Math Problem Statement

Complete the table by finding the unknown values for the given compound interest problems, and determine how much money is needed to deposit today at 15% annual interest compounded annually to have ₱1,500,000 in the account after 9 years.

Solution

To solve the problems from the compound interest table, we use the formula for compound interest:

F=P(1+r)tF = P(1 + r)^t Where:

  • FF is the future value (or maturity value),
  • PP is the principal,
  • rr is the interest rate (expressed as a decimal),
  • tt is the time (in years).

The compound interest earned, IcI_c, is given by: Ic=FPI_c = F - P

Let's work on filling the table step-by-step:

  1. First row:

    • P=10,000P = 10,000, r=9%r = 9\%, t=5t = 5
    • Future Value F=P(1+r)t=10,000(1+0.09)5F = P(1 + r)^t = 10,000(1 + 0.09)^5
    • Calculating: F=10,000(1.09)515,386.24F = 10,000(1.09)^5 \approx 15,386.24
    • Interest earned: Ic=FP=15,386.2410,000=5,386.24I_c = F - P = 15,386.24 - 10,000 = 5,386.24
    • So, Ic=5,386.24I_c = 5,386.24 and F=15,386.24F = 15,386.24.
  2. Second row:

    • P=25,000P = 25,000, r=3%r = 3\%, t=4t = 4
    • Future Value: F=25,000(1+0.03)4=25,000(1.03)428,246.36F = 25,000(1 + 0.03)^4 = 25,000(1.03)^4 \approx 28,246.36
    • Interest earned: Ic=FP=28,246.3625,000=3,246.36I_c = F - P = 28,246.36 - 25,000 = 3,246.36
    • So, Ic=3,246.36I_c = 3,246.36 and F=28,246.36F = 28,246.36.
  3. Third row:

    • P=?P = ?, r=4.5%r = 4.5\%, t=15t = 15, F=1,000,000F = 1,000,000
    • Using the future value formula: 1,000,000=P(1+0.045)151,000,000 = P(1 + 0.045)^{15} Simplifying: 1,000,000=P(1.045)15P×1.9311,000,000 = P(1.045)^{15} \approx P \times 1.931 Solving for PP: P=1,000,0001.931517,598.34P = \frac{1,000,000}{1.931} \approx 517,598.34
    • So, the principal PP is approximately 517,598.34.
    • To calculate the interest earned: Ic=1,000,000517,598.34=482,401.66I_c = 1,000,000 - 517,598.34 = 482,401.66
    • So, Ic=482,401.66I_c = 482,401.66.

Problem 6:

The problem asks how much money you would need to deposit today at 15% annual interest compounded annually to have ₱1,500,000 in 9 years.

We use the compound interest formula: F=P(1+r)tF = P(1 + r)^t Given:

  • F=1,500,000F = 1,500,000,
  • r=15%=0.15r = 15\% = 0.15,
  • t=9t = 9.

Solving for PP: 1,500,000=P(1+0.15)9=P(1.15)9P×3.0591,500,000 = P(1 + 0.15)^9 = P(1.15)^9 \approx P \times 3.059 P=1,500,0003.059490,383.21P = \frac{1,500,000}{3.059} \approx 490,383.21 So, the required deposit is approximately ₱490,383.21.


Would you like more details on any specific step?

Here are five questions for further exploration:

  1. How does the interest rate affect the compound interest earned?
  2. What happens if the time period is doubled?
  3. How does compound interest differ from simple interest?
  4. Can you derive the formula for compound interest from basic principles?
  5. What would happen if the interest is compounded monthly instead of annually?

Tip: Always express the interest rate as a decimal (i.e., 9% = 0.09) when using it in formulas.

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Math Problem Analysis

Mathematical Concepts

Compound Interest
Exponential Growth

Formulas

Future Value Formula: F = P(1 + r)^t
Compound Interest: Ic = F - P

Theorems

Properties of Exponential Functions

Suitable Grade Level

Grades 10-12