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04 Jun 2021, 02:15
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
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Question Stats:
83% (01:49) correct
17%
(02:38)
wrong
based on 122
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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
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
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
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.
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.
