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DS - Speed problem

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Manager
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DS - Speed problem [#permalink]

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New post 01 May 2009, 06:36
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A
B
C
D
E

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Please explain. TIA

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Re: DS - Speed problem [#permalink]

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New post 01 May 2009, 19:22
stmnt1 - if his speed is 17 feet / sec, then the distance traveled is less than 6 miles, if 19 feet/sec, then the distance is more than 6 miles.
nsf

stmnt2 - same reasoning as above: if his speed is 14 feet/sec, then the distance is less than 6 miles, but if the speed is 17.99 feet/sec, then the distance is more than 6 miles.
nsf

combining clearly does not help
IMO E.

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Re: DS - Speed problem [#permalink]

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New post 02 May 2009, 10:06
typhoidX wrote:
I think E.
0.5 hour = 1800 seconds; 6 miles = 5280*6 = 31680 feet.

1. From this we can find out the minimum distance traveled in 0.5 hours, but not the maximum, for all we know from this alone he could've been going 50 ft per sec. 1 alone isn't enough.

2. From this we can find out the maximum distance traveled, and whether that distance is bigger than 5280*6 feet. We find that the max distance possible is 32400 feet at 18/sec (which is greater than 31680 ft). It is possible that he traveled slightly greater than 6 miles, but for all we know he could've been going 1 ft/sec. 2 alone also isn't enough.

If we put the two speeds together. We get a lower distance limit of 28800 & an upper limit of 32400. So we're not really sure whether he traveled more or less than 31680 ft.

What I did took too much calculation, is there an easier algebraic way to do this?

EDIT: I'm a little unsure of this; are we supposed to assume his speed is a whole number (i.e. 16 ft/sec, 17ft/sec, etc.); or can we be open to the possibility that his speed is greater than 16 but lower than 18, but it could be 16.001 or 17.999?




Answer is clearly E.
In response to your doubt that whether to consider the possibility of "17.999999" kind of numbers in our calculation. YES, WE NEED TO, unless specifically mentioned.

To make the calculations easy, there is an easier approach.
The distance to exceed = (5280 x 6) ft/sec.

Statement 1 - Consider 16.00000001 16 ft/sec. (16 x 1800) ft/sec = (16 x 300 x 6) = (4800 x 6) < (5280 x 6). We only to compare not calculate.

Statement 2 - Consider 17.9999999 18 ft/sec. 18 x 1800 ft/sec = (18 x 300 x 6) = (5400 x 6) > (5280 x 6).


Combining still doesnt give us a specific answer. Therefore, E.

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Re: DS - Speed problem   [#permalink] 02 May 2009, 10:06
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