AAPL wrote:Magoosh
Every day at noon, a bus leaves for Townville and travels at a speed of x kilometers per hour. Today, the bus left 30 minutes late. If the driver drives 7/6 times as fast as usual, she will arrive in Townville at the regular time. If the distance to Townville is 280 kilometers, what is the value of x?
A. 66
B. 72
C. 80
D. 84
E. 90
\[? = x\]
Let´s use UNITS CONTROL, one of the most powerful tools of our method!
\[\frac{{30\,\,\,{\text{minutes}}\,\,{\text{saved}}}}{{280\,\,{\text{km}}}} = \,\,\frac{{\,\,\frac{3}{{28}}\,\,\,{\text{minutes}}\,\,{\text{saved}}\,}}{{1\,\,\,{\text{km}}}}\,\,\,\,\,\left( * \right)\]
\[\left. \begin{gathered}
x\,\,\frac{{{\text{km}}}}{{\text{h}}}\,\,\,\,:\,\,\,1\,{\text{km}}\,\,\left( {\frac{{1\,\,{\text{hour}}}}{{\,x\,\,{\text{km}}\,}}} \right)\left( {\frac{{60\,\,{\text{minutes}}\,}}{{\,1\,\,{\text{hour}}\,}}} \right)\,\,\,\, = \,\,\,\,\frac{{60}}{x}\,\,\,{\text{minutes}} \hfill \\
\frac{{7x}}{6}\,\,\frac{{{\text{km}}}}{{\text{h}}}\,\,\,\,:\,\,\,1\,{\text{km}}\,\,\left( {\frac{{6\,\,{\text{hour}}}}{{\,7x\,\,{\text{km}}\,}}} \right)\left( {\frac{{60\,\,{\text{minutes}}\,}}{{\,1\,\,{\text{hour}}\,}}} \right)\,\,\,\, = \,\,\,\,\frac{{6 \cdot 60}}{{7x}}\,\,\,{\text{minutes}}\,\,\, \hfill \\
\end{gathered} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{60}}{x} - \frac{{6 \cdot 60}}{{7x}}\,\,\,\mathop = \limits^{\left( * \right)} \,\,\frac{3}{{28}}\]
\[\frac{{60 \cdot \boxed7}}{{x \cdot \boxed7}} - \frac{{6 \cdot 60}}{{7x}}\, = \frac{3}{{28}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\frac{{60}}{{7x}} = \frac{3}{{28}}\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,? = x = \frac{{28 \cdot 60}}{{7 \cdot 3}} = 80\,\,\,\,\,\,\left[ {\,\frac{{{\text{km}}}}{{\text{h}}}\,} \right]\]
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio. 
