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Manager
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Average [#permalink] New post 11 Aug 2009, 08:22
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

100% (01:13) correct 0% (00:00) wrong based on 6 sessions
This is actually a SAT problem....

The average(arithmetic mean) of nine numbers is 9. When a tenth number is added the average of the ten numbers is also 9. What is the tenth number ?

A) 0
B) 9/10
c) 10/9
d) 9
e) 10
[Reveal] Spoiler: OA
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Re: Average [#permalink] New post 11 Aug 2009, 08:38
The average(arithmetic mean) of nine numbers is 9. When a tenth number is added the average of the ten numbers is also 9. What is the tenth number ?

A) 0
B) 9/10
c) 10/9
d) 9
e) 10


The answer is D.
The avg of 9 nos is 9 => Total of 9 nos = 9*9 = 81

For the avg to remain 9 when another number, say x, is added, the following equation should hold true: (81 + x)/10 = 9
=> x = 9
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Senior Manager
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Re: Average [#permalink] New post 11 Aug 2009, 08:59
Agree with D. For the answer to be A, the problem definition must be different.
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Re: Average [#permalink] New post 11 Aug 2009, 10:20
[quote="sdrandom1"]This is actually a SAT problem....

The average(arithmetic mean) of nine numbers is 9. When a tenth number is added the average of the ten numbers is also 9. What is the tenth number ?

A) 0
B) 9/10
c) 10/9
d) 9
e) 10

sum = 81
sum+x = 90 thus x = 9
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Re: Average [#permalink] New post 13 Aug 2009, 04:39
Well Sat or no Sat......x has to be 9 in this case......
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Re: Average [#permalink] New post 25 Aug 2009, 22:53
IMO D....
OA must be wrong....
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Re: Average [#permalink] New post 26 Aug 2009, 00:21
I also think OA must be wrong.

\frac{81 + X}{10} = 9
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Re: Average   [#permalink] 26 Aug 2009, 00:21
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