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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

DS questions on Arithmetic: search.php?search_id=tag&tag_id=30
PS questions on Arithmetic: search.php?search_id=tag&tag_id=51
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The rear wheels of a car crossed a certain line 0.5 second after the front wheels crossed the same line. If the centers of the front and rear wheels are 20 feet apart and the car traveled in a straight line at a constant speed, which of the following gives the speed of the car in miles per hour? (5280 feet = 1 mile)

The car traveled 20 feet in 1/2 second. This means it traveled 40 feet a second or 2400 feet/minute. This equates to .45 miles a minute or roughly 27 miles/hour.

20/5280 * (60^2) / .5
20/5280 * 3600/.5
20/5280 * 7200
300/11 = roughly 27

ANSWER: A
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Hi All,

Since this question is wordy and the answers are somewhat "crazy"-looking, it's possible that you might feel overwhelmed by this prompt. If you stay calm though and do the unit conversions properly, then you can use the answer choices to your advantage (and eliminate most of them for being too big or too small). Here's how:

The first step is to figure out approximately how fast the car was going. This will take a little bit of work, but the individual steps are not too hard.

The information in the first part of the question tells us that it basically takes .5 seconds to travel 20 feet. Since the question asks for a speed in MILES per HOUR, we have to convert these numbers....

.5 seconds = 20 feet
1 second = 40 feet
60 seconds = 2400 feet
1 hour = (2400)(60) = 144,000 feet

144,000 feet = about 28 miles

So the car was traveling about 28 miles per hour.

Now, by paying attention to how the answer choices are "structured", we can eliminate the wrong answers without having to calculate much...Remember that we're looking for an answer that is about 28.....

Let's start with Answers B and C, since they're the easiest to eliminate. Look at the NUMERATORS (relative to the DENOMINATORS)....

Answer B: 1200/2640. This is a FRACTION less than 1. Eliminate B.

Answer C: 10/(GIGANTIC PRODUCT). This is a REALLY small fraction. Eliminate C.

Of the remaining 3 answers, Answer E is reasonably easy to eliminate....

Answer E: (20)(5280)/30 This will be in the THOUSANDS. It's much too BIG. Eliminate E.

Between Answers A and D, notice how the "20" and the "0.5" are in the same relative positions.....

20/.05 = 40

So we're multiplying some fraction by 40. To get an answer that is equal to about 28, we need that fraction to be LESS than 1...

Answer A: 60^2/5280 is LESS than 1

Answer D: 5280/60^2 is GREATER than 1. Eliminate D.

Final Answer:
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Stiv
The rear wheels of a car crossed a certain line 0.5 second after the front wheels crossed the same line. If the centers of the front and rear wheels are 20 feet apart and the car traveled in a straight line at a constant speed, which of the following gives the speed of the car in miles per hour? (5280 feet = 1 mile)

A. \((\frac{20}{5280})(\frac{60^2}{0.5})\)

B. \((\frac{20}{5280})(\frac{60}{0.5})\)

C. \((\frac{20}{5280})(\frac{0.5}{60^2})\)

D. \(\frac{(20)(5280)}{(60^2)(0.5)}\)

E. \(\frac{(20)(5280)}{(60)(0.5)}\)

We see that the car drives 20 ft in 0.5 seconds. Since the car drives at a constant speed, its rate o is 20 ft/0.5 seconds.

Let’s convert 20 ft/0.5 sec into mph:

20 ft/0.5 sec x 1 mi/5280 ft x 3600 sec/1 hr

In the above expression, we see the units “seconds” and “feet” will cancel and we are left with:

(20 x 1 x 3600)/(0.5 x 5280 x 1) mi/hr

(20 x 3600)/(0.5 x 5280) mi/hr

Answer: A
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Stiv
The rear wheels of a car crossed a certain line 0.5 second after the front wheels crossed the same line. If the centers of the front and rear wheels are 20 feet apart and the car traveled in a straight line at a constant speed, which of the following gives the speed of the car in miles per hour? (5280 feet = 1 mile)

A. \((\frac{20}{5280})(\frac{60^2}{0.5})\)

B. \((\frac{20}{5280})(\frac{60}{0.5})\)

C. \((\frac{20}{5280})(\frac{0.5}{60^2})\)

D. \(\frac{(20)(5280)}{(60^2)(0.5)}\)

E. \(\frac{(20)(5280)}{(60)(0.5)}\)

A quick approach-

Eliminate D & E as 5280 needs to be a part of the denominator (as we are converting feet into miles).

Eliminate B & C as we need 60^2 in the numerator (as we are converting seconds into an hour).

Hence, A is the answer.

Aiena.
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Stiv
The rear wheels of a car crossed a certain line 0.5 second after the front wheels crossed the same line. If the centers of the front and rear wheels are 20 feet apart and the car traveled in a straight line at a constant speed, which of the following gives the speed of the car in miles per hour? (5280 feet = 1 mile)

A. \((\frac{20}{5280})(\frac{60^2}{0.5})\)

B. \((\frac{20}{5280})(\frac{60}{0.5})\)

C. \((\frac{20}{5280})(\frac{0.5}{60^2})\)

D. \(\frac{(20)(5280)}{(60^2)(0.5)}\)

E. \(\frac{(20)(5280)}{(60)(0.5)}\)
\(1\,\,{\rm{mile}}\,\,\, \leftrightarrow \,\,\,5280\,\,{\rm{feet}}\)

\(V\left( {{\rm{speed}}} \right) = {{\,20\,\,{\rm{feet }}} \over {0.5\,\,{\rm{s}}}}\,\, = \,\,\,?\,\,{\rm{mph}}\,\,\,\,\,\)

Perfect opportunity to use UNITS CONTROL, one of the most powerful tools of our course!

\(?\,\,\, = \,\,\,{{\,20\,\,{\rm{feet }}} \over {0.5\,\,{\rm{s}}}}\left( {{{1\,\,{\rm{mile}}} \over {5280\,\,{\rm{feet}}}}\matrix{\\
\nearrow \cr \\
\nearrow \cr \\
\\
} } \right)\,\left( {{{60\,\,{\rm{s}}} \over {1\,\,{\rm{min}}}}\matrix{\\
\nearrow \cr \\
\nearrow \cr \\
\\
} } \right)\left( {{{60\,\,{\rm{min}}} \over {1\,\,{\rm{h}}}}\matrix{\\
\nearrow \cr \\
\nearrow \cr \\
\\
} } \right)\,\,\, = \,\,\,{{\,20\,\, \cdot \,\,60\,\, \cdot \,\,60\,} \over {0.5\,\, \cdot \,\,5280}}\,\,\,\,\left[ {{\rm{mph}}} \right]\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( A \right)\)

Obs.: arrows indicate licit converters.


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

Regards,
Fabio.
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Stiv
The rear wheels of a car crossed a certain line 0.5 second after the front wheels crossed the same line. If the centers of the front and rear wheels are 20 feet apart and the car traveled in a straight line at a constant speed, which of the following gives the speed of the car in miles per hour? (5280 feet = 1 mile)

A. \((\frac{20}{5280})(\frac{60^2}{0.5})\)

B. \((\frac{20}{5280})(\frac{60}{0.5})\)

C. \((\frac{20}{5280})(\frac{0.5}{60^2})\)

D. \(\frac{(20)(5280)}{(60^2)(0.5)}\)

E. \(\frac{(20)(5280)}{(60)(0.5)}\)

Hi BrentGMATPrepNow

Question asked the rear wheels of a car crossed a certain line 0.5 second after the front wheels crossed the same line. So time for front wheel time is t and rear wheel is t+0.5s. Is my understanding correct or something wrong here as not seeing any answer choice used calculation of t+0.5s but only 0.5s? Could you help? Thanks
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Stiv
The rear wheels of a car crossed a certain line 0.5 second after the front wheels crossed the same line. If the centers of the front and rear wheels are 20 feet apart and the car traveled in a straight line at a constant speed, which of the following gives the speed of the car in miles per hour? (5280 feet = 1 mile)

A. \((\frac{20}{5280})(\frac{60^2}{0.5})\)

B. \((\frac{20}{5280})(\frac{60}{0.5})\)

C. \((\frac{20}{5280})(\frac{0.5}{60^2})\)

D. \(\frac{(20)(5280)}{(60^2)(0.5)}\)

E. \(\frac{(20)(5280)}{(60)(0.5)}\)

Hi BrentGMATPrepNow

Question asked the rear wheels of a car crossed a certain line 0.5 second after the front wheels crossed the same line. So time for front wheel time is t and rear wheel is t+0.5s. Is my understanding correct or something wrong here as not seeing any answer choice used calculation of t+0.5s but only 0.5s? Could you help? Thanks

Before you start using variables to create algebraic expressions, it's important to clearly define what those variables represent.
Here you are saying that t = "the front wheel time", which doesn't make any sense.

The given information tells us that the car travels 20 feet in 0.5 seconds.
So, we don't need to assign any variables at all.
All you need to do is convert 20 feet per 0.5 seconds into miles per hour (which the above posters have successfully done)
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Kimberly77
Stiv
The rear wheels of a car crossed a certain line 0.5 second after the front wheels crossed the same line. If the centers of the front and rear wheels are 20 feet apart and the car traveled in a straight line at a constant speed, which of the following gives the speed of the car in miles per hour? (5280 feet = 1 mile)

A. \((\frac{20}{5280})(\frac{60^2}{0.5})\)

B. \((\frac{20}{5280})(\frac{60}{0.5})\)

C. \((\frac{20}{5280})(\frac{0.5}{60^2})\)

D. \(\frac{(20)(5280)}{(60^2)(0.5)}\)

E. \(\frac{(20)(5280)}{(60)(0.5)}\)

Hi BrentGMATPrepNow

Question asked the rear wheels of a car crossed a certain line 0.5 second after the front wheels crossed the same line. So time for front wheel time is t and rear wheel is t+0.5s. Is my understanding correct or something wrong here as not seeing any answer choice used calculation of t+0.5s but only 0.5s? Could you help? Thanks

Before you start using variables to create algebraic expressions, it's important to clearly define what those variables represent.
Here you are saying that t = "the front wheel time", which doesn't make any sense.

The given information tells us that the car travels 20 feet in 0.5 seconds.
So, we don't need to assign any variables at all.
All you need to do is convert 20 feet per 0.5 seconds into miles per hour (which the above posters have successfully done)

Noted thanks BrentGMATPrepNow.
Got confused by the wording here that "The rear wheels of a car crossed a certain line 0.5 second after the front wheels crossed the same line" so thought there are 2 times associated with it :cry:
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