Math 103,
Spring 2011, Solutions to Chapter 3A homework
Problem 12:
The given table is completed by using the fact that each partner values the entire piece of land at
$400,000.
This implies that each
row
(not column!) must add up to $400,000:
s
1
s
2
s
3
s
4
Adams
$90,000
$96,000
$110,000
$104,000
Benson
$98,000
$100,000
$104,000
$98,000
Cagle
$106,000
$102,000
$94,000
$98,000
Duncan
$96,000
$96,000
$112,000
$96,000
For each partner, and share worth at least $100,000 is a fair share.
(a)
s
3
and s
4
are fair shares to Adams.
(b)
s
2
and s
3
are fair shares to Benson.
(c)
s
1
and s
2
are fair shares to Cagle.
(d)
s
3
is the only fair share to Duncan.
(e)
Using the parcels s
3
and s
4
as fair shares, the only possibility for Duncan is to receive s
3
.
The only possibility for s
4
is for it to go to Adams, since no other partner considers it a fair share.
This leaves s
1
and s
2
, of which only s
2
can be given to Benson. Cagle must therefore receive the
only remaining share, s
1
.
__________________________________________________________________
Problem 18:
(a)
Nick likes roast beef twice as much as ham, and likes ham and turkey equally.
That means that each inch of roast beef sandwich is worth twice as much as each inch of ham or
turkey sandwich.
If
x
is how much 1 inch of turkey sub is worth to Nick, then
x
is how much 1 inch of ham sub is worth to him, and
2
x
is how much 1 inch of roast beef sub is worth to him.
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View Full DocumentThus the whole sub is worth
(value of ham)+(value of turkey)+(value of roast beef) = 8
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 Spring '07
 Berkowitz
 Math, Inch, Betty, Jared

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