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27 Sep 2018, 00:55
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
59% (01:43) correct
41%
(01:19)
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What is the standard deviation of \(a, b\) and \(c\)?
(1) \(a^2+b^2+c^2 = 77\)
(2) \(a+b+c =15\)
(1) \(a^2+b^2+c^2 = 77\)
(2) \(a+b+c =15\)
27 Sep 2018, 16:23
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MathRevolution
Very nice problem, congrats Max!
\(? = \sigma \left( {a,b,c} \right)\)
\(\left( 1 \right)\,\,{a^2} + {b^2} + {c^2} = 77\,\,\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {a;b;c} \right) = \left( {\sqrt {\frac{{77}}{3}} \,;\sqrt {\frac{{77}}{3}} \,;\sqrt {\frac{{77}}{3}} \,} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{0}}\,\, \hfill \\\\
\,\,{\text{Take}}\,\,\left( {a;b;c} \right) = \left( {\sqrt {77} \,;0\,;0\,} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{\text{?}}\,\, \ne \,\,{\text{0}}\,\,\,\, \hfill \\ \\
\end{gathered} \right.\)
\(\left( 2 \right)\,\,a + b + c = 15\,\,\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {a;b;c} \right) = \left( {5\,;5\,;5\,} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{0}}\,\, \hfill \\\\
\,\,{\text{Take}}\,\,\left( {a;b;c} \right) = \left( {15\,;0\,;0\,} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{\text{?}}\,\, \ne \,\,{\text{0}}\,\,\,\, \hfill \\ \\
\end{gathered} \right.\)
\(\left( {1 + 2} \right)\,\,\,\,\mu = \frac{{a + b + c}}{3} = 5\)
\(? = \sqrt {\frac{{{{\left( {a - \mu } \right)}^2} + {{\left( {b - \mu } \right)}^2} + {{\left( {c - \mu } \right)}^2}}}{3}} \,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\boxed{\,\,\,? = {{\left( {a - 5} \right)}^2} + {{\left( {b - 5} \right)}^2} + {{\left( {c - 5} \right)}^2}\,\,\,}\)
\(?\,\,\, = \,\,\,{\left( {a - 5} \right)^2} + {\left( {b - 5} \right)^2} + {\left( {c - 5} \right)^2}\,\,\, = \,\,\,\,\underbrace {{a^2} + {b^2} + {c^2}}_{77} - 10\underbrace {\left( {a + b + c} \right)}_{15} + 3 \cdot 25\,\,\,\,\,{\text{unique}}\)
The correct answer is (C), indeed.
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
30 Sep 2018, 22:51
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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.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
The mean of \(a, b\) and \(c\) is \(m = \frac{( a + b + c )}{3}\). So, \(a + b + c = 3m.\)
The variance of a, b and c, that is, the square of the standard deviation of a, b and c, is given by
\(VAR = SD^2 = \frac{{ ( a – m )^2 + ( b – m )^2 + ( c – m )^2 }}{3}\)
\(= \frac{{ a^2 - 2am + m^2 + b^2 - 2bm + b^2 + c^2 - 2cm + m^2 }}{3}\)
\(= \frac{{ a^2 + b^2 + c^2 – 2m(a+b+c) + 3m^2 }}{3}\)
\(= \frac{{ a^2 + b^2 + c^2 – 2m*3m + 3m^2 }}{3}\)
\(= \frac{{ a^2 + b^2 + c^2 – 3m^2 }}{3}\)
\(= \frac{( a^2 + b^2 + c^2 )}{3} – m^2\)
\(= \frac{( a^2 + b^2 + c^2 )}{3} – { \frac{( a + b + c )}{3} }^2\)
This value can be calculated using the information given in conditions 1) and 2). Thus, both conditions together are sufficient.
Therefore, C is the answer.
Answer: C
When we have a sum of data values and a sum of the squares of data values, we can always calculate the standard deviation.
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.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
The mean of \(a, b\) and \(c\) is \(m = \frac{( a + b + c )}{3}\). So, \(a + b + c = 3m.\)
The variance of a, b and c, that is, the square of the standard deviation of a, b and c, is given by
\(VAR = SD^2 = \frac{{ ( a – m )^2 + ( b – m )^2 + ( c – m )^2 }}{3}\)
\(= \frac{{ a^2 - 2am + m^2 + b^2 - 2bm + b^2 + c^2 - 2cm + m^2 }}{3}\)
\(= \frac{{ a^2 + b^2 + c^2 – 2m(a+b+c) + 3m^2 }}{3}\)
\(= \frac{{ a^2 + b^2 + c^2 – 2m*3m + 3m^2 }}{3}\)
\(= \frac{{ a^2 + b^2 + c^2 – 3m^2 }}{3}\)
\(= \frac{( a^2 + b^2 + c^2 )}{3} – m^2\)
\(= \frac{( a^2 + b^2 + c^2 )}{3} – { \frac{( a + b + c )}{3} }^2\)
This value can be calculated using the information given in conditions 1) and 2). Thus, both conditions together are sufficient.
Therefore, C is the answer.
Answer: C
When we have a sum of data values and a sum of the squares of data values, we can always calculate the standard deviation.
