BTGmoderatorDC wrote:If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned, but if he buys 100 more chickens, he will run out of feed 15 days earlier than planned. If no chickens are sold or bought, the farmer will be exactly on schedule. How many chickens does the farmer have?
A. 60
B. 120
C. 240
D. 275
E. 300
Source: Manhattan Prep
$$\left( {\# {\rm{chicken}}} \right)\left( {\# days} \right) = {\rm{constant}}$$
$$? = x = \# {\rm{chicken}}$$
$$\left\{ \matrix{
\left( {x - 75} \right)\left( {d + 20} \right) = \left( {x + 100} \right)\left( {d - 15} \right)\,\,\,\,\left( {\rm{I}} \right) \hfill \cr
\left( {x - 75} \right)\left( {d + 20} \right) = xd\,\,\,\,\left( {{\rm{II}}} \right) \hfill \cr} \right.$$
$$\left( {\rm{I}} \right)\,\,\, \Rightarrow \,\,\,xd + 20x - 75d - 75 \cdot 20 = xd - 15x + 100d - 15 \cdot 100$$
$$ \Rightarrow \,\,\,35x - 175d = 20\left( {75 - 5 \cdot 15} \right) = 0\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,35} \,\,\,\,\,x = 5d\,\,\,\left( {{\rm{III}}} \right)$$
$$\left( {{\rm{II}}} \right)\,\,\, \Rightarrow \,\,\,xd + 20x - 75d - 75 \cdot 20 = xd\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {{\rm{III}}} \right)} \,\,\,\,\,20x - {{75 \cdot x} \over 5} = 75 \cdot 20$$
$$ \Rightarrow \,\,\,5x = 75 \cdot 20\,\,\,\,\, \Rightarrow \,\,\,\,\,x = 15 \cdot 20 = 300\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\rm{E}} \right)$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio. 
