document.write( "Question 1174215: dr e man invested $10,000, part at 8 1/2% per annum and the rest at 6 3/4% per annum in one year the amiunt earned at the 8 1/% was $14 more than twice the amount earned at the 6 3/4% how much was invested at each rate and how much interest was earned at each rate?
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Algebra.Com's Answer #799613 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Note that the calculations are straightforward (although a bit ugly) once you have set up the problem; by far the most important part of solving the problem is converting the given information into an appropriate equation. \n" ); document.write( "So pay close attention to that part of the solution below. \n" ); document.write( "The total investment is $10,000; let x be the amount invested at 8.5% and ($10,000-x) be the amount invested at 6.75%. Note that is a typical strategy for starting on a problem where the sum of two numbers is given -- one number is x, and the other is the given sum minus x. \n" ); document.write( "The interest from the first investment is 8.5% of x; the interest from the second is 6.75% of ($10,000-x). \n" ); document.write( "The interest from the first investment was $14 more than twice the interest from the second: \n" ); document.write( " \n" ); document.write( "Solve algebraically; or graph the two expressions on a graphing calculator and see that they intersect at x=6200. \n" ); document.write( "So $6200 was invested at 8.5%, and the other $3800 was invested at 6.75%. \n" ); document.write( "To check that.... \n" ); document.write( ".085(6200) = 527 \n" ); document.write( ".0675(3800) = 256.50 \n" ); document.write( "2(256.50)+14 = 513+14 = 527 \n" ); document.write( " \n" ); document.write( " |