document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1185556>Question 1185556</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A woman has a total of $7,000 to invest. She invests part of the money in an account that pays 8% per year and the rest in an account that pays 10% per year. If the interest earned in the first year is $620, how much did she invest in each account?  </SPAN>\n" );
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document.write( "(1) Using a traditional formal algebraic approach (good practice in solving problems using algebra)....<br><BR>\n" );
document.write( "x = amount invested at 8%<BR>\n" );
document.write( "7000=x = amount invested at 10%<br><BR>\n" );
document.write( "The total interest was $620:<br><BR>\n" );
document.write( ".08(x)+.10(7000-x)=620<br><BR>\n" );
document.write( "Solve using basic algebra -- probably start by multiplying everything by 100 to clear the decimals....<br><BR>\n" );
document.write( "(2) A quick and easy way to solve any 2-part \"mixture\" problem like this, if a formal algebraic solution is not required....<br><BR>\n" );
document.write( "$7000 all invested at 8% would yield $560 interest; all invested at 10% would yield $700 interest; the actual interest was $620.<br><BR>\n" );
document.write( "The amount invested at each rate is exactly determined by where the actual interest of $620 lies between the two extremes $560 and $700.<br><BR>\n" );
document.write( "(1) Picture the three interest amounts $560, $620, and $700 on a number line and observe/calculate that $620 is 60/140 = 3/7 of the way from $560 to $700.<BR>\n" );
document.write( "(2) That means 3/7 of the total was invested at the higher rate.<br><BR>\n" );
document.write( "ANSWER: 3/7 of the $7000, or $3000, was invested at 10%; the other $4000 at 8%.<br><BR>\n" );
document.write( "CHECK: .10(3000)+.08(4000)=300+320=620<br><BR>\n" );
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