Question 1193974: How can $69,000 be invested, part at 12% annual simple interest and the remainder at 11% annual simple interest, so that the interest earned by the two accounts will be equal?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52775)   (Show Source):
You can put this solution on YOUR website! .
How can $69,000 be invested, part at 12% annual simple interest and the remainder at 11% annual simple interest,
so that the interest earned by the two accounts will be equal?
~~~~~~~~~~~~~~
x dollars invested at 12%; the rest (69000-x) dollars invested at 11%.
Write equation as you read the problem
0.12x = 0.11*(69000-x).
Simplify and find x
0.12x = 0.11*69000 - 0.11x
0.12x + 0.11x = 0.11*69000
0.23x = 7590
x = 7590/0.23 = 33000.
ANSWER. $33000 invested at 12%; the rest 69000-33000 = 36000 invested at 11%.
CHECK. 0.12*33000 = 3960 dollars, the annual interest from the 12% investment.
0.11*36000 = 3960 dollars, the annual interest from the 11% investment.
the interest is the same, so the answer is correct.
Solved.
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
The solution from tutor @ikleyn is a typical formal algebraic solution; you should understand it and know how to do it.
But a bit of logical reasoning gives us an easier way to solve the problem.
If the amounts of interest from the investments at 12% and 11% are equal, then the amounts invested in the two accounts must be in the ratio 11:12. So
11x = amount invested at 12%
12x = amount invested at 11%
The total amount invested was $69,000:
11x+12x=69000
23x=69000
x=3000
ANSWER: The amounts of interest are equal if 11x=$33,000 is invested at 12% and 12x=$36,000 is invested at 11%.
|
|
|
