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03 Jun 2021, 05:15
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If the average (arithmetic mean) of k and k + 5 is b and if the average of k and k − 9 is c, what is the average of b and c?
A. k − 2
B. k − 1
C. k
D. k + 1 /2
E. 2k
A. k − 2
B. k − 1
C. k
D. k + 1 /2
E. 2k
03 Jun 2021, 07:00
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Asked: If the average (arithmetic mean) of k and k + 5 is b and if the average of k and k − 9 is c, what is the average of b and c?
k + k + 5 = 2b
k + k - 9 = 2c
2b + 2c = 4k - 4
Average of b & c = (b + c)/2 = k - 1
IMO B
k + k + 5 = 2b
k + k - 9 = 2c
2b + 2c = 4k - 4
Average of b & c = (b + c)/2 = k - 1
IMO B
04 Jun 2021, 02:03
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Method I:
Sum = Average * Total
=> 2k + 5 = 2b
=> 2k - 9 = 2c
Adding both: 4k - 4 = 2b + 2c
=> 2(b + c) = 4 (k - 1)
=> \(\frac{b + c }{ 2} = k - 1\)
Answer B
Method II: Take k = 9
=> \(b = \frac{9 + 14 }{ 2} = 11.5\)
=> \(c = \frac{9 + 0 }{ 2} = 4.5 \)
=> \(\frac{b + c }{ 2} = \frac{11.5 + 4.5 }{ 2} = \frac{16}{2} = 8 \)
=> 8 is 9 - 1 can be expressed as 'k - 1'.
Answer B
Sum = Average * Total
=> 2k + 5 = 2b
=> 2k - 9 = 2c
Adding both: 4k - 4 = 2b + 2c
=> 2(b + c) = 4 (k - 1)
=> \(\frac{b + c }{ 2} = k - 1\)
Answer B
Method II: Take k = 9
=> \(b = \frac{9 + 14 }{ 2} = 11.5\)
=> \(c = \frac{9 + 0 }{ 2} = 4.5 \)
=> \(\frac{b + c }{ 2} = \frac{11.5 + 4.5 }{ 2} = \frac{16}{2} = 8 \)
=> 8 is 9 - 1 can be expressed as 'k - 1'.
Answer B
