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The average (arithmetic mean) of k numbers is 23 and the average
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28 Apr 2021, 18:57
28 Apr 2021, 18:57
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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
A. 21
B. 22
C. 22.2
D. 22.4
E. 22.5
Re: The average (arithmetic mean) of k numbers is 23 and the average
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28 Apr 2021, 19:15
28 Apr 2021, 19:15
2
Avg of \(k\) nos = 23
Avg of \(n\) nos = 20
\(k = 4n\)
Take \(n = 1\) so \(k = 4\)
Total of \(k\) nos = 92 & Total of \(n\) nos = 20
Total of \(k+n\) nos= 92 + 20 = 112
Average = \(\frac{112}{5}\) = 22.4
Answer D
Avg of \(n\) nos = 20
\(k = 4n\)
Take \(n = 1\) so \(k = 4\)
Total of \(k\) nos = 92 & Total of \(n\) nos = 20
Total of \(k+n\) nos= 92 + 20 = 112
Average = \(\frac{112}{5}\) = 22.4
Answer D
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The average (arithmetic mean) of k numbers is 23 and the average
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05 May 2021, 00:33
05 May 2021, 00:33
wkdfkdus 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
A. 21
B. 22
C. 22.2
D. 22.4
E. 22.5
Given:
\(\frac{a_1 + a_2 + ........ + a_{4n}}{4n} = 23\)
i.e. \(a_1 + a_2 + ........ + a_{4n} = 92n\)
and, \(\frac{b_1 + b_2 + ........ + b_n}{n} = 20\)
i.e. \(b_1 + b_2 + ........ + b_n = 20n\)
Now, \(a_1 + a_2 + ........ + a_{4n} + b_1 + b_2 + ........ + b_n = 92n + 20n = 112n\)
Average of \(4n + n = 5n\) numbers will be = \(\frac{112n}{5n} = 22.4\)
Hence, option D
