I85PREP
⌘ K
PE Industrial study materials
PE Industrial study resources

PE Industrial practice questions

One question at a time. Build your PE Industrial understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Basic Engineering Practice

An engineer deposits $10,000 in an account that earns 6% per year compounded annually. What is the balance after 5 years?

  1. $13,000
  2. $13,382 Correct answer.
  3. $11,910
  4. $12,625

Explanation

Compound interest multiplies the principal by (1 + i)^n each year, so the future worth is $10,000(1.06)^5 = $13,382. Interest is earned on previously accumulated interest, which is what distinguishes this from simple interest.

Question 2Basic Engineering Practice

A 12 V source is connected across a 6 ohm resistor. How much power is dissipated in the resistor?

  1. 2 W
  2. 72 W
  3. 24 W Correct answer.
  4. 0.5 W

Explanation

For a resistive element, power equals V squared divided by R, so 144/6 = 24 W. The same result follows from the current of 2 A multiplied by the 12 V drop.

Question 3Basic Engineering Practice

What is the median of the data set 5, 9, 2, 7, 12, 9, 4?

  1. 5
  2. 7 Correct answer.
  3. 6
  4. 9

Explanation

Ordering the seven values gives 2, 4, 5, 7, 9, 9, 12; with an odd count the median is the fourth ordered value, which is 7. The median is a positional measure and is not affected by the magnitude of the extreme values.

Question 4Basic Engineering Practice

In hypothesis testing, a Type I error occurs when the analyst does which of the following?

  1. Fails to reject a true null hypothesis
  2. Rejects a false null hypothesis
  3. Rejects a true null hypothesis Correct answer.
  4. Fails to reject a false null hypothesis

Explanation

A Type I error is a false positive: rejecting the null hypothesis when it is actually true, and its probability is the significance level alpha. A Type II error is the opposite mistake, failing to reject a null hypothesis that is false.

Question 5Basic Engineering Practice

Equipment costing $50,000 has a salvage value of $5,000 and a depreciable life of 9 years. Using straight-line depreciation, what is the annual depreciation charge?

  1. $4,500
  2. $5,000 Correct answer.
  3. $5,556
  4. $6,000

Explanation

Straight-line depreciation spreads the depreciable base, cost minus salvage, evenly over the recovery period: (50,000 - 5,000)/9 = $5,000 per year. Salvage value must be removed before the division.

Question 6Basic Engineering Practice

A measured length is reported as 0.00450 m. How many significant figures does this value have?

  1. 2
  2. 3 Correct answer.
  3. 4
  4. 5

Explanation

Leading zeros only locate the decimal point and are never significant, while a trailing zero after the decimal point is significant. The significant digits are therefore 4, 5, and 0, giving three.

Question 7Basic Engineering Practice

A process produces defective items with probability 0.2, and 5 items are sampled independently. What is the probability that exactly 2 are defective?

  1. 0.0512
  2. 0.32768
  3. 0.4096
  4. 0.2048 Correct answer.

Explanation

The binomial probability is C(5,2)(0.2)^2(0.8)^3, which equals 10 times 0.04 times 0.512, or 0.2048. The combination factor counts the number of orderings in which the two defects can appear.

Question 8Basic Engineering Practice

A sample of 25 items has a mean of 50. The population standard deviation is known to be 10 and the data are normally distributed. What is the 95% confidence interval for the population mean?

  1. (46.08, 53.92) Correct answer.
  2. (48.04, 51.96)
  3. (30.40, 69.60)
  4. (44.00, 56.00)

Explanation

With a known standard deviation, the interval is the sample mean plus or minus 1.96 times the standard error, where the standard error is 10 divided by the square root of 25, or 2. The result is 50 plus or minus 3.92.

Question 9Basic Engineering Practice

A loan of $100,000 is to be repaid in 10 equal annual payments at an interest rate of 8% per year. What is the annual payment?

  1. $14,903 Correct answer.
  2. $12,500
  3. $20,000
  4. $10,000

Explanation

The capital recovery factor converts a present amount into an equivalent uniform annual series: A = P[i(1+i)^n]/[(1+i)^n - 1]. Substituting P = 100,000, i = 0.08, and n = 10 gives a factor of 0.14903, so the payment is about $14,903.

Question 10Basic Engineering Practice

A 2 m long steel rod with a 25 mm diameter carries an axial tensile load of 100 kN. Taking the modulus of elasticity as 200 GPa, what is the elastic elongation?

  1. 4.07 mm
  2. 8.15 mm
  3. 1.02 mm
  4. 2.04 mm Correct answer.

Explanation

Axial stress is load divided by area, giving 203.7 MPa for an area of 4.909 times 10 to the minus 4 square meters. Strain is stress divided by the modulus, 1.019 times 10 to the minus 3, and multiplying by the 2 m length gives about 2.04 mm.

Question 11Basic Engineering Practice

Water flows in a horizontal pipe that tapers from 100 mm diameter to 50 mm diameter. If the velocity in the 100 mm section is 2 m/s, what is the pressure drop between the two sections? Take the density of water as 1000 kg/m^3.

  1. 4 kPa
  2. 30 kPa Correct answer.
  3. 15 kPa
  4. 60 kPa

Explanation

Continuity requires the velocity to rise with the square of the diameter ratio, so the 50 mm section carries 8 m/s. Bernoulli's equation for a horizontal pipe then gives a pressure drop of one half times density times the difference of the squared velocities, or 30 kPa.

Question 12Basic Engineering Practice

A heat engine receives 500 kJ of heat from a hot reservoir and rejects 350 kJ to a cold reservoir each cycle. What is its thermal efficiency?

  1. 30% Correct answer.
  2. 25%
  3. 15%
  4. 70%

Explanation

The net work is the heat supplied minus the heat rejected, or 150 kJ, and thermal efficiency is net work divided by heat supplied. That gives 150/500 = 0.30, or 30%.

Question 13Basic Engineering Practice

A project network has three paths: A-B-D with durations 3, 4, and 5 days; A-C-D with durations 3, 6, and 5 days; and A-C-E with durations 3, 6, and 2 days. Which path is critical, and what is the project duration?

  1. A-C-E, 11 days
  2. A-B-D, 14 days
  3. A-B-D, 12 days
  4. A-C-D, 14 days Correct answer.

Explanation

The critical path is the longest path through the network, since it sets the earliest project completion. A-B-D totals 12 days, A-C-E totals 11 days, and A-C-D totals 14 days, so A-C-D governs the schedule.

Question 14Basic Engineering Practice

A product sells for $60 per unit and has a variable cost of $40 per unit. Annual fixed costs are $120,000. What is the break-even quantity?

  1. 2,000 units
  2. 3,000 units
  3. 12,000 units
  4. 6,000 units Correct answer.

Explanation

Break-even occurs where total contribution equals fixed cost, so the quantity is fixed cost divided by the unit contribution margin of $20. That gives 120,000/20 = 6,000 units.

Question 15Basic Engineering Practice

A normally distributed quality characteristic has a mean of 100 and a standard deviation of 15. What is the probability that a single measurement exceeds 130?

  1. 0.9772
  2. 0.1587
  3. 0.0228 Correct answer.
  4. 0.4772

Explanation

The standardized value is (130 - 100)/15 = 2.00, and the upper-tail area beyond z = 2.00 is 0.0228. Tables usually give the cumulative area 0.9772, which must be subtracted from one for an upper-tail question.

Question 16Basic Engineering Practice

A production process follows an 80% learning curve, and the first unit required 100 hours. What time should the fourth unit require?

  1. 40.9 h
  2. 80.0 h
  3. 64.0 h Correct answer.
  4. 51.2 h

Explanation

Unit time under a learning curve is the first-unit time multiplied by the unit number raised to the exponent b, where b equals the log of the learning rate divided by the log of 2, or -0.3219 for 80%. Thus T4 = 100 times 4 to the power -0.3219, which is about 64 hours.

Question 17Basic Engineering Practice

A simply supported beam spans 6 m and carries a uniformly distributed load of 10 kN/m. Taking E = 200 GPa and I = 50 x 10^-6 m^4, what is the maximum deflection?

  1. 4.2 mm
  2. 16.9 mm Correct answer.
  3. 33.8 mm
  4. 8.4 mm

Explanation

The maximum deflection of a simply supported beam under a uniform load is 5wL^4 divided by 384EI. Substituting 10,000 N/m, 6 m, 200 x 10^9 Pa, and 50 x 10^-6 m^4 gives 0.0169 m, or about 16.9 mm.

Question 18Basic Engineering Practice

A sample of 16 measurements has a mean of 54 and a standard deviation of 8. To test the null hypothesis that the population mean is 50 against a two-sided alternative at a significance level of 0.05, the critical t value is 2.131. What is the correct decision?

  1. Reject H0 because 2.0 exceeds 1.96
  2. Fail to reject H0 because 2.0 is less than 2.131 Correct answer.
  3. Reject H0 because 4 exceeds 2.131
  4. Fail to reject H0 because 2.0 is less than 1.753

Explanation

The one-sample t statistic is the difference of 4 divided by the standard error of 8 divided by the square root of 16, which is 2.0, with 15 degrees of freedom. Because 2.0 does not exceed the two-sided critical value of 2.131, the null hypothesis is not rejected.

Question 19Basic Engineering Practice

A process has specification limits of 49.0 and 51.0, an estimated mean of 50.2, and a standard deviation of 0.4. What is the process capability index Cpk?

  1. 0.67 Correct answer.
  2. 1.00
  3. 0.83
  4. 1.25

Explanation

Cpk is the smaller of the upper and lower distances from the mean to the specification limits, each divided by three standard deviations. The distances are 0.8 and 1.2, divided by 1.2 gives 0.67, showing the process is off-center toward the upper limit.

Question 20Engineering Economics

An engineer deposits $10,000 in a savings account that earns 5% per year compounded annually. No other deposits or withdrawals are made. What is the account balance at the end of three years?

  1. $12,155
  2. $11,025
  3. $11,576 Correct answer.
  4. $11,500

Explanation

Compound interest earns interest on previously accrued interest, so the balance grows by the factor (1+i)^n. Here F = $10,000(1.05)^3 = $11,576.25.

Question 21Engineering Economics

A $60,000 investment is expected to produce uniform net annual savings of $15,000. Ignoring the time value of money, what is the simple payback period?

  1. 6 years
  2. 4 years Correct answer.
  3. 5 years
  4. 3 years

Explanation

Simple payback is the initial cost divided by the uniform annual net cash flow: $60,000 / $15,000 = 4 years. It measures how long until cumulative undiscounted savings recover the investment.

Question 22Engineering Economics

Equipment costs $50,000, has an estimated salvage value of $5,000, and a depreciable life of 9 years. Using straight-line depreciation, what is the annual depreciation charge?

  1. $5,000 Correct answer.
  2. $6,111
  3. $5,556
  4. $4,500

Explanation

Straight-line depreciation spreads the depreciable basis evenly: D = (B - S)/N = ($50,000 - $5,000)/9 = $5,000 per year.

Question 23Engineering Economics

In an engineering economic analysis, the minimum attractive rate of return (MARR) is best described as which of the following?

  1. The interest rate a lender charges on borrowed capital
  2. The average return actually earned by a firm's past projects
  3. The lowest return a firm will accept on an investment Correct answer.
  4. The rate at which an asset's book value declines over time

Explanation

The MARR is the hurdle rate a firm uses to judge investments; a project is justified only if its rate of return meets or exceeds this minimum. It reflects the cost of capital plus an allowance for risk.

Question 24Engineering Economics

A firm spent $200,000 last year on preliminary design work for a proposed plant. The plant is now being re-evaluated as a new investment. How should the $200,000 be treated in the new analysis?

  1. As an opportunity cost of the new plant
  2. As a sunk cost excluded from the comparison Correct answer.
  3. As a deductible operating expense in year one
  4. As part of the plant's installed first cost

Explanation

Money already spent and not recoverable is a sunk cost, and sunk costs are identical across the alternatives, so they cannot influence the choice. Only future differences in cash flow belong in the analysis.

Question 25Engineering Economics

A public project has a conventional benefit-cost ratio of 1.4 computed at the agency's discount rate. What does this indicate?

  1. The project is acceptable at the stated discount rate Correct answer.
  2. The project should be rejected because the ratio is above 1.0
  3. The project's internal rate of return equals 1.4% per year
  4. The project returns $1.40 of cost for every $1.00 of benefit

Explanation

A conventional B/C ratio of 1.0 or greater means discounted benefits equal or exceed discounted costs at that discount rate, so the project is economically justified. The 1.4 value means $1.40 of benefit per $1.00 of cost.

Question 26Engineering Economics

Net revenue from a process improvement is expected to be $2,000 per year for 10 years. At an interest rate of 8% per year, what is the present worth of these revenues?

  1. $2,000
  2. $28,973
  3. $20,000
  4. $13,420 Correct answer.

Explanation

A uniform series is discounted with the P/A factor, (P/A,8%,10) = 6.7101. Multiplying, PW = $2,000 x 6.7101 = $13,420.

Question 27Engineering Economics

A bridge has an initial cost of $2,000,000 and annual maintenance costs of $50,000 that continue indefinitely. At an interest rate of 5% per year, what is the capitalized cost?

  1. $4,000,000
  2. $2,500,000
  3. $2,050,000
  4. $3,000,000 Correct answer.

Explanation

Capitalized cost adds the present worth of a perpetual annual cost, A/i, to the first cost: $2,000,000 + $50,000/0.05 = $3,000,000.

Question 28Engineering Economics

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

  1. 12.75%
  2. 12.55%
  3. 12.68% Correct answer.
  4. 12.00%

Explanation

The effective annual rate converts the monthly compounding to a yearly basis: i = (1 + 0.12/12)^12 - 1 = 0.1268, or 12.68% per year.

Question 29Engineering Economics

An asset costs $20,000, has a $2,000 salvage value, and a 5-year life. Using double declining balance depreciation, what is its book value at the end of year 2?

  1. $12,000
  2. $7,200 Correct answer.
  3. $8,000
  4. $4,800

Explanation

The DDB rate is 2/5 = 40% applied to the undepreciated book value. Book value after year 1 is $20,000(0.60) = $12,000, and after year 2 it is $12,000(0.60) = $7,200.

Question 30Engineering Economics

Alternative B requires $20,000 more in initial investment than alternative A and returns $5,000 more per year for 6 years. Neither alternative has a salvage value. What is the approximate incremental rate of return on the extra investment?

  1. 8%
  2. 15%
  3. 13% Correct answer.
  4. 10%

Explanation

Set the incremental investment equal to the incremental savings series: $20,000 = $5,000(P/A,i,6), so (P/A,i,6) = 4.0. That factor equals 3.998 at about 13% per year.

Question 31Engineering Economics

An investment earns a market (nominal) return of 9% per year while inflation runs at 3% per year. What is the real return?

  1. 3.0%
  2. 5.8% Correct answer.
  3. 6.0%
  4. 12.0%

Explanation

The Fisher relationship is (1 + i) = (1 + r)(1 + f), so r = 1.09/1.03 - 1 = 5.8% per year. The simple difference of 6.0% only approximates this result.

Question 32Engineering Economics

A product sells for $20 per unit, has a variable cost of $8 per unit, and incurs fixed costs of $120,000 per year. What is the break-even quantity per year?

  1. 4,000 units
  2. 15,000 units
  3. 10,000 units Correct answer.
  4. 6,000 units

Explanation

At break-even, revenue equals total cost: 20Q = 120,000 + 8Q, so Q = 120,000/12 = 10,000 units per year. The $12 per unit is the contribution margin.

Question 33Engineering Economics

A company must accumulate $50,000 in 10 years to replace a machine. What equal annual deposit at 6% per year, compounded annually, is required?

  1. $5,000
  2. $3,793 Correct answer.
  3. $6,793
  4. $27,920

Explanation

The sinking fund factor converts a future amount to an equivalent annual series: A = F(A/F,6%,10) = $50,000 x 0.06/[(1.06)^10 - 1] = $3,793.

Question 34Engineering Economics

The economic life of an asset is the ownership period that does which of the following?

  1. Maximizes the total undiscounted net revenue earned from the asset
  2. Minimizes the equivalent uniform annual cost of ownership Correct answer.
  3. Maximizes the resale or salvage value received at disposal time
  4. Matches the asset's depreciable life used for tax purposes

Explanation

Economic life is the retention period that minimizes the equivalent uniform annual cost of capital recovery plus operating and maintenance costs. It is generally shorter than the physical or depreciable life.

Question 35Engineering Economics

A proposed project's net cash flow series changes sign twice over its life. Which statement about its internal rate of return is correct?

  1. There may be as many positive IRRs as there are sign changes in the series Correct answer.
  2. Exactly two IRRs always exist and both should be averaged
  3. The IRR is undefined because the present worth equation has no real roots
  4. The IRR equals the MARR whenever the cumulative cash flow is positive

Explanation

By Descartes' rule of signs, the number of positive real roots of the present worth polynomial cannot exceed the number of sign changes in the cash flow series. With two sign changes the project may have zero, one, or two IRRs, so an external rate of return or a present worth test at the MARR is used instead.

Question 36Engineering Economics

A permanent fund must pay a $20,000 replacement cost every 10 years, with the first payment at the end of year 10 and payments continuing forever. At 8% per year, what amount must be deposited now to fund all future replacements?

  1. $250,000
  2. $17,257 Correct answer.
  3. $9,264
  4. $20,000

Explanation

Convert the periodic replacement to an equivalent annual amount with the A/F factor, then capitalize it: $20,000(0.06903) = $1,381 per year, and $1,381/0.08 = $17,257.

Question 37Engineering Economics

Two mutually exclusive machines have no salvage value. Machine A: first cost $30,000, annual operating cost $8,000, life 4 years. Machine B: first cost $50,000, annual operating cost $5,000, life 6 years. At a MARR of 10% per year, which is preferred and by how much on an annual worth basis?

  1. Machine B, because its annual cost is lower by about $984 Correct answer.
  2. Machine A, because its annual cost is lower by about $984
  3. Machine A, because its first cost is lower by $20,000
  4. The two machines are equivalent on an annual worth basis

Explanation

Annual worths may be compared directly under the repeatability assumption: AW(A) = -$30,000(A/P,10%,4) - $8,000 = -$17,464, while AW(B) = -$50,000(A/P,10%,6) - $5,000 = -$16,481. Machine B is cheaper by roughly $984 per year.

Question 38Engineering Economics

Annual maintenance for a system is $10,000 in year 1 and is expected to increase 4% per year thereafter. At an interest rate of 10% per year, what is the present worth of maintenance for 5 years?

  1. $65,687
  2. $37,908
  3. $40,785 Correct answer.
  4. $50,000

Explanation

A constant-percentage increase is a geometric gradient: P = A1[1 - (1+g)^n(1+i)^-n]/(i - g) = $10,000[1 - (1.04/1.10)^5]/(0.10 - 0.04) = $40,785.

Question 39Probability and Statistics

For the sample 12, 15, 11, 18, 14, what is the median?

  1. 14 Correct answer.
  2. 15
  3. 12
  4. 18

Explanation

The median is the middle value once the observations are arranged in order: 11, 12, 14, 15, 18. With five ordered values the third one is the median, so the median is 14.

Question 40Probability and Statistics

A sample of five measurements has a mean of 20. A sixth measurement of 30 is added to the sample. What is the new sample mean?

  1. 25.0
  2. 20.0
  3. 22.0
  4. 21.67 Correct answer.

Explanation

The original five values sum to 5 × 20 = 100. Adding 30 gives a total of 130 over six observations, so the new mean is 130 / 6 ≈ 21.67.

Question 41Probability and Statistics

Events A and B are mutually exclusive with P(A) = 0.3 and P(B) = 0.5. What is P(A or B)?

  1. 0.15
  2. 0.80 Correct answer.
  3. 0.35
  4. 0.20

Explanation

For mutually exclusive events the joint probability is zero, so the addition rule reduces to P(A or B) = P(A) + P(B). Adding 0.3 and 0.5 gives 0.80.

Question 42Probability and Statistics

Which measure of dispersion is expressed in the same units as the original observations?

  1. Coefficient of variation
  2. Standard deviation Correct answer.
  3. Variance
  4. Correlation coefficient

Explanation

The standard deviation is the square root of the variance, so squaring the original units is undone and the result carries the original measurement units. The variance carries squared units, while the coefficient of variation and correlation coefficient are unitless ratios.

Question 43Probability and Statistics

How many distinct ways can 3 items be chosen from 8 distinct items when order does not matter?

  1. 56 Correct answer.
  2. 24
  3. 336
  4. 112

Explanation

Selection without regard to order is counted by the combination formula C(8,3) = 8! / (3! 5!). This equals (8 × 7 × 6) / (3 × 2 × 1) = 56.

Question 44Probability and Statistics

A dimension is normally distributed with mean 170 mm and standard deviation 10 mm. What proportion of parts exceeds 185 mm?

  1. About 0.025
  2. About 0.159
  3. About 0.067 Correct answer.
  4. About 0.933

Explanation

The standardised value is z = (185 − 170) / 10 = 1.5. The standard normal table gives 0.9332 below z = 1.5, so the upper tail is 1 − 0.9332 ≈ 0.067.

Question 45Probability and Statistics

A process produces 10% defective parts independently. In a sample of 10 parts, what is the probability that exactly one is defective?

  1. 0.349
  2. 0.100
  3. 0.387 Correct answer.
  4. 0.651

Explanation

The count of defectives is binomial with n = 10 and p = 0.1, so P(X = 1) = C(10,1)(0.1)(0.9)^9. This gives 10 × 0.1 × 0.3874 ≈ 0.387.

Question 46Probability and Statistics

A 95% confidence interval for a population mean is calculated as 42.0 to 48.0. Which interpretation is correct?

  1. The sample mean has a 95% chance of equalling the midpoint of the interval
  2. There is a 95% probability that the population mean lies between 42.0 and 48.0
  3. About 95% of individual measurements lie between 42.0 and 48.0
  4. About 95% of samples would give intervals containing the true mean Correct answer.

Explanation

Confidence levels describe the long-run performance of the procedure: if the sampling were repeated many times, roughly 95% of intervals constructed this way would contain the fixed but unknown population mean. Once a single interval is computed, the mean is either inside it or not.

Question 47Probability and Statistics

In hypothesis testing, a Type I error occurs when which of the following happens?

  1. The significance level is set equal to 0.05
  2. A false null hypothesis is not rejected
  3. A true null hypothesis is rejected Correct answer.
  4. The test statistic falls inside the acceptance region

Explanation

A Type I error is rejecting the null hypothesis when it is in fact true, and its probability is controlled by the significance level α. Failing to reject a false null hypothesis is instead a Type II error.

Question 48Probability and Statistics

Defects along a cable occur as a Poisson process with an average of 3 defects per 100 m. What is the probability of exactly 2 defects in 100 m?

  1. 0.149
  2. 0.448
  3. 0.050
  4. 0.224 Correct answer.

Explanation

For a Poisson variable with mean λ = 3, P(X = 2) = e^(−3) × 3^2 / 2!. This equals 0.0498 × 9 / 2 ≈ 0.224.

Question 49Probability and Statistics

A discrete random variable X takes the values 0, 1 and 2 with probabilities 0.5, 0.3 and 0.2 respectively. What is E(X)?

  1. 0.70 Correct answer.
  2. 1.00
  3. 0.35
  4. 0.50

Explanation

The expected value is the probability-weighted sum of the outcomes: 0(0.5) + 1(0.3) + 2(0.2) = 0 + 0.3 + 0.4 = 0.70.

Question 50Probability and Statistics

A population is strongly right-skewed with mean 50 and standard deviation 12. A sample of 36 observations is taken. Which statement describes the sampling distribution of the sample mean?

  1. Right-skewed, mean 50, standard error 2
  2. Approximately normal, mean 50, standard error 12
  3. Approximately normal, mean 50, standard error 2 Correct answer.
  4. Exactly normal, mean 50, standard deviation 12

Explanation

By the central limit theorem, with n = 36 the distribution of the sample mean is approximately normal regardless of the shape of the population. Its mean equals the population mean 50 and its standard error is σ/√n = 12/6 = 2.