ENGR 301 Engineering Management Principles and
Economics TIPS QUESTIONS AND ANSWERS FOR FINAL
Concordia University
Exam Practice Questions – Solutions
1. At age 30 you invest $5,000 into a mutual fund. If the fund averages an 8% annual return, your
investment is worth how much at age 55?
(A) $23,300 (B) $34,240 (C) $50,310 (D) $344,570
n= 55 – 30 = 25
i= 8%
P= $5000
F= P(1+i)n = $5000(1+0.08)25 = $34,242
2. You are saving up for a big investment in six years. You estimate it will take $14,500 to secure this
investment. How much do you need to put into a savings account at the end of each year if the
savings account earns 4%? Neglect taxes.
(A) $2,185 (B) $2,375 (C) $2,415 (D) $2,485
F= $14,500
A=?
N=6
i=4%
A= F (A/F, 6,4%) = $14,500(0.1508) = $2186.60
3. You are buying your first car and need to borrow $16,000 over 5 years. If interest is 6%, what are
your monthly payments?
(A) $267 (B) $309 (C) $347 (D) $389
P=$16,000
N=5 years = 60 months
i=6%, imonthly = 6/12=0.5%
A=?
A= P(A/P,60,0.5%) = $16000(0.0193) = $308.80
4. You are considering investing in a 5-yr CD (certificate of deposit) with an annual yield of 6.5% and
monthly compounding. If you invest $5,000, your effective interest earned is most nearly:
(A) 6.5% (B) 6.6% (C) 6.7% (D) 6.8%
Annual rate = 6.5%
Compounding frequency = 12
Therefore, effective interest rate = ia = (1+ r/m)m--1 = (1+0.065/12)12 --1 = 6.7%
5. A lift station sewage pump initially costs $20,000. Annual maintenance costs are $300. The pump
salvage value is 10 percent of the initial cost in 20 years. Using 4% interest, the annual cost of the
pump is most nearly:
(A) $1,200 (B) $1,705 (C) $1,772 (D) $1,840
, Initial cost = $20,000
Annual Maintenance Cost = $300
Salvage value after 20yrs = $20,000 x 10%= $2000
i=4%
EUAC=?
Bring the salvage value to the present
2000(P/F, 4%,20) = $2000(0.4564) = $912.80
Therefore, adjusted cost at time zero = $20,000 – $912.80 = $19,087
Convert $19,087 to a uniform amount A over 20years and add it to $300
$19087(A/P,4%,20) = $19087(0.0736)=$1404.80 + $300 = $1704.80
6. A computerized wood lathe, costing $17,000, will be used to make ornamental parts for sale.
Receipts are estimated at $28,000 per year with costs running $25,000 per year. The salvage value is
$2,000 at the end of 10 years. If the MARR is 8%, what is the present worth of this investment?
(A) -$410 (B) $3,130 (C) $4,060 (D) $5,210
Initial cost = $17,000
Annual sales (benefit) =$28,000
Annual costs = $25,000
Salvage value at 10 years = $2000
PW=?
PW = --$17,000 + $28,000(P/A,8%,10) -- $25,000(P/A,8%,10) + $2000(P/F,8%,10)
= --$17,000 + $28,000(6.710) -- $25,000(6.710) + 2000(0.4632)
= $4056
7. Two alternatives are available for producing logos on sport shirts. Costs are shown below. Interest
is 4%.
Machine A Machine B
Initial Cost $54,000 $74,000
Salvage Value $8,100 $7,400
Operating Costs $2,100/yr $1,400/yr – 1st 10 years
$1,800/yr – 2nd 10 years
Life 15 years 20 years
Q1: The annual cost for machine A (ACA) is: (A) $6,350 (B) $6,550 (C) $6,750 (D) $6,950 Q2: The
annual cost for machine B (ACB) is: (A) $6,360 (B) $6,560 (C) $6,760 (D) $6,960
Annual cost for machine A = ($54,000 -- $8100(P/F,4%,15))(A/P,4%,15) + $2100 = $6,550
($74,000 – $7,400(P/F,4%,20))(A/P,4%,20) +
[$1400(P/A,4%,10) + $1800(P/A,4%,10)(P/F,4%,10)](A/P,4%,20)
Economics TIPS QUESTIONS AND ANSWERS FOR FINAL
Concordia University
Exam Practice Questions – Solutions
1. At age 30 you invest $5,000 into a mutual fund. If the fund averages an 8% annual return, your
investment is worth how much at age 55?
(A) $23,300 (B) $34,240 (C) $50,310 (D) $344,570
n= 55 – 30 = 25
i= 8%
P= $5000
F= P(1+i)n = $5000(1+0.08)25 = $34,242
2. You are saving up for a big investment in six years. You estimate it will take $14,500 to secure this
investment. How much do you need to put into a savings account at the end of each year if the
savings account earns 4%? Neglect taxes.
(A) $2,185 (B) $2,375 (C) $2,415 (D) $2,485
F= $14,500
A=?
N=6
i=4%
A= F (A/F, 6,4%) = $14,500(0.1508) = $2186.60
3. You are buying your first car and need to borrow $16,000 over 5 years. If interest is 6%, what are
your monthly payments?
(A) $267 (B) $309 (C) $347 (D) $389
P=$16,000
N=5 years = 60 months
i=6%, imonthly = 6/12=0.5%
A=?
A= P(A/P,60,0.5%) = $16000(0.0193) = $308.80
4. You are considering investing in a 5-yr CD (certificate of deposit) with an annual yield of 6.5% and
monthly compounding. If you invest $5,000, your effective interest earned is most nearly:
(A) 6.5% (B) 6.6% (C) 6.7% (D) 6.8%
Annual rate = 6.5%
Compounding frequency = 12
Therefore, effective interest rate = ia = (1+ r/m)m--1 = (1+0.065/12)12 --1 = 6.7%
5. A lift station sewage pump initially costs $20,000. Annual maintenance costs are $300. The pump
salvage value is 10 percent of the initial cost in 20 years. Using 4% interest, the annual cost of the
pump is most nearly:
(A) $1,200 (B) $1,705 (C) $1,772 (D) $1,840
, Initial cost = $20,000
Annual Maintenance Cost = $300
Salvage value after 20yrs = $20,000 x 10%= $2000
i=4%
EUAC=?
Bring the salvage value to the present
2000(P/F, 4%,20) = $2000(0.4564) = $912.80
Therefore, adjusted cost at time zero = $20,000 – $912.80 = $19,087
Convert $19,087 to a uniform amount A over 20years and add it to $300
$19087(A/P,4%,20) = $19087(0.0736)=$1404.80 + $300 = $1704.80
6. A computerized wood lathe, costing $17,000, will be used to make ornamental parts for sale.
Receipts are estimated at $28,000 per year with costs running $25,000 per year. The salvage value is
$2,000 at the end of 10 years. If the MARR is 8%, what is the present worth of this investment?
(A) -$410 (B) $3,130 (C) $4,060 (D) $5,210
Initial cost = $17,000
Annual sales (benefit) =$28,000
Annual costs = $25,000
Salvage value at 10 years = $2000
PW=?
PW = --$17,000 + $28,000(P/A,8%,10) -- $25,000(P/A,8%,10) + $2000(P/F,8%,10)
= --$17,000 + $28,000(6.710) -- $25,000(6.710) + 2000(0.4632)
= $4056
7. Two alternatives are available for producing logos on sport shirts. Costs are shown below. Interest
is 4%.
Machine A Machine B
Initial Cost $54,000 $74,000
Salvage Value $8,100 $7,400
Operating Costs $2,100/yr $1,400/yr – 1st 10 years
$1,800/yr – 2nd 10 years
Life 15 years 20 years
Q1: The annual cost for machine A (ACA) is: (A) $6,350 (B) $6,550 (C) $6,750 (D) $6,950 Q2: The
annual cost for machine B (ACB) is: (A) $6,360 (B) $6,560 (C) $6,760 (D) $6,960
Annual cost for machine A = ($54,000 -- $8100(P/F,4%,15))(A/P,4%,15) + $2100 = $6,550
($74,000 – $7,400(P/F,4%,20))(A/P,4%,20) +
[$1400(P/A,4%,10) + $1800(P/A,4%,10)(P/F,4%,10)](A/P,4%,20)
