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Re: The average of (n-2), (n-1),(n),(n+1),(n+2) is 20. [#permalink]
A more simple solution is as follows:

The above mentioned sequence is consecutive, so the mean equals the average of the sequence. In this case n is the median and hence, n equals 20 and 2n equals 40 and thus quantity A is greater.
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Re: The average of (n-2), (n-1),(n),(n+1),(n+2) is 20. [#permalink]
Not a relevant ans pls give explanation.

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Re: The average of (n-2), (n-1),(n),(n+1),(n+2) is 20. [#permalink]
Sonalika42 wrote:
The average of (n-2), (n-1),(n),(n+1),(n+2) is 20.

Quantity A
Quantity B
2n
16

the answer is B. As 20 is not average. it is the sum of the series. hence 5n=20 and n=4. so, 2n=2(4)=8. which is less than 18=6. Hence, answer B is correct.


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
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Re: The average of (n-2), (n-1),(n),(n+1),(n+2) is 20. [#permalink]
1
The answer is B.

We have the series (n-2), (n-1),(n),(n+1),(n+2) with SUM of 20.

So ((n−2)+(n−1)+(n)+(n+1)+(n+2))5=20((n−2)+(n−1)+(n)+(n+1)+(n+2))5=20


As 20 is not average. it is the sum of the series.
5n=20
n=4.
so, 2n=2(4)=8. which is less than 18=6. Hence, answer B is correct.
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Re: The average of (n-2), (n-1),(n),(n+1),(n+2) is 20. [#permalink]
1
This problem uses concept of mean and median. Here the numbers are consecutive and total number of terms in the sequence is odd.
So, mean is the mid point, which is n.

So n = 20, or 2n =40 > 16
hence A > B
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Re: The average of (n-2), (n-1),(n),(n+1),(n+2) is 20. [#permalink]
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