document.write( "Question 1206065: Ms. Burke invested 20,000 in two accounts, one yielding 7% interest and the other one yielding 11%. If she received a total of $1560 in interest at the end of the year, how much did she invest in each account?\r
\n" ); document.write( "\n" ); document.write( "The amount invested at 7% was
\n" ); document.write( "The amount invested at 11% was
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Algebra.Com's Answer #843259 by greenestamps(13209)\"\" \"About 
You can put this solution on YOUR website!


\n" ); document.write( "The standard setup for solving the problem algebraically....:

\n" ); document.write( "x = amount that yielded 7% interest
\n" ); document.write( "20000-x = amount that yielded 11% interest

\n" ); document.write( "The total interest earned was $1560:

\n" ); document.write( "\".07%28x%29%2B.11%2820000-x%29=1560\"

\n" ); document.write( "I'll leave it to you to finish solving the problem by that method, using basic algebra.

\n" ); document.write( "A quick informal method for solving any 2-part \"mixture\" problem like this, if a formal algebraic solution is not required....

\n" ); document.write( "(1) $1560 interest on an investment of $20,000 is a return of 1560/20000 = 0.078 = 7.8%.
\n" ); document.write( "(2) Using a number line if it helps, observe/calculate that 7.8% is 0.8/4.0 = 1/5 of the way from 7% to 11%.
\n" ); document.write( "(3) That means 1/5 of the total was invested at the higher rate.

\n" ); document.write( "1/5 of $20,000 is $4000.

\n" ); document.write( "ANSWER: $4000 was invested at 11%; the other $16,000 was invested at 7%.

\n" ); document.write( "CHECK: .11(4000)+.07(16000) = 440+1120 = 1560

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