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by hakyology » Mon Jul 07, 2008 7:11 am
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? (5,280 feet = 1 mile)

A) (20/5,280)(60^2/0.5)
B)(20/5,280)(60/0.5)
C)(20/5,280)(0.5/60^2)
D)(20*5,280)/(60^2)(0.5)
E)(20)*(5,280)/(60)(0.5)


The answer is A.


Please explain.

Thank u.
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method I used

by dsk » Mon Jul 07, 2008 1:59 pm
for the rear wheels to pass the object after the front wheels have passed them, the distance rear wheel has to cover = distance from the rim of the front wheel to the end rim of the rear wheel. So in effect it is the distance between the centres of the tow wheels. this is 20 ft or (20/5280) miles.
Time taken to cover pass the object = 0.5 secs or (0.5/3600) hours or (0.5/60^2) hours.

So the speed would be (20/5280)/(0.5/60^2) = (20/5280)(60^2/0.5) which is option A.
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Re: CompliCATed

by parallel_chase » Mon Jul 07, 2008 2:34 pm
hakyology 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? (5,280 feet = 1 mile)

A) (20/5,280)(60^2/0.5)
B)(20/5,280)(60/0.5)
C)(20/5,280)(0.5/60^2)
D)(20*5,280)/(60^2)(0.5)
E)(20)*(5,280)/(60)(0.5)


The answer is A.


Please explain.

Thank u.
You can easily pick the answer from the choices.

We are converting ft to mile therefore we have to divide. You can eliminate D and E.
You can also eliminate B because it is converting only in minutes not in hours.

Answer C is trap answer.

We know Speed= Distance/Time

(20/5280) / (0.5/60^2) => (20/5280) (60^2/0.5)

Hence A is the correct answer.
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Re: method I used

by Carol » Tue Jul 08, 2008 1:41 pm
I can't figure how they came up with the (60^2/0.5)


dsk you said:

"So the speed would be (20/5280)/(0.5/60^2) = (20/5280)(60^2/0.5) which is option A." which is the same as answer C = answer A


could someone explain, why it is (60^2/0.5) and not (0.5/60^2)? :?

thank you!
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Re: method I used

by parallel_chase » Tue Jul 08, 2008 2:31 pm
Carol wrote:I can't figure how they came up with the (60^2/0.5)


dsk you said:

"So the speed would be (20/5280)/(0.5/60^2) = (20/5280)(60^2/0.5) which is option A." which is the same as answer C = answer A


could someone explain, why it is (60^2/0.5) and not (0.5/60^2)? :?

thank you!
answer C does not have a fraction sign

(20/5280)/(0.5/60^2) = (20/5280)(60^2/0.5)

Hope this helps
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by Carol » Tue Jul 08, 2008 2:48 pm
that was it :oops:

thx! parallel_chase
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by mbaapplicant2008 » Thu Oct 30, 2008 8:31 pm
Thank you so much, parallel_chase
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by Jeff@TargetTestPrep » Tue Jan 02, 2018 10:50 am
hakyology 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? (5,280 feet = 1 mile)

A) (20/5,280)(60^2/0.5)
B)(20/5,280)(60/0.5)
C)(20/5,280)(0.5/60^2)
D)(20*5,280)/(60^2)(0.5)
E)(20)*(5,280)/(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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