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03 Jan 2019, 12:39
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
92% (01:40) correct
8%
(03:01)
wrong
based on 84
sessions
History
Date
Time
Result
Not Attempted Yet
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
A. 22456
B. 22457
C. 22458
D. 22459
E. 22460
03 Jan 2019, 12:53
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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.
Posted from my mobile device
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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03 Jan 2019, 14:48
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Bunuel
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
