During a certain production time, is the number of workers required to create w1 widgets at the rate of r1 widgets per minute per worker less than the number of workers required to create w2 widgets at the rate of r2 widgets per minute per worker?

(1) w1 is 20 less than w2.
(2) r1 is 20 less than r2.

Excellent opportunity for UNITS CONTROL, one of the most powerful tools of our method!

$$N \ge 1\,\,{\mathop{\rm int}} \,\,{\rm{workers}}\,\,\,\,\,\, \to \,\,\,\,\,{w_1}\,\,{\rm{widgets}}\,\,\,\left( {{{1\,\,{\rm{minute}}} \over {N \cdot {r_1}\,\,{\rm{widgets}}}}} \right)\,\,\,\,\, = \,\,\,\,{{{w_1}\,\,} \over {N \cdot {r_1}\,\,}}\,\,\,{\rm{minutes}}$$
$$M \ge 1\,\,{\mathop{\rm int}} \,\,{\rm{workers}}\,\,\,\,\,\, \to \,\,\,\,\,{w_2}\,\,{\rm{widgets}}\,\,\,\left( {{{1\,\,{\rm{minute}}} \over {M \cdot {r_2}\,\,{\rm{widgets}}}}} \right)\,\,\,\,\, = \,\,\,\,{{{w_2}\,\,} \over {M \cdot {r_2}\,\,}}\,\,\,{\rm{minutes}}$$
$${{{w_1}\,\,} \over {N \cdot {r_1}\,\,}} = {{{w_2}\,\,} \over {M \cdot {r_2}\,\,}}\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{{\rm{all}}\,\, \ne \,\,0} \,\,\,\,\,\,\,{N \over M} = {{{w_1} \cdot {r_2}} \over {{w_2} \cdot {r_1}}}$$
$$N\,\,\mathop < \limits^? \,\,M\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{M\, > \,0} \,\,\,\,\,\,\,{N \over M}\,\,\mathop < \limits^? \,\,1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,{{{w_1} \cdot {r_2}} \over {{w_2} \cdot {r_1}}}\,\,\mathop < \limits^? \,\,1\,\,\,$$

Let´s BIFURCATE both statement together at once, so that we are sure the correct answer is (E):

$$\left( {1 + 2} \right)\,\,\,\left\{ \matrix{
\,{w_2} = {w_1} + 20 \hfill \cr
\,\,{r_2} = {r_1} + 20 \hfill \cr} \right.\,\,\,\,\,$$
$$\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {{w_1},{w_2},{r_1},{r_2}} \right) = \left( {10,30,10,30} \right)\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {{w_1},{w_2},{r_1},{r_2}} \right) = \left( {1,21,10,30} \right)\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$


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

Regards,
Fabio.