Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

What is the standard deviation of a, b and c?

1) a^2+b^2+c^2 = 77
2) a+b+c =15

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 therefore [spoiler]__(C)_____[/spoiler] .

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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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 = ( 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 = { ( a - m )^2 + ( b - m )^2 + ( c - m )^2 } / 3
= { a^2 - 2am + m^2 + b^2 - 2bm + b^2 + c^2 - 2cm + m^2 } / 3
= { a^2 + b^2 + c^2 - 2m(a+b+c) + 3m^2 } / 3
= { a^2 + b^2 + c^2 - 2m*3m + 3m^2 } / 3
= { a^2 + b^2 + c^2 - 3m^2 } / 3
= ( a^2 + b^2 + c^2 ) / 3 - m^2
= ( a^2 + b^2 + c^2 ) / 3 - { ( 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.