CAIIB-ABM / BFM- NUMERICALS/CASE STUDIES 1. An annuity consists of monthly repayments of Rs. 600 made over 20 years and if rate is 14% monthly. What is the present value of the annuity? Explanation : Apply FV formula to get the Answer Here R = 14% / 12 = 0.01166 T = 20 × 12 = 240 PV = FV ÷ (1+r)t = 48278
2. Assume that you have a 6% Coupon console bond. The original face value is Rs. 1000 and the interest rate is 9%. Find the current value of this bond. Explanation : Current value of console bond = Coupon ÷ interest rate = 60 ÷ 0.09 = Rs. 667
3. A person invested Rs. 100000 in a bank FDR @ 6% p.a. for 1 year. If interest is compounded on half-yearly basis, the amount payable shall be ...... Explanation : Here, P = 100000 R = 6% half-yearly = 3%@ p.a. = 0.03 p.a. T = 1 yr = 2 half yrs FV = P * (1 + R)^T So, FV = 100000 * (1+0.03)^2 = 106090
4. A person invested Rs. 100000 in a bank FDR @ 6% p.a. for 1 year. If interest is compounded on quarterly basis, the amount payable shall be ...... Explanation : Here, P = 100000 R = 6% quarterly = 0.015% p.a. T = 1 yr = 4 quarters FV = P * (1 + R)^T So, FV = 100000 * (1+0.015)^4 = 106136
5. A person deposited Rs. 10000 in a post-office scheme @ 8% p.a. with quarterly compounding, for 2 years. What is the amount payable? Explanation : Here, P = 10000 R = 8% quarterly = 0.02% p.a. T = 2 Y = 8 quarters FV = P * (1 + R)^T So, FV = 10000 * (1+0.02)^8 = 11717
6. A person borrowed Rs. 10000 from the bank @ 12% p.a. for 1 year, payable on EMI basis. What is the amount of EMI? Explanation : Here, P = 10000 R = 12% yearly = 0.01% monthly T = 1 Y = 12 months EMI = P * R * [(1+R)^T/(1+R)^T-1)] So, EMI = 10000*0.01*(1+0.01)^12 ÷ {(1+0.01)^12 – 1} = 889
7. A person raised a house loan of Rs. 10 lac @ 12% roi repayable in 10 years. Calculate EMI. Explanation : Here, P = 1000000 R = 12% monthly = 0.01% p.a. T = 10 Y = 120 months EMI = P * R * [(1+R)^T/(1+R)^T-1)] So, EMI = 1000000*0.01*(1+0.01)^120 ÷ {(1+0.01)^120 – 1} = 14347
8. Mr. Raj is to invest Rs. 100000 by end of each year for 5 years @ 5% roi. How much amount he will receive? Explanation : Here, P = 1000000 R = 5% p.a. T = 5 Y FV = P / R * [(1+R)^T - 1] FV, if invested at end of each year, is: So, FV = (100000÷0.05) * {{1+0.05}^5 – 1} = 552563
9. Mr x is to receive Rs. 10000, as interest on bonds by end of each year for 5 years @ 5% roi. Calculate the present value of the amount he is to receive. Explanation : Here, P = 10000 R = 5% p.a. T = 5 Y PV = P / R * [(1+R)^T - 1]/(1+R)^T PV to be received, if the amount invested at end of each year: So, FV = (100000÷0.05) * {(1+0.05)^5 – 1} ÷ (1+0.05)^5 = 43295
10. Population of a town is 100000. The rate of change is 4% p.a. what it will be after 5 years? Explanation : Here, P = 100000 R = 4% T = 5 yrs FV = P*(1+R)^T So, FV = 10000*(1+0.04)^5 = 121665
11. Population of a town is 100000. The rate of change is 4% p.a. what is was 5 years ago? Explanation : Here, P = 100000 R = 4% T = 5 yrs FV = P*(1+R)^-T So, FV = 100000*(1+0.04)^-5 = 82193
12. Xyz purchased machinery of Rs. 100000. The rate of depreciation is 10%. At wdv method, what is the amount of depreciation for 4 years? Explanation : Here, P = 100000 R = 10% T = 5 yrs FV = P*(1-R)^T So, FV = 100000*(1-0.1)^4 = 65610 So, amount of depreciation = 100000 – 65610 = 34390
13. Xyz purchased machinery of Rs. 100000. The rate of depreciation is 10%. At wdv method, what is the average rate of depreciation for 4 years? Explanation : Here, P = 100000 R = 10% T = 5 yrs FV = P*(1-R)^T So, FV = 100000*(1-0.1)^4 = 65610 So, amount of depreciation = 100000 – 65610 = 34390 Average rate of depreciation = (34390 ÷100000) * (4÷10) % = 13.76%