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Manager
Joined: 30 May 2009
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11 Aug 2009, 09:22
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88% (01:14) correct 13% (00:00) wrong based on 8 sessions

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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
Manager
Joined: 25 Jul 2009
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Schools: NYU, NUS, ISB, DUKE, ROSS, DARDEN
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11 Aug 2009, 09:38
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
Joined: 20 Mar 2008
Posts: 452
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11 Aug 2009, 09:59
Agree with D. For the answer to be A, the problem definition must be different.
SVP
Joined: 05 Jul 2006
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11 Aug 2009, 11:20
[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
Intern
Joined: 11 Aug 2009
Posts: 2
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Kudos [?]: 0 [0], given: 1

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13 Aug 2009, 05:39
Well Sat or no Sat......x has to be 9 in this case......
Manager
Joined: 15 Jun 2009
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25 Aug 2009, 23:53
IMO D....
OA must be wrong....
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Joined: 13 Jul 2009
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Location: Barcelona
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26 Aug 2009, 01:21
I also think OA must be wrong.

$$\frac{81 + X}{10} = 9$$
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Re: Average   [#permalink] 26 Aug 2009, 01:21
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