Math Problem Statement

Solve the following problems regarding simple interest based on the provided details.

Solution

Let's go over each of the questions from the image and solve them step by step.

1. Determine the simple interest rate if an investment of ₱100,000 accumulates to ₱110,000 in 48 months.

  • Given:
    • Principal (P) = ₱100,000
    • Final amount = ₱110,000
    • Time (t) = 48 months = 48/12 = 4 years
    • Simple Interest (I) = Final amount - Principal = ₱110,000 - ₱100,000 = ₱10,000
  • Formula: I=P×r×tI = P \times r \times t Where:
    • I = Interest
    • P = Principal
    • r = Rate (to find)
    • t = Time
  • Calculation: 10,000=100,000×r×410,000 = 100,000 \times r \times 4 r=10,000100,000×4=0.025r = \frac{10,000}{100,000 \times 4} = 0.025 Therefore, the interest rate is 2.5% per year.

2. If the simple interest on ₱10,800 is ₱600 after 11 months, find the rate.

  • Given:
    • Principal (P) = ₱10,800
    • Interest (I) = ₱600
    • Time (t) = 11 months = 11/12 years
  • Formula: I=P×r×tI = P \times r \times t
  • Calculation: 600=10,800×r×1112600 = 10,800 \times r \times \frac{11}{12} Solving for rr: r=60010,800×1112=0.06r = \frac{600}{10,800 \times \frac{11}{12}} = 0.06 Therefore, the interest rate is 6% per year.

3. What is the rate of interest on an 18-month loan for ₱26,000 if the interest on the loan is ₱1,200?

  • Given:
    • Principal (P) = ₱26,000
    • Interest (I) = ₱1,200
    • Time (t) = 18 months = 1.5 years
  • Formula: I=P×r×tI = P \times r \times t
  • Calculation: 1,200=26,000×r×1.51,200 = 26,000 \times r \times 1.5 Solving for rr: r=1,20026,000×1.5=0.0308r = \frac{1,200}{26,000 \times 1.5} = 0.0308 Therefore, the interest rate is 3.08% per year.

4. Find the rate of interest if the interest earned was ₱300, the principal was ₱8,000, and the term was 11 months.

  • Given:
    • Principal (P) = ₱8,000
    • Interest (I) = ₱300
    • Time (t) = 11 months = 11/12 years
  • Formula: I=P×r×tI = P \times r \times t
  • Calculation: 300=8,000×r×1112300 = 8,000 \times r \times \frac{11}{12} Solving for rr: r=3008,000×1112=0.0409r = \frac{300}{8,000 \times \frac{11}{12}} = 0.0409 Therefore, the interest rate is 4.09% per year.

5. If ₱4,200 is the interest in investing ₱50,000 for 8 years, find the rate of interest.

  • Given:
    • Principal (P) = ₱50,000
    • Interest (I) = ₱4,200
    • Time (t) = 8 years
  • Formula: I=P×r×tI = P \times r \times t
  • Calculation: 4,200=50,000×r×84,200 = 50,000 \times r \times 8 Solving for rr: r=4,20050,000×8=0.0105r = \frac{4,200}{50,000 \times 8} = 0.0105 Therefore, the interest rate is 1.05% per year.

Would you like to see further breakdowns of any of these solutions?

Here are 5 related questions:

  1. How would you calculate the interest if the rate was compounded instead of simple?
  2. What happens to the rate if the time period is reduced by half but interest stays the same?
  3. How can you compare simple interest and compound interest over a period of 5 years?
  4. What if the principal changes over time? How would that affect the simple interest calculation?
  5. Can you explain how inflation might affect the value of money earned through interest?

Tip: Always convert months into years in simple interest calculations to keep the formula consistent.

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Math Problem Analysis

Mathematical Concepts

Simple Interest
Interest Rates
Percentage
Time Conversion

Formulas

Simple Interest Formula: I = P × r × t

Theorems

Simple Interest Calculation

Suitable Grade Level

Grade 9-10