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What is the average (arithmetic mean) of x1,x2,x3,x4,x5,x6 ?

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What is the average (arithmetic mean) of x1,x2,x3,x4,x5,x6 ?  [#permalink]

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What is the average (arithmetic mean) of \(x_1\), \(x_2\), \(x_3\), \(x_4\), \(x_5\), \(x_6\) ?

(1) The average of \(x_1\), \(x_2\), \(x_3\), \(x_4\) is 28.
(2) The average of \(x_4\), \(x_5\), \(x_6\) is 21.

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Originally posted by reto on 17 May 2015, 10:10.
Last edited by reto on 18 May 2015, 09:01, edited 2 times in total.
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Re: What is the average (arithmetic mean) of x1,x2,x3,x4,x5,x6 ?  [#permalink]

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New post 17 May 2015, 12:26
Hi reto,

The formatting of this question is unclear. Is the question referring to 6 different variables (using subscript) or is it referring to 1 variable with 6 different exponents?

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Re: What is the average (arithmetic mean) of x1,x2,x3,x4,x5,x6 ?  [#permalink]

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New post 17 May 2015, 20:57
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reto wrote:
What is the average (arithmetic mean) of x1,x2,x3,x4,x5,x6 ?

(1) The average of x1,x2,x3,x4 is 28.
(2) The average of x4,x5,x6 is 21.



Given 6 Numbers : \(x_1,x_2,x_3,x_4,x_5,x_6\) , what is average of these 6 numbers ?

(1) The average of \(x_1,x_2,x_3,x_4\) is 28.
no info about \(x_5,x_6\) . NS
(2) The average of \(x_4,x_5,x_6\) is 21.
no info about \(x_1,x_2,x_3\) . NS

Together :
Brute force approach(we know that \(x_4\) is not known so answer is E but lets see why still insufficient)
\(x_1+x_2+x_3+x_4 = 28*4=112\)
and
\(x_4+x_5+x_6=21*3=63\)

So, \(x_1+x_2+x_3+x_4 +x_5 +x_6 = 175-x_4\)
Average= \(\frac{x_1+x_2+x_3+x_4 +x_5 +x_6}{6} = \frac{175-x_4}{6}\)
we do not know \(x_4\) (it can be 0, or any other value) so the answer is E .
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Re: What is the average (arithmetic mean) of x1,x2,x3,x4,x5,x6 ?  [#permalink]

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New post 21 Dec 2016, 16:55
Here is my solution to this one =>

To get the mean => We need the sum=> x1+x2+...x6

Statement 1=>
We have x1+x2+x3+x4
But er don't know the values of x5 and x6.
Hence insufficient

Statement 2=>
x4+x5+x6
Again insufficient for similar reasons

Combing the two statements =>

x1+x2+x3+x4 = 112
x4+x5+x6
No clue of x4
Hence insufficient

Hence E

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Re: What is the average (arithmetic mean) of x1,x2,x3,x4,x5,x6 ?  [#permalink]

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New post 17 Dec 2017, 13:35
If I think of St 1) x1 ...x4 = 28*4 = 112

St 2) x4+x5..x6 = 21*3 = 63

If i combine and subtract one total from other - the difference (49) is x 4 (which is the overlap)

So we get: x1 + x2 + x3 + 49 + x5 + x6 -->now 112 - 49 = x1 + x2 + x3 and 63 -49 = x5 + x6

Thus we can divide the totals by 6.

Would like to know why E is the answer, and what is wrong with my logic
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Re: What is the average (arithmetic mean) of x1,x2,x3,x4,x5,x6 ?  [#permalink]

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New post 17 Dec 2017, 21:56
Madhavi1990 wrote:
If I think of St 1) x1 ...x4 = 28*4 = 112

St 2) x4+x5..x6 = 21*3 = 63

If i combine and subtract one total from other - the difference (49) is x 4 (which is the overlap)

So we get: x1 + x2 + x3 + 49 + x5 + x6 -->now 112 - 49 = x1 + x2 + x3 and 63 -49 = x5 + x6

Thus we can divide the totals by 6.

Would like to know why E is the answer, and what is wrong with my logic


Hi

So your first equation is: x1+x2+x3+x4 = 112
and your second equn is: x4+x5+x6 = 63

If you add the above two equations, you will get: x1+x2+x3+x4+x5+x6+x4 = 175 Or Total Sum+x4 = 175. Since we dont know x4, we cant find the total sum

Or instead of adding if we subtract the above two equations, x1+x2+x3-x5-x6 = 49. How does it give us the total sum? or how does it give us the value of x4? It doesnt. It only tells us that x4 will be subtracted and thus eliminated, and we are left with x1+x2+x3-x5-x6.. this doesnt help us to get the required sum at all.
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Re: What is the average (arithmetic mean) of x1,x2,x3,x4,x5,x6 ?  [#permalink]

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New post 28 May 2018, 23:46
amanvermagmat wrote:
Madhavi1990 wrote:
If I think of St 1) x1 ...x4 = 28*4 = 112

St 2) x4+x5..x6 = 21*3 = 63

If i combine and subtract one total from other - the difference (49) is x 4 (which is the overlap)

So we get: x1 + x2 + x3 + 49 + x5 + x6 -->now 112 - 49 = x1 + x2 + x3 and 63 -49 = x5 + x6

Thus we can divide the totals by 6.

Would like to know why E is the answer, and what is wrong with my logic


Hi

So your first equation is: x1+x2+x3+x4 = 112
and your second equn is: x4+x5+x6 = 63

If you add the above two equations, you will get: x1+x2+x3+x4+x5+x6+x4 = 175 Or Total Sum+x4 = 175. Since we dont know x4, we cant find the total sum

Or instead of adding if we subtract the above two equations, x1+x2+x3-x5-x6 = 49. How does it give us the total sum? or how does it give us the value of x4? It doesnt. It only tells us that x4 will be subtracted and thus eliminated, and we are left with x1+x2+x3-x5-x6.. this doesnt help us to get the required sum at all.


If we use both (i) and (ii), assuming x1 to x4, each the value is 28, so x4 = 28. If we know x4, then we can find x5 + x6 which is 35. Wouldnt this give us the total x1,x2,x3,x4,x5,x6 ?
Re: What is the average (arithmetic mean) of x1,x2,x3,x4,x5,x6 ? &nbs [#permalink] 28 May 2018, 23:46
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