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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Understanding the Problem: Compound Interest Growth
The question asks for the minimum number of years it takes for a sum of money to grow to be more than 100 times its original value when compounded annually at a rate of 20%.
This is a classic compound interest problem where we need to find the time period (\(n\)) given the principal amount (\(P\)), the final amount (\(A\)), and the interest rate (\(r\)). The condition is that the final amount must be strictly greater than 100 times the initial principal.
Compound Interest Formula
The formula for compound interest is:
\($A = P(1 + r)^n$\)
Where:
- \(A\) is the amount after \(n\) years.
- \(P\) is the principal amount (initial investment).
- \(r\) is the annual interest rate (as a decimal).
- \(n\) is the number of years.
Setting up the Inequality
We are given that the interest rate \(r = 20\%\), which is \(0.20\) as a decimal. The problem states that the amount \(A\) becomes more than 100 times the principal \(P\). Mathematically, this is expressed as:
\($A > 100P$\)
Substitute the compound interest formula for \(A\):
\($P(1 + r)^n > 100P$\)
Substitute the value of \(r = 0.20\):
\($P(1 + 0.20)^n > 100P$\)
\($P(1.2)^n > 100P$\)
Assuming the principal \(P\) is a positive value (which it must be for money), we can divide both sides of the inequality by \(P\) without changing the direction of the inequality sign:
\($ (1.2)^n > 100$\)
Solving for n using Logarithms
To solve for the exponent \(n\) in the inequality \( (1.2)^n > 100 \), we can use logarithms. Taking the base-10 logarithm of both sides (since the given log values are base 10):
\($\log_{10}((1.2)^n) > \log_{10}(100)$\)
Using the logarithm property \(\log(a^b) = b \log(a)\):
\($n \log_{10}(1.2) > \log_{10}(100)$\)
We know that \(\log_{10}(100) = \log_{10}(10^2) = 2\).
\($n \log_{10}(1.2) > 2$\)
Calculating \(\log_{10}(1.2)\)
We need to find the value of \(\log_{10}(1.2)\) using the given values \(\log_{10} 2 = 0.301\) and \(\log_{10} 3 = 0.477\).
\($\log_{10}(1.2) = \log_{10}\left(\frac{12}{10}\right) = \log_{10}\left(\frac{6}{5}\right)$\)
Using the logarithm property \(\log(\frac{a}{b}) = \log(a) - \log(b)\):
\($\log_{10}\left(\frac{6}{5}\right) = \log_{10}(6) - \log_{10}(5)$\)
Now calculate \(\log_{10}(6)\) and \(\log_{10}(5)\) using the given values:
\($\log_{10}(6) = \log_{10}(2 \times 3)$\)
Using the logarithm property \(\log(a \times b) = \log(a) + \log(b)\):
\($\log_{10}(6) = \log_{10}(2) + \log_{10}(3) = 0.301 + 0.477 = 0.778$\)
\($\log_{10}(5) = \log_{10}\left(\frac{10}{2}\right)$\)
Using the logarithm property \(\log(\frac{a}{b}) = \log(a) - \log(b)\):
\($\log_{10}(5) = \log_{10}(10) - \log_{10}(2)$\)
Since \(\log_{10}(10) = 1\):
\($\log_{10}(5) = 1 - 0.301 = 0.699$\)
Now, substitute these values back to find \(\log_{10}(1.2)\):
\($\log_{10}(1.2) = \log_{10}(6) - \log_{10}(5) = 0.778 - 0.699 = 0.079$\)
Finding the Value of n
Substitute the calculated value of \(\log_{10}(1.2)\) back into the inequality \(n \log_{10}(1.2) > 2\):
\($n \times 0.079 > 2$\)
Divide both sides by 0.079:
\($n > \frac{2}{0.079}$\)
Now, perform the division:
\($n > 25.316...\$\)
Since \(n\) represents the number of years and must be an integer, the least integer value of \(n\) that is greater than 25.316... is 26.
Therefore, it takes at least 26 years for the sum of money to become more than 100 times its original value at a 20% compound interest rate.
Conclusion
The least value of \(n\) for which the sum of money at 20% rate of compound interest per annum becomes more than 100 times is 26 years.
Revision Table: Compound Interest Calculation
| Concept | Description |
|---|---|
| Compound Interest Formula | \(A = P(1 + r)^n\) |
| Problem Condition | \(A > 100P\) |
| Resulting Inequality | \((1.2)^n > 100\) |
| Logarithm Property Used | \(\log(a^b) = b\log(a)\) |
| Log Calculation | \(\log_{10}(1.2) = \log_{10}(6/5) = \log_{10}(6) - \log_{10}(5)\) |
| Required Log Values | \(\log_{10} 2, \log_{10} 3, \log_{10} 5, \log_{10} 6, \log_{10} 1.2\) |
| Inequality with Log Values | \(n \times 0.079 > 2\) |
| Minimum Integer Years | \(n = 26\) |
Additional Information: Logarithm Properties in Finance
Logarithms are very useful in finance, especially when dealing with exponential growth or decay like compound interest. They allow us to solve for exponents (like the number of years or the interest rate) in the compound interest formula.
Key logarithm properties used in such calculations include:
- Product Rule: \(\log_b(xy) = \log_b(x) + \log_b(y)\)
- Quotient Rule: \(\log_b(\frac{x}{y}) = \log_b(x) - \log_b(y)\)
- Power Rule: \(\log_b(x^p) = p \log_b(x)\)
- Change of Base Formula: \(\log_b(x) = \frac{\log_k(x)}{\log_k(b)}\) (Though not directly used here as we stuck to base 10)
- Log of Base: \(\log_b(b) = 1\)
- Log of 1: \(\log_b(1) = 0\)
In this problem, breaking down \(\log_{10}(1.2)\) involved using the quotient rule (\(\log_{10}(12/10)\)) and then the product rule (\(\log_{10}(2 \times 6)\) or calculating \(\log_{10}(6)\) and \(\log_{10}(5)\) separately) and the fact that \(\log_{10}(10) = 1\).
Similar Questions
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Important Questions from Compound Interest
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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?
