document.write( "Question 963566: If 6 men take 3 days to dig 8 ditches, then how long would it take 4 men to dig 10 ditches? Assume that all the ditches are the same size and take equally long to dig, and that all the men work at the same steady rate.
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Algebra.Com's Answer #588693 by Edwin McCravy(20054)\"\" \"About 
You can put this solution on YOUR website!
If 6 men take 3 days to dig 8 ditches, then how long would it take 4 men to dig
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document.write( "You got this far:\r\n" );
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\n" ); document.write( "The rate of 1 ditch by the 6 men is 3/8 of A DITCH PER day. \r
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document.write( "So 4 men would take 1/4 of 2 1/4 days for 1 ditch.\r\n" );
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document.write( "That's (1/4)(2 1/4) = (1/4)(9/4) = (9/16) of a day for 1 ditch.\r\n" );
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document.write( "Therefore it would take them 10 times as long for 10 ditches.\r\n" );
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document.write( "(9/16)(10) = 90/16 = 45/8 = 5 5/8 days.\r\n" );
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document.write( "Another way to do it is:\r\n" );
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document.write( "The LCM of 6 men and 4 men is 12 men\r\n" );
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document.write( "Since 6 men take 3 days to dig 8 ditches, then \r\n" );
document.write( "12 men take 1 1/2 or 3/2 days to dig 8 ditches.\r\n" );
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document.write( "So 4 men would take 3 times as long or 9/2 days to dig 8 ditches\r\n" );
document.write( "So 4 men would take 1/8th of 9/2 days, or 9/16ths of a day to dig 1 ditch.\r\n" );
document.write( "So to dig 10 ditches, it would take them 10 times as long or  \r\n" );
document.write( "(9/16)(10) = 90/16 = 45/8 = 5 5/8 days.\r\n" );
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document.write( "Another way to do it is:\r\n" );
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document.write( "The LCM of 8 ditches and 10 ditches is 40 ditches\r\n" );
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document.write( "Since 6 men take 3 days to dig 8 ditches, then \r\n" );
document.write( "6 men take 15 days to dig 40 ditches.\r\n" );
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document.write( "So 6 men would take 15/4 days to dig 10 ditches\r\n" );
document.write( "So 2 men would take 3 times as long or 45/4 days to dig 10 ditches.\r\n" );
document.write( "So 4 men would take 1/2 of 45/4 days or 45/8 or 5 5/8 days to dig \r\n" );
document.write( "10 ditches.\r\n" );
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document.write( "The most straight-forward way is to use the worker-time-job formula, which is:\r\n" );
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document.write( "\"%28W%5B1%5DT%5B1%5D%29%2FJ%5B1%5D\"\"%22%22=%22%22\"\"%28W%5B2%5DT%5B2%5D%29%2FJ%5B2%5D\"\r\n" );
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document.write( "where\r\n" );
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document.write( "W1 = the number of workers in the first situation.\r\n" );
document.write( "T1 = the number of time units (days in this case) in the first situation.\r\n" );
document.write( "J1 = the number of jobs in the first situation.\r\n" );
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document.write( "W2 = the number of workers in the second situation.\r\n" );
document.write( "T2 = the number of time units (days in this case) in the second situation.\r\n" );
document.write( "J2 = the number of jobs in the second situation.\r\n" );
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document.write( "W1 =  6             W2 = 4     \r\n" );
document.write( "T1 =  3             T2 = the unknown quantity\r\n" );
document.write( "J1 =  8             J2 = 10\r\n" );
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document.write( "\"%28W%5B1%5DT%5B1%5D%29%2FJ%5B1%5D\"\"%22%22=%22%22\"\"%28W%5B2%5DT%5B2%5D%29%2FJ%5B2%5D\"\r\n" );
document.write( "\r\n" );
document.write( "\"%286%2A3%29%2F8\"\"%22%22=%22%22\"\"%284%2AT%5B2%5D%29%2F10\"\r\n" );
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document.write( "\"18%2F8\"\"%22%22=%22%22\"\"%284%2AT%5B2%5D%29%2F10\"\r\n" );
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document.write( "\"18%2F8\" reduces to \"9%2F4\" and \"4%2F10\" reduces to \"2%2F5\"\r\n" );
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document.write( "\"9%2F4\"\"%22%22=%22%22\"\"%282%2AT%5B2%5D%29%2F5\"\r\n" );
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document.write( "Cross-multiply\r\n" );
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document.write( "8T2 = 45\r\n" );
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document.write( "Divide both sides by 8\r\n" );
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document.write( "T2 = 45/8 or 5 5/8.\r\n" );
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document.write( "Answer: 5 5/8 days.\r\n" );
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document.write( "Edwin
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