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# Average

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Manager
Joined: 30 May 2009
Posts: 218
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Kudos [?]: 120 [0], given: 0

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11 Aug 2009, 08:22
00:00

Difficulty:

(N/A)

Question Stats:

100% (01:14) correct 0% (00:00) wrong based on 7 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
Posts: 116
Schools: NYU, NUS, ISB, DUKE, ROSS, DARDEN
Followers: 5

Kudos [?]: 229 [0], given: 17

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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 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: 454
Followers: 1

Kudos [?]: 119 [0], given: 5

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11 Aug 2009, 08:59
Agree with D. For the answer to be A, the problem definition must be different.
SVP
Joined: 05 Jul 2006
Posts: 1743
Followers: 6

Kudos [?]: 317 [0], given: 49

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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
Intern
Joined: 11 Aug 2009
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 1

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13 Aug 2009, 04:39
Well Sat or no Sat......x has to be 9 in this case......
Manager
Joined: 15 Jun 2009
Posts: 163
Followers: 2

Kudos [?]: 50 [0], given: 8

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25 Aug 2009, 22:53
IMO D....
OA must be wrong....
Current Student
Joined: 13 Jul 2009
Posts: 145
Location: Barcelona
Schools: SSE
Followers: 17

Kudos [?]: 372 [0], given: 22

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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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