A total of $12,000 is invested into two simple interest accounts. On one account the annual simple interest rate is 12%, while on the second account the annual interest rate is 18%. How much should be invested at 12% so that the interest earned is the same?
since 12% is 2/3 as much interest, 3/2 as much should be invested there.
Or, since the interest ratio is 2:3, the investment should be 3:2
So, 40% at 18% and 60% at 12%
.12 * 7200 = .18 * 4800 = 864
or, in the usual algebraic way,
.12x = .18(12000-x)
To determine how much should be invested at 12% so that the interest earned is the same, we need to set up an equation based on the given information.
Let's assume the amount to be invested at 12% is x, so the amount to be invested at 18% would be (12,000 - x), since the total investment is $12,000.
Now, let's calculate the interest earned on each account:
Interest earned on the account with 12% interest rate = x * 0.12
Interest earned on the account with 18% interest rate = (12,000 - x) * 0.18
Since we want the interest earned to be the same, we can set up the equation:
x * 0.12 = (12,000 - x) * 0.18
To solve for x, we can simplify the equation:
0.12x = 0.18(12,000 - x)
0.12x = 2,160 - 0.18x
0.12x + 0.18x = 2,160
0.30x = 2,160
Dividing both sides of the equation by 0.30:
x = 2,160 / 0.30
x = 7,200
Therefore, $7,200 should be invested at a 12% interest rate to earn the same interest as the account with an 18% interest rate.