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The average (arithmetic mean) of the 5 positive integers k, m, r, s, a
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22 Sep 2024, 23:48
22 Sep 2024, 23:48
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The average (arithmetic mean) of the 5 positive integers k, m, r, s, and t is 16, where, k < m < r < s < t. If t = 40, what is the greatest possible value of the median of these 5 integers?
A. 16
B. 18
C. 19
D. 20
E. 22
A. 16
B. 18
C. 19
D. 20
E. 22
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The average (arithmetic mean) of the 5 positive integers k, m, r, s, a
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23 Sep 2024, 13:24
23 Sep 2024, 13:24
1
If the average of the 5 numbers is 16, their sum will be 80
t = 40 the largest.
Using the principle of max/min we will need to reduce k and m to the least possible, therefore
k = 1
m = 2
Since t = 40, that leaves us with 37 for the two remaining values, r and s. Since there are odd number of terms, r will have to be the median so s = 19 and r = 18
Answer B
t = 40 the largest.
Using the principle of max/min we will need to reduce k and m to the least possible, therefore
k = 1
m = 2
Since t = 40, that leaves us with 37 for the two remaining values, r and s. Since there are odd number of terms, r will have to be the median so s = 19 and r = 18
Answer B
Re: The average (arithmetic mean) of the 5 positive integers k, m, r, s, a
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06 Oct 2024, 15:57
06 Oct 2024, 15:57
adewale223 wrote:
If the average of the 5 numbers is 16, their sum will be 80
t = 40 the largest.
Using the principle of max/min we will need to reduce k and m to the least possible, therefore
k = 1
m = 2
Since t = 40, that leaves us with 37 for the two remaining values, r and s. Since there are odd number of terms, r will have to be the median so s = 19 and r = 18
Answer B
t = 40 the largest.
Using the principle of max/min we will need to reduce k and m to the least possible, therefore
k = 1
m = 2
Since t = 40, that leaves us with 37 for the two remaining values, r and s. Since there are odd number of terms, r will have to be the median so s = 19 and r = 18
Answer B
What is principle of max/min ??Can you please Elaborate??
