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The average of (n-2), (n-1),(n),(n+1),(n+2) is 20.
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Updated on: 22 Sep 2016, 20:43
Updated on: 22 Sep 2016, 20:43
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26% (00:31) wrong based on 65 sessions
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The average of (n-2), (n-1),(n),(n+1),(n+2) is 20.
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.
Quantity A |
Quantity B |
2n |
16 |
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.
Originally posted by Sonalika42 on 22 Sep 2016, 10:20.
Last edited by Sonalika42 on 22 Sep 2016, 20:43, edited 1 time in total.
Last edited by Sonalika42 on 22 Sep 2016, 20:43, edited 1 time in total.
The average of (n-2), (n-1),(n),(n+1),(n+2) is 20.
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27 Jul 2021, 00:30
27 Jul 2021, 00:30
Hey,
Please read the stem :
The average of (n-2), (n-1),(n),(n+1),(n+2) is 20.
\(\frac{n - 2 + n - 1 + n + n + 1 + n + 2}{5} = 20\)
\(5n = 100\)
\(n = 20\)
Answer A
Please read the stem :
The average of (n-2), (n-1),(n),(n+1),(n+2) is 20.
\(\frac{n - 2 + n - 1 + n + n + 1 + n + 2}{5} = 20\)
\(5n = 100\)
\(n = 20\)
Answer A
Asim41000 wrote:
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.
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.
General Discussion
Re: The average of (n-2), (n-1),(n),(n+1),(n+2) is 20.
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22 Sep 2016, 11:16
22 Sep 2016, 11:16
2
Expert Reply
We have the series (n-2), (n-1),(n),(n+1),(n+2) with average of 20.
So \(\frac{((n-2)+(n-1)+(n)+(n+1)+(n+2))}{5}=20\)
or \(\frac{5n}{5}=20\) or \(n=20\).
So \(2n=40\).
Hence Quantity A is greater.
So \(\frac{((n-2)+(n-1)+(n)+(n+1)+(n+2))}{5}=20\)
or \(\frac{5n}{5}=20\) or \(n=20\).
So \(2n=40\).
Hence Quantity A is greater.
