BTGmoderatorDC wrote:Set P consists of all the multiples of 4 from 12 to 52, inclusive. Set Q consists of 9 different integers drawn from set P. What is the average (arithmetic mean) of the integers in set Q?
(1) Set Q contains at most 4 consecutive multiples of 4.
(2) Set Q contains exactly 1 perfect square.
Source: Manhattan Prep
$$P = \left\{ {12,16,20,24, \ldots ,52} \right\}\,\,\,\,\,\,\left[ {16,36\,\,{\rm{perfect}}\,\,{\rm{squares}}} \right]$$
$$Q \subset P\,\,\,,\,\,\,\# Q = 9$$
$$? = {{\sum\nolimits_Q {} } \over 9}\,\,\,\,\, \Leftrightarrow \,\,\,\,\,? = \sum\nolimits_Q {} $$
$$\left( {1 + 2} \right)\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,{{\rm{Q}}_{\rm{1}}} = \left\{ {12,16,20,24} \right\} \cup \left\{ {32} \right\} \cup \left\{ {40,44,48,52} \right\}\,\,\,\, \Rightarrow \,\,\,\,? = \sum\nolimits_{{Q_1}} {} \hfill \cr
\,{\rm{Take}}\,\,{{\rm{Q}}_{\rm{2}}} = \left\{ {12,16} \right\} \cup \left\{ {24,28,32} \right\} \cup \left\{ {40, 44,48,52} \right\}\,\,\,\, \Rightarrow \,\,\,\,? = \sum\nolimits_{{Q_2}} \ne \sum\nolimits_{{Q_1}} {} \hfill \cr} \right.$$
The correct answer is therefore (E).
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio. 
