right|\,\,\,\,\,\left( * \right)\)
\(\left. \matrix{\\
{\left( {x - 2} \right)^2} \ge 0 \hfill \cr \\
- \left| {y - 3} \right| \le 0\,\, \hfill \cr} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\left\{ \matrix{\\
\,x - 2 = 0 \hfill \cr \\
\,y - 3 = 0 \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {x,y} \right) = \left( {2,3} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}{\rm{.}}\)
\(\left( 2 \right)\,\,\,\left| {x - 2} \right| + \sqrt {y - 3} \,\, = 0\,\,\,\,\,\,\left( {**} \right)\)
\(\left. \matrix{\\
\left| {x - 2} \right| \ge 0 \hfill \cr \\
\sqrt {y - 3} \ge 0\,\, \hfill \cr} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\left\{ \matrix{\\
\,x - 2 = 0 \hfill \cr \\
\,y - 3 = 0 \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {x,y} \right) = \left( {2,3} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}{\rm{.}}\)
The correct answer is (D).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
