[FREE] End of Year Math Assessments (Grade 4 and Grade 5)

The assessments cover a range of topics to assess your students' math progress and help prepare them for state assessments.

Here you will learn about **simple interest**, including how to calculate simple interest for increasing and decreasing values, and set-up, solve and interpret growth and decay problems.

Students will first learn about **simple interest** as part of Ratios and Proportional Thinking in 7th grade.

**Simple interest **is calculated by finding a percent of the principal (original) amount and multiplying by the time period of the investment. The **final value **of an investment can then be found by **adding/subtracting **the **simple interest **to the **principal amount**.

You can simply this by using the simple interest formulas.

Where:

- I = simple interest
- A = total amount,
- P = principal (original amount),
- r = rate of interest (written as a decimal),
- t = time period (number of intervals)
- ex. Annual percentage rate (once per year): t=1
- ex. Monthly interest rate (once per month): t=12

For example, calculate the interest earned on \$3,000 with a **simple interest** rate of 5\% over 2 years.

Using the formula I=Prt:

- P=3,000
- r=0.05 (remember to write 5\% as a decimal)
- t=2

To find the final value of the investment you can now add the interest to the principal amount.

\begin{aligned} A & =3,000+300 \\\\ & =\$ 3,300 \end{aligned}You could have calculated this directly using the formula A=P\left( 1+rt \right)

\begin{aligned} A & =3,000(1+0.05 \times 2) \\\\ & =\$ 3,300 \end{aligned}How does this relate to 7th grade math?

**7th Grade: Ratios and Proportional Relationships (7.RP.A.3)**Use proportional relationships to solve multistep ratio and percent problems.

Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, percent error.

In order to calculate simple interest:

**Identify the value of each known variable in**\textbf{I = Prt, A = P(1 + rt)}**or**\textbf{A = P(1 − rt)}.**Substitute the values into the formula.****Solve the equation.**

Use this quiz to check your grade 6 to 7 students’ understanding of percents. 10+ questions with answers covering a range of 6th and 7th grade topics to identify areas of strength and support!

DOWNLOAD FREEUse this quiz to check your grade 6 to 7 students’ understanding of percents. 10+ questions with answers covering a range of 6th and 7th grade topics to identify areas of strength and support!

DOWNLOAD FREE\$2,100 is invested for 3 years at an annual percentage rate of 2\% per year simple interest. Find the interest earned on the investment in that time?

**Identify the value of each known variable in**\textbf{I = Prt, A = P(1 + rt)}**or**\textbf{A = P(1 − rt)}.

To find the interest, you use I=Prt.

- P=2,100
- r=0.02
- t=3

2**Substitute the values into the formula.**

Substituting these values into the simple interest formula I=Prt, you get:

I=2,100\times 0.02\times 33**Solve the equation.**

\$126 was earned on the investment.

An investment of \$1,500 is made at a simple interest rate of 5\% per year for 4 years. What is the value of the investment after this time?

**Identify the value of each known variable in** \textbf{I = Prt, A = P(1 + rt)} **or** \textbf{A = P(1 − rt)}.

To find the total after the increase, you use A=P(1+rt).

- P=1,500
- r=0.05
- t=4

**Substitute the values into the formula.**

Substituting these values into the simple interest formula A=P(1+rt), you get:

A=1,500(1+0.05\times{4})

**Solve the equation.**

\begin{aligned}
A&=1,500(1+0.20)\\\\
A&=1,500(1.20)\\\\
A&=1,500\times{1.2}\\\\
A&=\$1,800
\end{aligned}

A car is bought for \$10,000 and loses 9\% of its value per year, simple interest. What is the value of the car after 8 years?

**Identify the value of each known variable in** \textbf{I = Prt, A = P(1 + rt)} **or** \textbf{A = P(1 − rt)}.

To find the total after the decrease, you use A=P(1-rt).

- P=10,000
- r=0.09
- t=8

**Substitute the values into the formula.**

Substituting these values into the simple interest formula A=P(1-rt), you get:

A=10,000(1-0.09\times{8})

**Solve the equation.**

\begin{aligned}
A&=10,000(1-0.72)\\\\
A&=10,000(0.28)\\\\
A&=10,000\times{0.28}\\\\
A&=\$2,800
\end{aligned}

\$7,600 is borrowed for 2 years on a credit card. The cost of borrowing is a 1\% interest payment per month simple interest for the life of the loan. What is the total cost to pay off after this time?

**Identify the value of each known variable in** \textbf{I = Prt, A = P(1 + rt)} **or** \textbf{A = P(1 − rt)}.

To find the total after the increase, you use A=P(1+rt).

- P=7,600
- r=0.01
- t=2\times{12}=24 (remember there are 12 months in 1 year)

**Substitute the values into the formula.**

Substituting these values into the simple interest formula A=P(1+rt), you get:

A=7,600(1+0.01\times{24})

**Solve the equation.**

\begin{aligned}
A&=7,600(1+0.24)\\\\
A&=7,600(1.24)\\\\
A&=7,600\times{1.24}\\\\
A&=\$9,424
\end{aligned}

A house is currently valued at \$175,000. For the first 3 years, the value of the house increases by the rate of simple interest of 0.2\% per year.

For the following 4 years, the value of the house decreases in value by a simple interest rate of 0.18\% per annum. Calculate the value of the house after these 7 years.

**Identify the value of each known variable in** \textbf{I = Prt, A = P(1 + rt)} **or** \textbf{A = P(1 − rt)}.

To find the total after the increase, you use A=P\left(1+r_{1} t_{1}-r_{2} t_{2}\right).

- P=175,000
- r_1=0.002 (the first interest rate)
- t_1=3 (the first time period)
- r_2=0.0018 (the second interest rate)
- t_2=4 (the second time period)

**Substitute the values into the formula.**

Substituting these values into the simple interest formula A=P\left(1+r_{1} t_{1}-r_{2} t_{2}\right), you get:

A=175,000(1+0.002\times{3}-0.0018\times{4})

**Solve the equation.**

\begin{aligned}
A&=175,000(1+0.006-0.0072)\\\\
A&=175,000(0.9988)\\\\
A&=175,000\times{0.9988}\\\\
A&=\$174,790
\end{aligned}

- Assign students a personal finance project that requires them to use what they are learning about simple interest in the real world.

- Let students spend time using a simple interest calculator, trying different values for each part of the equation and writing or talking about what patterns they notice.

- Although students will not learn the compound interest formula until algebra, spend time explaining the differences so that students understand what simple interest does.

**Applying the incorrect formula to the question**

Pay attention to whether the question is asking for the interest or the total after an increase or a loss. This will determine which of the following formulas you need to use: I=Prt, \; A=P(1+rt) or A=P(1-rt).

Additionally, these formulas cannot be used to calculate compound interest or other problems that do not involve simple interest.

**Inputting the incorrect time scale**

Example 3: \$7,600 is invested for 2 years at 1\% per month simple interest.

What is the value of the investment after this time?”

The interest rate is monthly, but the period of time on the investment is years.

It is important to convert the 2 years into months ( 2 \times 12=24 months) before multiplying by the principal balance. This will show you the correct amount for monthly payments.

**Not converting the percent to a decimal**

Using the whole number from the percent as the value for r (therefore not converting the percent to a decimal). For example, when using simple interest to increase \$100 by 2\% for 5 years, this calculation is incorrectly made:

A=100(1+2\times{5})

A=100\times{11}

A=\$1,100

Notice that the value increased from \$100 to \$1,100. This is an extremely large increase, and clearly does not show 2\%.

- Percent
- Percent of a number
- Percent decrease
- Percent increase
- Percent change
- Percent increase and decrease
- Percent error

1. Freya invests \$6,700 for 2 years. The simple interest rate is 1.2\% per year. Which calculation below works out the total value after 2 years?

\$6,700\times1.012\times2

\$6,700\times1.2\times2

\$6,700\times1.012^2

\$6,700 \times 1.024

Since the investment is an increase, use the equation I=P(1 +rt):

\begin{aligned} P&=\$6,700 \\ r&=0.012 \\ t&=2 \end{aligned}

\begin{aligned} & \$6,700\times(1+0.012\times2) \\\\ & =\$6,700\times(1+0.024) \\\\ & =\$6,700\times1.024 \end{aligned}

2. A technology store has a back to school offer: save 20\% on all full price laptops. Paula buys a laptop that was \$689 full price. After 3 years, the value of the purchased laptop has decreased by 4\% per year, simple interest. What is the value of the laptop after these 3 years?

\$485.06

\$551.20

\$485.06

\$529.15

The original price was \$689. The sale price was 80\% of this:

\$689\times0.80=\$551.20

Since the value decreased, use the equation I=P(1-rt):

\begin{aligned} P&=\$551.20 \\ r&=0.04 \\ t&=3 \end{aligned}

\$551.20 \; (1-0.04\times3)=\$485.056

\$485.056 \, rounds to \, \$485.06

3. \$7,342 is invested in a savings account with a 0.4\% simple interest rate per month. What is the total interest earned after 4 years?

\$7,459.47

\$1,694.42

\$1,409.66

\$8,751.66

Use the equation I=Prt:

\begin{aligned} P&=\$7,342 \\ r&=0.004 \\ t&=4 \text{ years} \times12=48 \text{ months} \end{aligned}

\$7,342 \; (0.004\times48)=\$1,409.664

\$1,409.664 \, rounds to \, \$1,409.66

4. A boat is valued at \$365,500. The value of the boat decreases by an average of 0.25\% per year, simple interest. How much is the boat worth after 12 years?

\$255,850

\$376,465

\$354,535

\$233,920

Since the value is decreasing, use the equation I=P(1-rt):

\begin{aligned} P&=\$365,500 \\ r&=0.0025 \\ t&=12 \end{aligned}

\$365,500 \; (1-0.0025\times12)=\$354,535

5. To buy a new car, Jeff gets an auto loan of \$22,000 that he will pay off over five years. On the loan, the lender charges a 0.5\% simple interest rate per month. How much will Jeff pay for the loan in total?

\$28,600

\$15,400

\$37,400

\$22,550

Since the interest is being added to the original loan amount, use the equation I=P(1+rt):

\begin{aligned} P&=\$22,000 \\ r&=0.005 \\ t&=5 \text{ years} \times12=60 \text{ months} \end{aligned}

\$22,000 \; (1+0.005\times60)=\$28,600

Simple interest is used to calculate growth or decay, in terms of money. For example, you can use it to calculate the interest charges based on a loan amount or it can be used to calculate the amount of interest you can earn if you invest your money.

- Converting fractions, decimals, and percentages
- Exponents
- Algebraic expressions

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