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The average (arithmetic mean) of k numbers is 23 and the average (arit

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Bunuel
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The average (arithmetic mean) of k numbers is 23 and the average (arit [#permalink] New post   04 Jun 2021, 02:15
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The average (arithmetic mean) of k numbers is 23 and the average (arithmetic mean) of n numbers is 20. If k=4n, what is the average (arithmetic mean) of k+n numbers?

A. 21
B. 22
C. 22.2
D. 22.4
E. 22.5
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D
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The average (arithmetic mean) of k numbers is 23 and the average (arit [#permalink] New post   04 Jun 2021, 02:57
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(D) is the answer

Sum of k numbers =23k
sum of n numbers = 20n
sum of (K+n) numbers = 23k + 20n
average (arithmetic mean) of k+n numbers =\(\frac{ sum_of_ (K+n) _ numbers }{(k+n)}\)

=\(\frac{23k + 20n }{(k+ n)}\)
k=4n
=\(\frac{112}{5}\) = 22.4

Bunuel wrote:
The average (arithmetic mean) of k numbers is 23 and the average (arithmetic mean) of n numbers is 20. If k=4n, what is the average (arithmetic mean) of k+n numbers?

A. 21
B. 22
C. 22.2
D. 22.4
E. 22.5
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Re: The average (arithmetic mean) of k numbers is 23 and the average (arit [#permalink] New post   04 Jun 2021, 03:16
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Solution:

It is given that the average of \(k \)numbers is \(23\)
So, the sum of \(k \)numbers = \(23k\)

Similarly, It is given that the average of \(n \)numbers is \(20\)
So, the sum of \(n \)numbers = \(20n\)

The average of k+n numbers = \(\frac{sum of k numbers + sum of n numbers}{k+n}\)
⇒ \(\frac{23k+20n}{k+n}\)
⇒ \(\frac{23(4n)+20n}{4n+n}\) (since \(k=4n\))
⇒ \(\frac{112n}{5n}\)
⇒ \(22.4\)

Hence the right answer is Option D.
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Re: The average (arithmetic mean) of k numbers is 23 and the average (arit [#permalink] New post   04 Jun 2021, 03:23
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Sum of sequence k: 23*k = 23k which then becomes 23k*4 to get it in terms of n 23*4 = 92n

Sum of sequence n: 20*n = 20n

Sum of both sequences: 20n + 92n = 112n

n+k total numbers in sequence: if k = 4n then k+n = 5n

Mean of n+k: 112n/5n = 22 2/5 or 22.4
Answer D
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Re: The average (arithmetic mean) of k numbers is 23 and the average (arit [#permalink] New post   04 Jun 2021, 21:48
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Sum of k nos. = 23k, Sum of n nos. = 20n

Sum of (k+n) nos. = 23k + 20n
mean of (k+n) nos. = \(\frac{23k + 20n}{k+n} = \frac{23*4n + 20n}{4n+n}\) = 18.4 +4 =22.4

So, I think D. :)
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Re: The average (arithmetic mean) of k numbers is 23 and the average (arit [#permalink] New post   05 Jun 2021, 01:51
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Sum = Average * Total

=> The average (arithmetic mean) of k numbers is 23: Sum(k) = 23 * k
=> The average (arithmetic mean) of n numbers is 20: Sum(n) = 20 * n

=> Total sum(k + n) = 23k + 20n

=> k = 4n so total numbers: k + n = 4n +n = 5n
=> Total sum(k + n) = 23(4)n + 20n = 112n

Average: \(\frac{112n }{ 5n} = 22.4\)

Answer D
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Re: The average (arithmetic mean) of k numbers is 23 and the average (arit [#permalink]
05 Jun 2021, 01:51
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