Question

Hattie had $1200 to invest and wants to earn 4.6% interest per year. She will put some of the money into an account that earns 3.6% per year and the rest into an account that earns 4.8% per year. How much money should she put into each account?

a) $720 at 3.6%, $480 at 4.8%
b) $480 at 3.6%, $720 at 4.8%
c) $600 at 3.6%, $600 at 4.8%
d) $800 at 3.6%, $400 at 4.8%

Answer 1

Hattie should invest $200 at 3.6% and $1000 at 4.8%.

Step-by-step explanation:

To determine how much money Hattie should put into each account, we can set up a system of equations based on the amount of money and the interest rates. Let's call the amount of money Hattie puts into the account that earns 3.6% 'x', and the amount she puts into the account that earns 4.8% 'y'. The interest earned from the first account would be 0.036x, and the interest earned from the second account would be 0.048y. We know that the total interest earned should be 4.6% of $1200, which is 0.046(1200). So we can set up the equation 0.036x + 0.048y = 0.046(1200) which simplifies to 0.036x + 0.048y = 55.2. We also know that the sum of the amounts invested should be $1200, so we can set up the equation x + y = 1200.

To solve this system of equations, we can use the substitution method. Rearrange the second equation to solve for x: x = 1200 - y. Substitute this expression for x into the first equation: 0.036(1200 - y) + 0.048y = 55.2. Distribute the 0.036: 43.2 - 0.036y + 0.048y = 55.2. Combine like terms: 0.012y = 12, and divide both sides by 0.012: y = 1000. Substitute this value for y back into the equation x = 1200 - y: x = 1200 - 1000 = 200. Therefore, Hattie should invest $200 at 3.6% and $1000 at 4.8%.