Question

You invested $6000 between two accounts paying 3% and 9% annual interest, respectively. If the total interest earned for the year was $480, how much was invested at each rate?

Answer 1

By setting up equations for the interest earned from each account and the total interest earned, you can solve for the amounts invested at each rate. In this case, $1000 was invested at a 3% interest rate, and $5000 was invested at a 9% interest rate.

Step-by-step explanation:

To solve this problem, we can use a system of equations. Let's say you invested x dollars in the account with a 3% interest rate, and (6000 - x) dollars in the account with a 9% interest rate. The interest earned from the first account would be 0.03x, and the interest earned from the second account would be 0.09(6000 - x). The total interest earned is given as $480, so we have the equation:

0.03x + 0.09(6000 - x) = 480

Simplify and solve for x:

1. 0.03x + 0.09(6000 - x) = 480
2. 0.03x + 540 - 0.09x = 480
3. -0.06x + 540 = 480
4. -0.06x = -60
5. x = -60 / -0.06 = 1000

So you invested $1000 at a 3% interest rate, and $5000 at a 9% interest rate.

Answer 2

$1,000 and $5,000

Step-by-step explanation:

let the 2 amounts be x and y (which add to $6,000)

i.e,

x + y = 6000 -----------(eq1)

assume x is invested at 3% and y is invested at 9%

Assuming simple interest, the following formula is applicable.

I = Prt

where

P = principal ( i.e either $x or $y)

r = rate (i.e 3% or 9%)

t = time elapsed (given as 1 year)

since t = 1 year, the equation reduces to

I = Pr

For $x invested at 3% (i.e 0.03) ---> I = 0.03x

For $y invested at 9% (i.e 0.09) ---> I = 0.09y

given that the total interest for the year = $480,

0.03x + 0.09y = 480 (multiplying both sides by 100)

3x + 9y = 48,000 ---------(eq2)

solving the system of equations made up of eq 1 and eq 2 by elimination,

eq 1 x 3

3x + 3y = 18,000 ------(eq 3)

eq2 - eq3

9y - 3y = 48,000 - 18,000

6y = 30,000

y = 30,00 / 6 = $5,000 (answer)

substitute y = 5,000 into eq 1.

x + 5000 = 6000

x = $1,000 (answer)