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sydc8
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EIV (4th edition) Ch.12 Problem 5 on Page 187

by sydc8 Wed Feb 24, 2010 4:19 am

A cyclist travels the length of a bike path that is 225 miles long, rounded to the nearest mile. If the trip took him 5 hours, rounded to the nearest hour, then his average speed must be between:

a) 38 and 50 MPH
b) 40 and 50 MPH
c) 40 and 51 MPH
d) 41 and 50 MPH
e) 41 and 51 MPH

The answer goes something like this.

Max speed = 225.5/4.5
Min speed = 224.5/5.5

My question is why max speed is calculated with 225.5 which is 226 miles (rounded)
and min speed uses 5.5 seconds which is 6 seconds (rounded)

Thanks,
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Re: EIV (4th edition) Ch.12 Problem 5 on Page 187

by Ben Ku Thu Feb 25, 2010 1:17 am

sydc8 Wrote:A cyclist travels the length of a bike path that is 225 miles long, rounded to the nearest mile. If the trip took him 5 hours, rounded to the nearest hour, then his average speed must be between:

a) 38 and 50 MPH
b) 40 and 50 MPH
c) 40 and 51 MPH
d) 41 and 50 MPH
e) 41 and 51 MPH

The answer goes something like this.

Max speed = 225.5/4.5
Min speed = 224.5/5.5

My question is why max speed is calculated with 225.5 which is 226 miles (rounded)
and min speed uses 5.5 seconds which is 6 seconds (rounded)

Thanks,


The distance is 225 rounded to the nearest mile, which means the max distance must be 225.5 ≤ d < 225.5. Basically, it has to be less than 225.5, but not equal to it. We can think of 225.5 as our max limit. I hope that makes sense.
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Re: EIV (4th edition) Ch.12 Problem 5 on Page 187

by itzdeepika Sat Mar 27, 2010 7:33 pm

Hi Ben,

I think the limit mentioned should be 224.5 ≤ d < 225.4.
Hence the

Max speed = 225.4/4.5
Min speed = 224.5/5.4

Please correct me if am missing somethg.

Regards
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Re: EIV (4th edition) Ch.12 Problem 5 on Page 187

by tim Tue Apr 06, 2010 6:44 pm

itzdeepika Wrote:Hi Ben,

I think the limit mentioned should be 224.5 ≤ d < 225.4.
Hence the

Max speed = 225.4/4.5
Min speed = 224.5/5.4

Please correct me if am missing somethg.

Regards


You are missing something. :) d could be 225.49, which is outside the range you allowed for but inside Ben's. d<225.5 is definitely correct.
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Re: EIV (4th edition) Ch.12 Problem 5 on Page 187

by hurshid.narmatov Sat Nov 12, 2011 4:27 pm

Tim, thanks for previous replies.

I've got another question regarding the same problem.
So, according to the solution:
Max speed = 50(1/9) and Min speed = 40(9/11), and therefore cyclist's speed must be between 40 and 51 mph.

I don't understand though, how 40(9/11) equals to 40, when it is closer to 41, and 50(1/9) is closer to 50 rather than 51?

I mean the range should be between 41 and 50?

Thanks!
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Re: EIV (4th edition) Ch.12 Problem 5 on Page 187

by jnelson0612 Tue Nov 22, 2011 6:20 pm

hurshid.narmatov Wrote:Tim, thanks for previous replies.

I've got another question regarding the same problem.
So, according to the solution:
Max speed = 50(1/9) and Min speed = 40(9/11), and therefore cyclist's speed must be between 40 and 51 mph.

I don't understand though, how 40(9/11) equals to 40, when it is closer to 41, and 50(1/9) is closer to 50 rather than 51?

I mean the range should be between 41 and 50?

Thanks!


Okay, let's check this out again! As you stated above: "Max speed = 50(1/9) and Min speed = 40(9/11)".

Thus, the slowest the cyclist could go is 40 9/11 and the fastest the cyclist could go is 50 1/9.

Which of the answer choices contains every possible speed? If I say 41 to 50, as you suggest, I leave out the possibilities of 40 9/11 and 50 1/9. Thus, to include those possibilities, I have to say 40 to 51. 40 to 51 includes EVERY possible speed.

Let me know if this doesn't make sense. :-)
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Re: EIV (4th edition) Ch.12 Problem 5 on Page 187

by aps_asks Sun Nov 27, 2011 4:47 am

For the same question , i.e

A cyclist travels the length of a bike path that is 225 miles long, rounded to the nearest mile. If the trip took him 5 hours, rounded to the nearest hour, then his average speed must be between:

a) 38 and 50 MPH
b) 40 and 50 MPH
c) 40 and 51 MPH
d) 41 and 50 MPH
e) 41 and 51 MPH



According to me the answer should be
Average speed should be between & ( inclusive ) of the below values

Max Speed = 225.5/4.6
Min Speed = 224.6/5.4

But in the MGMAT Book we have considered ,
Max Speed = 225.4/4.5
Min Speed = 224.5/5.5

Did we make the adjustments to the decimal above as the Question asked his average speed must be between ---- and ---- ?

I hope i am clear
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Re: EIV (4th edition) Ch.12 Problem 5 on Page 187

by munique2 Tue Dec 13, 2011 1:48 am

Hello,

I'm also interested in this type of question because there is a similar one in the OG Guide as well.

I do get the concept of the upper limit. We take 225.5 to make sure the quotient will be inbetween the boundaries. So far, so good.
What I'm struggling with is the lower limit. As 225.5 is the next value to be rounded up to 226, we choose this one because it can't be the actual number and embraces the other possible solutions. However, 224.5 is rounded up to 225. Therefore, 224.5 could be an actual number that fulfills the requirements opposed to 225.5. Why don't we choose 224.4 for the lower limit to make sure the average speed is between the numbers we are looking for?

Do I misunderstand a concept wrong here or is the meaning of "between" different in terms of GMAT? I would assume when something is considered to be between, let's say 3 and 8, both numbers are not included in the open interval. Here it seems to be equivalent to a half-open interval, including 3 bot not 8.
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Re: EIV (4th edition) Ch.12 Problem 5 on Page 187

by tim Tue Dec 13, 2011 2:43 am

The MGMAT book never uses a decimal ending in .4 for this problem. We use .5 everywhere for all the reasons discussed above. Please read the prior discussion and let us know if you still have questions..
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Re: EIV (4th edition) Ch.12 Problem 5 on Page 187

by munique2 Tue Dec 13, 2011 4:14 am

alright, I think I found an explanation for the problem regarding my thought process. Just in case somebody is going through the same:
If the numerator is already adjusted we don't need to adjust the denominator any more and the result will be more exact or narrowed down and still meet the criteria of "between".
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Re: EIV (4th edition) Ch.12 Problem 5 on Page 187

by jnelson0612 Mon Dec 26, 2011 12:16 am

Thanks munique. :-)
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Re: EIV (4th edition) Ch.12 Problem 5 on Page 187

by aps_asks Sat Jan 28, 2012 5:40 pm

Hi Guys ,

I have found out one explanation.

The Question gives the following details :
Distance : 225 miles which means
( Considering d as the distance
224.6 <=d <=225.4)
Hours required for Trip : 5 Hours which means
( Considering t as the hours
4.6 <= t <= 5.4)

But wait ,
The Question asks average speed to be between

Hence we will consider
(224.6 <=d <=225.4) Equal to 224.5 <d < 225.5
and
(4.6 <= t <= 5.4) Equal to 4.5 <t<5.5


So the answer is
Max speed = 225.5/4.5
Min speed = 224.5/5.5

Please let me know your comments
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Re: EIV (4th edition) Ch.12 Problem 5 on Page 187

by tim Thu Feb 02, 2012 7:40 pm

every part of your work that involved something that ended in .4 or .6 was incorrect, but fortunately you could strip all that out of your solution and everything else is correct..

PLEASE understand everyone, if x rounds down to 225, that does not mean that x <= 225.4. This is NOT good enough, because 225.499 rounds down to 225 as well. all we can say is that x < 225..
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Re: EIV (4th edition) Ch.12 Problem 5 on Page 187

by aps_asks Sat Feb 04, 2012 9:50 am

Tim , Do i need to understand like this ???---

If the rounded values of the distance ( d) is 225 miles ,rounded to the nearest mile

Then 224.5<d<225.5
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Re: EIV (4th edition) Ch.12 Problem 5 on Page 187

by tim Mon Feb 06, 2012 3:11 pm

very close:

224.5 <= d < 225.5
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