document.write( "Question 1018957: sheryl won $60000 on slot machine in oxford. She invested part of the money at 2% and the rest at 3%. in one year she earned a total of $1600 in interest. How much was invested at each rate? \n" ); document.write( "

Algebra.Com's Answer #634932 by Marz157(7) $\"\"$ $\"About$

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Lets call the percent of the money invested in the 2% account \"x\" which means the percent invested in the 3% account is \"1-x\" (this makes sense because if we add together x and 1-x, we just get 1 meaning the total 100% of the money).\r
\n" ); document.write( "\n" ); document.write( "Depending on different values of p, we get different amounts of money at the end. If we find the value of p that makes $1600 we solve the problem.\r
\n" ); document.write( "\n" ); document.write( "If we invest $A at a interest rate of B per year, we make A*B money in interest. \r
\n" ); document.write( "\n" ); document.write( "So, the formula for the money we make in interest is\r
\n" ); document.write( "\n" ); document.write( " $\"60000%2Ax%2A.02+%2B+60000%2A%281-x%29%2A.03+=+1600\"$ \r
\n" ); document.write( "\n" ); document.write( "Solving this for x can be done like this\r
\n" ); document.write( "\n" ); document.write( " $\"x%2A.02+%2B+%281-x%29%2A.03+=+1600+%2F+60000\"$ Dividing by 60000
\n" ); document.write( " $\"x%2A.02+%2B+.03+-+.03+%2A+x+=+.0266666\"$ Expand (1-x) * .03 by multiplying both by .03
\n" ); document.write( " $\"-.01+%2A+x+=+-.003333\"$ Add the x terms together and subract the .03 from the other side.
\n" ); document.write( " $\"x=0.33333\"$ divide both sides by -.01\r
\n" ); document.write( "\n" ); document.write( "So 33.33% (one third) of the money was invested in the 2% interest account. To double check, if we plug this number in for x in the first equation we should see that we get $1600.\r
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