Compound Interest Calculation: Sum Growth Over Time

This problem involves understanding how compound interest works and how the principal amount grows exponentially over time. We are given information about the growth of a sum over a specific period and asked to find the time required for a much larger growth under the same compound interest rate.

Understanding Compound Interest

Compound interest is calculated on the initial principal and also on the accumulated interest from previous periods. The formula for the amount (A) after 't' years with principal (P) and annual interest rate (R) compounded annually is:

$$A = P(1 + R)^t$$

Here, $(1 + R)$ represents the growth factor per year.

Analyzing the Given Information

We are told that a sum becomes 5 times of itself in 3 years. Let the principal amount be P. After 3 years, the amount becomes 5P. Using the compound interest formula, we can write:

$$5P = P(1 + R)^3$$

Dividing both sides by P, we get:

$$5 = (1 + R)^3$$

This equation tells us that the growth factor raised to the power of 3 is equal to 5. This is a crucial relationship.

Calculating Time for 125 Times Growth

We want to find out in how many years, say $t'$, the sum will become 125 times of itself. So, the amount after $t'$ years will be 125P. Using the compound interest formula again:

$$125P = P(1 + R)^{t'}$$

Dividing both sides by P, we get:

$$125 = (1 + R)^{t'}$$

Using the Relationship to Find Time

We know from the first part that $5 = (1 + R)^3$. We also know that 125 can be expressed in terms of 5:

$$125 = 5 \times 5 \times 5 = 5^3$$

Now, we can substitute $5^3$ for 125 in the second equation:

$$5^3 = (1 + R)^{t'}$$

Since we know $5 = (1 + R)^3$, we can substitute $(1 + R)^3$ for 5 on the left side:

$$((1 + R)^3)^3 = (1 + R)^{t'}$$

Using the exponent rule $(a^m)^n = a^{m \times n}$, we simplify the left side:

$$(1 + R)^{3 \times 3} = (1 + R)^{t'}$$

$$(1 + R)^9 = (1 + R)^{t'}$$

Now, since the bases are the same and positive (as $1+R$ must be > 1 for growth), we can equate the exponents:

$$t' = 9$$

Therefore, the sum will become 125 times of itself in 9 years.

Summary of Steps

• Identify the compound interest formula: $A = P(1 + R)^t$.
• Use the first condition ($A=5P, t=3$) to find the relationship: $5 = (1+R)^3$.
• Use the second condition ($A=125P, t=t'$) to set up the equation: $125 = (1+R)^{t'}$.
• Express 125 as a power of 5: $125 = 5^3$.
• Substitute the relationship from step 2 into the equation from step 3: $5^3 = ((1+R)^3)^3$.
• Simplify the exponents: $(1+R)^9 = (1+R)^{t'}$.
• Equate the exponents to find $t' = 9$ years.

The sum becomes 125 times of itself in 9 years.

Revision Table: Key Concepts in Compound Interest

Additional Information on Compound Interest Growth

Compound interest leads to exponential growth because the interest earned in each period is added to the principal for the next period's calculation. This means the base on which interest is calculated keeps increasing.

• In this problem, the growth factor over 3 years is 5.
• To find the growth over 6 years (which is $3 \times 2$ years), the growth would be $5 \times 5 = 5^2 = 25$ times. This is because the initial 3-year period's growth factor (5) acts as a new multiplier over the *next* 3 years.
• Similarly, over 9 years (which is $3 \times 3$ years), the growth would be $5 \times 5 \times 5 = 5^3 = 125$ times.
• This pattern holds true: if a sum becomes 'x' times in 'y' years at compound interest, it will become $x^n$ times in $n \times y$ years. Here, $x=5$, $y=3$, and $x^n = 125 = 5^3$, so $n=3$. The time is $n \times y = 3 \times 3 = 9$ years.

This exponential nature is what makes compound interest powerful for investments over longer periods.