Question

Your cousin borrowed $6000, some on a home-equity loan at an interest rate of 9.5% and the rest on a consumer loan at an interest rate of 11%. Her total interest paid was $645. How much did she borrow at each rate?

Answer 1

The student's cousin borrowed $1000 at a 9.5% interest rate and $5000 at an 11% interest rate, calculated by setting up a system of linear equations based on the total loan amount and interest paid.

Step-by-step explanation:

The question involves solving a system of linear equations based on given conditions for two types of loans. We are told the total loan amount ($6000) and the total interest paid ($645). We need to determine how much was borrowed at each interest rate. Let's denote the amount borrowed for the home-equity loan as x and the amount borrowed for the consumer loan as y. The system of linear equations can be set up as follows:

x + y = 6000 (total amount borrowed)

0.095x + 0.11y = 645 (total interest paid)

To solve for x and y, you can use methods such as substitution or elimination. Here's a step-by-step explanation:

Solve the first equation for y: y = 6000 - x.

Substitute y in the second equation: 0.095x + 0.11(6000 - x) = 645.

Solve for x: 0.095x + 660 - 0.11x = 645.

This simplifies to -0.015x = -15, so x = 1000.

Substitute x back into y = 6000 - x to find y: y = 6000 - 1000, so y = 5000.

The amounts borrowed are $1000 at 9.5% and $5000 at 11%.

Answer 2

I=PRT 9.5% in decimal form is .095=11 times rate times tax but we don't know what the tax is.