Question

Question 10

A bank loaned out $29,000, part of it at the rate of 13% annual interest, and the rest at 4% annual interest.

The total interest earned for both loans was $2,195.00. How much was loaned at each rate?

Answer 1

Answer:

$11,500 was invested at 13%.

$17,500 was invested at 4%

Step-by-step explanation:

This is a simple interest problem.

The simple interest formula is given by:

$E = P*I*t$

In which E is the amount of interest earned, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.

After t years, the total amount of money is:

$T = E + P$

In this question:

Loans totaling 29,000.

P was invested at 13%

29000 - P was invested at 4%.

First investment:

Principal P.

Interest 13% = 0.13.

One year, so t = 1.

So

$E_{1} = P*0.13*1$

$E_{1} = 0.13P$

Second investment:

Principal 29000 - P.

Interest 4% = 0.04.

One year, so t = 1.

So

$E_{2} = (29000-P)*0.04$

The total interest earned for both loans was $2,195.00.

This means that $E_{1} + E_{2} = 2195$

So

$E_{2} = 2195 - E_{1}$

So we solve the following system:

$E_{1} = 0.13P$

$E_{2} = (29000-P)*0.04$

$2195 - E_{1} = (29000-P)*0.04$

$2195 - 0.13P = 1160 - 0.04P$

$0.09P = 2195 - 1160$

$P = \frac{2195 - 1160}{0.09}$

$P = 11500$

$11,500 was invested at 13%.

29000 - 11500 = 17500

$17,500 was invested at 4%