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# The average (arithmetic mean) of a, b, c, d, e is 7. What is

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The average (arithmetic mean) of a, b, c, d, e is 7. What is [#permalink]

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28 Jul 2012, 15:28
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67% (02:33) correct 33% (00:37) wrong based on 96 sessions

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The average (arithmetic mean) of a, b, c, d, e is 7. What is the average of a, b, c, d, e, f ?

(1) The average of b, c, d, e, f is 9.

(2) The average of b, c, d, e is 11.
[Reveal] Spoiler: OA

Last edited by Bunuel on 29 Jul 2012, 00:34, edited 1 time in total.
Edited the question.
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Re: The average (arithmetic mean) of a,b,c,d,e is 7. What is the [#permalink]

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28 Jul 2012, 20:58
superpus07 wrote:
The average (arithmetic mean) of a,b,c,d,e is 7. What is the average of a,b,c,d,e,f ?

(1) The average of b,c,d,e,f is 9.

(2) The average of b,c,d,e is 11.

Hi

Q says (a+b+c+d+e)/5=7
=>a+b+c+d+e=35----1)

From startement 1:
(b+c+d+e+f)/5=9;
=> b+c+d+e+f=45---- 2)

from 1) and 2) we can't solve for F hence INSUFFICIENT

from statement 2
(b+c+d+e)/4=11
b+c+d+e=44--3)
from 1) and 3) we can't solve for F hence INSUFFICIENT

combining 1 & 2 statements
from 2) and 3) 23 can easily solve for F

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Re: The average (arithmetic mean) of a, b, c, d, e is 7. What is [#permalink]

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29 Jul 2012, 00:43
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The average (arithmetic mean) of a, b, c, d, e is 7. What is the average of a, b, c, d, e, f ?

It's almost always better to express the average in terms of the sum.
Given: $$a+b+c+d+e=5*7$$.
Question: $$a+b+c+d+e+f=?$$ --> $$5*7+f=?$$ So basically we are asked to find the value of $$f$$.

(1) The average of b, c, d, e, f is 9 --> $$b+c+d+e+f=5*9$$. Not sufficient.

(2) The average of b, c, d, e is 11 --> $$b+c+d+e=4*11$$. Not sufficient.

(1)+(2) Subtract (2) from (1): $$f=5*9-4*11=1$$. Sufficient.

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Re: The average (arithmetic mean) of a, b, c, d, e is 7. What is [#permalink]

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21 Dec 2016, 20:27
Nice one.
Here is what i did in this one =>
Mean =35+f/6
We need the value of f

Statement 1=>
b+c+d+e+f=45
No clue of a => no clue of f=>not sufficient

Hence not sufficient

Statement 2-->
No clue of f => Not sufficient

Combing them => f=1
Hence mean = 35+1/6 = 6
Hence sufficient
Hence C

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Re: The average (arithmetic mean) of a, b, c, d, e is 7. What is   [#permalink] 21 Dec 2016, 20:27
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