Math Problem Statement

Benjamin borrowed £1800 at a simple interest rate of 8% per year. After a certain number of years, he owes a total of £2952 on this loan. How many years have passed since he took out the loan?

Solution

This is a simple interest problem. The formula for simple interest is:

A=P(1+rt)A = P(1 + rt)

Where:

  • AA is the total amount after interest is applied.
  • PP is the principal amount (initial loan).
  • rr is the annual interest rate in decimal form.
  • tt is the time in years.

Given:

  • P=1800P = 1800 pounds
  • r=8%=0.08r = 8\% = 0.08
  • A=2952A = 2952 pounds

We need to solve for tt, the number of years.

First, let's plug the values into the formula:

2952=1800(1+0.08t)2952 = 1800(1 + 0.08t)

Now, solve for tt:

  1. Divide both sides by 1800: 29521800=1+0.08t\frac{2952}{1800} = 1 + 0.08t 1.64=1+0.08t1.64 = 1 + 0.08t

  2. Subtract 1 from both sides: 0.64=0.08t0.64 = 0.08t

  3. Divide by 0.08: t=0.640.08=8t = \frac{0.64}{0.08} = 8

So, 8 years have passed since Benjamin took out the loan.

Would you like more details or explanations? Here are five related questions for further exploration:

  1. How would the time change if the interest rate were higher or lower?
  2. How is compound interest different from simple interest in terms of loan growth?
  3. What is the total interest paid by Benjamin over the 8 years?
  4. How would the total amount owed change if the loan was for a different principal?
  5. How does inflation affect the real value of interest payments over time?

Tip: When working with percentages, always convert them to decimals by dividing by 100 before using them in equations.

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Math Problem Analysis

Mathematical Concepts

Simple Interest
Algebra

Formulas

Simple Interest Formula: A = P(1 + rt)

Theorems

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Suitable Grade Level

Grades 8-10