document.write( "Question 1127979: Tamara invests 8000 in two different accounts. The first account has a simple interest rate of 3% and the second has a simple interest rate of 2%. How much did she invest in each account if the interest earned in them is the same at the end of 1 year? \n" ); document.write( "
Algebra.Com's Answer #744463 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Algebraically, we can let x be the amount invested at 2%; then (8000-x) is the amount invested at 3%. Then since the amounts of interest are the same, \n" ); document.write( " \n" ); document.write( "That equation is easily solved using basic algebra. I'll let you finish. \n" ); document.write( "Here is a different way to solve the problem informally, using ratios and logical reasoning. \n" ); document.write( "Since the ratio of the interest rates is 2:3 and the amounts of interest are the same, the money must be invested in the ratio 3:2. The ratio 3:2 means 3/5 of the money is invested at the lower rate and 2/5 of it at the higher rate. \n" ); document.write( "ANSWER: 3/5 of $8000 = $4800 at 2%; 2/5 o $8000 = $3200 at 3%. \n" ); document.write( " |