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The rear wheels of a car crossed a certain line 0.5 second

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The rear wheels of a car crossed a certain line 0.5 second  [#permalink]

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New post 14 Jul 2012, 00:53
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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)

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

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Re: The rear wheels of a car crossed a certain line 0.5 second  [#permalink]

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New post 14 Jul 2012, 01:57
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Stiv wrote:
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 need to find the speed of the car in miles per hour. So, we should convert feet in miles and seconds in hours.

20 feet is \(\frac{20}{5280}\) miles;
0.5 second is \(\frac{0.5}{3600}=\frac{0.5}{60^2}\) hours;

The speed of the car therefore is \(\frac{distance}{time}=\frac{(\frac{20}{5280})}{(\frac{0.5}{60^2})}=(\frac{20}{5280})*(\frac{60^2}{0.5})\).

Answer: A.

Similar question to practice: a-train-traveling-at-a-constant-speed-down-a-straight-track-87124.html

Hope it helps.
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Re: The rear wheels of a car crossed a certain line 0.5 second  [#permalink]

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New post 02 Jul 2013, 00:18
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: The rear wheels of a car crossed a certain line 0.5 second  [#permalink]

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New post 02 Jul 2013, 00:18
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Other conversion problems to practice:
if-an-object-travels-100-feet-in-2-seconds-what-is-the-103364.html
the-rear-wheels-of-a-car-crossed-a-certain-line-0-5-second-135753.html
a-train-traveling-at-a-constant-speed-down-a-straight-track-87124.html
a-welder-received-an-order-to-make-a-1-million-liter-126154.html
a-car-traveling-at-a-certain-constant-speed-takes-2-seconds-126658.html
if-an-automobile-averaged-22-5-miles-per-gallon-of-gasoline-109083.html
if-1-kilometer-is-approximately-0-6-mile-which-of-the-140731.html
measuring-standards-conversion-106516.html
if-the-speed-of-x-meters-per-second-is-equivalent-to-the-127005.html
if-10-millimeters-equal-1-centimeter-how-many-square-48538.html
the-speed-of-light-is-approximately-149492.html

Hope it helps
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Re: The rear wheels of a car crossed a certain line 0.5 second  [#permalink]

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New post 05 Aug 2013, 12:31
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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Re: The rear wheels of a car crossed a certain line 0.5 second  [#permalink]

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New post 25 Feb 2015, 22:29
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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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Re: The rear wheels of a car crossed a certain line 0.5 second  [#permalink]

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New post 01 Sep 2017, 11:05
Stiv wrote:
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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Re: The rear wheels of a car crossed a certain line 0.5 second  [#permalink]

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New post 11 Feb 2018, 04:11
Stiv wrote:
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.

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The rear wheels of a car crossed a certain line 0.5 second  [#permalink]

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New post 18 Oct 2018, 04:56
1
Stiv wrote:
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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