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Let's solve the two problems related to annuities.

### Problem 1: Kenyo's Loan

Kenyo is paying Php 3,000 every 3 months for 4 years at an interest rate of 8% compounded quarterly. We need to find out how much Kenyo borrowed (present value of the loan).

This is a present value of an ordinary annuity problem. The formula for the present value \( PV \) of an ordinary annuity is:

\[
PV = P \times \left( \dfrac{1 - (1 + i)^{-n}}{i} \right)
\]

Where:
- \( P \) = periodic payment (Php 3,000)
- \( i \) = interest rate per period
- \( n \) = total number of periods

**Step 1: Identify the values**

- Interest rate per year = 8% or 0.08
- Since the interest is compounded quarterly, the interest rate per period is \( i = \dfrac{0.08}{4} = 0.02 \).
- Kenyo is making payments every 3 months for 4 years, so the total number of periods is \( n = 4 \times 4 = 16 \).

**Step 2: Plug the values into the formula**

\[
PV = 3000 \times \left( \dfrac{1 - (1 + 0.02)^{-16}}{0.02} \right)
\]

Now, calculating:

\[
PV = 3000 \times \left( \dfrac{1 - (1.02)^{-16}}{0.02} \right)
\]

\[
PV = 3000 \times \left( \dfrac{1 - 0.73503}{0.02} \right)
\]

\[
PV = 3000 \times \left( \dfrac{0.26497}{0.02} \right)
\]

\[
PV = 3000 \times 13.2485
\]

\[
PV \approx 39,745.50
\]

So, Kenyo borrowed approximately **Php 39,745.50**.

---

### Problem 2: Sasa's Savings

Sasa is saving Php 5,000 every 6 months for 10 years in a cooperative that pays 0.25% compounded monthly. We need to calculate her total savings (future value of the annuity).

This is a future value of an ordinary annuity problem. The formula for the future value \( FV \) of an ordinary annuity is:

\[
FV = P \times \left( \dfrac{(1 + i)^n - 1}{i} \right)
\]

Where:
- \( P \) = periodic payment (Php 5,000)
- \( i \) = interest rate per period
- \( n \) = total number of periods

**Step 1: Identify the values**

- Interest rate per month = 0.25% or 0.0025
- Since payments are made every 6 months (equivalent to 6 periods of compounding), we must adjust this. The effective interest rate per 6 months is:
  \[
  i = (1 + 0.0025)^6 - 1 \approx 0.0150375
  \]
- Sasa is saving for 10 years, which means \( n = \dfrac{10 \times 12}{6} = 20 \) periods of saving.

**Step 2: Plug the values into the formula**

\[
FV = 5000 \times \left( \dfrac{(1 + 0.0150375)^{20} - 1}{0.0150375} \right)
\]

Now, calculating:

\[
FV = 5000 \times \left( \dfrac{(1.0150375)^{20} - 1}{0.0150375} \right)
\]

\[
FV = 5000 \times \left( \dfrac{1.346855 - 1}{0.0150375} \right)
\]

\[
FV = 5000 \times \left( \dfrac{0.346855}{0.0150375} \right)
\]

\[
FV = 5000 \times 23.0658
\]

\[
FV \approx 115,329.00
\]

So, Sasa’s savings after 10 years will be approximately **Php 115,329.00**.

---

Would you like more details or clarification on these problems? 

Here are some related questions:
1. How would the solution change if Kenyo's loan had a different interest rate?
2. What if Kenyo made monthly payments instead of quarterly?
3. How would Sasa’s savings change if the cooperative compounded interest yearly?
4. What happens if Sasa increases her periodic payments after 5 years?
5. How is the future value formula used in planning for retirement?

**Tip**: Always remember that the number of compounding periods can significantly affect the total interest, especially in long-term financial planning!

1. Kenyo is paying Php 3,000 every 3 months for the amount he borrowed at an interest rate of 8% compounded quarterly. How much did he borrow if he agreed that the loan will be paid in 4 years? 2. Sasa, an accountant saves Php 5,000 every 6 months in a cooperative that pays 0.25% compounded monthly. How much will be her savings after 10 years?

Solve two annuity problems: Kenyo's loan repayment of Php 3,000 every quarter for 4 years with 8% interest compounded quarterly, and Sasa's savings of Php 5,000 every 6 months for 10 years with 0.25% interest compounded monthly. Learn about present and future value calculations with step-by-step solutions.

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