IF YOU CAN SOLVE THESE PROBLEMS IN 1 HR, THEN YOU ARE A CAL TECH EXPERT PROBLEMS TAKEN FROM 1ST MEETING REFRESHER AT TOLENTINO AND ASSOCIATES.. 1. A deposit of P110, 000 was made for 31 days. The net interest after deducting 20% withholding tax is P890.36. Find the rate of return annually. 2. P5, 000 is borrowed for 75 days at 16% per annum simple interest. How much will be due at the end of 75 days? S-1 Mr. J de la Cruz borrowed money from a bank. He received from the bank P1, 340.00 and promised to pay P1, 500.00 at the end of 9 months. Determine the following 3. Simple interest rate. 4. The corresponding discount rate or often referred to as the “Bankers Discount”. S-2 John borrowed P50, 000.00 from the bank at 25% compounded semi-annually. 5. What is the equivalent effective rate of interest? 6. How much must he pay if he pay after 5 years? 7. When will the money double ? 8. If he pay 10,000 after 1 year and 10,000 on the next year , how much must he pay on the 3rd year to wipe out his loan? 9. Convert the interest rate to compounded quarterly 10. Convert the interest rate to compounded monthly. 11. Convert the interest rate to compounded continuously . 12. Find the present worth of a future payment of P100, 000 to be made in 10 years with an interest of 12% compounded quarterly. 13. If P5, 000.00 shall accumulate for 10 years at 8% compounded quarterly. Find the compounded interest at the end of 10 years. 14. By the condition of a will, the sum of P2, 000 is left to a girl to be held in trust fund by her guardian until it amounts to P50, 000, when will the girl received the money if the fund is invested at 8% compounded quarterly? 15. A Mechanical Engineer wishes to accumulate a total of P10, 000 in a savings account at the end of 10 years. If the bank pays only 4% compounded quarterly, what should be the initial deposit? Mr X borrowed P 100,000 at en effective interest of 8% in a bank. He will pay the loan monthly for 5 years 16. Convert interest to compounded monthly 17. What should be the monthly payment? 18. If he decides to pay on a quarterly basis, convert the interest rate to compounded quarterly. 19. What must be the quarterly payment? 20. If he decides to pay on a semiannual basis, convert the interest rate to compounded seminannually. 21. What should be the semiannual payment? 22. Convert the interest rate to compounded continuously. Mr Y deposited P 1,000 at the end of every month in an account that earns 12% effective interest. 23. What is the monthly interest rate? 24. What will be the amount of money after 5 years? 25. If Mr Y deposited P 1,000 at the beginning of each year, what will be the amount in 5 years. (There will be only 60 deposits )
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S -5 Mr Z borrowed P 300,000 from a bank. The bank charges 16% effective interest. 26. What is the equivalent monthly interest rate? 27. What is the monthly amortization if he will pay at the end of each month for 5 years? 28. What will be the monthly amortization if he delays his payment by 6 months (his 1st payment happened at the end of 6 months ) . 29. What is the monthly amortization if he pays at the beginning of each month for 5 years. 30. Which is the biggest interest rate? a. 12% compounded quarterly b. 11.88% compounded seminually c. 11.99 % compounded monthly d. 12.02% compounded continuously S-6 Given the series of payments at the end of each year: Year 1 2000 Year 2 5000 Year 3 8000 and so on until year 20, If i = 12% per annum 31. The present worth of all payments. (year 0 ) 32. The future worth of all payments ( year 20) 33. The equivalent equal annual worth of all the payments. S–7 Given the first cost of a machine FC = 500,000 Salvage Value is 150,000 Economic Life = 12 years 34. USING straight line method , the annual depreciation is 35. USING Straight line method, the book value after 5 years is 36. USING SOYD, the book value after 3 years is 37 USING SOYD, the maximum depreciation in any year is 38. USING SOYD, the total depreciation after 6 years is 39. USING MATHESONS METHOD, the Matheson’s constant is 40. The sunk cost after 3 years using MATHESONS formula is 41. The total depreciation after 7 years is (MATHESONS FORMULA ) 42. USING DOUBLE DECLINING BALANCE METHOD , the book value after 3 years is 43. USING SINKING FUND METHOD , with i = 8% , the annual depreciation is 44. The book value after 6 years is (SINKING FUND METHOD ) 45. The total depreciation after 5 years is 46. The depreciation on the 4th year is S-8 Find the present value of a perpetuity of P 15,000 payable semiannually if money is worth 8% compounded quarterly 47. The interest rate every six months is 48. The present value of the given perpetuity is 2
49. A man invest in a medium scale business that cost him P 57,000.00. The net annual return is estimated at P 14,000 for each of the next 8 years. Find the benefit cost ratio if annual interest rate is 8%. 50. Compute the benefit cost ratio of the given project Project Cost P 80,000 Gross Income 25,000 per year Operating Cost P 6,000 per year Salvage Value 0 Life of Project 10 years Rate of Interest 12% 51. If the nominal interest rate is 3%, how much is P 5,000 worth in 10 years in a continuous compounded account? 52. P 6,000 is deposited each year into a savings account that pays 6% nominal interest compounded continuously. How much will be the account at the end of 8 years? 53. A man borrows P 100,000 at a rate of 6% compounded continuously for 5 years. How much money must he pay annually? 54. At 6% , find the capitalized cost of a bridge whose cost is 250 M and life is 20 years if the bridge must be partially rebuilt at a cost of 100 M every 20 years. S-9 A machine cost P 150,000 and will have a scrap value of P 10,000 when retired at the end of 15 years. If money is worth 4% 55. Find the annual investment 56. The Capitalized Cost is S-10 57. The cost of producing a certain commodity consist of P 45 per unit for labor and material cost and P 15 per unit for other variable cost. The fixed cost per month is P 450,000. If the commodity is sold at P 250 per unit 57. What is the break even quantity ? 58. What is the profit/loss if the number of items sold is 3000? 59. A certain copier machine cost P 150,000 with a trade in value of P 1.50,000 after making 800,000 copies. Using the declining balance method, what is the book value of the machine when it had made 300,000 copies? 60. What is the present worth of two P 100.00 payments at the end of the 3rd and the 4th year is interest rate is 8%? 61. The nominal rate of interest is 4%. How much is P 10,000 worth in 10 years in a continuously compounded account? 62. If P 500,000 is deposited at a rate of 11.25% compounded monthly, what will be the amount after 7 years and 9 months? 63. In year zero, you invest P 10,000 in a 15% security for 5 years. During the time, the average inflation is 6%. How much in terms of year zero pesos will be the account at the maturity?
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S – 11 A man inherited a regular endowment of P 100,000 every end of 3 months for 10 years. However he may elect to get a single lump sum after 4 years. 64 . How much is the lump sum if the cost of money is 14% compounded quarterly? 65. What is the present worth of the endowment? 66. A debt of x pesos with interest rate of 7% compounded annually will be retired at the end of 10 years through the accumulation of deposit in a sinking fund invested at 6% compounded semiannually. The deposit in the sinking fund every end of the six months is P 21,962.68. The value of x is 67. A man paid 10% downpayment of P 200,000 for a house and lot and agreed to pay the balance on a monthly installments for 60 months at an interest rate of 15% compounded monthly. Determine the required monthly payment. S -12 Determine the break even point in terms of the number of units produced per month. Cost are in pesos per unit Selling Price Per unit P 600 Labor Cost P 115 Cost of Materials P 76 Other Cost P 2.32 Total Monthly Overhead Cost 428,000 68. The break even point is 69. Determine the profit/loss if it sells 1000 units. S -13 Given the bank deposits in an account that earns 3% per annum. End of : Year 1 5,000 Year 2 6,000 Year 3 7,200 70. The deposit at year 20 is 71. The total deposits made without interest after 20 years. 72. The Present Worth of 20 deposits. 73. The Future worth of the 20 deposits 74. The equivalent equal annual deposits for the 20 deposits. 75. An amount of 1,000 becomes 3500 after 6 years. If money is x% compounded quarterly , x = ? S – 14 A 3 deg simple curve is used to connect two highways whose bearings are S 110 W and S 470 W . Use arc basis. 76. The radius of the curve is 77. The length of the curve is 78. The external distance is 79. The length of the middle ordinate is 4
80. The tangent distance is S- 15 A simple curve has a degree of 6 deg and length of 140 m. What is the central angle of the curve. Use Chord Basis. 81. The radius of the curve is 82. The central angle is S – 16 A 40 simple curve have tangents with bearings of N 200 E and N 600 E. The stationing of PI is 3 + 980 83. The radius of the curve is 84. The tangent distance is 85. The stationing of PC is 86. The length of the curve is 87. The stationing of the PT is S -17 A reversed curve consists of two curves of equal radius has the following properties D1 = 3 deg I1 = 18 deg D2 = 2 deg I2 = 24 deg Stationing Of PT is 32 + 121 Use Chord Basis 88. The radius of curve 1 is 89. The radius of the 2nd curve is 90. The stationing of PC is S – 18 A vertical sag parabolic curve is to connect a -2% grade to a 3% grade. The change in grade per meter station is 0.2%. The PT of of the curve is at station 25 + 325 and elevation 25.42. 91. The length of the curve is 92. The stationing of the PC of the curve is 93. The stationing of the lowest point of the curve is 94. The elevation of the PC is at 95. The elevation of the lowest point of the curve is at 96. The elevation of PI is at 97. The stationing of the PI is at 98. The elevation at station 25 + 075 is S – 19 Given the following cross section notes for a road in cut x +2.5
0 +c
7.8 +y
S – 20
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The road has a width of 10 m and have a slope of 2: 1 (horizontal: vertical ) . The area of the cross section is 47.4 m2 99. The value of x is 100. The value of y is 101. The value of c is a. 4.23 m
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