SOLUTION: 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 di
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-> SOLUTION: 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 di
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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?
The amount invested at 7% was
The amount invested at 11% was Answer by greenestamps(13203) (Show Source):
The standard setup for solving the problem algebraically....:
x = amount that yielded 7% interest
20000-x = amount that yielded 11% interest
The total interest earned was $1560:
I'll leave it to you to finish solving the problem by that method, using basic algebra.
A quick informal method for solving any 2-part "mixture" problem like this, if a formal algebraic solution is not required....
(1) $1560 interest on an investment of $20,000 is a return of 1560/20000 = 0.078 = 7.8%.
(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%.
(3) That means 1/5 of the total was invested at the higher rate.
1/5 of $20,000 is $4000.
ANSWER: $4000 was invested at 11%; the other $16,000 was invested at 7%.
