__BANK FINANCIAL MANAGEMENT NUMERICALS : 6__

ABM/BFM NUMERICALS

ABM/BFM NUMERICALS

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%