103s11chap3Asolutions

103s11chap3Asolutions - Math 103, Spring 2011, Solutions to...

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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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Thus the whole sub is worth (value of ham)+(value of turkey)+(value of roast beef) = 8
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This note was uploaded on 01/20/2012 for the course MATH 103 taught by Professor Berkowitz during the Spring '07 term at Rutgers.

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103s11chap3Asolutions - Math 103, Spring 2011, Solutions to...

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