SCHOOL OF COMMECE BACHELOR OF BUSINESS MANAGEMENT AND ENTREPRENUERSHIP (BBME) OR BACHELOR OF APPLIED ACCOUNTING AUDITING AND INFORMATION SYSTEM SUBJECT: I.B.M 114 SEMESTER: ONE ASSGNMENT: ONE STUDENT NO: KA1227 PRESENTED TO: MUSSI DUE DATE OF SUBMISSION: 9th September 2019 Question one: (a) i. to calculate the value of first machine 1 January 2017. Solution Cost of first machine = k 350,000.00 Scrap value = k 25,000.00 Useful life = 10 years Depreciation charge = cost of first machine – scrap value Useful life = k350,000.00- k25,000.00 10 years = k 325,000.00 10 = k 32,500.00/year Value of first machine = depreciation charge * time used Total depreciation = k 32,500.00 * 7 = k 227,500.00 Value of machine on I January 2017 = = k 350 000.00 – 227,500.00 k122,500.00 Question one: ii. To calculate profit or loss on the sale of first machine Solution amount sold for – value of machine Profit or loss = k 150, 000.00 – k 122,500.00 = = k27,500.00 There was a profit of 27,500.00 on the sale of the first machine Question one: b. To calculate the value, the new machine on 1 January 2021 Solution cost of second machine (B)= k 400,000.00 deprecation rate (i) = 20% useful life = 4 years future value = B(1-i)n = k 400,000.00(1- 0.2)4 = k 400,000.00(0.8)4 = k 400,000.00 (0.4096) Value of second machine on 1 January 1017 was = k163,840.00 Question two: (b) to find time within which Sandra’s deposit will earn an interest of k 50,000.00 at 4.5% per annum simple interest. Solution Simple interest = P*R*T 100 Where P (principal amount) = k 220,000.00 R (interest rate) = 4.5% T (Period of investment) = ? Simple interest = k 50,000.00 K 50,000.00 = 220,000.00*4.5*T 100 K 50,000.00 = 990,000.00T 100 K 50,000.00 = K 9900 9900T 9900 Sandra’s deposit will take 5 years to yield a simple interest of k 50, 000.00 Question two ii. Option 1 To calculate interest earned if the amount is deposited for four years at 4.5% per annum compound interest directly credited to her account at the end of each year. Solution compound interest = p(1 + i)n where p (principal amount) = n (time of investment) k 220,000.00 = 4 years I (interest rate) = 4.5% Compound interest = k 220,000.00(1 + 4.5/100)4 = k 220,000.00(1 + 0.045)4 = k 220,000.00(1.045)4 = k220,000.00(1.1925) Compound interest will be k 262,354.09 To calculate option (b) To calculate compound interest if the amount is deposited for four years at 4%per compound interest compounded every six months. Solution Compound interest p (1 + i)n = Where p (principal amount) = k 220,000.00 I (interest rate as proportion) = 4%/2 N (period of investment) = 2*4=8 Compound interest = k 220,000.00(1 + 2%/100)8 = k 220,000.00(1+0.02)8 = k 220,000.00(1.02)8 = k 220,000.00(0.1716) = k 257, 765.06 The best option will = compound interest under option 1 – option 2 As such the best option would be option one as it’s giving a higher interest . Question four: (a) To calculate length and width of Asha’s garden Solution Area = length * width Area = 60ft2 Let width = x Length = 60ft2= x+4 x*(x+4) X2 + 4x – 60 ft2 = 0 X2 + 10x – 6x – 60ft2 = 0 X (x + 10) – 6 (x – 10) =0 (X + 10) (x – 6) = 0 X = -10 X= 6 Width of the garden = 6ft Length = = 6ft + 4ft 10ft Question four (b): To find price an adults meal and a child’s meal. Solution Let adults price = y Childs price = x Sue and mike plus their two children 2y + 2x = k 55 Ahmed and his three children Y + 3x = k48 In equation 1. When x = 0 2y + 2(0) = k 55 2y = k 55 Y = 27.5 And in equation 1 When y = 0 2(0) + 2x = k55 2x = k55 X=27.5 Therefore, when x = 0, y is 27.5 and when y = 0, x = 27.5 In equation two Y + 3x = k48 When y= 0 0 +3x = k48 3x=k48 X=16 And when x =0 Y + 3 (0) = k48 Y = k48 Therefore: 2y + 2x = k55 x y 0 27.5 27.5 0 Y + 3x =k48 X y 0 48 16 0 Question three: Solution Book value of machine = Scrap value = k 1,000,000.00 150,000.00 Useful life = 5 year book value of machine – scrap value Net cash outflows = = k 1000,000.00 – k 150,000.00 Yea r Cash in flow Cash out flow Net cash flow 0 - K (850,000.00 ) 1 K 200,000.0 0 K 350,000.0 0 K (850,000.00 ) K 200,000.00 2 Discoun t factor (18%) 1.00 Present value Discoun t factor (24%) 1.00 Present value 0.847 K 169,400.00 0.806 0.718 K 251,300.00 0.650 K 161,20 0 K 217,50 0 K 350,000.00 (850,000.00 ) 3 K 350,00.00 K 350,000.00 0.609 K 213,150.00 0.524 4 K 350,000.0 0 K 350,000.0 0 K 350,000.00 0.516 K 180,600.00 0.423 K 350,000.00 0.437 K 152,950 0.341 5 (850,0 00.00) K 183,40 0.00 K148,0 50.00 119.35 0 Total present value 1 = = Total present value 2 = (k 850,000.00) + 967400 k 117 400 (k 850,000.00) + k 839, 500.00 (k 10, 500) Internal rate of return = NPVi1 – NPVi2 NPV1 – NPV2 = k 117,400(18%) – (k 10,500*24) 117,400 – (10,500) = K 2, 113,200.00 – (k252,000.00) K 117, 400 – (k10,500) = K 2, 365,700.00 K 127, 900.00 = Discount factor= 20.07% 1 (1+i)n Internal rate return falls between 18% and 24% as such the project is viable. Question two: to calculate the net amount paid by Sandra to nur and udoka in January. Solution Rate/hour = 10$ Normal hours= 150 hours Statutory deductions= 10% Other deductions= Nurs pay 5% = normal hours × rate /hour+ ( over time × 1.5) = 150 × 10$ (12 × 1.5) = 1500$ + 180 $ = 1680 $ Stator deduction Other deductions = (1689$ - 1400 $) ×10%÷100 = 280 ×0.1 = 28 $ = 280 $- 28 $ = 252$ = 252$ × 5%÷100 = 252 × 0.05 = 12.6$ Total deduction = Total nurs pay Udoka pay 280$ - ( 28$ + 12.6 ) = 239.4 = 1400 $ + 239.4 = 1639.4$ = 135 hours × 10 $/hour = 1350 Total money payed to nur and udoka by Sandra = 1350$ + 1639.4$ = 2989.4$