document.write( "Question 117100: Janet invested 10,000, part at 15% and part at 7%. If the total interest at the end of the year is $1,260, how much did she invest @ 15%?
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document.write( "a) 3,000
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document.write( "b) 8,000
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document.write( "c) 7,000
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document.write( "d) 6,000\r
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document.write( "Show all work \n" );
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Algebra.Com's Answer #85203 by ptaylor(2198)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Interest(I) equals Principal(P) times Rate(R) times Time(T) or I=PRT \n" ); document.write( "Let x=amount invested at 15% \n" ); document.write( "Then 10,000-x=amount invested at 7% \n" ); document.write( "Interest at 15% =x*0.15*1 or 0.15x \n" ); document.write( "Interest at 7%=(10,000-x)*0.07*1=0.07(10,000-x)\r \n" ); document.write( "\n" ); document.write( "Now we are told that the total interest for the year was $1260 so our equation to solve is:\r \n" ); document.write( "\n" ); document.write( "0.15x+0.07(10,000-x)=1260 get rid of parens\r \n" ); document.write( "\n" ); document.write( "0.15x+700-0.07x=1260 subtract 700 from both sides \n" ); document.write( "0.15x+700-700-0.07x=1260-700 collect like terms\r \n" ); document.write( "\n" ); document.write( "0.08x=560 divide both sides by 0.08 \n" ); document.write( "x=$7000----amount invested at 15% \n" ); document.write( "10,000-x=10,000-7000=$3000---------------amount invested at 7%\r \n" ); document.write( "\n" ); document.write( "CK\r \n" ); document.write( "\n" ); document.write( "7000*0.15+3000*0.07=1260 \n" ); document.write( "1050+210=1260 \n" ); document.write( "1260=1260\r \n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( " \n" ); document.write( " |