A sum of money compounded annually doubles itself in 5 years. In how many years will it become four times of itself ?
10 years
Solving Compound Interest Problems: Doubling and Quadrupling Time
This problem involves understanding how money grows under annual compound interest. Compound interest means that the interest earned each year is added to the principal, and the next year's interest is calculated on this new, larger amount. This leads to exponential growth.
Understanding the Compound Interest Formula
The formula for the amount \(A\) after \(t\) years when a principal \(P\) is invested at an annual interest rate \(r\) compounded annually is:
\[ A = P(1 + r)^t \]Here:
- \(A\) is the amount after \(t\) years.
- \(P\) is the principal amount.
- \(r\) is the annual interest rate (as a decimal).
- \(t\) is the time in years.
Applying the Formula to the Problem Conditions
We are given two pieces of information:
- The sum of money doubles itself in 5 years.
- We need to find the time it takes for the money to become four times itself.
Condition 1: Money Doubles in 5 Years
Let the principal amount be \(P\). After 5 years, the amount becomes \(2P\). Using the compound interest formula:
\[ 2P = P(1 + r)^5 \]We can divide both sides by \(P\) (assuming \(P > 0\)):
\[ 2 = (1 + r)^5 \]This equation tells us the relationship between the interest rate \(r\) and the time it takes for the money to double. We don't need to calculate \(r\) itself, just keep this relationship in mind.
Condition 2: Money Becomes Four Times Itself
We want to find the time, let's call it \(T\) years, when the amount becomes \(4P\). Using the compound interest formula again:
\[ 4P = P(1 + r)^T \]Divide both sides by \(P\):
\[ 4 = (1 + r)^T \]Finding the Time to Become Four Times
Now we have two equations:
- \( 2 = (1 + r)^5 \)
- \( 4 = (1 + r)^T \)
We know that \(4\) is the square of \(2\), i.e., \(4 = 2^2\). Let's substitute this into the second equation:
\[ 2^2 = (1 + r)^T \]Now, we can substitute the expression for \(2\) from the first equation (\(2 = (1 + r)^5\)) into this equation:
\[ ((1 + r)^5)^2 = (1 + r)^T \]Using the exponent rule \((a^m)^n = a^{m \times n}\), the left side becomes:
\[ (1 + r)^{5 \times 2} = (1 + r)^{10} \]So, the equation becomes:
\[ (1 + r)^{10} = (1 + r)^T \]For this equality to hold, the exponents must be equal (assuming \(1+r \neq 1\) and \(1+r \neq -1\), which is true for any reasonable positive interest rate):
\[ T = 10 \]Thus, it will take 10 years for the sum of money to become four times itself.
Summary of the Steps
- Identify the compound interest formula: \( A = P(1 + r)^t \).
- Use the first condition (doubling in 5 years) to find the relationship \( 2 = (1 + r)^5 \).
- Use the second condition (four times) to set up the equation \( 4 = (1 + r)^T \).
- Recognize that \( 4 = 2^2 \).
- Substitute the relationship from step 2 into the equation from step 3.
- Solve for \(T\) using exponent properties.
The time taken for the sum to become four times is 10 years.
Revision Table: Compound Interest Growth
| Multiple of Principal | Time Taken (Years) | Formula Relationship |
|---|---|---|
| 1 (Starting) | 0 | \(1 = (1+r)^0\) |
| 2 (Doubles) | 5 | \(2 = (1+r)^5\) |
| 4 (Four Times) | 10 | \(4 = (1+r)^{10}\) (Since \(4=2^2 = ((1+r)^5)^2 = (1+r)^{10}\)) |
| 8 (Eight Times) | 15 | \(8 = (1+r)^{15}\) (Since \(8=2^3 = ((1+r)^5)^3 = (1+r)^{15}\)) |
Additional Information: Compound Interest and Doubling Time
This problem highlights a useful property of compound interest, especially when the question involves geometric progression of the amount (like doubling, quadrupling, eight times, etc.). If an investment doubles in \(t_d\) years at a constant compound interest rate, then it will become \(2^n\) times the original amount in \(n \times t_d\) years. In this case, the doubling time \(t_d = 5\) years.
- To become \(2^1 = 2\) times, it takes \(1 \times 5 = 5\) years.
- To become \(2^2 = 4\) times, it takes \(2 \times 5 = 10\) years.
- To become \(2^3 = 8\) times, it takes \(3 \times 5 = 15\) years.
- To become \(2^4 = 16\) times, it takes \(4 \times 5 = 20\) years, and so on.
This pattern holds true specifically for compound interest. Simple interest does not follow this pattern.
Understanding this relationship can quickly solve problems where the growth is described as doubling or other powers of the doubling factor.
Similar Questions
A sum of money at 20% rate of compound interest per annum becomes more than 100 times in n years. What is the least value of n? (Use log10 2 = 0.301, log10 3 = 0.477)
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What is the least number of complete years in which a sum of money put out at 40% annual compound interest will be more than tripled?
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Important Questions from Compound Interest
A person borrowed Rs. 10000 on compound interest at the rate of 40 percent per annum. If the interest is compounded half yearly, then what will be the amount to be paid after 1.5 years?
The difference between the compound interest (compounding annually) and the simple interest on a sum of money at the rate of 40 per cent per annum for 2 years is Rs. 2400. What is the amount?
In how many years will a sum of Rs.1875 amount to Rs.2187 at 8 percent p.a. compound interest?
A sum of money has increased by 45% in 9 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?
At a certain rate of compound interest a certain sum amounts to Rs. 64800 in 4 years and Rs. 93312 in 6 years. What is the compound interest earned in fifth year?
