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The average (arithmetic mean) population in town X was recorded as 22,

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The average (arithmetic mean) population in town X was recorded as 22,  [#permalink]

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New post 03 Jan 2019, 12:39
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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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The average (arithmetic mean) population in town X was recorded as 22,  [#permalink]

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New post 03 Jan 2019, 12:53
Answer is B.
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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The average (arithmetic mean) population in town X was recorded as 22,  [#permalink]

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New post 03 Jan 2019, 14:48
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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
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The average (arithmetic mean) population in town X was recorded as 22,  [#permalink]

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New post 04 Jan 2019, 03:11

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

Answer: B

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The average (arithmetic mean) population in town X was recorded as 22,  [#permalink]

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New post 04 Jan 2019, 04:42
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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.

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