document.write( "Question 1124862: A hardware supplier manufactures three kinds of clamps, types A, B, and C. Production restrictions force it to make 10 more type C clamps than the total of the other types and twice as many type B clamps as type A clamps. The shop must produce 310 clamps per day. How many of each type are made per day? \n" ); document.write( "

Algebra.Com's Answer #741144 by ikleyn(52866) $\"\"$ $\"About$

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document.write( "Take off for a minute 10 clamps of type C.\r\n" );
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document.write( "Then the rest amount of clamps C will be exactly equal to sum of types A and B,\r\n" );
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document.write( "and altogether (C-10), A and B will comprise 300 clamps.\r\n" );
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document.write( "Hence,  C = 150 + 10, while A + B = 150.\r\n" );
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document.write( "We also are given that the number of clamps of the type B is twice as many as that of type A.\r\n" );
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document.write( "Hence, we just get the\r\n" );
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document.write( "Answer.  A = 50,  B = 100  and  C = 160.\r\n" );
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\n" ); document.write( "\n" ); document.write( "The problem is SO SIMPLE, that equations are not needed.\r
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