Question

Jose has placed his money into two simple interest accounts. He allocates twice the amount in an account yielding a 12% interest rate compared to what he invests in an account with a 7% interest rate. If the total interest earned from both accounts after one year amounts to $124.00, what is the total amount he invested?

A) $600.00
B) $800.00
C) $1,000.00
D) $1,200.00

Answer 1

By setting up a system of equations based on the simple interest formula, we can determine that Jose invested a total of $1200.00, where $400 was placed in the 7% account and $800 in the 12% account.

Step-by-step explanation:

To solve the problem, we need to set up a system of equations based on the information given. Let's denote the amount invested in the account with a 7% interest rate as x, then the amount invested in the account with a 12% interest rate would be twice the amount, which is 2x.

Using the formula for simple interest, I = Prt (where I is the interest earned, P is the principal amount, r is the rate of interest, and t is the time in years), we can create the following equations:

• For the 7% account: Interest I1 = x * 0.07 * 1
• For the 12% account: Interest I2 = 2x * 0.12 * 1

The total interest earned from both accounts after one year is $124.00, so we can write:

I1 + I2 = $124

Substituting the expressions for I1 and I2 from the equations above, we get:

x * 0.07 + 2x * 0.12 = $124

Simplifying this gives us:

0.07x + 0.24x = $124

0.31x = $124

Dividing both sides by 0.31, we find that:

x = $124 / 0.31

x = $400

Since x represents the amount invested at 7%, the amount invested at 12% is 2x, which equals $800.

Therefore, the total amount Jose invested is x + 2x, which is $400 + $800 = $1,200.00.

The correct answer is D) $1,200.00.