document.write( "Question 1196875: Hi
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document.write( "Boxes A and B contained 150 marbles altogether. When 1/3 of the marbles in box A were transferred to box B and 12 marbles were taken out from B there were twice as many marbles in B than in A.
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document.write( "How many were in B in the beginning.
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document.write( "Thanks \n" );
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Algebra.Com's Answer #829908 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "12 of the 150 marbles were removed from B at the end, leaving 138 marbles in the two boxes. \n" ); document.write( "At that point, there were twice as many in B than in A -- i.e., 2/3 of the 138 marbles were in B and 1/3 of them in A. That makes 46 in A and 92 in B. \n" ); document.write( "Prior to that, 1/3 of the marbles in A had been moved to B, so the number of marbles remaining in A was 2/3 of the number originally in A. We know that number of marbles was 46; so the number of marbles originally in A was 46*(3/2) = 69. \n" ); document.write( "(Algebraically, that is \n" ); document.write( "And that means the number originally in B was 150-69 = 81. \n" ); document.write( "ANSWER: 81 marbles in B originally \n" ); document.write( "CHECK: \n" ); document.write( "start: A = 69; B = 81 \n" ); document.write( "move 1/3 of the marbles in A (23) to B: A = 69-23 = 46; B = 81+23 = 104 \n" ); document.write( "remove 12 from B: A = 46; B = 104-12 = 92. \n" ); document.write( "92 = 2(46) \n" ); document.write( " \n" ); document.write( " |