BTGmoderatorDC wrote:Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?
A. 1/2
B. 2
C. 3
D. 5
E. 6
Source: Manhattan Prep
$${A_{{\text{usual}}}}\,\,\, - \,\,\,1\,\,Job\,\,\, - \,\,\,\boxed{? = A}\,\,{\text{h}}$$
$${{\text{A}}_{{\text{doubled}}}} - 1\,\,Job\,\,\, - \,\,\,\frac{A}{2}\,\,{\text{h}}$$
$${B_{{\text{usual}}}}\,\,\, - \,\,\,1\,\,Job\,\,\, - \,\,\,B\,\,{\text{h}}$$
Let´s use the "work/rate shortcut":
$${1 \over {{T_{x \cup y}}}} = {1 \over {{T_x}}} + {1 \over {{T_y}}}$$
$$\left\{ \matrix{
\,\left( {\rm{I}} \right)\,\,\,\,{1 \over 3} = {1 \over A} + {1 \over B} \hfill \cr
\,\left( {{\rm{II}}} \right)\,\,\,{1 \over 2} = {2 \over A} + {1 \over B} \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {{\rm{II}}} \right) - \left( {\rm{I}} \right)} \,\,\,\,\,{1 \over 6} = {1 \over 2} - {1 \over 3} = {2 \over A} - {1 \over A} = {1 \over A}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = A = 6\,\,\,\,\left[ {\rm{h}} \right]$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
