Question

assessment 1

1. find the simple interest and maturity value if ₱155 000 pesos is invested in a bank at an interest rate of 6% for a) 5 years and b) 9 months.
2. suppose you save ₱12 500.00 of your salary and deposit it into an account that earns simple interest. after 4 years, the balance is ₱15 000. what is the annual interest rate?
3. if ₱100 000 is invested at 6% simple interest, how long will it take to grow to ₱112 000?
4. how much is the principal amount or the present value borrowed if maturity value is ₱96 500 at 5% simple interest for 5 years?
5. jane borrowed a money from her friend and agreed to pay the simple interest of 5, 400 pesos at 12% simple interest for 6 months. how much is face value or the principal amount that she borrowed?

Explanation:

Problem 1a: Simple Interest and Maturity Value (5 years)

The formula for simple interest is \( I = P \times r \times t \), where \( P \) is the principal, \( r \) is the annual interest rate (in decimal), and \( t \) is the time in years. The maturity value \( A \) is \( A = P + I \).

Step 1: Identify given values

\( P = 155000 \) pesos, \( r = 6\% = 0.06 \), \( t = 5 \) years.

Step 2: Calculate simple interest (\( I \))

Using the formula \( I = P \times r \times t \):
\( I = 155000 \times 0.06 \times 5 \)
\( I = 155000 \times 0.3 \)
\( I = 46500 \) pesos.

Step 3: Calculate maturity value (\( A \))

Using \( A = P + I \):
\( A = 155000 + 46500 \)
\( A = 201500 \) pesos.

Problem 1b: Simple Interest and Maturity Value (9 months)

Time \( t \) in years: \( 9 \) months \( = \frac{9}{12} = 0.75 \) years. Use the same formulas for \( I \) and \( A \).

Step 1: Identify given values (adjust \( t \))

\( P = 155000 \) pesos, \( r = 0.06 \), \( t = 0.75 \) years.

Step 2: Calculate simple interest (\( I \))

\( I = 155000 \times 0.06 \times 0.75 \)
\( I = 155000 \times 0.045 \)
\( I = 6975 \) pesos.

Step 3: Calculate maturity value (\( A \))

\( A = 155000 + 6975 \)
\( A = 161975 \) pesos.

Problem 2: Annual Interest Rate

Maturity value \( A = P + I \), so \( I = A - P \). Then use \( I = P \times r \times t \) to solve for \( r \).

Step 1: Find interest (\( I \))

\( A = 15000 \), \( P = 12500 \), so \( I = 15000 - 12500 = 2500 \) pesos.
\( t = 4 \) years.

Step 2: Solve for \( r \)

From \( I = P \times r \times t \), rearrange: \( r = \frac{I}{P \times t} \)
\( r = \frac{2500}{12500 \times 4} \)
\( r = \frac{2500}{50000} \)
\( r = 0.05 \) or \( 5\% \).

Problem 3: Time to Grow to ₱112,000

Answer:

Simple interest \( I = P \times r \times t \). Time \( t = 6 \) months \( = \frac{6}{12} = 0.5 \) years. Solve for \( P \): \( P = \frac{I}{r \times t} \).

Step 1: Identify given values

\( I = 5400 \) pesos, \( r = 12\% = 0.12 \), \( t = 0.5 \) years.

Step 2: Solve for \( P \)

\( P = \frac{5400}{0.12 \times 0.5} \)
\( P = \frac{5400}{0.06} \)
\( P = 90000 \) pesos.

Final Answers:

1a. Simple Interest: \( \boldsymbol{46500} \) pesos; Maturity Value: \( \boldsymbol{201500} \) pesos.
1b. Simple Interest: \( \boldsymbol{6975} \) pesos; Maturity Value: \( \boldsymbol{161975} \) pesos.

1. Annual Interest Rate: \( \boldsymbol{5\%} \).
2. Time: \( \boldsymbol{2} \) years.
3. Principal Amount: \( \boldsymbol{77200} \) pesos.
4. Principal Amount (Borrowed): \( \boldsymbol{90000} \) pesos.