document.write( "Question 22826: suppose five bales of hay are weighed two at a time in all possible ways. the weights in pounds are 110, 112, 113, 114, 115, 116, 117, 118, 120, and 121. how much does each bale weigh \n" ); document.write( "
Algebra.Com's Answer #11216 by khwang(438)\"\" \"About 
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suppose five bales of hay are weighed two at a time in all possible ways. the weights in pounds are 110, 112, 113, 114, 115, 116, 117, 118, 120, and 121. how much does each bale weigh\r
\n" ); document.write( "\n" ); document.write( " Let them be A,B,C,D & E. (in increasing order)
\n" ); document.write( " Since each of them appeared in the list of partial sums 4 times(why ?)
\n" ); document.write( " we have 4(A+B+C+D+E)=\"SIGMA\" 110, 112, 113, 114, 115, 116, 117, 118, 120, and 121 = 1156.\r
\n" ); document.write( "\n" ); document.write( " So, A+B+C+D+E = 289.\r
\n" ); document.write( "\n" ); document.write( " We see that A+B = 110 & D+E = 121, so C = 289-(110+121) = 58.
\n" ); document.write( " Also, since A+C = 112, we get A = 112-58= 54,B = 110 - 54 = 56.
\n" ); document.write( " Similarly, C+E = 120, E = 120 - 58 = 62, D = 121 - 62 = 59.\r
\n" ); document.write( "\n" ); document.write( " Answer: 54, 56, 58, 59, 62.
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\n" ); document.write( " Kenny \r
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