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03 Jun 2021, 05:00
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The average (arithmetic mean) of 17 numbers is j. If two of the numbers are k and m, what is the average of the remaining 15 numbers in terms of j, k and m?
A. \(\frac{k+m}{17}\)
B. \(17j+k+m\)
C. \(\frac{16j-k-m}{17}\)
D. \(\frac{17j-k-m}{15}\)
E. \(\frac{17(k-m)-j}{15}\)
A. \(\frac{k+m}{17}\)
B. \(17j+k+m\)
C. \(\frac{16j-k-m}{17}\)
D. \(\frac{17j-k-m}{15}\)
E. \(\frac{17(k-m)-j}{15}\)
03 Jun 2021, 05:10
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Bunuel
The average (arithmetic mean) of 17 numbers is j. If two of the numbers are k and m, what is the average of the remaining 15 numbers in terms of j, k and m?
A. \(\frac{k+m}{17}\)
B. \(17j+k+m\)
C. \(\frac{16j-k-m}{17}\)
D. \(\frac{17j-k-m}{15}\)
E. \(\frac{17(k-m)-j}{15}\)
A. \(\frac{k+m}{17}\)
B. \(17j+k+m\)
C. \(\frac{16j-k-m}{17}\)
D. \(\frac{17j-k-m}{15}\)
E. \(\frac{17(k-m)-j}{15}\)
Let F = the sum of the 15 numbers that don't include k and m
So, the average of the 15 numbers \(= \frac{F}{15}\)
The average (arithmetic mean) of 17 numbers is j.
We can write: \(\frac{F + k + m}{17} = j\)
What is the average of the remaining 15 numbers in terms of j, k and m?
Take: \(\frac{F + k + m}{17} = j\)
Multiply both sides of the equation by 17 to get: \(F + k + m= 17j\)
Subtract k and m from both sides: \(F= 17j - k - m\)
Divide both sides by 15: \(\frac{F}{15}= \frac{17j - k - m}{15}\)
Answer: D
03 Jun 2021, 10:10
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Bunuel
The average (arithmetic mean) of 17 numbers is j. If two of the numbers are k and m, what is the average of the remaining 15 numbers in terms of j, k and m?
A. \(\frac{k+m}{17}\)
B. \(17j+k+m\)
C. \(\frac{16j-k-m}{17}\)
D. \(\frac{17j-k-m}{15}\)
E. \(\frac{17(k-m)-j}{15}\)
A. \(\frac{k+m}{17}\)
B. \(17j+k+m\)
C. \(\frac{16j-k-m}{17}\)
D. \(\frac{17j-k-m}{15}\)
E. \(\frac{17(k-m)-j}{15}\)
Average of 17 numbers = j
Sum of 17 numbers = 17j
Sum of other 2 numbers = k + m
Sum of the 15 numbers = 17j - k - m
Average of 15 numbers = \(\frac{17j - k - m}{15}\) -(D), IMO!
