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Question
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?
Tk.400
Tk.300
TK. 250
Tk. 200
ANSWER : 3
Descrption
<p>This is a compound interest problem, and we'll use the compound interest formula:</p><p>\[A = P(1 + r/n)^(nt)\]</p><p>Where:<br>- \(A\) is the final amount after \(t\) years.<br>- \(P\) is the principal amount (the initial amount).<br>- \(r\) is the annual interest rate (decimal).<br>- \(n\) is the number of times interest is compounded per year.<br>- \(t\) is the time the money is invested for (in years).</p><p>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 (\(P\)).</p><p>Let's denote the initial amount as \(P\), and the final amount as \(A\). We're given that \(A - P = 52.50\).</p><p>\(A = P(1 + r/n)^(nt)\)</p><p>\(A = P(1 + 0.1/1)^(1*2)\) (Since it's compounded annually, \(n = 1\))</p><p>\(A = P(1.1)^2\)</p><p>\(A = 1.21P\)</p><p>We know that \(A - P = 52.50\), so:</p><p>\(1.21P - P = 52.50\)</p><p>\(0.21P = 52.50\)</p><p>\(P = \frac{52.50}{0.21}\)</p><p>\(P ≈ 250\)</p><p>So, the initial amount invested was approximately Tk. 250.</p>
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