EXAMINATION-SET A 1. Find the interest on P 5,000 at 5% simple interest for 3 years. 2. What is the simple interest rate on P 60,000 for 6½ years if money earns P 5,000? 3. If P 3,000 is the interest at 5% after 5 months, how much was borrowed? 4. How long will it take for P 2,000 to become P3,500 at 5% simple interest? 5. Accumulate P10,000 at 10% for 5 years.

Let's solve these one by one using the simple interest formula:

The formula for simple interest is:

\[
I = P \times r \times t
\]

Where:
- \( I \) is the interest,
- \( P \) is the principal (initial amount),
- \( r \) is the annual interest rate (in decimal form),
- \( t \) is the time in years.

---

### 1. **Find the interest on P 5,000 at 5% simple interest for 3 years.**

Using the formula:

\[
I = P \times r \times t
\]

\[
I = 5,000 \times 0.05 \times 3 = 750
\]

So, the interest is **P 750**.

---

### 2. **What is the simple interest rate on P 60,000 for 6½ years if money earns P 5,000?**

We are given:
- \( I = 5,000 \),
- \( P = 60,000 \),
- \( t = 6.5 \text{ years} \).

We need to find the interest rate \( r \). Rearranging the simple interest formula to solve for \( r \):

\[
r = \frac{I}{P \times t}
\]

Substitute the given values:

\[
r = \frac{5,000}{60,000 \times 6.5} = \frac{5,000}{390,000} \approx 0.0128
\]

Convert to percentage:

\[
r = 0.0128 \times 100 = 1.28\%
\]

So, the interest rate is **1.28%**.

---

### 3. **If P 3,000 is the interest at 5% after 5 months, how much was borrowed?**

We are given:
- \( I = 3,000 \),
- \( r = 5\% = 0.05 \),
- \( t = \frac{5}{12} \) years (since 5 months is \( \frac{5}{12} \) of a year).

We need to find the principal \( P \). Using the formula:

\[
P = \frac{I}{r \times t}
\]

Substitute the values:

\[
P = \frac{3,000}{0.05 \times \frac{5}{12}} = \frac{3,000}{0.02083} \approx 144,000
\]

So, the amount borrowed was **P 144,000**.

---

### 4. **How long will it take for P 2,000 to become P 3,500 at 5% simple interest?**

We are given:
- \( P = 2,000 \),
- \( A = 3,500 \),
- \( r = 5\% = 0.05 \).

First, we need to calculate the interest \( I \):

\[
I = A - P = 3,500 - 2,000 = 1,500
\]

Now, using the formula for time \( t \):

\[
t = \frac{I}{P \times r} = \frac{1,500}{2,000 \times 0.05} = \frac{1,500}{100} = 15 \text{ years}
\]

So, it will take **15 years**.

---

### 5. **Accumulate P 10,000 at 10% for 5 years.**

To accumulate means to find the total amount \( A \), which is:

\[
A = P + I
\]

First, calculate the interest:

\[
I = P \times r \times t = 10,000 \times 0.10 \times 5 = 5,000
\]

Now, add the interest to the principal:

\[
A = 10,000 + 5,000 = 15,000
\]

So, the accumulated amount is **P 15,000**.

---

### Do you want details or have any questions?

Here are 5 related questions to expand on this topic:

1. What would the interest be if the rate were 6% instead of 5% for question 1?
2. How would the total amount change in question 5 if the rate were compounded annually?
3. Can you explain the difference between simple and compound interest in terms of growth?
4. What would be the time required to double the amount for question 4 using compound interest?
5. How does inflation affect the real value of the accumulated amounts over time?

**Tip:** Always convert the time into years when using the simple interest formula, especially when time is given in months or days.

EXAMINATION-SET A 1. Find the interest on P 5,000 at 5% simple interest for 3 years. 2. What is the simple interest rate on P 60,000 for 6½ years if money earns P 5,000? 3. If P 3,000 is the interest at 5% after 5 months, how much was borrowed? 4. How long will it take for P 2,000 to become P 3,500 at 5% simple interest? 5. Accumulate P 10,000 at 10% for 5 years.

Learn how to solve simple interest problems, including finding the interest on a loan, calculating the rate, determining the principal, and calculating accumulation for different scenarios. This problem set includes examples such as calculating the interest on P 5,000 at 5% for 3 years and determining the amount borrowed when P 3,000 is earned as interest in 5 months.

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