Question

A person invested ​$7700 for 1​ year, part at 7​%, part at 11​%, and the remainder at 12​%. the total annual income from these investments was ​$832. the amount of money invested at 12​% was ​$1300 more than the amounts invested at 7​% and 11​% combined. find the amount invested at each rate.

Answer 1

The problem is solved by establishing a system of equations based on the given information about investments and their returns. After solving, it's found that $1,500 was invested at 7%, $1,900 at 11%, and $3,300 at 12%.

Step-by-step explanation:

To solve this problem, we need to set up a system of equations based on the information given:

• Total investment: $7,700
• Total annual income: $832
• Amount at 12% is $1,300 more than the sum of the amounts at 7% and 11%

Let x be the amount invested at 7%, y be the amount invested at 11%, and z be the amount invested at 12%. We can now formulate the following equations:

1. x + y + z = $7,700 (total investment)
2. 0.07x + 0.11y + 0.12z = $832 (total income)
3. z = x + y + $1,300 (relationship between investments)

Using substitution or elimination methods, we will solve these equations to find the values of x, y, and z. After doing the calculations, we find that x = $1,500, y = $1,900, and z = $3,300 as the amounts invested at 7%, 11%, and 12% respectively.

The amount invested at 7% is $1,500, the amount invested at 11% is $1,900, and the amount invested at 12% is $3,300, which is $1,300 more than the combined amounts invested at 7% and 11%.