Calculating Simple Interest Rate: Money Tripling Problem

This question asks for the rate of simple interest at which a sum of money grows to three times its original value over a period of 36 years. To solve this, we need to understand the concept of simple interest and its formula.

Understanding Simple Interest

Simple interest is calculated only on the principal amount, or on that portion of the principal amount that remains unpaid. It does not compound, meaning the interest earned in previous periods does not earn interest itself.

The formula for simple interest is:

\( SI = \frac{P \times R \times T}{100} \)

Where:

• \( SI \) is the Simple Interest
• \( P \) is the Principal amount (the initial sum of money)
• \( R \) is the Rate of interest per annum (what we need to find)
• \( T \) is the Time period in years

The total amount (\( A \)) after \( T \) years is the Principal plus the Simple Interest:

\( A = P + SI \)

So, \( A = P + \frac{P \times R \times T}{100} \)

Solving the Simple Interest Rate Problem

Let's break down the information given in the question:

• The sum of money (Principal) becomes three times itself. This means the Amount (\( A \)) is 3 times the Principal (\( P \)). So, \( A = 3P \).
• The time period (\( T \)) is 36 years.
• We need to find the Rate of simple interest (\( R \)).

First, let's find the Simple Interest (\( SI \)) earned over 36 years. Since \( A = P + SI \) and \( A = 3P \), we have:

\( 3P = P + SI \)

\( SI = 3P - P \)

\( SI = 2P \)

So, the simple interest earned is equal to twice the principal amount.

Now, we can plug the values into the simple interest formula \( SI = \frac{P \times R \times T}{100} \):

\( 2P = \frac{P \times R \times 36}{100} \)

We want to solve for \( R \). Notice that the Principal (\( P \)) appears on both sides of the equation. Assuming \( P \) is not zero (which it must be for there to be a sum of money), we can cancel \( P \) from both sides:

\( 2 = \frac{R \times 36}{100} \)

Now, rearrange the equation to isolate \( R \):

\( R = \frac{2 \times 100}{36} \)

\( R = \frac{200}{36} \)

Simplify the fraction:

\( R = \frac{50}{9} \)

Now, calculate the decimal value of \( \frac{50}{9} \):

\( \frac{50}{9} = 5.555... \)

This value represents the rate of simple interest per annum. Therefore, the rate is approximately 5.55 percent.

Verification of Simple Interest Rate Calculation

Let's check our answer. Assume a principal \( P = 100 \) and the rate \( R = 5.55\% \) (or \( \frac{50}{9} \)). The time is \( T = 36 \) years.

Simple Interest \( SI = \frac{P \times R \times T}{100} \)

\( SI = \frac{100 \times \frac{50}{9} \times 36}{100} \)

\( SI = \frac{100 \times 50 \times 36}{9 \times 100} \)

\( SI = \frac{50 \times 36}{9} \)

\( SI = 50 \times 4 \)

\( SI = 200 \)

The amount after 36 years would be \( A = P + SI = 100 + 200 = 300 \). Since the principal was 100, the amount is 300, which is three times the principal. This confirms our calculated rate is correct.

Using the rearranged formula for rate:

\( R = \frac{SI \times 100}{P \times T} \)

\( R = \frac{2P \times 100}{P \times 36} \)

\( R = \frac{2 \times 100}{36} \)

\( R = \frac{200}{36} = \frac{50}{9} \approx 5.55 \)

The rate of simple interest is approximately 5.55 percent per annum.

Revision Table: Simple Interest Concepts

Additional Information on Simple Interest and Growth

Understanding how money grows under simple interest is different from compound interest. In simple interest, the growth is linear because only the initial principal earns interest. If the money triples, it means the interest earned is exactly twice the principal.

• If money doubles, \( A=2P \), so \( SI = P \). Then \( P = \frac{P \times R \times T}{100} \), which simplifies to \( 1 = \frac{R \times T}{100} \), or \( RT = 100 \).
• If money triples, \( A=3P \), so \( SI = 2P \). Then \( 2P = \frac{P \times R \times T}{100} \), which simplifies to \( 2 = \frac{R \times T}{100} \), or \( RT = 200 \).
• If money becomes 'n' times, \( A=nP \), so \( SI = (n-1)P \). Then \( (n-1)P = \frac{P \times R \times T}{100} \), which simplifies to \( (n-1) = \frac{R \times T}{100} \), or \( RT = 100(n-1) \).

In this specific problem, money triples (n=3), so \( RT = 100(3-1) = 200 \). Given \( T = 36 \) years, we have \( R \times 36 = 200 \), which means \( R = \frac{200}{36} = \frac{50}{9} \approx 5.55\% \). This confirms the general relationship for simple interest growth.