document.write( "Question 1203293: Brandon invested $20 000. He invested part of the money in a term deposit paying 4% annual interest, three times as much in a government bond paying 5% annual interest, and the rest in a second mortgage paying 7% annual interest. If he earned a total of $1130 interest in one year, how much did he invest at each rate? \n" ); document.write( "

Algebra.Com's Answer #838742 by greenestamps(13198) $\"\"$ $\"About$

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\n" ); document.write( "Here is a very non-standard solution without algebra. With practice, many similar problems can be solved using the ideas demonstrated in this solution.

\n" ); document.write( "The interest of $1130 on a $20,000 investment represents and average rate of return of 1130/20000, which is 5.65%.

\n" ); document.write( "Since he invested 3 times as much at 5% as at 4%, 3/4 of the total invested at those two rates was at the higher rate. That means the average rate of return for those two investments was 3/4 of the way from 4% to 5%; that is 4.75%.

\n" ); document.write( "So we can view the investment as one amount at 4.75% and the other amount at 7%.

\n" ); document.write( "Now look at the three rates of 4.75%, 5.65%, and 7.00% (on a number line, if it helps), and observe/calculate that 5.65% is 0.90/2.25 = 2/5 of the way from 4.75% to 7%. That means 2/5 of the total of $20,000 was invested at the higher rate of 7%. 2/5 of $20,000 is $8000, so $8000 was invested at 7%.

\n" ); document.write( "That leaves $12,000 invested at 4% or 5%; with 3 times as much invested at 5% as at 4%, simple mental calculations show that $3000 was invested at 4% and $9000 at 5%.

\n" ); document.write( "ANSWERS: $3000 at 4%, $9000 at 5%, and $8000 at 7%

\n" ); document.write( "CHECK: .04(3000) + .05(9000) + .07(8000) = 120 + 450 + 560 = 1130

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