The question asks us to find the factor (\(n\)) by which a sum of money grows in 20 years, given that it triples itself in 8 years at simple interest. Simple interest means the interest earned each year is constant and calculated only on the initial principal amount.
Let the principal amount be \(P\). The problem states that the sum of money triples itself in 8 years. This means the final amount (\(A\)) after 8 years is \(3P\). The formula for the final amount (\(A\)) in simple interest is \(A = P + SI\), where \(SI\) is the simple interest earned.
So, the simple interest earned in 8 years is:
\( SI = A - P = 3P - P = 2P \)
Now, we use the simple interest formula:
\( SI = \frac{P \times R \times T}{100} \)
Here, \(SI = 2P\) and \(T = 8\) years. Plugging these values in:
\( 2P = \frac{P \times R \times 8}{100} \)
We can cancel \(P\) from both sides (assuming \(P \ne 0\)):
\( 2 = \frac{R \times 8}{100} \)
Now, we solve for the rate (\(R\)):
\( R = \frac{2 \times 100}{8} \)
\( R = \frac{200}{8} \)
\( R = 25\% \)
So, the rate of simple interest is 25% per year.
Next, we need to find out how many times (\(n\)) the principal amount becomes in 20 years at this rate (\(R = 25\%\)).
Let the final amount after 20 years be \(nP\). The simple interest earned in 20 years (\(T = 20\)) is:
\( SI = nP - P = (n-1)P \)
Using the simple interest formula again with \(R = 25\%\) and \(T = 20\):
\( SI = \frac{P \times R \times T}{100} \)
\( (n-1)P = \frac{P \times 25 \times 20}{100} \)
Cancel \(P\) from both sides:
\( n-1 = \frac{25 \times 20}{100} \)
\( n-1 = \frac{500}{100} \)
\( n-1 = 5 \)
Solving for \(n\):
\( n = 5 + 1 \)
\( n = 6 \)
Therefore, the sum of money becomes 6 times itself in 20 years.
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