document.write( "Question 1008375: If 5 men and 3 boys can reap 23 acres in 4 days and 3 men and 2 boys can reap 7 acres in 2 days , how many boys must assist 7 men in order that they may reap 45 acres in 6 days?
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Algebra.Com's Answer #624096 by ikleyn(52908)\"\" \"About 
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\n" ); document.write( "If 5 men and 3 boys can reap 23 acres in 4 days and 3 men and 2 boys can reap 7 acres in 2 days , how many boys must assist 7 men in order that they may reap 45 acres in 6 days?
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document.write( "Let m be (an unknown) rate of work for one man, measured in \"acres%2Fday\", \r\n" );
document.write( "and let b be (an unknown) rate of work for one boy, measured in same units.\r\n" );
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document.write( "Then we have a system of two linear equations in two unknowns m and b\r\n" );
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document.write( "4*(5m + 3b) = 23,   (1)\r\n" );
document.write( "2*(3m + 2b) =  7.   (2)\r\n" );
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document.write( "Or\r\n" );
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document.write( "20m + 12b = 23,     (1')\r\n" );
document.write( " 6m +  4b =  7.     (2')\r\n" );
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document.write( "To solve it, multiply (2') by 3 and then distract it from (1'). You will get\r\n" );
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document.write( "20m - 18m = 23 - 3*7, or\r\n" );
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document.write( "2m = 2, hence, m = 1. \r\n" );
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document.write( "Substitute it into (2'), and you will get 4b = 7 - 6 = 1, hence, b = \"1%2F4\".\r\n" );
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document.write( "Thus you obtained that 1 man can reap 1 acr in 1 day, while 1 boy can reap \"1%2F4\" of acr in 1 day.\r\n" );
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document.write( "Then 7 men in 6 days can reap 7*6*1 = 42 acres.\r\n" );
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document.write( "How many boys must assist to reap remaining 45-42 = 3 acres acres in 6 days?\r\n" );
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document.write( "Their number is \"3%2F%28%281%2F4%29%2A6%29\" = \"4%2F2\" = 2.\r\n" );
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document.write( "Answer. Two boys.\r\n" );
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