# An easy Problem(or is it?)

• Oct 14th 2009, 11:23 AM
ruthvik
An easy Problem(or is it?)
7 non-negative numbers have a average of 24. What is the largest possible value for the median of the numbers?

I know the sum has to be 168. Anyone know where to go from there?

Help is appreciated
• Oct 14th 2009, 11:50 AM
aidan
Quote:

Originally Posted by ruthvik
7 non-negative numbers have a average of 24. What is the largest possible value for the median of the numbers?

I know the sum has to be 168. Anyone know where to go from there?

Help is appreciated

Quote:

7 non-negative numbers

This can include zero.
Quote:

What is the largest possible value for the median of the numbers?
The median in the MIDDLE value when the values are sorted ascending/descending.
There are 7 numbers; the middle or median value will be the 4th number.
Divide the total (7*24 = 168) by 4 to get the four highest values possible.
Spoiler:

0 0 0 42 42 42 42

If zero is not allowed:
1 1 1 41 41 41 42

.
• Oct 15th 2009, 04:38 AM
swanz
Quote:

Originally Posted by ruthvik
7 non-negative numbers have a average of 24. What is the largest possible value for the median of the numbers?

I know the sum has to be 168. Anyone know where to go from there?

Help is appreciated

this is assuming that all the 7 numbers are different
\$\displaystyle 7*24=168\$

choose the lowest 3 non-negative numbers as first half that is 1,2,3
that gives a total of 6
168-6 = 162
now to give the largest median the four remaining numbers should add up to 162, the way in which the median (4th number) would be the largest would be when all the four remaining numbers are consecutive which ultimately means:
\$\displaystyle n+n+1+n+2+n+3=162\$
\$\displaystyle 4n+6=162\$
\$\displaystyle 4n=156\$
\$\displaystyle n=39\$
ans is 39