BTGmoderatorDC wrote:At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Source: Official Guide
$$? = b\,\,\left( {{\rm{back}}} \right)$$
$$10 \to 10 - b\,\,{\rm{fruits}}\,\,\,\left\{ \matrix{
\,{\rm{apples}}\,\, = a\,\,{\rm{units}}\,\,,\,\,40\,\,{\rm{cents}}\,\,{\rm{each}} \hfill \cr
\,{\rm{oranges}}\,\,{\rm{ = }}\,\,10 - a\,\,\, \to \,\,\,{\rm{10}} - b - a\,\,{\rm{units}}\,\,,\,\,60\,\,{\rm{cents}}\,\,{\rm{each}} \hfill \cr} \right.$$
$$\sum\nolimits_{10} {\,\, = \,\,\,10 \cdot 56 = 560\,\,\,{\rm{cents}}\,\,\,\,\,\,\left( {{\rm{homogeneity}}\,\,{\rm{nature}}\,\,{\rm{of}}\,\,{\rm{the}}\,\,{\rm{average}}} \right)} $$
$$\sum\nolimits_{10 - b} {\,\, = \,\,\,\left( {10 - b} \right) \cdot 52\,\,\,{\rm{cents}}\,\,\,\,\,\,\left( {{\rm{homogeneity}}\,\,{\rm{nature}}\,\,{\rm{of}}\,\,{\rm{the}}\,\,{\rm{average}}} \right)} $$
$$\sum\nolimits_{10 - b} {\,\, + b \cdot 60\,\,\, = } \,\,\,\,\sum\nolimits_{10} {\,\,\,\,\, \Rightarrow \,\,\,\,\,\,52\left( {10 - b} \right) + 60b = 560\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\, \ldots \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = b = 5} $$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
