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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # The average (arithmetic mean) population in town X was recorded as 22,

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Math Expert V
Joined: 02 Sep 2009
Posts: 56257
The average (arithmetic mean) population in town X was recorded as 22,  [#permalink]

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Difficulty:   15% (low)

Question Stats: 88% (01:45) correct 13% (03:08) wrong based on 39 sessions

### HideShow timer Statistics 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?

A. 22456
B. 22457
C. 22458
D. 22459
E. 22460

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Intern  B
Joined: 26 Dec 2018
Posts: 3
The average (arithmetic mean) population in town X was recorded as 22,  [#permalink]

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Step 1: Multiply 11 years by the original mean (which is 22,455) to get the Total of 11 years.
Step 2: Subtract the erroneous value from the Total.
Step 3: Add the corrected value to the Total.
Step 4: Divide the Total of Step 3 by 11 to get the updated mean.

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VP  P
Joined: 07 Dec 2014
Posts: 1206
The average (arithmetic mean) population in town X was recorded as 22,  [#permalink]

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1
Bunuel wrote:
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?

A. 22456
B. 22457
C. 22458
D. 22459
E. 22460

units digit of erroneous total 11*22,455=5
correct figure (22,500) minus incorrect figure (22,478)=error of 22
5+22 gives corrected total a units digit of 7
so only 22,457 will work when corrected total is divided by 11
B
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 2943
The average (arithmetic mean) population in town X was recorded as 22,  [#permalink]

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Solution

Given:
• The average population in town X during the years 2000-2010 is 22,455
• The figure for the year 2009 was erroneously recorded as 22,478.
• It should have been correctly recorded as 22,500

To find:
• The average population in town X during the years 2000-2010, after correction

Approach and Working:
• Sum of all the recordings = 22,455 * 11
• Correct sum = 22,455 * 11 – 22,478 + 22,500 = 22,455 * 11 + 22

Therefore, average population during 2000-2010, after correction = $$\frac{(22,455 * 11 + 22)}{11} = 22,455 + 2 = 22,457$$

Hence, the correct answer is Option B

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Intern  B
Joined: 20 Nov 2018
Posts: 1
The average (arithmetic mean) population in town X was recorded as 22,  [#permalink]

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2
Above processes are long, we know the error count is 22 and years are 11 so on an average 2 people will be increased per year. Add 2 to 22455, you get the solution in 10 seconds.

Posted from my mobile device The average (arithmetic mean) population in town X was recorded as 22,   [#permalink] 04 Jan 2019, 04:42
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# The average (arithmetic mean) population in town X was recorded as 22,  