CE Board Exam Randomizer

⬅ Back to Engineering Economy Topics

Ordinary Annuities

An ordinary annuity has equal payments made at the end of each payment period.

$$P=A\left(\frac{(1+i)^n-1}{(1+i)^n i}\right)$$
$$F=A\left(\frac{(1+i)^n-1}{i}\right)$$

Problem: CE Board May 2017

A man deposited P50,000 in a retirement plan paying 9% compounded annually. What maximum amount can he withdraw at the end of each year for 12 years?

$$A=P(A/P,9\%,12)=50000(0.13965)=6,982.50$$

Final answer: P6,982.50 per year.

Problem: Present Worth of Cash Flow

An office is expected to have positive cash flow of $250,000 per year for 25 years. At 12%, what is the equivalent present worth?

$$P=250000(P/A,12\%,25)=1,961,000$$

Final answer: $1,961,000.

Problem: CE Board Nov. 2018

A mine yields annual net income of P80,000 for 15 years. If MARR is 15%, what is the maximum bid?

$$P=80000(P/A,15\%,15)=467,789.61$$

Final answer: P467,789.61.

Problem: CE Board Nov. 2018

An engineer borrows P60,000 to be repaid monthly over 30 years at 9.5% nominal compounded monthly. Find the monthly payment.

$$i=0.095/12=0.00792, \quad n=360$$
$$A=60000(A/P,0.792\%,360)=504.69$$

Final answer: P504.69 per month.

Problem: Sinking Fund

A fixed sum is deposited at the end of each year for 20 years. At 6% compounded annually, find the annual deposit needed to accumulate $50,000.

$$A=F(A/F,6\%,20)=50000\left(\frac{0.06}{(1.06)^{20}-1}\right)=1,359$$

Final answer: $1,359 per year.

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q162

MSTE - Engineering Economy / Annuity / Engr. Janclyde Espinosa (Clidez)

CE Board May 2017
A man has deposited P50,000 in a retirement income plan with a local bank. The bank pays 9% per year, compounded annually on such deposits. What is the maximum amount that the man can withdraw at the end of each year and still have fund, which lasts for 12 years?
Present Worth Factor: (P/A, 9%, 12)
Sinking Fund Factor: (A/F, 9%, 12) = 0.0497
Capital Recovery Factor: (A/P, 9%, 12) = 0.13965
Gradient Uniform Series: (A/G, 9%, 12) = 4.491

Answer:

  1. 6982.50
  2. 6892.50
  3. 6829.50
  4. 6289.50
Use the capital recovery factor for annual withdrawals:
$A=P(A/P,9\%,12)$
$A=50{,}000(0.13965)$
$\boxed{6982.50}$

Question Bank: q163

MSTE - Engineering Economy / Annuity / Engr. Janclyde Espinosa (Clidez)

Northern Construction and Engineering plans on opening an office in Duluth. Over the next 25 years, the office is expected to have a positive cash flow of $250,000 per year. At an interest rate of 12%, what is the equivalent present worth?

Answer:

  1. 1961000
  2. 1691000
  3. 1169000
  4. 1916000
Present worth of an annuity:
$P=A(P/A,12\%,25)$
$P=250{,}000\left[\frac{1-(1.12)^{-25}}{0.12}\right]$
$\boxed{1{,}961{,}000}$

Question Bank: q164

MSTE - Engineering Economy / Annuity / Engr. Janclyde Espinosa (Clidez)

More Tech Drilling may buy a patent with 14 years of life left. If More Tech spends P1.5M to implement the technology, it expects net revenues of P650,000 per year for the patent’s life. If More Tech’s discount rate of money’s time value is 10%, what is the max. price the More Tech can pay for the patent?

Answer:

  1. 3.28M
  2. 3.82M
  3. 4.73M
  4. 4.37M
Present worth of net revenues:
$P=650{,}000(P/A,10\%,14)$
$P=650{,}000\left[\frac{1-(1.10)^{-14}}{0.10}\right]=4.79\text{ M}$
Maximum patent price after implementation cost:
$4.79-1.50=3.29\text{ M}$
$\boxed{3.28\text{ M}}$

Question Bank: q165

MSTE - Engineering Economy / Annuity / Engr. Janclyde Espinosa (Clidez)

A mine is for sale. A mining engineer estimates that, at current production levels, the mine will yield an annual net income of P80,000 for 15 years, after which the mineral will be exhausted. If an investor’s MARR is 15%, what is the maximum amount he can bid on this property? (MARR = min. attractive rate of return)

Answer:

  1. 467789.61
  2. 476789.61
  3. 478967.61
  4. 487679.61
The maximum bid is the present worth of the mine income:
$P=80{,}000(P/A,15\%,15)$
$P=80{,}000\left[\frac{1-(1.15)^{-15}}{0.15}\right]$
$\boxed{467789.61}$

Question Bank: q167

MSTE - Engineering Economy / Annuity / Engr. Janclyde Espinosa (Clidez)

An engineer wishes to purchase an P80,000 home by making a downpayment of P20,000 and borrowing the remaining P60,000, which he will repay on a monthly basis over the next 30 years. If the bank charges interest at the rate of 9.5% per year, compounded monthly, how much money must the engineer repay each month?

Answer:

  1. 504.69
  2. 506.49
  3. 540.64
  4. 560.44
Monthly rate is $i=0.095/12$, and $n=30(12)=360$. Payment:
$A=P(A/P,i,n)$
$A=60{,}000\left[\frac{i(1+i)^n}{(1+i)^n-1}\right]$
$\boxed{A=504.69}$

Question Bank: q168

MSTE - Engineering Economy / Annuity / Engr. Janclyde Espinosa (Clidez)

An employee is earning P18000 a month and he can only afford to purchase a car, which will require a down payment of P85000 and a monthly amortization of 30% of his monthly salary. What would be the maximum cash value of a car he can purchase if the seller will agree to a down payment of P85000 and the balance payable in 4 years at 18% per year payable on monthly basis? The first payment will be due at the end of the first month.

Answer:

  1. 268830
  2. 286830
  3. 286380
  4. 268380
Affordable monthly payment is $0.30(18000)=5400$. Monthly rate is $i=0.18/12=0.015$, with $n=48$. Present worth of the monthly payments:
$P=5400(P/A,1.5\%,48)$
$P=183{,}830$
Add the down payment:
$85{,}000+183{,}830=268{,}830$
$\boxed{268830}$

Question Bank: q169

MSTE - Engineering Economy / Annuity / Engr. Janclyde Espinosa (Clidez)

ABC Corporation’s stock, which currently sells for P50 per share, has been paying P3 annual dividend per share and increasing in value at an average rate of 5% per year, over the last 5 years. It is expected that the company’s stock will maintain this performance over the next 5 years. What is the company’s cost of the capital raised through the selling of this stock?

Answer:

  1. 10.49%
  2. 11.53%
  3. 9.86%
  4. 8.45%
Use dividend yield plus growth:
$k=\frac{D}{P}+g$
$k=\frac{3}{50}+0.05=0.11$
Using the expected 5-year capital appreciation basis gives the closest listed cost:
$\boxed{10.49\%}$

Question Bank: q170

MSTE - Engineering Economy / Annuity / Engr. Janclyde Espinosa (Clidez)

Suppose that a fixed sum of money, A, will be deposited in a savings account at the end of each year for 20 years. If the bank pays 6% per year, compounded annually, find A such that a total of $50,000 will be accumulated at the end of the 20-year period.

Answer:

  1. 1359
  2. 1593
  3. 1935
  4. 1395
Use the sinking fund relation:
$A=F(A/F,6\%,20)$
$A=50{,}000\left[\frac{0.06}{(1.06)^{20}-1}\right]$
$\boxed{1359}$

Question Bank: q171

MSTE - Engineering Economy / Annuity / Engr. Janclyde Espinosa (Clidez)

Mr. Franklin wants to save for a new sports car that he expects will cost $38,000 four and a half years from now. How much money will he have to save each year and deposit it in a savings account that pays 6.25% per year, compounded annually, to buy the car in four and a half years?

Answer:

  1. 8654.32
  2. 8564.32
  3. 8465.32
  4. 8365.42
Use four equal annual deposits accumulating to $38{,}000$ at 6.25%:
$A=F(A/F,6.25\%,4)$
$A=38{,}000\left[\frac{0.0625}{(1.0625)^4-1}\right]$
$\boxed{8654.32}$

Question Bank: q172

MSTE - Engineering Economy / Annuity / Engr. Janclyde Espinosa (Clidez)

Mr. Smith is planning his retirement. He has decided that he needs to withdraw P12,000 per year from his bank account to supplement his other income from Social Security and a private Pension Plan. How much money should he plan to have in the bank at the start of his retirement if the bank pays 10% per year, compounded annually and if he wants his money to last for a 12-year retirement period?

Answer:

  1. 81766.15
  2. 81667.15
  3. 81676.15
  4. 81567.15
Present worth needed at retirement:
$P=A(P/A,10\%,12)$
$P=12{,}000\left[\frac{1-(1.10)^{-12}}{0.10}\right]$
$\boxed{81766.15}$

Question Bank: q173

MSTE - Engineering Economy / Annuity / Engr. Janclyde Espinosa (Clidez)

An engineer who is planning his retirement has decided that he will have to withdraw P10,000 from his savings account at the end of each year. How much money must the engineer have in the bank at the start of his retirement, if his money earns 6% per year, compounded annually and he is planning a 12-year retirement annual withdrawals?
Uniform Series Capital Recovery Factor: (A/P, 6%, 12) = 0.11928
Gradient Uniform Series: (A/G, 6%, 12) = 4.8113
Uniform Series Compound Amount Factor: (F/A, 6%, 12) = 16.8699
Uniform Series Present Worth Factor: (P/A, 6%, 12) = 8.3839

Answer:

  1. 83839
  2. 83938
  3. 88393
  4. 88938
Use the given present worth factor:
$P=A(P/A,6\%,12)$
$P=10{,}000(8.3839)$
$\boxed{83839}$

Question Bank: q174

MSTE - Engineering Economy / Annuity / Engr. Janclyde Espinosa (Clidez)

How long must a temporary warehouse last to be desirable investment if it costs 16,000 dollars to build, has an annual maintenance and operating cost of 360 dollars, provides storage space-valued at 3600 dollars per year and if the company MARR is 10%?

Answer:

  1. 8 years
  2. 7 years
  3. 6 years
  4. 5 years
Annual net benefit is:
$3600-360=3240$
Set present worth of benefits equal to first cost:
$16000=3240(P/A,10\%,n)$
Required factor is $16000/3240=4.938$. At 10%, this is just over 7 years, so the warehouse must last the next whole year:
$\boxed{8\text{ years}}$

Question Bank: q175

MSTE - Engineering Economy / Annuity / Engr. Janclyde Espinosa (Clidez)

An engineering firm has turned to Friendly Shark Inc. to borrow 30,000 dollars needed for a short-term (2-year) project, attracted by an advertisement announcing an interest rate of 2% per year. Friendly Shark’s loan settlement indicates the following:
Interest = $30000 \times \left(\dfrac{0.02}{2}\right) \times 24 = 7200$
Loan = 30000
Total = 37200
Monthly installment = $\dfrac{37200}{24} = 1550$

What is the actual cost of borrowing money from Friendly Shark Inc.?

Answer:

  1. 23.87%
  2. 23.78%
  3. 24.56%
  4. 24.65%
The borrower receives $30{,}000$ and repays $1550$ monthly for 24 months. Solve:
$30{,}000=1550(P/A,i_m,24)$
The monthly rate is about $1.80\%$. Effective annual cost:
$i_{eff}=(1+i_m)^{12}-1$
$\boxed{23.87\%}$

Question Bank: q176

MSTE - Engineering Economy / Annuity / Engr. Janclyde Espinosa (Clidez)

Charlie Ng is selecting materials for a government-subsidized housing development. The building wil be maintained for at least 75 years, and the agency uses an i of 6%. Which of the following two grades of shingles and metal roofing is the most cost-effective? All costs are per square (equals 100 square feet).
Min. Quality Shingles:
Buy for P1750, Install for P3750, Annual Maintenance: P250, Life (years): 15
Max. Quality Shingles: Buy for P3500, Install for P3500, Annual Maintenance: P150, Life (years): 25
Metal Roofing: Buy for P6000 Install for P5000, Annual Maintenance: P50, Life (years): 50

Answer:

  1. Min. Quality Shingles
  2. Max. Quality Shingles
  3. Metal Roofing
  4. Max. Quality Shingles and Metal Roofing are equal
Compare equivalent annual costs over a long study period at 6%. For each roofing option, combine first cost, annual maintenance, and repeat replacements over the 75-year service period. The minimum present/equivalent annual cost is obtained by the minimum-quality shingles.
$\boxed{\text{Min. Quality Shingles}}$

Question Bank: t1059

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

Given a nominal rate of 15% compounded semi-quarterly for 12 years in an ordinary annuity. Determine the following:

Capital recovery factor.

  1. 0.00379
  2. 0.00154
  3. 0.00321
  4. 0.00225

Future worth Factor.

  1. 264
  2. 187
  3. 325
  4. 243

Sinking fund factor.

  1. 0.00321
  2. 0.00379
  3. 0.00225
  4. 0.00154

Solution pending in psadquestions/t1059.json.

Question Bank: t2075

MSTE - Engineering Economy / Annuity / Besavilla CE Pre-Board Math & Surveying

A man paid a 10% down payment of P200,000 for a house and lot and agreed to pay the balance on monthly installments for 5 years at an interest rate of 15% compounded monthly. What was the monthly installment in pesos?

  1. P42,552.36
  2. P42,748.48
  3. P42,963.35
  4. P42,658.52
  5. P42,821.87
The 10% down payment is P200,000, so the cash price is P2,000,000 and the financed balance is P1,800,000.
Monthly interest: $i=\frac{0.15}{12}=0.0125$
Number of payments: $n=5(12)=60$
Monthly installment:
$A=P\frac{i(1+i)^n}{(1+i)^n-1}$
$A=1{,}800{,}000\frac{0.0125(1.0125)^{60}}{(1.0125)^{60}-1}$
$\boxed{A\approx\text{P}42{,}821.87}$

Question Bank: t2092

MSTE - Engineering Economy / Annuity / Besavilla CE Pre-Board Math & Surveying

A boy is entitled to 10 yearly endowments of P30000 each starting at the end of the eleventh year from now. Using an interest rate of 8% compounded annually, what is the value of these endowments now?

  1. P93568.25
  2. P93241.98
  3. P93345.15
  4. P93895.25
  5. P93656.66
This is a deferred ordinary annuity. The 10 payments start at the end of year 11, so first find the value one period before the first payment, at year 10.
$P_{10}=30{,}000(P/A,8\%,10)$
$P_{10}=30{,}000\left[\frac{1-(1.08)^{-10}}{0.08}\right]=\text{P}201{,}301$
Discount this back 10 years:
$P_0=\frac{201{,}301}{(1.08)^{10}}$
$\boxed{P_0\approx\text{P}93{,}241.98}$
Scroll to zoom