## Related questions with answers

Mom found an open box of her children's favorite candy bars. She decides to apportion the candy bars among her three youngest children according to the number of minutes each child spend doing homework during the week. (a) Suppose that there were 10 candy bars in the box. Given that Bob did homework for a total of 54 minutes, Peter did homework for a total of 243 minutes, and Ron did homework for a total of 703 minutes, apportion the 10 candy bars among the children using Hamilton's method. (b) Suppose that just before she hands out the candy bars, mom finds one extra candy bar. Using the same total minutes as in (a), apportion now the 11 candy bars among the children using Hamilton's method. (c) Two results of (a) and (b) illustrate one of the paradoxes of Hamilton's method. Which one? Explain.

Solution

Verified**(a)**

We will determine the total number of hours spent on homework as follows:

$54+243+703=1000 \textrm{ minutes}$

We are given of $10$ candy bars. Hence,

$\textrm{Standard Divisor}=\dfrac{1000}{10}=100$

Below is apportionment under Hamilton's method:

$\small{\begin{array}{|c|c|c|c|c|c|c|} \hline \text{Child} & \begin{array}{c}\text{Minutes}\\ \text{spent}\\\text{ on}\\ \text{homework}\end{array} & \begin{array}{c}\text{Standard}\\\text{ Quota}\end{array} & \begin{array}{c}\text{Lower}\\ \text{quota}\end{array} & \text{Residue} & \begin{array}{c}\text{Order of}\\\text{ Surplus}\end{array} & \begin{array}{c}\text{Apportion}\\\text{-ment}\end{array} \\ \hline & & & & & & \\ \text{Bob} & 54 & \dfrac{56}{100}=0.54 & 0 & 0.54 & \text{First} & 1 \\ & & & & & & \\ \hline & & & & & & \\ \text{Peter} & 243 & \dfrac{243}{100}=2.43 & 2 & 0.43 & & 2 \\ & & & & & & \\ \hline & & & & & & \\ \text{Ron} & 703 & \dfrac{703}{100}=7.03 & 7 & 0.03 & & 7 \\ & & & & & & \\ \hline \text{Total} & 1000 & 10 & 9 & 1 & & 10 \\ \hline \end{array}}$

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