Question

A bank loaned out $14000, part of it at the rate of 9% per year and the rest at 17% per year. If the interest received in one year totaled $2000, how much was loaned at 9%?

Answer 1

Using simple interest, it is found that $13,872 was loaned at 9%.

Simple Interest

The amount of money after t years in simple interest is modeled by:

$A(t) = P(1 + rt)$

In which:

• P is the initial amount.

• r is the interest rate, as a decimal.

The interest earned is:

$I(t) = Prt$

A bank loaned out $14000, in two parts, hence:

$P_2 = 14000 - P_1$

Part of it at the rate of 9% per year and the rest at 17% per year, hence:

$r_1 = 0.09, r_2 = 0.17$

The interest received in one year totaled $2000, hence:

$I_2 = 2000 - I_1$

Then:

$I_1 = 0.09P_1$

$I_2 = 0.17I_2$

$14000 - P_1 = 0.17(2000 - I_1)$

Isolating $I_1$ as a function of $P_1$, then we can replace on the equation for the first interest.

$14000 - P_1 = 340 - 0.17I_1$

$0.17I_1 = -13660 + P_1$

$I_1 = \frac{P_1 - 13660}{0.17}$

Then:

$I_1 = 0.09P_1$

$\frac{P_1 - 13660}{0.17} = 0.09P_1$

$P_1 - 13660 = 0.0153P_1$

$0.9847P_1 = 13660$

$P_1 = \frac{13660}{0.9847}$

$P_1 = 13872$

$13,872 was loaned at 9%.

To learn more about simple interest, you can take a look at brainly.com/question/25296782