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# Question No : 3 If you deposit 10 \$ in an account, that pays 5% interest,...

Question No : 3

1. If you deposit 10 \$ in an account, that pays 5% interest, compounded annually, how much you will have at the end of 10 years? 50 years and 100 years
2. How much will be in account at the end of 5 years the amount deposited today is 10,000 and interest is 8% per year, compounded semiannually?
3. How much would I have to deposit in an account today that pays 12% interest, compounded quarterly, so that I have a balance of 20,000 in the account at the end of 10 years?
4. Suppose you deposit 1,00,000 in an account today that pays 6% interest compounded annually. How long does it take before the balance in your account is 5,00,000?
5. At what interest rate an investment of 12,000 be compounded annually to become 28,000 in 12 years?

 a. Deposit Amount= 10 Interest Rate = 5% Annual Compounding Value of deposits at the end of - 10 Years = 10x(1+0.05)^10 = 16.29 50 Years = 10x(1+0.05)^50 = 114.67 100 Years = 10x(1+0.05)^100 = 1,315.01 b. Deposit Amount= 10,000 Interest Rate = 8% Semi-Annual Compounding Value of deposits at the end of - 5 Years = 10000 x (1+(0.08/2))^(5x2) = 10000 x 1.04^10 14802.44 c. Deposit Amount= A Interest Rate = 12% Quarterly Compounding Value of deposits at the end of - 10 Years = 20000 20000= A x (1+(0.12/4))^(10x4) = 20000= A x 1.03^40 = 20000= A x 3.2620 = A = 20000/3.2620 A = 6131.14 d. Deposit Amount= 1,00,000 Interest Rate = 6% Annual Compounding Value of deposits at the end of x years = 500000 100000 x (1+0.06)^X = 500000 1.06^x = 500000/100000 = 5 At x = 27 4.82 At x = 28 5.11 Therefore it will take 28 years to reach 500000 e. Deposit = 12000 Rate = r Annual Compounding Value at the end of 12 years = 28000 12000 x (1+r)^12 = 28000 (1+r)^12 = 28000/12000 (1+r)^12 = 2.33 Using Linear Interpolation - 7% 2.25 r 2.33 6% 2.01 (r-6)/(7-6) = (2.33-2.01)/(2.25-2.01) r-6 = 0.33809748 r= 6.34

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