Question

1. $1,760 at 7% for 2 years
2. $35 at 2.9% for \\(\frac{1}{4}\\) years
3. $37,000 at 1.2% for \\(2\frac{3}{4}\\) years
4. $1,200 at 1% for \\(4\frac{1}{4}\\) years
5. $13,700 at 5% for 3 years
6. $650 at 3.6% for \\(5\frac{1}{4}\\) years

Explanation:

Let's solve problem 30 first. The formula for simple interest is \( I = P \times r \times t \), where \( P \) is the principal amount, \( r \) is the annual interest rate (in decimal), and \( t \) is the time in years.

Step 1: Identify values for problem 30

For problem 30: \( P = \$1760 \), \( r = 7\% = 0.07 \), \( t = 2 \) years.

Step 2: Apply the simple interest formula

Substitute the values into the formula: \( I = 1760 \times 0.07 \times 2 \)
First, calculate \( 1760 \times 0.07 = 123.2 \)
Then, \( 123.2 \times 2 = 246.4 \)

Now, let's solve problem 32.

Step 1: Identify values for problem 32

For problem 32: \( P = \$35 \), \( r = 2.9\% = 0.029 \), \( t = \frac{1}{4} = 0.25 \) years.

Step 2: Apply the simple interest formula

Substitute the values: \( I = 35 \times 0.029 \times 0.25 \)
First, \( 35 \times 0.029 = 1.015 \)
Then, \( 1.015 \times 0.25 = 0.25375 \)

For problem 34:

Step 1: Identify values for problem 34

\( P = \$37000 \), \( r = 1.2\% = 0.012 \), \( t = 2\frac{3}{4} = 2.75 \) years.

Step 2: Apply the simple interest formula

\( I = 37000 \times 0.012 \times 2.75 \)
First, \( 37000 \times 0.012 = 444 \)
Then, \( 444 \times 2.75 = 1221 \)

For problem 36:

Step 1: Identify values for problem 36

\( P = \$1200 \), \( r = 1\% = 0.01 \), \( t = 4\frac{1}{4} = 4.25 \) years.

Step 2: Apply the simple interest formula

\( I = 1200 \times 0.01 \times 4.25 \)
First, \( 1200 \times 0.01 = 12 \)
Then, \( 12 \times 4.25 = 51 \)

For problem 38:

Step 1: Identify values for problem 38

\( P = \$13700 \), \( r = 5\% = 0.05 \), \( t = 3 \) years.

Step 2: Apply the simple interest formula

\( I = 13700 \times 0.05 \times 3 \)
First, \( 13700 \times 0.05 = 685 \)
Then, \( 685 \times 3 = 2055 \)

For problem 40:

Step 1: Identify values for problem 40

\( P = \$650 \), \( r = 3.6\% = 0.036 \), \( t = 5\frac{1}{4} = 5.25 \) years.

Step 2: Apply the simple interest formula

\( I = 650 \times 0.036 \times 5.25 \)
First, \( 650 \times 0.036 = 23.4 \)
Then, \( 23.4 \times 5.25 = 122.85 \)

Answer:

s:

• Problem 30: \(\$246.40\)
• Problem 32: \(\$0.25375\) (or \(\$0.25\) if rounded to the nearest cent)
• Problem 34: \(\$1221\)
• Problem 36: \(\$51\)
• Problem 38: \(\$2055\)
• Problem 40: \(\$122.85\)