document.write( "Question 769376: a famer has $100 doallrs to spend on 100 animals.
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document.write( "he has to end up with 100 animals.
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document.write( "pigs cost $10 each, turkeys cost .50cents and chickens cost 12.5cents
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document.write( "how many of each can he have for his $100? \n" );
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Algebra.Com's Answer #468856 by KMST(5328)![]() ![]() You can put this solution on YOUR website! 1 pig - $10 \n" ); document.write( "1 turkey - 50 cents \n" ); document.write( "1 chicken - 12.5 cents \n" ); document.write( "I am assuming the farmer has to get exactly 100 animals and spend exactly $100 dollars. Otherwise there would be too many possible solutions. \n" ); document.write( " \n" ); document.write( "The easiest strategy seems to be casework (checking the possibilities case by case) with a little thinking to reduce the number of cases. \n" ); document.write( "If each pig costs $10, the farmer could get 0, 1, 2, 3, ....7, 8, 9, 10 pigs. \n" ); document.write( " \n" ); document.write( "Low numbers of pigs will not work. \n" ); document.write( "If the farmer buys only 5 pigs, he would spend $50 om the pigs and would have $50 left to spend, which would get him at least 100 more animals (if he spends the remaining $50 on just turkeys). \n" ); document.write( " \n" ); document.write( "High numbers of pigs will not work either. \n" ); document.write( "Ten pigs would use up the $100 and he would be getting only 10 animals. \n" ); document.write( "Nine pigs would use up $90, and leave $10 left, which at most would get him 80 animals (if he spends it all on chickens). \n" ); document.write( " \n" ); document.write( "That leaves 6, 7, or 8 pigs as the only possibilities. \n" ); document.write( " \n" ); document.write( "Buying 6 pigs uses $60, leaving $40 to buy another 100-6=94 animals. \n" ); document.write( "If the farmer buys 6 pigs, he needs to also buy 94 birds (between turkeys and chickens). \n" ); document.write( "$40 buys only 80 turkeys, so to get 94 birds, the farmer would need some chickens. \n" ); document.write( "At 50 cents=4x(12.5 cents), each turkey cost the same as 4 chickens. \n" ); document.write( "For $40, instead of 80 turkeys, the farmer could get: \n" ); document.write( "80-1=79 turkeys and 4 chickens, for a total of 79+4=83 birds, or \n" ); document.write( "80-2=78 turkeys and 2x4=8 chickens, for a total of 78+8=86 birds, or \n" ); document.write( "80-3=77 turkeys and 3x4=12 chickens, for a total of 77+12=89 birds, or \n" ); document.write( "80-4=76 turkeys and 4x4=16 chickens, for a total of 76+16=92 birds, or \n" ); document.write( "80-5=75 turkeys and 5x4=20 chickens, for a total of 75+20=95 birds. \n" ); document.write( "The number of birds that can be bought for $40 (or for any fixed amount of many varies by 3's (86, 89, 92, 95). \n" ); document.write( "With 6 pigs, there is no way of buying the 94 birds needed. \n" ); document.write( " \n" ); document.write( "Buying 8 pigs does not work either. \n" ); document.write( " \n" ); document.write( "Buying 7 pigs uses $70, leaving $30 to buy another 100-7=93 animals. \n" ); document.write( "If the farmer buys 7 pigs, he needs to also buy 93 birds (between turkeys and chickens). \n" ); document.write( "$30 buys only 60 turkeys, so to get 93 birds, the farmer would need quite a few chickens. \n" ); document.write( "Again, trading each turkey for 4 chickens, the number of birds the farmer can buy for $30 varies by 3's. \n" ); document.write( "He can get 60, 63, 66, 69, ..... \n" ); document.write( "He can numbers that are multiples of 3. \n" ); document.write( "Each time he reduces the number of turkeys by 1, the total number of birds increases by 3. \n" ); document.write( "He needs \n" ); document.write( "He can get 93 birds using 11 less turkeys (60-11=49)and 4x11=44 more chickens. \n" ); document.write( "He could get |