Question

1. $7,600 at 15% for 2 years
2. $17,300 at 1% for $1\frac{1}{2}$ years
3. $48,200 at 2.9% for 2 years
4. $1,940 at 5.6% for 9 years
5. $44,600 at 1.3% for 5 years

Explanation:

Assuming we are calculating simple interest for each problem, the formula for simple interest is \( I = P \times r \times t \), where \( I \) is the interest, \( P \) is the principal amount, \( r \) is the annual interest rate (in decimal form), and \( t \) is the time in years.

Problem 31:

Step 1: Identify values

\( P = 7600 \), \( r = 15\% = 0.15 \), \( t = 2 \)

Step 2: Apply formula

\( I = 7600 \times 0.15 \times 2 \)
\( I = 7600 \times 0.3 \)
\( I = 2280 \)

Problem 33:

Step 1: Identify values

\( P = 17300 \), \( r = 1\% = 0.01 \), \( t = 1.5 \) (since \( 1\frac{1}{2}=1.5 \))

Step 2: Apply formula

\( I = 17300 \times 0.01 \times 1.5 \)
\( I = 17300 \times 0.015 \)
\( I = 259.5 \)

Problem 35:

Step 1: Identify values

\( P = 48200 \), \( r = 2.9\% = 0.029 \), \( t = 2 \)

Step 2: Apply formula

\( I = 48200 \times 0.029 \times 2 \)
\( I = 48200 \times 0.058 \)
\( I = 2795.6 \)

Problem 37:

Step 1: Identify values

\( P = 1940 \), \( r = 5.6\% = 0.056 \), \( t = 9 \)

Step 2: Apply formula

\( I = 1940 \times 0.056 \times 9 \)
First, \( 1940 \times 0.056 = 108.64 \)
Then, \( 108.64 \times 9 = 977.76 \)
So, \( I = 977.76 \)

Problem 39:

Step 1: Identify values

\( P = 44600 \), \( r = 1.3\% = 0.013 \), \( t = 5 \)

Step 2: Apply formula

\( I = 44600 \times 0.013 \times 5 \)
\( I = 44600 \times 0.065 \)
\( I = 2899 \)

Answer:

s:

• 31) Interest = $\$2280$
• 33) Interest = $\$259.5$
• 35) Interest = $\$2795.6$
• 37) Interest = $\$977.76$
• 39) Interest = $\$2899$