dflorez wrote:

List R contains five numbers that have an average value of 55. If the median of the numbers in the list is equal to the mean and the largest number is equal to 20 more than two times the smallest number, what is the smallest possible value in the list?

a. 35

b. 30

c. 25

d. 20

e. 15

:

Any suggestions about how to approach maximization/minimization problems? OA: CAverage of List R\(=\)Median of List R \(=55\)

Given: List R :\([x,A,55,B,2x+20]\)

Average\(=\frac{x+A+55+B+(2x+20)}{5}\)

\(55=\frac{x+A+55+B+(2x+20)}{5}\)

\(x =\frac{55*5 -55-20-A-B}{3}= \frac{200-A-B}{3}\).........(1)

For \(x\) to be minimum, \(A\) and \(B\) have to maximum

There are following constraints on the value of \(A\) and \(B\)

\(x≤A≤55\) and \(55≤B≤2x+20\)

The maximum value of \(A\) can be \(55\) and \(B\) can be \(2x+20\). putting these in (1), we get

\(x= \frac{200-55-(2x+20)}{3}=\frac{125-2x}{3}\)

\(3x=125-2x\)

\(5x=125,x=\frac{125}{5}=25\)

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