FE Other Disciplines (NCEES Fundamentals of Engineering) — All Questions

What is the future worth of $1,000 invested for 5 years at 6% annual compound interest?

• a.$1,060
• b.$1,500
• c.$1,300
• d.$1,338✓

Future worth F = P(1 + i)^n = 1000(1.06)^5. Since 1.06^5 = 1.3382, F = $1,338. This single-payment compound-amount factor is a core time-value-of-money relation.

What is the present worth of $5,000 to be received 10 years from now at an 8% annual discount rate?

• a.$2,316✓
• b.$2,000
• c.$2,500
• d.$4,630

Present worth P = F / (1 + i)^n = 5000 / (1.08)^10. Since 1.08^10 = 2.159, P = 5000 / 2.159 = $2,316. Discounting converts a future cash flow to its value today.

What is the future worth of $2,000 invested for 10 years at 5% annual compound interest?

• a.$2,600
• b.$4,000
• c.$3,000
• d.$3,258✓

Future worth F = P(1 + i)^n = 2000(1.05)^10. Since 1.05^10 = 1.6289, F = $3,258. This is the single-payment compound-amount factor (F/P).

What is the present worth of $10,000 to be received 5 years from now at a 7% annual discount rate?

• a.$14,026
• b.$7,000
• c.$6,500
• d.$7,130✓

Present worth P = F/(1 + i)^n = 10,000/(1.07)^5. Since 1.07^5 = 1.4026, P = 10,000/1.4026 = $7,130. Discounting shrinks a future amount to today's value.

What is the total amount owed on a $5,000 loan after 3 years at 8% SIMPLE interest per year?

• a.$6,300
• b.$5,400
• c.$1,200
• d.$6,200✓

Simple interest is I = P·i·n = 5000(0.08)(3) = $1,200, so the total is 5000 + 1200 = $6,200. Simple interest accrues only on the original principal, not on accumulated interest.

A nominal annual interest rate of 12% is compounded monthly. What is the effective annual interest rate?

• a.12.00%
• b.12.68%✓
• c.12.55%
• d.12.36%

Effective rate = (1 + r/m)^m - 1 = (1 + 0.12/12)^12 - 1 = (1.01)^12 - 1 = 0.1268 = 12.68%. More frequent compounding raises the effective rate above the nominal rate.

What is the future worth of a uniform series of $1,000 deposited at the end of each year for 5 years at 6% annual interest?

• a.$5,637✓
• b.$5,000
• c.$6,000
• d.$5,975

The uniform-series compound-amount factor gives F = A·[((1+i)^n - 1)/i] = 1000·[(1.3382 - 1)/0.06] = 1000(5.637) = $5,637. This is the (F/A) factor.

What is the present worth of a uniform series of $2,000 received at the end of each year for 10 years at 8% annual interest?

• a.$20,000
• b.$16,000
• c.$12,000
• d.$13,420✓

The uniform-series present-worth factor gives P = A·[(1 - (1+i)^-n)/i] = 2000·[(1 - 0.4632)/0.08] = 2000(6.710) = $13,420. This is the (P/A) factor.

A $100,000 loan is repaid with equal annual payments over 20 years at 10% interest. What is the annual payment?

• a.$10,000
• b.$6,727
• c.$11,746✓
• d.$5,000

The capital-recovery factor gives A = P·[i(1+i)^n/((1+i)^n - 1)] = 100,000·[0.1(6.7275)/(6.7275 - 1)] = 100,000(0.11746) = $11,746. This is the (A/P) factor.

How much must be deposited at the end of each year to accumulate $50,000 in 10 years at 6% annual interest?

• a.$4,193
• b.$5,000
• c.$3,793✓
• d.$3,000

The sinking-fund factor gives A = F·[i/((1+i)^n - 1)] = 50,000·[0.06/(1.7908 - 1)] = 50,000(0.07587) = $3,793. This is the (A/F) factor.

What is the present worth of a perpetuity paying $5,000 per year forever at an 8% interest rate?

• a.$40,000
• b.$50,000
• c.$625,000
• d.$62,500✓

The present worth of a perpetuity (capitalized cost) is P = A/i = 5,000/0.08 = $62,500. As the number of periods approaches infinity, the present-worth factor approaches 1/i.

An investment of $10,000 returns $12,000 after one year. What is the rate of return?

• a.16.7%
• b.12%
• c.20%✓
• d.10%

Rate of return = (gain)/(investment) = (12,000 - 10,000)/10,000 = 2,000/10,000 = 20%. The rate of return measures profit relative to the amount invested.

A project costs $50,000 up front and produces $10,000 in annual savings. What is the simple payback period?

• a.10 years
• b.5 years✓
• c.4 years
• d.6 years

Simple payback = initial cost/annual savings = 50,000/10,000 = 5 years. The simple payback method ignores the time value of money.

An asset costs $100,000 with a $10,000 salvage value and a 9-year life. What is the annual straight-line depreciation?

• a.$12,222
• b.$10,000✓
• c.$9,000
• d.$11,111

Straight-line depreciation = (cost - salvage)/life = (100,000 - 10,000)/9 = 90,000/9 = $10,000 per year. The salvage value is subtracted before dividing by the life.

An asset costs $60,000 with zero salvage and a 10-year life (straight-line). What is its book value after 3 years?

• a.$54,000
• b.$18,000
• c.$48,000
• d.$42,000✓

Annual depreciation = 60,000/10 = $6,000. After 3 years accumulated depreciation is $18,000, so book value = 60,000 - 18,000 = $42,000. Book value is cost minus accumulated depreciation.

An asset costing $20,000 with a 5-year life is depreciated by the double-declining-balance method. What is the first-year depreciation?

• a.$8,000✓
• b.$6,000
• c.$4,000
• d.$12,000

The double-declining rate is 2/life = 2/5 = 40%. First-year depreciation = 0.40 x 20,000 = $8,000. DDB applies the rate to the current book value, ignoring salvage in the early years.

A public project has present-worth benefits of $500,000 and present-worth costs of $400,000. What is the benefit-cost ratio?

• a.1.10
• b.1.50
• c.0.80
• d.1.25✓

The benefit-cost ratio is B/C = benefits/costs = 500,000/400,000 = 1.25. A ratio above 1.0 indicates an economically justified project.

If the nominal (market) interest rate is 10% and inflation is 4%, what is the approximate real interest rate?

• a.5.77%✓
• b.14.00%
• c.6.00%
• d.5.50%

The exact real rate is (1 + i_nominal)/(1 + f) - 1 = (1.10/1.04) - 1 = 0.0577 = 5.77%. A rough estimate simply subtracts (10% - 4% = 6%), which slightly overstates the real rate.

A present cost of $20,000 is to be expressed as an equivalent uniform annual cost over 5 years at 10% interest. What is the annual amount?

• a.$5,276✓
• b.$6,000
• c.$3,000
• d.$4,000

Multiply by the capital-recovery factor (A/P, 10%, 5) = 0.2638: A = 20,000(0.2638) = $5,276. This spreads a lump sum into equal annual payments including interest.

Using the rule of 72, approximately how long does it take money to double at an 8% annual interest rate?

• a.12 years
• b.8 years
• c.9 years✓
• d.6 years

The rule of 72 estimates the doubling time as 72/interest rate = 72/8 = 9 years. It is a quick approximation of the exact compound-interest doubling time.

A nominal annual rate of 8% is compounded semiannually. What is the effective annual rate?

• a.8.00%
• b.8.24%
• c.16.00%
• d.8.16%✓

Effective rate = (1 + r/m)^m - 1 = (1 + 0.08/2)^2 - 1 = (1.04)^2 - 1 = 0.0816 = 8.16%. Semiannual compounding earns interest on interest twice per year.

A project requires $10,000 now and returns $5,000 at the end of each of the next 3 years. At 10% interest, what is the net present value?

• a.$12,435
• b.$2,434✓
• c.$5,000
• d.$3,000

Present worth of the annuity = 5,000·(P/A, 10%, 3) = 5,000(2.4869) = $12,435. NPV = 12,435 - 10,000 = $2,434. A positive NPV means the project earns more than the 10% rate.

What is the future worth of $1,000 after 5 years at 6% interest compounded continuously?

• a.$1,350✓
• b.$1,360
• c.$1,338
• d.$1,300

For continuous compounding F = P·e^(r·t) = 1000·e^(0.06·5) = 1000·e^0.3 = 1000(1.3499) = $1,350. Continuous compounding yields slightly more than annual compounding ($1,338).

A product has fixed costs of $10,000, a selling price of $50 per unit, and a variable cost of $30 per unit. What is the break-even quantity?

• a.1,000 units
• b.500 units✓
• c.200 units
• d.333 units

Break-even quantity = fixed cost/(price - variable cost) = 10,000/(50 - 30) = 10,000/20 = 500 units. The $20 contribution margin per unit covers fixed costs.

How much interest is earned on $8,000 invested for 2 years at 5% compounded annually?

• a.$820✓
• b.$800
• c.$840
• d.$1,025

Total value F = 8000(1.05)^2 = 8000(1.1025) = $8,820, so interest earned = 8,820 - 8,000 = $820. Simple interest would give only $800; the extra $20 is interest on interest.

What is the present worth of receiving $1,000 at the end of year 1 and $1,000 at the end of year 2 at 10% interest?

• a.$1,810
• b.$1,735✓
• c.$1,653
• d.$2,000

Discount each cash flow: 1000/1.10 + 1000/(1.10)^2 = 909.09 + 826.45 = $1,735. Each future amount is discounted by its own number of periods.

An investment of $1,000 grows to $1,331 after 3 years. What is the annual rate of return?

• a.10%✓
• b.11%
• c.33%
• d.9%

Solve (1 + i)^3 = 1,331/1,000 = 1.331, so 1 + i = 1.331^(1/3) = 1.10 and i = 10%. This is the internal rate of return for a single cash flow.

A structure has a first cost of $100,000 and annual maintenance of $5,000 forever. At 10% interest, what is its capitalized cost?

• a.$150,000✓
• b.$100,000
• c.$550,000
• d.$105,000

Capitalized cost = first cost + annual cost/i = 100,000 + 5,000/0.10 = 100,000 + 50,000 = $150,000. The perpetual maintenance is treated as a perpetuity worth A/i.

What is the present worth of a $20,000 salvage value to be received 10 years from now at 8% interest?

• a.$9,264✓
• b.$8,000
• c.$43,157
• d.$10,000

Present worth P = F/(1 + i)^n = 20,000/(1.08)^10 = 20,000/2.1589 = $9,264. A future salvage value is discounted back like any other future cash flow.

Machine B costs $6,000 now and has a $1,500 salvage value at the end of year 5. At 10% interest, what is its net present cost?

• a.$5,931
• b.$6,000
• c.$5,069✓
• d.$4,500

Net present cost = first cost - present worth of salvage = 6,000 - 1,500/(1.10)^5 = 6,000 - 1,500/1.6105 = 6,000 - 931 = $5,069. Salvage reduces the effective cost.

A single deposit of $10,000 is made now. What is the equivalent uniform annual worth over 10 years at 6% interest?

• a.$1,000
• b.$600
• c.$1,600
• d.$1,359✓

Multiply by the capital-recovery factor (A/P, 6%, 10) = 0.13587: A = 10,000(0.13587) = $1,359. This converts a present lump sum to an equivalent annual series.

A nominal annual rate of 12% is compounded continuously. What is the effective annual rate?

• a.12.68%
• b.12.75%✓
• c.13.00%
• d.12.00%

For continuous compounding the effective rate is e^r - 1 = e^0.12 - 1 = 1.1275 - 1 = 0.1275 = 12.75%. This is the maximum effective rate for a given nominal rate.