Susan invests 3 times as much money at
9% as she does at 6%. If her total interest after 1 year is $1980, how much does she have invested at each rate?
To solve this problem, let's denote the amount Susan invested at 6% as "x" dollars.
According to the problem, Susan invested 3 times as much money at 9% as she did at 6%. Therefore, the amount she invested at 9% would be 3x dollars.
Next, let's calculate the interest earned at each rate.
The interest earned at 6% can be calculated using the formula:
interest_6 = (principal_6 * rate_6 * time) / 100
Since Susan invested x dollars at 6%, the interest earned at 6% would be:
interest_6 = (x * 6 * 1) / 100 = 6x / 100
Similarly, the interest earned at 9% can be calculated using the formula:
interest_9 = (principal_9 * rate_9 * time) / 100
Since Susan invested 3x dollars at 9%, the interest earned at 9% would be:
interest_9 = (3x * 9 * 1) / 100 = 27x / 100
According to the problem, the total interest earned after one year is $1980. Therefore, we can write the equation:
interest_6 + interest_9 = 1980
Substituting the expressions for interest_6 and interest_9, we have:
6x / 100 + 27x / 100 = 1980
Combining like terms, we get:
33x / 100 = 1980
To isolate x, we'll multiply both sides of the equation by 100 / 33:
x = (1980 * 100) / 33
Calculating this expression, we find that x is approximately $6000.
Therefore, Susan has invested $6000 at 6% and 3 times that, which is $18000, at 9%.
Just write the info in math. If $x is invested at 6%, then
.09*3x + .06x = 1980