BTGmoderatorLU wrote:Source: Princeton Review

If Albert can travel 200 miles in 4 hours, how many hours will it take Albert, travelling at the same constant rate, to travel 350 miles?

A. 5
B. 6
C. 7
D. 8
E. 10

\[200\,\,{\text{miles}}\,\,\,\, \leftrightarrow \,\,\,\,{\text{4}}\,\,{\text{h}}\]
\[{\text{350}}\,\,\,{\text{miles}}\,\,\, \leftrightarrow \,\,\,{\text{?}}\,\,{\text{h}}\]
Same constant rate (speed) implies that the ratio miles/time is constant (and different from zero): we have direct proportionality.
\[\frac{{\text{?}}}{4} = \frac{{350}}{{200}}\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\frac{{{\text{?}} \cdot \boxed{50}}}{{4 \cdot \boxed{50}}} = \frac{{350}}{{200}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = \frac{{350}}{{50}} = 7\,\,\,\left[ {\text{h}} \right]\,\,\]
(*) This is what we call the "Bruce Lee" technique, because he is known to have said: "Punch when you have to punch. Kick when you have to kick." ... and he certainly thought about "denominator equal to the denominator, imply numerator equal to the numerator"... got it?

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

Regards,
Fabio.

BTGmoderatorLU wrote:Source: Princeton Review

If Albert can travel 200 miles in 4 hours, how many hours will it take Albert, traveling at the same constant rate, to travel 350 miles?

A. 5
B. 6
C. 7
D. 8
E. 10

And now a solution that our advanced students do as "second nature":
\[350\,{\text{miles}}\,\,\,\left( {\frac{{4\,\,{\text{h}}}}{{200\,\,{\text{miles}}}}\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}\,\,} \right)\,\,\, = \frac{{35 \cdot 4}}{{20}} = 7\,\,\,\,\left[ {\text{h}} \right]\]
Obs.: arrows indicate licit converter.

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

Regards,
Fabio.

BTGmoderatorLU wrote:Source: Princeton Review

If Albert can travel 200 miles in 4 hours, how many hours will it take Albert, traveling at the same constant rate, to travel 350 miles?

A. 5
B. 6
C. 7
D. 8
E. 10

And let´s offer a third possibility.

This one is dedicated to Julia, my 10-years old niece:
\[200\,\,{\text{miles}}\,\,\,\, \leftrightarrow \,\,\,\,{\text{4}}\,\,{\text{h}}\]
\[{\text{50}}\,\,\,{\text{miles}}\,\,\,\, \leftrightarrow \,\,\,\,{\text{1}}\,\,{\text{h}}\]
\[{\text{350}}\,\,\,{\text{miles}}\,\,\, \leftrightarrow \,\,\,{\text{?}}\,{\text{ = }}\,{\text{7}}\,\,\,{\text{h}}\]

Regards,
Fabio.

BTGmoderatorLU wrote:Source: Princeton Review

If Albert can travel 200 miles in 4 hours, how many hours will it take Albert, traveling at the same constant rate, to travel 350 miles?

A. 5
B. 6
C. 7
D. 8
E. 10

Since Albert travels 200 miles in 4 hours, his rate = 200/4 = 50 mph.
When the correct answer is divided into 350 miles, the resulting rate must be 50 mph.
Only C works:
350/7 = 50.

The correct answer is C.