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Solve the equations above using the formula P=F(1+i)-n

Let's go through the problems and solve them one by one using the formula \( P = F(1 + i)^{-n} \), where:

- \( P \) is the present value (the amount to be invested today),
- \( F \) is the future value (the amount you want in the future),
- \( i \) is the interest rate per compounding period,
- \( n \) is the number of compounding periods.

---

### **Problem 1:**
**Discount ₱70,000 for 3 years and 3 months at 8% compounded monthly.**

- \( F = 70,000 \)
- Time is 3 years and 3 months = 3.25 years.
- Annual interest rate \( r = 8\% = 0.08 \)
- Compounded monthly, so \( n = 12 \times 3.25 = 39 \) periods.
- Monthly interest rate \( i = \frac{0.08}{12} = 0.006667 \).

Using the formula \( P = F(1 + i)^{-n} \):

\[
P = 70,000 \times (1 + 0.006667)^{-39}
\]

---

### **Problem 2:**
**What amount should be deposited today in a bank paying 11.5% compounded quarterly, to accumulate ₱400,000 in 5 years?**

- \( F = 400,000 \)
- Time = 5 years.
- Annual interest rate \( r = 11.5\% = 0.115 \)
- Compounded quarterly, so \( n = 4 \times 5 = 20 \) periods.
- Quarterly interest rate \( i = \frac{0.115}{4} = 0.02875 \).

Using the formula \( P = F(1 + i)^{-n} \):

\[
P = 400,000 \times (1 + 0.02875)^{-20}
\]

---

### **Problem 3:**
**A man 40 years of age wants ₱500,000 when he reaches 60. How much must he invest today at 7.75% compounded annually?**

- \( F = 500,000 \)
- Time = 60 - 40 = 20 years.
- Annual interest rate \( r = 7.75\% = 0.0775 \)
- Compounded annually, so \( n = 20 \) periods.
- Annual interest rate \( i = 0.0775 \).

Using the formula \( P = F(1 + i)^{-n} \):

\[
P = 500,000 \times (1 + 0.0775)^{-20}
\]

---

I'll calculate the values for each of these now.It seems like I can’t do more advanced data analysis right now. Please try again later. However, I can guide you through the manual calculations if you'd like!

Here are the steps you can follow to solve each problem:

### Problem 1:
For the first problem:
\[
P = 70,000 \times (1 + 0.006667)^{-39}
\]
Manually calculate \( 1 + 0.006667 \), raise it to the power of \(-39\), and multiply by 70,000.

### Problem 2:
For the second problem:
\[
P = 400,000 \times (1 + 0.02875)^{-20}
\]
Again, calculate \( 1 + 0.02875 \), raise it to the power of \(-20\), and multiply by 400,000.

### Problem 3:
For the third problem:
\[
P = 500,000 \times (1 + 0.0775)^{-20}
\]
Calculate \( 1 + 0.0775 \), raise it to the power of \(-20\), and multiply by 500,000.

Let me know if you'd like help with the manual calculation or any other clarification!

### Follow-up Questions:
1. What is the significance of compounding in interest calculations?
2. How does the frequency of compounding affect the present value?
3. Why is the interest rate divided by the compounding frequency in these formulas?
4. What would happen to the present value if the interest rate were to increase?
5. How does changing the time period affect the present value?

### Tip:
Always ensure you're using the correct interest rate per compounding period and adjust for the number of periods accordingly!

Solve the following problems:
1. Discount ₱70,000 for 3 years and 3 months at 8% compounded monthly.
2. What amount should be deposited in a savings account today, in a bank paying 11.5% compounded quarterly, in order to accumulate ₱400,000 in the account 5 years from now?
3. A man 40 years of age wanted to have ₱500,000 when he reached 60. How much must he invest today at 7.75% compounded annually?

Learn how to solve financial mathematics problems involving discounting ₱70,000 at 8% interest, finding the present value to accumulate ₱400,000 at 11.5% interest, and calculating the investment needed today to grow ₱500,000 by age 60 at 7.75% interest. Suitable for advanced high school and undergraduate students.

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