(a) Suppose that there were 11 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 11 candy bars among the children using Hamilton’s method. (b) Suppose that before mom hands out the candy bars, the children decide to spend a “little” extra time on homework. Bob puts in an extra 2 minutes (for a total of 56 minutes), Peter an extra 12 minutes (for a total of 255 minutes), and Ron an extra 86 minutes (for a total of 789 minutes). Using these new totals, apportion the 11 candy bars among the children using Hamilton’s method. (c) The results of (a) and (b) illustrate one of the paradoxes of Hamilton’s method. Which one? Explain.

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(a) Suppose that there were 11 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 11 candy bars among the children using
Hamilton’s method.
(b) Suppose that before mom hands out the candy bars,
the children decide to spend a “little” extra time on
homework. Bob puts in an extra 2 minutes (for a total
of 56 minutes), Peter an extra 12 minutes (for a total
of 255 minutes), and Ron an extra 86 minutes (for a total
of 789 minutes). Using these new totals, apportion the
11 candy bars among the children using Hamilton’s
method.
(c) The results of (a) and (b) illustrate one of the paradoxes
of Hamilton’s method. Which one? Explain.

Expert Solution
Step 1

a). Total number of hours spent on homework:

54+243+703=1000 minutes

Number of candy bars=10

Standard Divisor:

100010=100

The table according to Hamilton's method:

Child Minutes spent on homework Standard Quota Lower quota Residue Order of surplus Apportionment
Bob 54 54100=0.54 0 0.54 First 1
Peter 243 243100=2.43 2 0.43   2
Ron 703 703100=7.03 7 0.03   7
Total 1000 10 9 1   10
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