Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
7. Exponential Growth and Decay
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Exercise 41 Page 466

Practice makes perfect
a We know that we are charged 1.8 % monthly interest on the account balance when we do not pay the entire balance after one month. The suit costs $250, and we can make a $30 payment each month.

250-30= 220 After the first monthly payment, the balance is $ 220.

b The remaining balance after our first payment is $220. Now we are charged 1.8 % monthly interest on the remaining balance. To do this, we can multiply 220 by 1.8 %.
220 * 1.8 %
Simplify
220 * 1.8/100
220* 1.8/100
396/100
3.96
We have to pay $3.96 more due to interest.
c The balance just before we make the second payment is equal to the sum of the balance remaining after the first payment and the interest charged on it.

220+3.96=223.96

d We know that the balance just before the second payment is $223.96. We can make another $30 payment this month. After the second payment, the balance is equal to $ 193.96.

223.96-30= 193.96

e We will make a table of values to find the number of months required to finish paying the entire bill. Basically, we will repeat what we did in the previous parts for each month. To do so, we first need to write a formula to find the balance just before the x^(th) payment. \begin{gathered} \text{Balance just before the ${\color{#0000FF}{x}}^{\text{th}}$ payment} \\ = \\ {\color{#009600}{\text{Balance after the $({\color{#0000FF}{x-1}})^\text{th}$ payment}}} \times \ ({\color{#FF0000}{1+ 0.018}} ) \end{gathered} Here, 0.018 is the monthly interest rate. For example, to find the balance just before the 4^\text{th} payment, we multiply the balance after the 3^\text{rd} payment by 1.018.

To find the balance after the x^(th) payment, we will subtract 30 from the balance just before the x^(th) payment.

\begin{gathered} {\color{#009600}{\text{Balance after the ${\color{#0000FF}{x}}^\text{th}$ payment}}} \\ = \\ \text{Balance just before the ${\color{#0000FF}{x}}^{\text{th}}$ payment} - \textcolor{darkviolet}{30} \end{gathered} Let's show them in a table.
x Balance just before the x^(th) payment Balance after the x^(th) payment
1 250 250-30= 220
2 220* 1.018=223.96 223.96-30= 193.96
3 193.96* 1.018≈ 197.45 197.45-30= 167.45
4 167.45* 1.018≈ 170.47 170.47-30= 140.47
5 140.47* 1.018≈ 142.99 142.99-30= 112.99
6 112.99* 1.018≈ 115.03 115.03-30= 85.03
7 85.03* 1.018≈ 86.56 86.56-30= 56.56
8 56.56 * 1.018 ≈ 57.58 57.58-30= 27.58
9 27.58 * 1.018≈ 28.07 28.07-28.07= 0

We see that the balance just before the ninth payment is $28.07, less than the amount we can pay each month. Therefore, we can finish paying the debt in 9 months.

f We will calculate the interest for each month after the first month. To find the interest for the {\color{#0000FF}{x}}^\text{th} month, we will multiply the balance after the ({\color{#0000FF}{x-1}})^\text{th} by the monthly interest rate.

\begin{gathered} \textcolor{darkorange}{\text{Interest for the ${\color{#0000FF}{x}}^\text{th}$ month }} \\ = \\ {\color{#009600}{\text{Balance after the $({\color{#0000FF}{x-1}})^\text{th}$ payment }}} \times \ 0.018 \end{gathered} For example, to find the interest for the 4^\text{th} month, we multiply the balance after the 3^\text{rd} payment by the monthly interest rate, 0.018. Let's show them in a table.

x Balance after the (x-1)^(th) payment Interest for the x^(th) month
2 220 220* 0.018=3.96
3 193.96 193.96* 0.018≈ 3.49
4 167.45 167.45* 0.018≈ 3.01
5 140.47 140.47 * 0.018≈ 2.53
6 112.99 112.99 * 0.018≈ 2.03
7 85.03 85.03* 0.018≈ 1.53
8 56.56 56.56 * 0.018 ≈ 1.02
9 27.58 27.58* 0.018≈ 0.50
Sum:18.07

The sum of the interests is 18.07. We have paid about $18.07 interest.