A set of five different natural numbers has a mean of 6 and a median of 6. What is the largest possible number in the set?
Answers
Answer:
Note that the numbers in the set have to be distinct.
If 6 numbers have a mean of 13, then their sum is 6∗13=786∗13=78.
If the median of the set is 10, then the mean of the center two numbers has to be 10. So the sum of the center two numbers is 2∗10=202∗10=20.
To get the largest possible number in the set, you want to start of with a 1 and 2, because you need larger numbers to balance out the equation:
1,2,w,x,y,z1,2,w,x,y,z
Since the sum of the center two numbers is 20, we can merge the center two numbers into one:
1,2,20,y,z1,2,20,y,z
For the “y” in the set, we can’t make it smaller than any of the two numbers, or else the median will change. This means to make the “z” as large as possible, we need “w” to be 9, “x” to be 11, and “y” to be 12:
1,1,9,11,12,z1,1,9,11,12,z
If we do the math, “z” should be 78−12−11−9−2−1=78−35=4378−12−11−9−2−1=78−35=43.
So “z” is 43, which is our answer
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Answer:
Note that the numbers in the set have to be distinct.
If 6 numbers have a mean of 13, then their sum is 6∗13=78 .
If the median of the set is 10, then the mean of the center two numbers has to be 10. So the sum of the center two numbers is 2∗10=20 .
To get the largest possible number in the set, you want to start of with a 1 and 2, because you need larger numbers to balance out the equation:
1,2,w,x,y,z
Since the sum of the center two numbers is 20, we can merge the center two numbers into one:
1,2,20,y,z
For the “y” in the set, we can’t make it smaller than any of the two numbers, or else the median will change. This means to make the “z” as large as possible, we need “w” to be 9, “x” to be 11, and “y” to be 12:
1,1,9,11,12,z
If we do the math, “z” should be 78−12−11−9−2−1=78−35=43 .
So “z” is 43, which is our answer.