fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 5)

Brenda will test-drive a vintage motor car on a circular track with radius 300 meters, at a constant speed of 10 meters per second. If her husband Jim is at rest, at the center of this track, to film the entire test-drive, which of the following values is closest to the approximate angle Jim will rotate his head during her 1-minute car ride?

(A) 75 degrees
(B) 90 degrees
(C) 115 degrees
(D) 140 degrees
(E) 155 degrees

$$?\,\,\,:\,\,\,{\rm{approx}}.\,\,{\rm{degrees - rotation}}\,\,\,{\rm{in}}\,\,\,{\rm{1}}\,\,{\rm{min}}$$
$$360\,\,{\rm{degrees - rotation}}\,\,\,\, \leftrightarrow \,\,\,\,2 \cdot \pi \cdot 300\,\,{\rm{meters}}\,\,\,\,\,\,\left( {1\,\,{\rm{lap}}} \right)$$
$$10\,\,{\rm{meters}}\,\,\,\, \leftrightarrow \,\,\,\,1\,\,\sec \,\,\,\,\left( {{\rm{speed}}} \right)$$

Let´s use UNITS CONTROL, one of the most powerful tools of our method!

$$?\,\,\, = \,\,\,1\,\,\min \,\,\,\left( {{{60\,\,\sec } \over {1\,\,\min }}} \right)\,\,\,\left( {{{10\,\,{\rm{meters}}} \over {1\,\,{\rm{sec}}}}} \right)\,\,\,\left( {{{360\,\,{\rm{degrees - rotation}}} \over {2 \cdot \pi \cdot 300\,\,{\rm{meters}}}}} \right)\,\,\,\,\,\left[ {{\rm{degrees - rotation}}} \right]$$
$$\pi \cong {{22} \over 7}\,\,\,\, \Rightarrow \,\,\,\,? \cong {{60 \cdot 10 \cdot 360} \over {2 \cdot {{22} \over 7} \cdot 300}}\,\, = \,\,{7 \over {22}} \cdot {{30 \cdot 10 \cdot 6} \over 5}\,\, = \,\,{7 \over {22}} \cdot 360$$
$$? \cong {7 \over {22}} \cdot 360 = {7 \over {11}} \cdot \left( {110 + 66 + 4} \right) = 7\left( {16 + {4 \over {11}}} \right) = 112 + 2 + {6 \over {11}} \cong 115^\circ $$

The correct answer is (C).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

P.S.: there is an immediate alternate way: (60*10)metres/300metres = 2 radians (1 radian is 180/pi degrees and it´s done), but "radians" is out-of-GMAT´s scope...

Radius of circular track = 300 meters
$$Circumference=2\pi r=2\cdot\frac{22}{7}\cdot300\approx1886$$

Speed = 10m/s
Distance in 1min (60s) =10*60 = 600meters
Angle that jim will rotate
$$1886=360$$ $$600=\frac{\left(600\cdot360\right)}{1880}=\frac{216000}{1886}$$
$$=114.5\approx115\deg ree$$

$$answer\ is\ Option\ C$$