Math Problem Statement

1. Rommel takes a loan of ₱5,000 to buy a used hand tractor at the rate of 9.5% simple interest. Calculate the annual interest to be paid for the loan amount. 2. Mary borrowed ₱800,000 from a bank to buy a car at 11.5%. If she paid ₱9,500 interest while clearing the loan, find the time for which the loan was given. 3. If Dianne paid an interest amounting to ₱4,450 for a loan in 1½ years at 15% per annum, what was the original loan? 4. Darryl borrowed ₱156,300 from a bank charging 9% simple interest with a promise that he would pay the principal and interest at the agreed term. If he paid ₱4,500 at the end of the specified term, how long did he use the money? 5. At what rate of interest did Annie invest the amount of ₱23,500 for it to earn a simple interest of ₱3,525 for four years?

Solution

Let’s solve these problems one by one using the formula for Simple Interest:

Formula:

I=P×r×tI = P \times r \times t Where:

  • II: Simple interest
  • PP: Principal amount
  • rr: Annual interest rate (in decimal form)
  • tt: Time (in years)

Problem 1:

Rommel takes a loan of ₱5,000\text{₱5,000} at 9.5% simple interest. Calculate the annual interest.

Given:

  • P=5000P = 5000
  • r=9.5%=0.095r = 9.5\% = 0.095
  • t=1t = 1 year

I=P×r×t=5000×0.095×1=₱475I = P \times r \times t = 5000 \times 0.095 \times 1 = \text{₱475}

Answer: The annual interest is ₱475.


Problem 2:

Mary borrowed ₱800,000\text{₱800,000} at 11.5% interest and paid ₱9,500\text{₱9,500} interest. Find the time.

Given:

  • P=800,000P = 800,000
  • r=11.5%=0.115r = 11.5\% = 0.115
  • I=9,500I = 9,500

Rearrange the formula to find tt: t=IP×r=9500800,000×0.115t = \frac{I}{P \times r} = \frac{9500}{800,000 \times 0.115}

t=950092,000=0.1033years(1.24months)t = \frac{9500}{92,000} = 0.1033 \, \text{years} \, (\approx 1.24 \, \text{months})

Answer: The loan term is approximately 0.1033 years or 1.24 months.


Problem 3:

Dianne paid ₱4,450\text{₱4,450} interest for 1.5 years at 15% per annum. What was the original loan?

Given:

  • I=4,450I = 4,450
  • r=15%=0.15r = 15\% = 0.15
  • t=1.5t = 1.5

Rearrange the formula to find PP: P=Ir×t=44500.15×1.5P = \frac{I}{r \times t} = \frac{4450}{0.15 \times 1.5}

P=44500.225=₱19,778P = \frac{4450}{0.225} = \text{₱19,778}

Answer: The original loan was ₱19,778.


Problem 4:

Darryl borrowed ₱156,300\text{₱156,300} at 9% interest and paid ₱4,500\text{₱4,500} interest. How long did he use the money?

Given:

  • P=156,300P = 156,300
  • r=9%=0.09r = 9\% = 0.09
  • I=4,500I = 4,500

Rearrange the formula to find tt: t=IP×r=4500156,300×0.09t = \frac{I}{P \times r} = \frac{4500}{156,300 \times 0.09}

t=450014,0670.32years or 3.84 monthst = \frac{4500}{14,067} \approx 0.32 \, \text{years or 3.84 months}

Answer: He used the money for approximately 0.32 years or 3.84 months.


Problem 5:

At what rate of interest did Annie invest ₱23,500\text{₱23,500} to earn ₱3,525\text{₱3,525} in 4 years?

Given:

  • P=23,500P = 23,500
  • I=3,525I = 3,525
  • t=4t = 4

Rearrange the formula to find rr: r=IP×t=352523,500×4r = \frac{I}{P \times t} = \frac{3525}{23,500 \times 4}

r=352594,000=0.0375=3.75%r = \frac{3525}{94,000} = 0.0375 = 3.75\%

Answer: The rate of interest was 3.75%.


Do you need further clarifications or detailed steps for any problem? Here are related questions for you:

  1. How can you convert months into years when solving simple interest problems?
  2. How does compound interest differ from simple interest?
  3. Can you calculate the total amount paid after adding principal and interest for these cases?
  4. How do you compute interest if the rate changes during the loan period?
  5. What would happen if the time or interest rate were fractions?

Tip: Always double-check the formula rearrangement when solving for a specific variable.

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Math Problem Analysis

Mathematical Concepts

Simple Interest
Algebra
Interest Calculation

Formulas

I = P × r × t
t = I / (P × r)
r = I / (P × t)
P = I / (r × t)

Theorems

Simple Interest Formula

Suitable Grade Level

Grades 8-10