The problem asks for the time it takes for an investment to grow to 9 times its initial value, given that it triples in 3 years under compound interest.
Let the principal amount be \(P\). The formula for the amount \(A\) after \(t\) years with compound interest is \(A = P(1+r)^t\), where \(r\) is the annual interest rate.
We are given that the sum triples itself in 3 years. This means:
\( 3P = P(1+r)^3 \)Dividing both sides by \(P\), we get:
\( 3 = (1+r)^3 \)This equation tells us the growth factor over a 3-year period is 3.
We need to find the time \(T\) when the amount becomes 9 times the principal, i.e., \(A = 9P\).
\( 9P = P(1+r)^T \)Dividing by \(P\) gives:
\( 9 = (1+r)^T \)We know that \(9\) can be expressed as \(3^2\). So, the equation becomes:
\( 3^2 = (1+r)^T \)Now, substitute the value of \(3\) from the first condition, which is \(3 = (1+r)^3\):
\( ((1+r)^3)^2 = (1+r)^T \)Using the power rule \((a^m)^n = a^{mn}\), we simplify the left side:
\( (1+r)^{3 \times 2} = (1+r)^T \) \( (1+r)^6 = (1+r)^T \)By comparing the exponents, since the base \((1+r)\) is the same and greater than 1 (as it represents growth), the time periods must be equal:
\( T = 6 \)Therefore, the sum will amount to 9 times itself in 6 years.
The certain sum amounts to Rs. 9,982.50 in \(2\frac{1}{2}\) years at 12% p.a., interest compounded 10-monthly. The sum (in Rs.) is:
The difference between the simple interest and the compound interest compounded annually on a certain sum of money for 2 years at a rate of 8% per annum is Rs. 16.80. Find the principle amount.
If a sum of ₹ 2000 is lent at 10% p.a. compound interest, what is the interest for the second year?
A sum becomes 5 times of itself in 3 years. at compound interest (interest is compounded annually). In how many years. will the sum becomes 125 times of itself?
If the compound interest on a certain sum of money for two years at 9% p.a. is Rs. 3,762, then the sum is: