document.write( "Question 1163681: There are two boxes of balls. The 1st box has 15 more than the 2nd box. All balls in the 2nd box are red. 2/5 of the first box are red balls. There are 69 red balls all together. How many balls in total from the 2 boxes? \n" ); document.write( "

Algebra.Com's Answer #787848 by ikleyn(52915) $\"\"$ $\"About$

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document.write( "    0.4x + y = 69     (1)   (counting all red balls:  69 red balls altogether)\r\n" );
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document.write( "       x - y = 15     (2)   (The 1st box has 15 more balls than the 2nd box.)\r\n" );
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document.write( "--------------------------------   Add the equations\r\n" );
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document.write( "    1.4x     = 69 + 15 = 84\r\n" );
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document.write( "       x               = 84/1.4 = 60.\r\n" );
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document.write( "Thus, first box has 60 balls.\r\n" );
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document.write( "The second box has  60 - 15 = 45 balls  (and they all are red, according to the condition).\r\n" );
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document.write( "The first box has  0.4*60 = 24 red balls.\r\n" );
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document.write( "The total number of red balls is  24 + 45 = 69.   <<<---===  It is THE CHECK (!) and it is CORRECT (!)\r\n" );
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document.write( "ANSWER.  There are 60 + 45 = 105 balls, in all.\r\n" );
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\n" ); document.write( "\n" ); document.write( "This problem has an underwater stone.\r
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\n" ); document.write( "\n" ); document.write( "One condition is EXCESSIVE, but FORTUNATELY, as I checked it, this excessive condition is CONSISTENT
\n" ); document.write( "with other two basic conditions,\r
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\n" ); document.write( "\n" ); document.write( "making the problem CORRECTLY POSED.\r
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\n" ); document.write( "\n" ); document.write( "When the problem has an excessive condition, the check becomes not only desired: it becomes NECESSARY.\r
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\n" ); document.write( "\n" ); document.write( "FORTUNATELY (again), as I checked it, this excessive condition is CONSISTENT with other two basic conditions,\r
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\n" ); document.write( "\n" ); document.write( "making the problem CORRECTLY POSED.\r
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