Last visit was: 22 Dec 2024, 02:28 It is currently 22 Dec 2024, 02:28

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
User avatar
Retired Moderator
Joined: 07 Jun 2014
Posts: 4815
Own Kudos [?]: 11267 [3]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
avatar
Intern
Intern
Joined: 10 Aug 2018
Posts: 29
Own Kudos [?]: 10 [0]
Given Kudos: 0
Send PM
avatar
Intern
Intern
Joined: 02 Sep 2018
Posts: 1
Own Kudos [?]: 2 [0]
Given Kudos: 0
Send PM
avatar
Intern
Intern
Joined: 22 Jul 2018
Posts: 39
Own Kudos [?]: 69 [3]
Given Kudos: 0
Send PM
Re: The average (arithmetic mean) population in town X was recor [#permalink]
3
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
11-year 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
Manager
Joined: 02 Sep 2019
Posts: 181
Own Kudos [?]: 144 [0]
Given Kudos: 94
Concentration: Finance
GRE 1: Q151 V148
GPA: 3.14
Send PM
Re: The average (arithmetic mean) population in town X was recor [#permalink]
1
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.
247005-22478+22500=‭247027‬ ==> this is the new sum after the edit.
finally the new average =247027/11=22457
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5088
Own Kudos [?]: 76 [0]
Given Kudos: 0
Send PM
Re: The average (arithmetic mean) population in town X was recor [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Prep Club for GRE Bot
Re: The average (arithmetic mean) population in town X was recor [#permalink]
Moderators:
GRE Instructor
88 posts
GRE Forum Moderator
37 posts
Moderator
1115 posts
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne