A sum of money doubles itself in 10 years on compound interest. In how many years will it become four times?
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Solving Compound Interest Problems: Doubling and Quadrupling Time
Understanding how compound interest works is key to solving problems involving the growth of money over time. In this specific problem, we are told that a sum of money doubles itself in a certain number of years under compound interest and we need to find out how many years it will take to become four times the original amount.
Understanding Compound Interest Growth
Compound interest means that the interest earned in each period is added to the principal, and then the next period's interest is calculated on the new, larger principal. This leads to exponential growth. The formula for compound interest is:
\[A = P\left(1 + \frac{r}{n}\right)^{nt}\]Where:
- \(A\) is the amount after \(t\) years.
- \(P\) is the principal amount.
- \(r\) is the annual interest rate (as a decimal).
- \(n\) is the number of times interest is compounded per year.
- \(t\) is the number of years.
In many problems where the compounding frequency (\(n\)) is not explicitly mentioned and the time period is in years, it's common to assume annual compounding, meaning \(n=1\). The formula simplifies to:
\[A = P(1 + r)^t\]Analyzing the Given Information
We are given two key pieces of information:
- Information 1: A sum of money doubles itself in 10 years on compound interest.
- Information 2: We need to find the number of years it takes for the same sum to become four times its original value.
Step-by-Step Solution using the Formula
Let the initial sum of money be \(P\). The interest rate is \(r\) per annum, compounded annually (assuming \(n=1\)).
Step 1: Use the doubling information.
After 10 years, the amount (\(A\)) is twice the principal (\(2P\)). Using the formula \(A = P(1 + r)^t\):
\[2P = P(1 + r)^{10}\]Divide both sides by \(P\):
\[2 = (1 + r)^{10}\]This equation relates the interest rate \(r\) to the doubling time. We don't need to find \(r\) explicitly, just the value of \((1 + r)\).
Step 2: Use the quadrupling information.
We want to find the time (\(t'\)) when the amount becomes four times the principal (\(4P\)). Using the formula again:
\[4P = P(1 + r)^{t'}\]Divide both sides by \(P\):
\[4 = (1 + r)^{t'}\]Step 3: Connect the two equations.
We have \(2 = (1 + r)^{10}\) and \(4 = (1 + r)^{t'}\).
Notice that \(4 = 2^2\). We can substitute the first equation into the second one:
\[4 = (1 + r)^{t'}\] \[2^2 = (1 + r)^{t'}\]Substitute \(2\) with \((1 + r)^{10}\):
\[\left((1 + r)^{10}\right)^2 = (1 + r)^{t'}\]Using the exponent rule \((a^m)^n = a^{mn}\):
\[(1 + r)^{10 \times 2} = (1 + r)^{t'}\] \[(1 + r)^{20} = (1 + r)^{t'}\]Since the bases \((1 + r)\) are equal and positive (as interest rate \(r\) must be positive for growth), the exponents must be equal:
\[t' = 20\]So, it will take 20 years for the money to become four times its original value.
Alternative Explanation using Doubling Periods
For compound interest, if money doubles in a certain period, it will double again in the same period. This is because the growth is exponential.
- Starting Amount: \(P\)
- After 10 years, the amount is \(2P\) (doubled).
- To reach \(4P\), the amount \(2P\) needs to double again.
- Since the compounding is consistent, doubling from \(2P\) to \(4P\) (which is also a doubling) will take the same amount of time as doubling from \(P\) to \(2P\).
- Time taken to go from \(P\) to \(2P\) = 10 years.
- Time taken to go from \(2P\) to \(4P\) = 10 years.
Total time taken to go from \(P\) to \(4P\) = Time (\(P \to 2P\)) + Time (\(2P \to 4P\))
Total time = 10 years + 10 years = 20 years.
This confirms the result obtained using the formula.
Statement-wise Analysis and Conclusion
- Statement: A sum of money doubles itself in 10 years on compound interest. This establishes the growth factor \((1+r)^{10} = 2\).
- Statement: In how many years will it become four times? We need to find \(t'\) such that \((1+r)^{t'} = 4\).
- Since \(4 = 2^2\), we can substitute the known growth factor: \((1+r)^{t'} = ( (1+r)^{10} )^2\).
- Comparing the exponents, \(t' = 10 \times 2 = 20\).
The time taken for the money to become four times is indeed 20 years.
| Initial Amount | Time Elapsed (Years) | Amount with Compound Interest | Growth Factor |
|---|---|---|---|
| \(P\) | 0 | \(P\) | 1 |
| \(P\) | 10 | \(2P\) | 2 |
| \(P\) | 20 | \(4P\) | 4 (\(=2^2\)) |
Final Answer
Based on both the compound interest formula and the concept of doubling periods, the sum of money will become four times its original value in 20 years.
Compound Interest Revision Table
| Concept | Description | Formula (Annual Compounding) |
|---|---|---|
| Principal (P) | The initial amount invested or borrowed. | N/A |
| Amount (A) | The total value after interest is added. | \(A = P(1 + r)^t\) |
| Interest Rate (r) | The percentage rate per year (as a decimal). | N/A |
| Time (t) | The duration of the investment/loan in years. | N/A |
| Compound Growth | Interest added to principal, leading to exponential growth. | \(A = P(1 + r)^t\) |
| Doubling Time | Time for an investment to double in value. | If \(A=2P\), \(2 = (1+r)^t\) |
| Quadrupling Time | Time for an investment to become four times its value. | If \(A=4P\), \(4 = (1+r)^t\) |
Additional Information on Compound Interest and Investment Growth
Compound interest is a powerful concept often called the "eighth wonder of the world". It's fundamental to understanding investments, loans, and savings growth. Here are some related points:
- Rule of 72: A simple way to estimate doubling time. Divide 72 by the annual interest rate percentage to get an approximate number of years for an investment to double. For example, at 7.2% interest, it takes roughly 72 / 7.2 = 10 years to double. (Note: This is an approximation, especially for higher rates). In our problem, the doubling time is 10 years, which implies an interest rate of approximately 7.2%.
- Exponential Nature: The key takeaway from this problem is the exponential nature of compound interest. If it takes \(T\) years to double, it takes \(2T\) years to quadruple (\(2^2\) times), \(3T\) years to become eight times (\(2^3\) times), and so on. It takes \(k \times T\) years to become \(2^k\) times the original amount.
- Effect of Compounding Frequency (n): If interest is compounded more frequently (e.g., semi-annually \(n=2\), quarterly \(n=4\), monthly \(n=12\)), the effective annual rate is slightly higher, leading to faster growth. The formula \(A = P(1 + r/n)^{nt}\) accounts for this. In this problem, assuming \(n=1\) was sufficient because we worked with growth factors over the stated time period.
- Applications: Compound interest calculations are vital in personal finance for understanding mortgages, car loans, retirement savings (like 401k or IRA), and long-term investments.
Understanding the relationship between time and growth factor, especially for doubling periods, can simplify many compound interest problems significantly without needing to calculate the exact interest rate.
Important Questions from Simple and Compound Interest
Simple interest on a certain sum at 8% per annum for 5 years is ₹2400. What will be the compound interest on the same sum at the same rate for 2 years?
Find the compound interest on ₹64000 at 10% per annum for 9 months when interest is compounded quarterly.
A man invested on simple interest, 1/4 of his capital at 7% p.a., another 1/4 of the capital at 8% p.a. and the remaining capital at 10% p.a. He earned Rs. 700 as interest in one year. Find his total capital invested.
A sum of ₹1500 is invested at 5% p.a. simple interest. If the interest is added to the principal after every 10 years, the amount will become ₹3000 after:
Find the compound interest on ₹42000 for 1½ years at 10% p.a. compounded annually.
