Nov 28 08:00 PM PST  09:00 PM PST Magoosh is excited to offer you a free GRE practice test with video answers and explanations. If you’re thinking about taking the GRE or want to see how effective your GRE test prep has been, pinpoint your strengths and weaknesses with this quiz! Nov 30 08:00 PM PST  09:00 PM PST Take 20% off the plan of your choice, now through midnight on 11/30 Dec 01 01:00 PM EST  02:00 PM EST Save up to 37% this weekend! $340.00 off our 6month Genius plan using promo code: GREBlackFridayGen $220.00 off our 3month Premium plan using promo code: GREBlackFridayPre Dec 02 07:30 AM PST  08:30 AM PST This webinar will focus on evaluating reading comprehension questions on the GRE and GMAT. This 60 minute class will be a mix of "presentation" and Q&A where students can get specific questions answered. Dec 04 10:00 PM PST  11:00 PM PST Regardless of whether you choose to study with Greenlight Test Prep, I believe you'll benefit from my many free resources. Dec 07 08:00 PM PST  09:00 PM PST This admissions guide will help you plan your best route to a PhD by helping you choose the best programs your goals, secure strong letters of recommendation, strengthen your candidacy, and apply successfully.
Author 
Message 
TAGS:


Retired Moderator
Joined: 07 Jun 2014
Posts: 4803
WE: Business Development (Energy and Utilities)
Followers: 175
Kudos [?]:
3036
[0], given: 394

The average (arithmetic mean) population in town X was recor [#permalink]
29 Aug 2018, 16:29
Question Stats:
87% (02:39) correct
12% (01:53) wrong based on 24 sessions
The average (arithmetic mean) population in town X was recorded as 22,455 during the years 2000–2010, inclusive. However, an error was later uncovered: the figure for 2009 was erroneously recorded as 22,478, but should have been correctly recorded as 22,500. What was the average population in town X during the years 2000–2010, inclusive, once the error was corrected?
_________________
Sandy If you found this post useful, please let me know by pressing the Kudos Button
Try our free Online GRE Test




Intern
Joined: 10 Aug 2018
Posts: 29
Followers: 1
Kudos [?]:
8
[0], given: 2

Re: The average (arithmetic mean) population in town X was recor [#permalink]
31 Aug 2018, 20:59
Please tell the method of solving it.



Intern
Joined: 02 Sep 2018
Posts: 1
Followers: 0
Kudos [?]:
2
[2]
, given: 6

Re: The average (arithmetic mean) population in town X was recor [#permalink]
02 Sep 2018, 05:05
2
This post received KUDOS
total sum for 11 year (2000  2010) inclusive = 11 * 22455 = X(say)
Now 22478 needs to be replace with 22500 , since their difference is 22
or
X  22478 + 22500 = Sum after correction
Divide X  22 by 11 years = 22457.



Intern
Joined: 22 Jul 2018
Posts: 39
Followers: 0
Kudos [?]:
20
[2]
, given: 5

Re: The average (arithmetic mean) population in town X was recor [#permalink]
15 Oct 2018, 10:15
2
This post received KUDOS
22,457. There is a simple shortcut for a change to an average. The figure for 2009 was recorded as 22,478, but actually should have been recorded as 22,500, meaning 22 people in that year were not counted. Thus, the sum should have been 22 higher when the average was originally calculated. 2000–2010, inclusive, is 11 years (subtract low from high and then add 1 to count an inclusive list of consecutive numbers). When taking an average, divide the sum by the number of things being averaged (in this case, 11). So the shortcut is to take the change to the sum and “spread it out” over all of the values being averaged by dividing the change by the number of things being averaged. Divide 22 by 11 to get 2. The average should have been 2 greater. Thus, the correct average for the 11year period is 22,457. Alternatively, use the traditional method: 22,455 × 11 years = 247,005, the sum of all 11 years’ recorded populations. Add the 22 uncounted people, making the corrected sum 247,027. Divide by 11 to get the corrected average: 22,457. (Note that while the traditional method is faster to explain, the shortcut is faster to actually execute!)



Manager
Joined: 02 Sep 2019
Posts: 69
Concentration: Finance
GPA: 3.14
Followers: 1
Kudos [?]:
39
[1]
, given: 21

Re: The average (arithmetic mean) population in town X was recor [#permalink]
12 Dec 2019, 01:28
1
This post received KUDOS
This question is tricky. First you need to work with the sum. it says there was an error in one of the year 2009 was 22478 and should have been 25000. Second pull out the average formula 1 average before the edit was 22455 = (Sum/11)= and Sum= 11*22455= 247,005 We know the sum of the years is 247005 and should add change to it. 24700522478+22500=247027 ==> this is the new sum after the edit. finally the new average =247027/11=22457




Re: The average (arithmetic mean) population in town X was recor
[#permalink]
12 Dec 2019, 01:28





