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If the eleven consecutive integers are listed from least to greatest,

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If the eleven consecutive integers are listed from least to greatest,  [#permalink]

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New post 01 Jul 2018, 19:45
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

71% (00:58) correct 29% (01:40) wrong based on 38 sessions

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If the eleven consecutive integers are listed from least to greatest, what is the average (arithmetic mean) of eleven integers?

(1) The average of the first nine integers is 7
(2) The average of the last nine integers is 9

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Re: If the eleven consecutive integers are listed from least to greatest,  [#permalink]

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New post 01 Jul 2018, 19:59
1
IMO Option D.

Let integers be n, n+1, n+2, ...., n +10
Average of 11 integers = (1st + last term)/2 = (n + n + 10)/ 2. Hence question just needs the value of n to be found.

Stmt 1: n + (n+1) + ... (n+8) = 7*9 = 63 => n can be found. Hence Sufficient
Stmt 2: (n+2) + (n+ 3) + ... (n+10) = 9*9 = 81 => n can be found. hence Sufficient.

Hence D.
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Re: If the eleven consecutive integers are listed from least to greatest, &nbs [#permalink] 01 Jul 2018, 19:59
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If the eleven consecutive integers are listed from least to greatest,

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