Every day at noon, a bus leaves for Townville and travels at

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

Time and rate have a RECIPROCAL RELATIONSHIP:
2 times as fast implies 1/2 the time.
3 times as fast implies 1/3 the time.

Let t = the normal time.
Since the bus travels today at 7/6 its normal rate, the time today is equal to 6/7 of the normal time:
(6/7)t.
Since the bus leaves 1/2 hour late today and arrives on time, the time for today's trip is 1/2 hour less than the normal time:
t - 1/2.
Since the expressions in blue both represent today's time, they must be EQUAL:
t - 1/2 = (6/7)t
(1/7)t = 1/2
t = 7/2.

Since the normal time = 7/2 hours, the normal rate for the 280-kilometer trip = 280/(7/2) = 80 kph.

The correct answer is C.

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.

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

Alternate approach:

7/6 of speed implies 6/7 of time (for any given distance).

The difference of time (30min) is therefore 1/7 (=1-6/7) of the original time, therefore the original time was 7*(1/2 h) = 7/2 hours.

It´s done: x = (280 km) / (7/2 h) = 80 km/h


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

Regards,
Fabio.