document.write( "Question 276055: A farmer spends $4,000 to obtain 100 head of livestock.Prices are: calves – $120 each; lambs – $50 each; piglets – $25 each. If he purchased at least one animal of each type and fewer than 10 calves,how many lambs did he buy?\r
\n" ); document.write( "\n" ); document.write( "a 44 b 46 c 48 d 50 e None of the above
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Algebra.Com's Answer #201325 by ankor@dixie-net.com(22740)\"\" \"About 
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A farmer spends $4,000 to obtain 100 head of livestock.Prices are: calves – $120 each; lambs – $50 each; piglets – $25 each.
\n" ); document.write( "If he purchased at least one animal of each type and fewer than 10 calves,how many lambs did he buy?
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\n" ); document.write( "Total animal equation:
\n" ); document.write( "C + L + P = 100
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\n" ); document.write( "Total cost equation
\n" ); document.write( "120C + 50L + 25P = 4000
\n" ); document.write( "Simplify, divide by 5
\n" ); document.write( "24C + 10L + 5P = 800
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\n" ); document.write( "Multiply the 1st equation by 5 subtract from the above, get rid of the pigs!
\n" ); document.write( "24C + 10L + 5P = 800
\n" ); document.write( "5C + 5L + 5P = 500
\n" ); document.write( "-----------------------Subtraction, eliminates p
\n" ); document.write( "19C + 5L = 300
\n" ); document.write( "5L = 300 - 19C
\n" ); document.write( "L = \"300%2F5\" - \"19%2F5\"C
\n" ); document.write( "L = 60 - \"19%2F5\"C
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\n" ); document.write( "We know L has to be an integer, therefore C has to be a multiple of 5
\n" ); document.write( "they tell his their are fewer than 10 Calves so obviously C = 5
\n" ); document.write( "L = 60 - \"19%2F5\"5
\n" ); document.write( "L = 60 - 19
\n" ); document.write( "L = 41 lambs
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\n" ); document.write( "Find the pigs
\n" ); document.write( "100 - 5 - 41 = 54 pigs
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\n" ); document.write( "Check solution by finding the total cost
\n" ); document.write( "120(5) + 41(50) + 25(54) =
\n" ); document.write( "600 + 2050 + 1350 = 4000
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