An amount of money is invested in a savings account for two years. It increased by Tk. 52.50 in two years. after annual compounding at the rate of 10% per year.

An amount of money is invested in a savings account for two years. It increased by Tk. 52.50 in two years. after annual compounding at the rate of 10% per year. What is the amount invested initially?

An amount of money is invested in a savings account for two years. It increased by Tk. 52.50 in two years. after annual compounding at the rate of 10% per year. What is the amount invested initially? ?

Created: 5 days ago | Updated: 5 days ago | Created By:

Imran Ahmed

Tk.400

Tk.300

TK. 250

Tk. 200

0 1

Best Answer Explained in Below-:

This is a compound interest problem, and we'll use the compound interest formula:\[A = P(1 + r/n)^(nt)\]Where:- \(A\) is the final amount after \(t\) years.- \(P\) is the principal amount (the initial amount).- \(r\) is the annual interest rate (decimal).- \(n\) is the number of times interest is compounded per year.- \(t\) is the time the money is invested for (in years).In this case, we know that \(r = 10\% = 0.1\), \(t = 2\) years, and the final amount increased by Tk. 52.50. We're trying to find the initial amount (\(P\)).Let's denote the initial amount as \(P\), and the final amount as \(A\). We're given that \(A - P = 52.50\).\(A = P(1 + r/n)^(nt)\)\(A = P(1 + 0.1/1)^(1*2)\) (Since it's compounded annually, \(n = 1\))\(A = P(1.1)^2\)\(A = 1.21P\)We know that \(A - P = 52.50\), so:\(1.21P - P = 52.50\)\(0.21P = 52.50\)\(P = \frac{52.50}{0.21}\)\(P ≈ 250\)So, the initial amount invested was approximately Tk. 250.

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An amount of money is invested in a savings account for two years. It increased by Tk. 52.50 in two years. after annual compounding at the rate of 10% per year. What is the amount invested initially? ?

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