FE exam · common sections · checked against the July 2020 specifications

FE practice test: the sections all FE exams share

Q: Does NCEES have an FE practice exam?

Yes. For each of the five disciplines here NCEES sells two 50-question practice exams at $50 each (store price as of October 2026), built on the July 2020 specifications. This page is not one of them and uses none of their items.

Q: Are these real FE exam questions?

No. Each one was set for this site against a section of the July 2020 specification and solved again by a second reviewer without the key; NCEES keeps its exam items confidential.

Q: How much of the FE is math?

It depends on the discipline: 6–9 questions on FE Mechanical, 11–17 on FE Electrical and Computer, and 8–12 on FE Civil, where statistics is included.

This FE practice test covers the part of the exam that every discipline shares: mathematics, probability and statistics, and ethics, in 70 problems with worked solutions. Engineering economics, also common to all of them, has its own page. Discipline problems are on each discipline’s practice test, linked below.

Updated Oct 29, 2026 · Not affiliated with NCEES.

problems on this page70
shared sections3
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Pick your discipline

The shared sections take a different number of questions in each specification. The rest of each exam is on its own practice test:

FE Civil practice test: mathematics and statistics are one section, 8–12 questions; ethics 4–6.
FE Mechanical: mathematics 6–9, probability and statistics 4–6, ethics 4–6.
FE Electrical and Computer: mathematics 11–17, statistics 4–6, ethics 4–6.
FE Environmental: mathematics 5–8, statistics 4–6, ethics 5–8.
FE Chemical: mathematics 6–9, statistics 4–6, ethics 3–5.
Coming: FE Industrial and Systems (statistics alone 10–15) and FE Other Disciplines (mathematics 8–12).

Each discipline test carries its own small set of shared-section problems, drawn from the same pool as this page but never the same problems, so working both gives you new ones.

70 problems, filter by section

0 / 70 answered · 0 correct

Question 1 of 70 · Mathematics

Classify the conic section defined by the equation 3x² − 4xy + 2y² − 5x + 6y − 8 = 0.

Given	Equation 3x² − 4xy + 2y² − 5x + 6y − 8 = 0
Find	Type of conic section

Solution

For Ax² + Bxy + Cy² + Dx + Ey + F = 0 the sign of B² − 4AC classifies the curve (FE Reference Handbook, Mathematics — Conic Sections).

B² − 4AC = (−4)² − 4(3)(2) = 16 − 24 = −8 < 0

Negative → ellipse; the 3 × 3 determinant (−25.5) is nonzero, so it is not degenerate.

Degenerate conics occur only when the full 3 × 3 determinant of the quadratic form is zero; here it is −25.5, so the curve does not split into lines.

A parabola needs B² − 4AC exactly zero; here the discriminant is −8.

A circle needs B = 0 and A = C; this equation has an xy term and A ≠ C, so it is only the ellipse family.

A hyperbola needs B² − 4AC > 0; this answer comes from reading −8 as positive or forgetting the minus in −4AC.

With A = 3, B = −4, C = 2 the discriminant B² − 4AC = 16 − 24 = −8 is negative, and the curve is a real, non-degenerate ellipse.

Question 2 of 70 · Mathematics

Use two Newton–Raphson iterations, starting from x₀ = 2, to approximate the root of f(x) = x³ − 2x − 5 nearest 2. What is x₂ to four significant figures?

Given	Function f(x) = x³ − 2x − 5 Initial guess x₀ = 2 Iterations: 2
Find	x₂ to four significant figures

Solution

Newton’s method (FE Reference Handbook, Mathematics — Numerical Methods):

x_n+1 = x_n − f(x_n)/f′(x_n)

x₁ = 2 − (−1)/10 = 2.1

x₂ = 2.1 − 0.061/11.23 = 2.09457 ≈ 2.095

2.083 is x₁ computed with f′(x) = 3x²: the −2 of the derivative was dropped, so the step 1/12 is too large.

2.093 is x₂ after dropping the −2 from f′(x) in both iterations; the derivative must be 3x² − 2.

x₁ = 2 − (−1)/10 = 2.1, then x₂ = 2.1 − 0.061/11.23 = 2.09457, which rounds to 2.095.

2.100 is x₁; the question asks for the second iterate, so one more correction step is missing.

2.105 adds the second correction instead of subtracting it: 2.1 + 0.061/11.23. The update is x_n − f/f′.

Question 3 of 70 · Mathematics

The polynomial P(x) = x³ − 5x² + 11x − 15 has real coefficients and a known complex root 1 + 2i. What is its remaining real root?

Given	P(x) = x³ − 5x² + 11x − 15 Known root 1 + 2i
Find	Remaining real root

Solution

Complex roots of a real polynomial come in conjugate pairs, so 1 − 2i is also a root. Sum of roots = −b/a:

r + (1 + 2i) + (1 − 2i) = 5 → r = 3

Check with the product of roots = −d/a: 3 × (1 + 2i)(1 − 2i) = 3 × 5 = 15. P(3) = 27 − 45 + 33 − 15 = 0.

This is the whole root sum −b/a; the two complex roots (sum 2) were not subtracted.

This is only the real part of the complex root; the conjugate pair 1 ± 2i contributes 2 to the sum.

A sign slip in the product of the roots: r·(1 + 2i)(1 − 2i) = 5r must equal −d/a = 15, so r = 3; taking −15 gives −3.

This is the sum of the conjugate pair (1 + 2i) + (1 − 2i), not the third root.

The conjugate 1 − 2i is also a root, so r + 2 = 5 gives r = 3, and P(3) = 27 − 45 + 33 − 15 = 0.

Question 4 of 70 · Mathematics

The integral of f(x) over [0, 10] is computed with the trapezoidal rule. The largest value of |f″(x)| on the interval is 0.6. What is the minimum number of subintervals N that guarantees an absolute error less than 0.001?

Given	Interval [0, 10] max \|f″(x)\| = 0.6 Required error < 0.001
Find	Minimum number of subintervals N

Solution

Trapezoidal-rule error bound (FE Reference Handbook, Mathematics — Numerical Integration):

|E| ≤ (b − a)³·max|f″|/(12N²) = 1000 × 0.6/(12N²) = 50/N²

50/N² < 0.001 → N > √50,000 = 223.6 → N = 224

50000 is N², not N: the square root of 50,000 was never taken.

2236 solves the bound for a tolerance of 10⁻⁵ instead of 10⁻³.

775 drops the 12 from the denominator of the error bound, which inflates the requirement to N > 774.6.

The bound 50/N² < 0.001 needs N > 223.6, so the smallest whole N is 224.

This truncates 223.6 instead of rounding up; at N = 223 the error bound is 50/223² = 0.00101, still above 0.001.

Question 5 of 70 · Mathematics

A closed cylindrical tank of fixed volume V is to be built with the least total surface area (top, bottom and wall cost the same per unit area). What relationship between the height h and the radius r minimizes the surface area?

Solution

Substitute h = V/(πr²) and differentiate (FE Reference Handbook, Mathematics — Differential Calculus):

A(r) = 2πr² + 2V/r

dA/dr = 4πr − 2V/r² = 0 → V = 2πr³

h = V/(πr²) = 2r

This counts each end disk twice (the optimum when end material costs double); here all surfaces cost the same.

Minimizing A = 2πr² + 2πrh with πr²h = V gives h = 2r, the height equal to the diameter.

Fixing V sets the size, not the shape: dA/dr = 0 gives V = 2πr³, and h/r = 2 for every volume.

This inverts the result: a squat tank with r = 2h has far more end area than needed.

This counts only one end disk, which is the optimum for an open-top tank; a closed tank has two.

Question 6 of 70 · Mathematics

The integral of f(x) = x⁴ over [0, 1] is computed with Simpson’s 1/3 rule. The error bound is |E| ≤ (b − a)h⁴·max|f⁽⁴⁾|/180. What is the minimum number of subintervals n that keeps the error at or below 10⁻⁵?

Given	f(x) = x⁴ Interval [0, 1] Tolerance \|E\| ≤ 10⁻⁵
Find	Minimum number of subintervals n

Solution

Simpson’s 1/3 rule error bound (FE Reference Handbook, Mathematics — Numerical Integration):

|E| ≤ (1)(1/n)⁴ × 24/180 ≤ 10⁻⁵

n⁴ ≥ 13,333 → n ≥ 10.75

Simpson’s rule needs an even n, so n = 12.

With n = 4 the bound is 24/(180 × 256) = 5.2 × 10⁻⁴, fifty times the tolerance.

With n = 6 the bound is 1.0 × 10⁻⁴, still ten times the tolerance.

This is the even number just below 10.75; n = 10 gives 1.33 × 10⁻⁵, which exceeds 10⁻⁵.

11 rounds 10.75 up correctly but ignores that Simpson’s 1/3 rule needs an even number of subintervals.

n⁴ ≥ 24/(180 × 10⁻⁵) = 13,333 gives n ≥ 10.75, and the next even integer is 12.

Question 7 of 70 · Mathematics

The concentration of a ground contaminant is modeled by C(x, y) = 100 − x² + 2x − 2y² + 4y. According to the second-derivative test, how is the critical point of this function classified?

Given	C(x, y) = 100 − x² + 2x − 2y² + 4y
Find	Location and type of the critical point

Solution

Set the gradient to zero, then test D = C_xxC_yy − C_xy² (FE Reference Handbook, Mathematics — Differential Calculus):

C_x = −2x + 2 = 0 → x = 1; C_y = −4y + 4 = 0 → y = 1

D = (−2)(−4) − 0 = 8 > 0, C_xx = −2 < 0 → local maximum

A saddle needs D < 0; the mixed partial is zero and the pure second partials have the same sign, so D = 8.

Both the location (the linear coefficients) and the type are wrong; there is only one critical point, at (1, 1).

This takes the linear coefficients 2 and 4 as coordinates instead of solving C_x = 0 and C_y = 0.

The point is right, but a minimum needs C_xx > 0; here C_xx = −2.

C_x = −2x + 2 and C_y = −4y + 4 vanish at (1, 1), and D = (−2)(−4) − 0 = 8 > 0 with C_xx < 0.

Question 8 of 70 · Mathematics

The temperature in a bioreactor over a 9-hour cycle is T(t) = t³ − 12t² + 36t (°C, t in hours) for 0 ≤ t ≤ 9. What is the maximum temperature reached during this interval?

Given	T(t) = t³ − 12t² + 36t Interval 0 ≤ t ≤ 9 h
Find	Maximum temperature on the interval (°C)

Solution

On a closed interval compare critical points and endpoints (FE Reference Handbook, Mathematics — Differential Calculus):

T′(t) = 3t² − 24t + 36 = 3(t − 2)(t − 6) = 0 → t = 2, 6

T(0) = 0, T(2) = 32, T(6) = 0, T(9) = 729 − 972 + 324 = 81

Maximum = 81 °C at t = 9 h.

This is the local minimum at t = 6 (and the start value T(0)); it is the lowest, not the highest, temperature.

16 °C is T(4), the inflection point where T″ = 0; an inflection point is not an extremum.

32 °C is the interior local maximum at t = 2; the endpoint t = 9 must also be checked and is higher.

63 °C is T′(9), the slope at the endpoint, not the temperature there.

The candidates are T(0) = 0, T(2) = 32, T(6) = 0 and T(9) = 81, so the absolute maximum is at the endpoint.

Question 9 of 70 · Mathematics

A conical sludge hopper (vertex down) has a top diameter of 4 m and a depth of 4 m. Sludge enters at a constant 0.5 m³/min. How fast is the sludge level rising when the depth is 2 m?

Given	Top diameter 4 m, depth 4 m Inflow dV/dt = 0.5 m³/min Depth h = 2 m
Find	Rate of level rise dh/dt (m/min)

Solution

Similar triangles give r = h/2; differentiate the cone volume with respect to time (FE Reference Handbook, Mathematics — Differential Calculus):

V = (1/3)πr²h = πh³/12

dV/dt = (πh²/4)·dh/dt

dh/dt = 0.5/(π × 2²/4) = 0.5/π = 0.159 m/min

0.040 m/min takes r = h (the 4 m diameter used as the radius at full depth), which quadruples the surface area.

0.053 m/min drops the 1/3 of the cone volume, treating the hopper like a cylinder of the same radius profile.

r = h/2 gives V = πh³/12, so dV/dt = (πh²/4)·dh/dt and dh/dt = 0.5/π = 0.159 m/min.

0.477 m/min differentiates V = πh³/12 without the chain-rule factor 3, i.e. dV/dt = (πh²/12)·dh/dt.

0.637 m/min takes r = h/4 (the ratio diameter/depth halved twice), which shrinks the surface area fourfold.

Question 10 of 70 · Mathematics

Using the excerpt from a superheated steam table below, apply linear interpolation to find the specific enthalpy of steam at 1.2 MPa and 275 °C.

Pressure (MPa)	Temperature (°C)	Enthalpy (kJ/kg)
1.0	250	2943.1
1.0	300	3051.6
1.5	250	2923.5
1.5	300	3038.9

Given	P = 1.2 MPa, T = 275 °C h(1.0 MPa, 250 °C) = 2943.1 kJ/kg h(1.0 MPa, 300 °C) = 3051.6 kJ/kg h(1.5 MPa, 250 °C) = 2923.5 kJ/kg h(1.5 MPa, 300 °C) = 3038.9 kJ/kg
Find	Specific enthalpy h (kJ/kg)

Solution

Interpolate in temperature on each pressure line, then in pressure (FE Reference Handbook, Mathematics — Numerical Methods, linear interpolation):

h(1.0, 275) = 2943.1 + 0.5(3051.6 − 2943.1) = 2997.35 kJ/kg

h(1.5, 275) = 2923.5 + 0.5(3038.9 − 2923.5) = 2981.20 kJ/kg

h(1.2, 275) = 2997.35 + 0.4(2981.20 − 2997.35) = 2990.9 kJ/kg

2935.3 kJ/kg interpolates only in pressure, on the 250 °C line: 2943.1 + 0.4(2923.5 − 2943.1); the temperature step to 275 °C is missing.

2981.2 kJ/kg is the 275 °C value on the 1.5 MPa line only; the pressure interpolation to 1.2 MPa was skipped.

2987.7 kJ/kg applies the pressure weight backwards (0.6 instead of 0.4 from the 1.0 MPa line), landing too close to 1.5 MPa.

2989.3 kJ/kg averages the two pressure lines with weight 0.5, as if 1.2 MPa were halfway between 1.0 and 1.5 MPa.

2997.35 kJ/kg at 1.0 MPa and 2981.20 kJ/kg at 1.5 MPa, then 2997.35 + 0.4(2981.20 − 2997.35) = 2990.9 kJ/kg.

Question 11 of 70 · Mathematics

A process instrument’s dynamic response is modeled by d²y/dt² + 4·dy/dt + 4y = 3e^−2t. What is the correct form of the particular solution y_p(t) for the method of undetermined coefficients?

Solution

Characteristic equation (FE Reference Handbook, Mathematics — Differential Equations):

r² + 4r + 4 = (r + 2)² = 0 → r = −2 (double)

The forcing e^−2t matches a root of multiplicity 2, so the trial is multiplied by t²:

y_p = At²e^−2t (substituting gives 2A = 3, A = 1.5)

One factor of t is not enough for a double root, and the constant B answers a forcing term that is not there.

The characteristic root r = −2 is repeated, so both e^−2t and te^−2t solve the homogeneous equation and the trial must be multiplied by t².

This is exactly the homogeneous solution; substituted into the left side it gives zero, so it cannot produce 3e^−2t.

Ae^−2t is a homogeneous solution and Be^2t matches no term of the forcing function.

Sine and cosine trials answer a sinusoidal forcing term or complex roots; this forcing is a real exponential.

Question 12 of 70 · Mathematics

What is the solution of the differential equation dy/dx + 2y = 0 with the initial condition y(0) = 3?

Given	dy/dx + 2y = 0 y(0) = 3
Find	y(x)

Solution

Separate the variables and integrate (FE Reference Handbook, Mathematics — Differential Equations):

dy/y = −2dx → ln|y| = −2x + C

y = Ce^−2x; y(0) = C = 3 → y = 3e^−2x

Separating gives dy/y = −2dx, so y = Ce^−2x, and y(0) = 3 sets C = 3.

This is a straight line with the right start value; it treats dy/dx as the constant −2 instead of −2y.

The decay rate is right but the sign of the constant is not: at x = 0 this gives −3, not 3.

This loses the minus sign when moving 2y to the right side, so the solution grows instead of decaying.

Question 13 of 70 · Mathematics

Three survey points on a construction site have coordinates A(2, 3), B(5, 7) and C(8, 4). What is the area of triangle ABC?

Given	A(2, 3) B(5, 7) C(8, 4)
Find	Area of triangle ABC (square units)

Solution

Shoelace (determinant) formula:

Area = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|

= ½|2(3) + 5(1) + 8(−4)| = ½|−21| = 10.5

½|2(7 − 4) + 5(4 − 3) + 8(3 − 7)| = ½|6 + 5 − 32| = ½(21) = 10.5.

12.0 square units is ½ × 6 × 4 using the x-span and y-span of the points as base and height; the triangle is not right-angled on those axes.

21.0 square units is the determinant sum without the factor ½, i.e. the area of the parallelogram on AB and AC.

21.5 square units drops the minus sign of the last term (+32 instead of −32) before halving.

Question 14 of 70 · Mathematics

The velocity of water is described by the vector field v = (2x²y)i + (x³)j. What is the divergence of this field?

Given	v = (2x²y)i + (x³)j
Find	Divergence ∇·v

Solution

Divergence (FE Reference Handbook, Mathematics — Vectors):

∇·v = ∂v_x/∂x + ∂v_y/∂y = ∂(2x²y)/∂x + ∂(x³)/∂y = 4xy + 0 = 4xy

∇·v = ∂(2x²y)/∂x + ∂(x³)/∂y = 4xy + 0.

A zero divergence would mean incompressible flow; ∂(2x²y)/∂x = 4xy is not zero here.

This takes ∂/∂y of the i-component, a cross derivative that belongs to the curl, not the divergence.

This adds the components themselves; the divergence adds their partial derivatives.

Question 15 of 70 · Mathematics

Evaluate the definite integral ∫₀¹ x√(1 − x²) dx.

Given	Integrand x√(1 − x²) Limits x = 0 to 1
Find	Value of the integral

Solution

Substitute u = 1 − x² and change the limits (x = 0 → u = 1, x = 1 → u = 0) (FE Reference Handbook, Mathematics — Integral Calculus):

∫₀¹ x√(1 − x²) dx = ½∫₀¹ u^1/2 du = ½ × (2/3) = 1/3

4/3 multiplies by 2 instead of ½ when replacing x dx with du (du = −2x dx gives x dx = −du/2): 2 × 2/3.

This is ∫₀¹ u^1/2 du without the ½ that comes from du = −2x dx.

This keeps the ½ from du = −2x dx but integrates u^1/2 as if it were 1.

With u = 1 − x², du = −2x dx, the integral becomes ½∫₀¹ u^1/2 du = ½ × 2/3 = 1/3.

Question 16 of 70 · Mathematics

What is the derivative of f(x) = e^3x·sin(2x) at x = 0?

Given	f(x) = e^3x·sin(2x) x = 0
Find	f′(0)

Solution

Product rule (FE Reference Handbook, Mathematics — Differential Calculus):

f′(x) = 3e^3xsin(2x) + 2e^3xcos(2x)

f′(0) = 3(1)(0) + 2(1)(1) = 2

5 adds the inner derivatives 3 + 2, as if both terms of the product rule were nonzero at x = 0.

This is the derivative of e^3x alone at x = 0; the sine factor was ignored instead of differentiated by the product rule.

f′(x) = 3e^3xsin(2x) + 2e^3xcos(2x), so f′(0) = 0 + 2(1)(1) = 2.

This differentiates only the exponential factor: 3e^3xsin(2x) is zero at x = 0, but the second product-rule term is missing.

Question 17 of 70 · Mathematics

Solve for x: log₁₀(x) + log₁₀(x − 3) = 1.

Given	log₁₀(x) + log₁₀(x − 3) = 1
Find	x

Solution

Combine the logarithms and solve the quadratic:

log₁₀[x(x − 3)] = 1 → x² − 3x − 10 = 0 → (x − 5)(x + 2) = 0

The domain requires x > 3, so x = 5.

This sets x = 10¹ from the first term alone and ignores log₁₀(x − 3).

x(x − 3) = 10 gives (x − 5)(x + 2) = 0, and only x = 5 keeps both arguments positive.

At x = 3 the term log₁₀(x − 3) = log₁₀(0) is undefined; 3 is only the domain boundary.

This is the second root of x² − 3x − 10 = 0; it must be rejected because the logarithm of a negative number is undefined.

Question 18 of 70 · Mathematics

For A = i + 2j + 3k and B = 2i + j + k, what is the k-component of the cross product A × B?

Given	A = i + 2j + 3k B = 2i + j + k
Find	k-component of A × B

Solution

Determinant form of the cross product (FE Reference Handbook, Mathematics — Vectors):

A × B = (A_yB_z − A_zB_y)i − (A_xB_z − A_zB_x)j + (A_xB_y − A_yB_x)k

k: (1)(1) − (2)(2) = −3 (full result −i + 5j − 3k)

The k-component is A_xB_y − A_yB_x = (1)(1) − (2)(2) = −3.

This is the i-component with its sign flipped: A_yB_z − A_zB_y = 2 − 3 = −1.

3 reverses the order of the products (B × A), which flips the sign of every component.

5 is the j-component of A × B, −(A_xB_z − A_zB_x) = −(1 − 6); the question asks for the k-component.

Question 19 of 70 · Mathematics

What is the equation of the line tangent to the curve y = x³ − 2x at the point (1, −1)?

Given	y = x³ − 2x Point (1, −1)
Find	Tangent line equation

Solution

Slope from the derivative, then point-slope form:

dy/dx = 3x² − 2 = 1 at x = 1

y − (−1) = 1(x − 1) → y = x − 2

This uses slope 3 from 3x² and drops the −2 of the derivative.

dy/dx = 3x² − 2 = 1 at x = 1, so y + 1 = 1(x − 1), i.e. y = x − 2.

The slope is right but the intercept has the wrong sign; this line does not pass through (1, −1).

This line passes through (1, −1) but has slope −1; the slope is 3(1)² − 2 = +1.

Question 20 of 70 · Mathematics

What is the general solution of the differential equation y″ − 5y′ + 6y = 0?

Given	y″ − 5y′ + 6y = 0
Find	General solution y(x)

Solution

Characteristic equation (FE Reference Handbook, Mathematics — Differential Equations):

r² − 5r + 6 = (r − 2)(r − 3) = 0 → r = 2, 3

y = C₁e^2x + C₂e^3x

This flips the signs of the roots, as if the equation were y″ + 5y′ + 6y = 0.

This uses the coefficients 5 and 6 as roots instead of solving the characteristic equation.

Sines and cosines come from complex roots; these roots are real, and a real pair would never mix two frequencies.

r² − 5r + 6 = (r − 2)(r − 3) = 0 gives the distinct real roots 2 and 3.

Question 21 of 70 · Mathematics

What are the eigenvalues of the matrix A = [5 2; 2 5]?

Given	A = [5 2; 2 5]
Find	Eigenvalues of A

Solution

Characteristic equation (FE Reference Handbook, Mathematics — Matrices):

det(A − λI) = (5 − λ)² − 4 = 0 → λ = 5 ± 2 = 3, 7

Check: λ₁ + λ₂ = 10 = trace, λ₁λ₂ = 21 = det A.

This reverses the sign in det(A − λI), i.e. solves det(A + λI) = 0.

This solves (5 − λ)² = 16 instead of 4, i.e. skips the square root of the off-diagonal product (2 × 2 = 4, not 4²).

(5 − λ)² − 4 = 0 gives 5 − λ = ±2, so λ = 3 and λ = 7 (sum 10 = trace, product 21 = det).

These are the diagonal entries; they are the eigenvalues only of a diagonal or triangular matrix.

Question 22 of 70 · Mathematics

Using the trapezoidal rule with n = 2 subintervals, approximate the integral of f(x) = x² from 0 to 2.

Given	f(x) = x² Interval 0 to 2 n = 2 subintervals
Find	Trapezoidal estimate of the integral

Solution

Trapezoidal rule (FE Reference Handbook, Mathematics — Numerical Integration):

∫ ≈ (h/2)[f₀ + 2f₁ + f₂] = (1/2)[0 + 2(1) + 4] = 3.00

(exact value 8/3 = 2.67)

4.00 is a single trapezoid over [0, 2]: (2/2)(0 + 4) = 4; the question asks for two subintervals.

h = 1, so (h/2)[f(0) + 2f(1) + f(2)] = ½[0 + 2 + 4] = 3.00.

2.67 is the exact value 8/3 (or Simpson’s rule); the trapezoidal estimate overshoots for a convex function.

2.50 forgets to double the interior ordinate: ½[0 + 1 + 4] = 2.5.

Question 23 of 70 · Mathematics

Using implicit differentiation, find dy/dx for x² + y² = 25.

Given	x² + y² = 25
Find	dy/dx

Solution

Differentiate both sides with respect to x:

2x + 2y·dy/dx = 0 → dy/dx = −x/y

2x + 2y·dy/dx = 0, so dy/dx = −x/y (the slope of a circle is perpendicular to its radius).

This inverts the ratio: the x-term gives 2x and the y-term gives 2y·dy/dx, so x belongs in the numerator.

This loses the minus sign when moving 2x across the equation.

This both inverts the ratio and drops the sign; it is the slope of the radius, not of the tangent.

Question 24 of 70 · Mathematics

What is the shortest distance from the point P(2, 2, 2) to the line through A(1, 0, 0) and B(0, 1, 0)?

Given	P(2, 2, 2) A(1, 0, 0) B(0, 1, 0)
Find	Shortest distance from P to line AB

Solution

Distance from a point to a line (FE Reference Handbook, Mathematics — Vectors):

d = |w × v|/|v|, w = P − A = (1, 2, 2), v = B − A = (−1, 1, 0)

w × v = (−2, −2, 3), |w × v| = √17 = 4.12

d = 4.12/√2 = 2.92

0.71 is 1/√2, the reciprocal of |v|; the cross-product magnitude was never computed.

1.41 is |v| = √2, the length of the direction vector AB, not a distance from P.

w = AP = (1, 2, 2), v = AB = (−1, 1, 0), |w × v| = √17, so d = √17/√2 = 2.92.

3.00 is |AP|, the distance from P to the point A, not the perpendicular distance to the line.

4.12 is |w × v| = √17 before dividing by |v| = √2.

Question 25 of 70 · Mathematics

What is the volume of the parallelepiped defined by the vectors A = (1, 1, 0), B = (0, 2, 1) and C = (1, 0, 3)?

Given	A = (1, 1, 0) B = (0, 2, 1) C = (1, 0, 3)
Find	Volume of the parallelepiped

Solution

Scalar triple product (FE Reference Handbook, Mathematics — Vectors):

B × C = (2·3 − 1·0, 1·1 − 0·3, 0·0 − 2·1) = (6, 1, −2)

V = |A·(B × C)| = |6 + 1 + 0| = 7.00

1.17 is the tetrahedron on the same three edges, one sixth of the parallelepiped.

2.45 is |A × B| = |(1, −1, 2)| = √6, the area of one face; the third vector C never enters.

3.50 is half the triple product, the volume of the triangular prism on the edges.

6.40 is |B × C| = √41 alone; the dot product with A was never taken.

B × C = (6, 1, −2) and A·(B × C) = 6 + 1 + 0 = 7.00.

Question 26 of 70 · Mathematics

What is the area bounded by the curves y₁ = −x² + 2x + 3 and y₂ = x² − 4x + 3?

Given	y₁ = −x² + 2x + 3 y₂ = x² − 4x + 3
Find	Area between the curves

Solution

Intersections and upper-minus-lower integral (FE Reference Handbook, Mathematics — Integral Calculus):

−x² + 2x + 3 = x² − 4x + 3 → 2x(x − 3) = 0 → x = 0, 3

A = ∫₀³(−2x² + 6x) dx = [−(2/3)x³ + 3x²]₀³ = −18 + 27 = 9.0

45.0 adds the x² terms instead of subtracting them, integrating 2x² + 6x: 18 + 27.

36.0 integrates 6x as 6x² (forgets to halve the coefficient), giving −18 + 54.

27.0 keeps only the linear term, ∫₀³6x dx = 27, and drops the −2x² term.

18.0 keeps only the x² terms, ∫₀³2x² dx = 18, and loses the 6x term.

The curves meet at x = 0 and x = 3, and ∫₀³(−2x² + 6x) dx = −18 + 27 = 9.0.

Question 27 of 70 · Probability and Statistics

Seven turbidity measurements (NTU) are taken from a water source: 5, 8, 9, 12, 14, 15, 30. Using the interquartile-range method with fences at Q₁ − 1.5·IQR and Q₃ + 1.5·IQR, which value(s) are outliers?

Given	Data 5, 8, 9, 12, 14, 15, 30 NTU
Find	Outlier(s) by the 1.5·IQR rule

Solution

Quartiles as medians of the lower and upper halves:

Q₁ = 8, Q₃ = 15, IQR = 7

fences 8 − 10.5 = −2.5 and 15 + 10.5 = 25.5 NTU

Only 30 NTU lies outside. (With interpolated quartiles 8.5 and 14.5 the upper fence is 23.5 and the answer is the same.)

15 is Q₃ itself, well inside the fences −2.5 to 25.5 NTU.

The lower fence is −2.5 NTU, so the smallest value 5 is not unusual; only the high end has an outlier.

The upper fence is 25.5 NTU, so 30 NTU lies outside it.

15 is the upper quartile, not an extreme value; only 30 exceeds the upper fence.

Q₁ = 8 and Q₃ = 15 give IQR = 7 and fences −2.5 and 25.5 NTU; only 30 falls outside (it stays the only one with any common quartile convention).

Question 28 of 70 · Probability and Statistics

Two independent water-quality samples are taken from a lake. Sample 1 has n = 6 and s = 5; sample 2 has n = 16 and s = 2. Assuming equal population variances, what is the pooled variance s_p²?

Given	n₁ = 6, s₁ = 5 n₂ = 16, s₂ = 2
Find	Pooled variance s_p²

Solution

Pooled variance (FE Reference Handbook, Probability and Statistics — Test Statistics):

s_p² = [(n₁ − 1)s₁² + (n₂ − 1)s₂²]/(n₁ + n₂ − 2)

= (5 × 25 + 15 × 4)/20 = 185/20 = 9.25

3.04 is the pooled standard deviation √9.25, not the pooled variance the question asks for.

3.50 is the plain average of the two standard deviations, (5 + 2)/2; variances must be pooled, not standard deviations.

[(6 − 1)(25) + (16 − 1)(4)]/(6 + 16 − 2) = 185/20 = 9.25.

9.73 weights the variances by n instead of n − 1: (6 × 25 + 16 × 4)/22.

14.50 is the unweighted mean of the variances, (25 + 4)/2, which ignores the very different sample sizes.

Question 29 of 70 · Probability and Statistics

A hydrologist fits a log-normal distribution to annual streamflow data grouped into 12 bins. The distribution’s mean and standard deviation are estimated from the same data. How many degrees of freedom does the chi-square goodness-of-fit test have?

Given	Bins k = 12 Mean and standard deviation estimated from the data
Find	Degrees of freedom

Solution

Chi-square goodness of fit:

df = k − 1 − p = 12 − 1 − 2 = 9

12 uses the number of bins directly; the bin counts must sum to n, which costs one degree of freedom before any estimation.

11 is k − 1, which applies only when no parameters are estimated from the sample.

10 subtracts only one estimated parameter; both the mean and the standard deviation came from the data.

Df = k − 1 − p = 12 − 1 − 2 = 9.

8 subtracts one parameter too many; only two (the mean and the standard deviation) were estimated.

Question 30 of 70 · Probability and Statistics

An engineer fits the power law y = ax^b to hydraulic data by linear regression on Y = ln y versus X = ln x. The regression slope is 1.0, the mean of X is 2.0 and the mean of Y is 5.0. What is the coefficient a?

Given	Model y = ax^b Regression of ln y on ln x: slope 1.0 X̄ = 2.0, Ȳ = 5.0
Find	Coefficient a

Solution

Least-squares line through the centroid (FE Reference Handbook, Probability and Statistics — Least Squares):

Ȳ = ln a + b·X̄ → ln a = 5.0 − 1.0 × 2.0 = 3.0

a = e^3.0 = 20.1

The fitted line passes through (X̄, Ȳ), so ln a = 5.0 − 1.0 × 2.0 = 3.0 and a = e³ = 20.1.

7.39 is e², the mean of X back-transformed; X̄ enters the intercept only through b·X̄.

3.00 is ln a, the intercept of the transformed line, before back-transforming with e^x.

2.72 is e¹, the slope b back-transformed; b is already the exponent and needs no conversion.

1.10 is ln 3, the logarithm of the intercept taken again instead of back-transforming it with e^x.

Question 31 of 70 · Probability and Statistics

The product concentration of a chemical process is normally distributed with a mean of 95.0% and a standard deviation of 1.2%. What is the probability that a randomly selected batch has a concentration above 97.0%?

Given	μ = 95.0% σ = 1.2% x = 97.0%
Find	P(concentration > 97.0%)

Solution

Standardize and read the unit normal table (FE Reference Handbook, Probability and Statistics — Normal Distribution):

z = (x − μ)/σ = (97.0 − 95.0)/1.2 = 1.67

P(Z > 1.67) = 1 − 0.9525 = 0.0475 = 4.75%

z = (97 − 95)/1.2 = 1.67, and the upper-tail area beyond z = 1.67 is 0.0475, i.e. 4.75%.

9.50% is the two-tailed area beyond ±1.67; the question asks only about the high side.

45.25% is the area between the mean and z = 1.67, not the tail beyond it.

95.25% is Φ(1.67), the probability of a concentration below 97.0%.

Question 32 of 70 · Probability and Statistics

A production line fills bottles to a target of 500 mL. An X-bar chart with limits at μ ± 3σ/√n is used, with μ = 500 mL, σ = 4 mL and samples of n = 4 taken hourly. What are the lower and upper control limits?

Given	μ = 500 mL σ = 4 mL n = 4
Find	LCL and UCL

Solution

X-bar chart limits:

UCL, LCL = μ ± 3σ/√n = 500 ± 3(4)/√4 = 500 ± 6 → 494 mL and 506 mL

488 mL and 512 mL is μ ± 3σ, the limits for individual bottles; sample means vary less, by the factor √n.

492 mL and 508 mL is μ ± 2σ for individual bottles: both the √n and the 3-sigma width are wrong.

3σ/√n = 3 × 4/2 = 6 mL, so the limits are 494 mL and 506 mL.

496 mL and 504 mL is μ ± 2σ/√n, two-sigma limits instead of the three-sigma limits the chart uses.

Question 33 of 70 · Probability and Statistics

Defects in a reactor product occur randomly at an average rate of 0.002 per unit volume. If 1,500 volume units are processed and defects follow a Poisson distribution, what is the probability of exactly 3 defects?

Given	Rate 0.002 per unit volume Volume 1,500 units x = 3
Find	P(X = 3)

Solution

Scale λ to the interval, then apply the Poisson pmf (FE Reference Handbook, Probability and Statistics — Poisson Distribution):

λ = 1,500 × 0.002 = 3

P(X = 3) = e⁻³·3³/3! = 0.0498 × 27/6 = 0.224

64.7% is the cumulative P(X ≤ 3), the chance of three or fewer defects, not exactly three.

λ = 1,500 × 0.002 = 3, and P(X = 3) = e⁻³·3³/3! = 0.224, i.e. 22.4%.

16.8% is P(X = 4), one count too many in the λ^x/x! term.

4.98% is e⁻³, the probability of zero defects, P(X = 0).

Question 34 of 70 · Probability and Statistics

The standard deviation of concrete cylinder strengths on a project is 3.5 MPa. If 100 cylinders are tested, what is the standard error of the mean?

Given	σ = 3.5 MPa n = 100
Find	Standard error of the mean

Solution

Standard error of the mean (FE Reference Handbook, Probability and Statistics):

SE = σ/√n = 3.5/10 = 0.35 MPa

0.035 MPa divides by n = 100 instead of √n = 10.

SE = σ/√n = 3.5/√100 = 0.35 MPa.

0.70 MPa is about two standard errors (the half-width of a 95% interval), not the standard error itself.

3.5 MPa is the standard deviation of single cylinders; the mean of 100 cylinders varies far less.

Question 35 of 70 · Probability and Statistics

A sample of n = 49 has a mean of 30 and a variance of 16. Using z = 2.576, what is the 99% confidence interval for the true mean?

Given	n = 49 x̄ = 30 Variance 16 z = 2.576
Find	99% confidence interval

Solution

Confidence interval on a mean with known σ (FE Reference Handbook, Probability and Statistics — Confidence Intervals):

x̄ ± zσ/√n = 30 ± 2.576 × 4/7 = 30 ± 1.47

28.53 to 31.47

This divides σ by n = 49 instead of √n = 7.

This uses the variance 16 as if it were the standard deviation; σ = √16 = 4.

This uses z = 1.96, the 95% value, instead of 2.576.

σ = 4, SE = 4/7 = 0.571, and 2.576 × 0.571 = 1.47 around the mean of 30.

Question 36 of 70 · Probability and Statistics

A digital link has a bit-error probability of 0.01. If 8 bits are sent independently, what is the probability that exactly one bit is received in error?

Given	p = 0.01 n = 8 k = 1
Find	P(exactly one error)

Solution

Binomial distribution (FE Reference Handbook, Probability and Statistics — Binomial Distribution):

P(1) = C(8, 1)p¹(1 − p)⁷ = 8 × 0.01 × 0.99⁷ = 0.0746 = 7.46%

92.27% is (0.99)⁸, the probability that no bit is in error.

8.00% is n·p = 8 × 0.01; it leaves out the (0.99)⁷ chance that the other seven bits are correct.

7.73% is 1 − 0.99⁸, the probability of at least one error, which also counts two or more errors.

C(8, 1)(0.01)(0.99)⁷ = 8 × 0.01 × 0.932 = 0.0746, i.e. 7.46%.

Question 37 of 70 · Probability and Statistics

A signal-processing system has two independent stages with reliabilities 0.95 and 0.90. Both stages must work for the system to work. What is the system reliability?

Given	R₁ = 0.95 R₂ = 0.90 Both stages must work
Find	System reliability

Solution

Independent components in series:

R_s = R₁R₂ = 0.95 × 0.90 = 0.855 = 85.5%

For components in series R = R₁R₂ = 0.95 × 0.90 = 0.855 = 85.5%.

90.0% takes the weakest stage as the system value; in series the system is always less reliable than its weakest part.

92.5% averages the two reliabilities; reliabilities of independent series stages multiply.

99.5% is the parallel (redundant) result 1 − (0.05)(0.10), where only one stage needs to work.

Question 38 of 70 · Probability and Statistics

A sample of 6 power measurements gives 100, 105, 110, 115, 120 and 130 W. What is the sample standard deviation, rounded to the nearest watt?

Given	Data 100, 105, 110, 115, 120, 130 W n = 6
Find	Sample standard deviation s (W)

Solution

Sample standard deviation (FE Reference Handbook, Probability and Statistics — Dispersion):

x̄ = 680/6 = 113.33 W

Σ(x − x̄)² = 177.8 + 69.4 + 11.1 + 2.8 + 44.4 + 277.8 = 583.3 W²

s = √(583.3/5) = √116.7 = 10.8 ≈ 11 W

4 W is the standard error of the mean, s/√n = 10.8/√6 = 4.4 ≈ 4 W, not the spread of single readings.

This is the mean absolute deviation (50/6 = 8.3 W), not the standard deviation.

10 W divides the sum of squares by n = 6 (population formula) instead of n − 1 = 5.

The mean is 113.33 W, the squared deviations sum to 583.3 W², and s = √(583.3/5) = 10.8 W.

Question 39 of 70 · Probability and Statistics

In a plant, 60% of products come from Machine A and 40% from Machine B. Machine A produces 2% defective items and Machine B produces 5%. What is the probability that a randomly selected defective item came from Machine A?

Given	P(A) = 0.60, P(B) = 0.40 P(D\|A) = 0.02, P(D\|B) = 0.05
Find	P(A\|D)

Solution

Bayes’ theorem (FE Reference Handbook, Probability and Statistics — Probability Laws):

P(D) = 0.60(0.02) + 0.40(0.05) = 0.032

P(A|D) = 0.012/0.032 = 0.375 = 37.5%

28.6% ignores the production shares and compares only the defect rates, 0.02/(0.02 + 0.05).

P(D) = 0.60 × 0.02 + 0.40 × 0.05 = 0.032, and P(A|D) = 0.012/0.032 = 0.375 = 37.5%.

60.0% is the prior share of Machine A; it does not update for the information that the item is defective.

62.5% is P(B|D) = 0.020/0.032, the posterior for the other machine.

Question 40 of 70 · Probability and Statistics

A quality engineer measures 25 resistors: the mean is 470 Ω and the sample standard deviation is 10 Ω. For a 95% confidence interval with t = 2.064, what is the margin of error?

Given	n = 25 s = 10 Ω t = 2.064
Find	Margin of error (Ω)

Solution

Confidence interval on a mean with unknown σ (FE Reference Handbook, Probability and Statistics — Confidence Intervals):

E = t·s/√n = 2.064 × 10/√25 = 4.13 Ω

20.6 Ω multiplies t by s without dividing by √n.

10.0 Ω is the sample standard deviation, the spread of single resistors.

8.3 Ω is the full width of the interval, 2 × 2.064 × 10/5; the margin of error is half of it.

E = t·s/√n = 2.064 × 10/5 = 4.13 Ω.

Question 41 of 70 · Probability and Statistics

Two independent sensors read X ~ N(10.0 V, (0.5 V)²) and Y ~ N(9.8 V, (0.4 V)²). What is the probability that the reading of X exceeds that of Y?

Given	X ~ N(10.0 V, (0.5 V)²) Y ~ N(9.8 V, (0.4 V)²)
Find	P(X > Y)

Solution

Difference of independent normals (FE Reference Handbook, Probability and Statistics — Normal Distribution):

D = X − Y ~ N(0.2, 0.41), σ_D = 0.640 V

P(D > 0) = P(Z > −0.2/0.640) = Φ(0.31) = 0.623

74.8% subtracts the variances (0.25 − 0.16); variances of a difference add, they never cancel.

D = X − Y has mean 0.2 V and σ = √(0.25 + 0.16) = 0.640 V, so P(D > 0) = Φ(0.31) = 0.623 = 62.3%.

58.8% adds the standard deviations (0.5 + 0.4 = 0.9 V); for independent variables the variances add.

37.7% is P(X < Y), the complement of the asked probability.

Question 42 of 70 · Probability and Statistics

A discrete random variable X has the probability mass function f(x) = kx² for x = 1, 2, 3, 4. What are the normalization constant k and the variance of X?

Given	f(x) = kx², x = 1, 2, 3, 4
Find	k and Var(X)

Solution

Normalize, then use Var = E[X²] − (E[X])² (FE Reference Handbook, Probability and Statistics — Expected Values):

k(1 + 4 + 9 + 16) = 1 → k = 1/30

E[X] = Σx³/30 = 100/30 = 10/3; E[X²] = Σx⁴/30 = 354/30 = 59/5

Var(X) = 59/5 − 100/9 = 31/45 = 0.689

Σx² = 30 gives k = 1/30, E[X] = 100/30 = 10/3, E[X²] = 354/30 = 59/5, and Var = 59/5 − 100/9 = 31/45.

The constant is right, but 10/3 is the mean E[X], not the variance.

This normalizes f(x) = kx (Σx = 10) instead of kx²; the variance 1 follows from that wrong pmf.

The constant is right, but 59/5 is E[X²]; the squared mean (10/3)² was never subtracted.

Question 43 of 70 · Probability and Statistics

A batch of 100 integrated circuits contains 5 defective units. If two circuits are selected at random without replacement, what is the probability that both are defective?

Given	N = 100, defective = 5 2 drawn without replacement
Find	P(both defective)

Solution

Multiply the conditional probabilities:

P = (5/100) × (4/99) = 0.00202 = 0.20%

10.00% adds the two draws’ probabilities; for “both” the conditional probabilities multiply.

5.00% is only the first draw; the second must also be defective.

0.25% is (5/100)², sampling with replacement; after one defective is drawn only 4 of 99 remain.

(5/100)(4/99) = 20/9,900 = 0.0020 = 0.20%.

Question 44 of 70 · Probability and Statistics

How many combinations are there of 6 items taken 2 at a time?

Given	n = 6 r = 2
Find	C(6, 2)

Solution

Combinations (FE Reference Handbook, Probability and Statistics — Permutations and Combinations):

C(n, r) = n!/[r!(n − r)!] = 6!/(2!·4!) = 15

30 is the number of permutations P(6, 2) = 6 × 5, which counts each pair twice (AB and BA).

C(6, 2) = 6!/(2!·4!) = 30/2 = 15.

12 multiplies 6 × 2, which does not count pairs.

6 is the number of items, not the number of pairs.

Question 45 of 70 · Ethics and Professional Practice

An engineer is offered a gift worth $500 by a vendor who is bidding on a project the engineer is evaluating. What should the engineer do?

Solution

The NSPE Code (II.4.c, II.5.b) forbids soliciting or accepting valuable consideration from parties connected with work the engineer is responsible for; the safe course is to decline and tell the employer.

Equal gifts from all bidders are still consideration from interested parties; they do not cancel each other out.

What the engineer does with the money afterwards does not change that a bidder gave value to the evaluator.

Disclosure after the decision does nothing for the fairness of the evaluation; the gift is improper when it is accepted.

Codes of ethics bar accepting valuable gifts from parties seeking work the engineer influences, and reporting the offer protects the evaluation.

Question 46 of 70 · Ethics and Professional Practice

An engineer recommends a filtration vendor for a municipal project, and the municipality will select the vendor based on this recommendation. The engineer owns stock in the vendor’s parent company and has told their employer about it. What additional step do the conflict-of-interest rules require?

Solution

NSPE Code II.4.a: engineers shall disclose all known or potential conflicts of interest that could influence or appear to influence their judgment or the quality of their services. The disclosure must reach the party relying on that judgment: the municipality chooses the vendor on the engineer’s recommendation, so telling only the employer leaves the decision-maker uninformed.

Divesting later does not cure the recommendation already made, and the rules require disclosure, not a sale.

A statement from the interested vendor is not disclosure to the client and adds no independence.

The employer is only one affected party; the client deciding on the vendor has not been told.

Giving away the profit leaves the client uninformed; the duty is to disclose the conflict, not to forgo the gain.

The conflict must be disclosed to every party whose interests it may affect, and the municipality is the client relying on the recommendation.

Question 47 of 70 · Ethics and Professional Practice

During foundation excavation a contractor hits rock at elevation −12 ft, although the contract boring logs show soil to elevation −20 ft. The contractor keeps daily logs and photos from the day of discovery but sends no written notice to the engineer until 3 weeks later. Which element of a differing-site-conditions claim is weakest, and what would most strengthen it?

Solution

A differing-site-conditions claim rests on a material difference from the contract indications, unforeseeability, causation of cost or delay, and timely written notice. Logs and photos cover the facts; the 3-week delay in written notice is the weak link, so notice should be given at once with the records attached.

The daily logs and photos from the day of discovery are already contemporaneous; later affidavits add little.

A schedule analysis quantifies impact later, but nothing in the facts shows causation is in doubt; the missing notice is the gap.

Most contracts make prompt written notice a condition of the claim; the 3-week gap is the one element with a documented failure, and notice with the contemporaneous records limits it.

Rock 8 ft above the logged soil depth is a clear material difference; surveying adds detail, not entitlement.

The boring logs showed soil to −20 ft, so the rock contradicts the contract indications; foreseeability is the strong part of this claim.

Question 48 of 70 · Ethics and Professional Practice

A company has developed a new chemical formulation that it wants to keep secret indefinitely. Which form of intellectual property protection fits best?

Solution

Trade secrets protect confidential business information for as long as secrecy is maintained; patents trade a 20-year exclusive right for full public disclosure.

A trade secret lasts as long as the information stays confidential and is protected by reasonable secrecy measures, with no fixed term.

A trademark protects a brand name or logo that identifies the source of goods, not the composition of the product.

A utility patent requires public disclosure of the invention and expires 20 years after filing, which defeats indefinite secrecy.

Question 49 of 70 · Ethics and Professional Practice

An employer asks an engineer to certify a design that the engineer did not prepare and has not reviewed. What should the engineer do?

Solution

NSPE Code II.2.b: engineers shall not affix their signatures to any plans or documents not prepared under their direction and control (NCEES Model Rules: responsible charge). A management directive cannot replace that; the engineer declines to certify work they have not examined and taken responsibility for.

A disclaimer does not undo a certification; signing still represents that the engineer stands behind work they have not examined.

Delegating the signature shifts the same problem to someone with even less basis for it.

A certification asserts the engineer’s own knowledge of the design; without examining it and taking responsible charge, the engineer cannot sign.

A management directive does not supply the knowledge a certification asserts; the engineer’s professional judgment cannot be overruled into a signature.

Question 50 of 70 · Ethics and Professional Practice

Under which circumstance may a chemical engineer ethically disclose confidential information about a client’s process without the client’s consent?

Solution

The NSPE Code (II.1.c) lets engineers reveal confidential information without consent only as authorized or required by law or by the Code, and public safety is paramount (I.1).

Research value does not override confidentiality; publication needs the client’s consent.

Personal gain is never a ground for breaking confidentiality; this is exactly what the duty forbids.

A payment dispute is settled by contract remedies, not by revealing the client’s process.

Confidentiality yields when disclosure is required by law or to protect public health and safety.

Question 51 of 70 · Ethics and Professional Practice

Software code that an employee engineer writes during working hours as part of the job, using company resources, is usually protected as:

Solution

Under U.S. copyright law (17 U.S.C. 101) a work prepared by an employee within the scope of employment is a work made for hire, and the employer is the author and owner.

Patents are not automatic and code is not patented by default; ownership would also follow the employment agreement.

Code is a work of authorship; written within the scope of employment it is a work made for hire, so the employer owns the copyright.

A trademark identifies the source of goods (a name or logo); it does not protect code.

Code may be kept confidential, but there is no default joint ownership between employee and employer.

Question 52 of 70 · Ethics and Professional Practice

How long does a U.S. utility patent on a new integrated-circuit design normally provide protection?

Solution

35 U.S.C. 154: a utility patent term ends 20 years from the earliest effective U.S. filing date (subject to term adjustments). Design patents: 15 years from grant; copyright: life + 70 years.

Only trademarks can be renewed indefinitely while in use; maintenance fees cannot extend a patent past its term.

A utility patent normally runs 20 years from the filing date of the application, provided maintenance fees are paid.

This is the term of a design patent (for applications filed since May 2015), not a utility patent.

Life plus 70 years is the copyright term for works by individual authors.

Question 53 of 70 · Ethics and Professional Practice

An engineer finds that a colleague is taking credit for work others did on a technical paper. What is the most appropriate first step?

Solution

The NSPE Code (III.9) requires engineers to give credit to those to whom it is due; the usual first step on a credit dispute is to raise it with the person involved, then escalate if it is not corrected.

Giving credit where due is a duty, and a private conversation lets the colleague correct the record before it is escalated.

Escalating before any attempt to resolve an authorship dispute directly is premature; a board report may follow if the colleague persists.

Misrepresenting credit is an ethics issue regardless of how common it is; ignoring it condones it.

Public accusations are unprofessional and can injure a colleague’s reputation without due process.

Question 54 of 70 · Ethics and Professional Practice

Which of the following is NOT typically covered by a non-disclosure agreement (NDA) on a chemical engineering project?

Solution

A project NDA protects the confidential technical and business information exchanged for the project (formulations, design data, know-how); employee compensation is governed by employment agreements.

Pay of team members is an employment matter, not the proprietary technical information a project NDA covers.

Formulations are core proprietary information and the usual subject of a project NDA.

Know-how about how a product is made is exactly what an NDA is written to protect.

Operating conditions and design parameters are confidential technical data an NDA protects.

Question 55 of 70 · Ethics and Professional Practice

A chemical engineer discovers that a colleague has falsified quality-control data to meet a production deadline. The product will ship to customers this week. What is the most appropriate action?

Solution

NSPE Code II.1.f: engineers having knowledge of an alleged violation shall report it to appropriate bodies; II.3.a: reports and data must be objective and truthful. With the product about to ship, the falsification goes at once to the person responsible for quality, so the product can be held.

Knowledge of a violation creates a duty to report it; staying silent lets the product ship.

Waiting for field failures puts the public at the very risk the QC data exist to prevent.

Falsified QC data can put an unsafe product in customers’ hands; engineers who know of such a violation must report it so the product can be held and the data corrected.

Altering records without telling anyone is itself falsification and leaves the shipped product unchecked.

Question 56 of 70 · Ethics and Professional Practice

A chemical engineer is asked to give expert testimony in a lawsuit about a process similar to one they designed for a former client. What is the primary ethical consideration?

Solution

NSPE Code III.4: engineers shall not disclose confidential information about the business affairs or technical processes of any present or former client without consent; testimony must also be objective and truthful (II.3.a).

Personal risk management is not an ethical obligation owed to anyone else.

The duty of confidentiality to a former client survives the engagement and limits what the engineer may use or reveal in court.

Expert testimony must be objective and truthful regardless of who pays for it.

The fee is a business term; it is not the ethical issue the prior engagement raises.

Question 57 of 70 · Ethics and Professional Practice

Which statement best describes how membership in a professional society relates to ethical obligations?

Solution

Professional society codes (for example the NSPE Code of Ethics) are binding on members; licensing boards enforce their own rules of professional conduct separately on licensees.

Joining a society adds obligations; it never releases anyone from licensing-board rules or the law.

Societies can censure, suspend or expel members who violate their codes.

A member accepts the society’s code by joining, so it applies to licensed and unlicensed members alike.

Licensure is a state legal status; society codes bind members by membership, not by license.

Question 58 of 70 · Ethics and Professional Practice

The owner asks a project engineer to certify a monthly pay application that includes concrete the engineer saw had visible honeycomb voids. The contractor has not yet repaired the defect, and the owner says certification is needed to maintain cash flow. What best meets the engineer’s professional duties?

Solution

Certification of a pay application represents that the work covered conforms to the contract. The engineer withholds the value of the defective work, documents it in writing to owner and contractor, and certifies the rest.

Legal comfort for the owner does not change the engineer’s duty to certify accurately.

Certifying payment for work known to be defective misrepresents its condition, whatever is noted in the file.

A bond covers the money, not the truth of the certificate; the work is still defective today.

The owner’s acceptance does not make the certification truthful; the engineer would still be certifying defective work.

The engineer certifies only conforming work and documents the defect; the withheld amount is released once the repair is accepted.

Question 59 of 70 · Ethics and Professional Practice

An engineer reviewing project documents finds a known safety hazard. A superior instructs the engineer to remove references to the hazard so the project gets approved. What should the engineer do?

Solution

NSPE Code I.1, II.3.a and III.3: engineers hold public safety paramount and are objective and truthful in reports; they may not omit material facts to obtain approval.

A hidden note does not correct the approved documents; reviewers would rely on a false record.

Hoping for a later fix leaves the hazard unknown to those approving the work.

Schedule pressure never justifies hiding a safety hazard.

Reports must be truthful and complete, and public safety comes first; omitting a known hazard to win approval is misrepresentation.

Question 60 of 70 · Ethics and Professional Practice

During a peer review, an engineer notices misleading safety claims in a project report. What is the most appropriate action?

Solution

NSPE Code II.3.a: engineers are objective and truthful in professional reports; a reviewer who finds misleading safety claims records them and asks for independent verification.

A reviewer must be objective: record the discrepancies and have the safety claims checked before the report is relied on.

Approval would put the reviewer’s name behind claims they know are misleading.

Cosmetic edits leave the misleading safety claims in place.

Safety claims are within any engineering review; ignoring them fails the reviewer’s duty.

Question 61 of 70 · Ethics and Professional Practice

Under competitive pressure to finish a safety report quickly, an engineer considers skipping some verification steps. What should the engineer do?

Solution

NSPE Code I.1 and II.1: public safety is paramount over economic or schedule pressure; the engineer completes and documents all verification steps.

Unsupervised delegation leaves no one responsible for the check the report depends on.

No verification step of a safety report is optional because of a deadline; “minor” is a judgment made after checking, not before.

Old data may not fit the current design; reusing it unchecked is the same as skipping verification.

Public safety outranks schedule and market pressure, so the report must be fully verified and its basis recorded.

Question 62 of 70 · Ethics and Professional Practice

Because of an email mistake, an engineer receives a competitor’s proprietary design information. What should the engineer do with it?

Solution

NSPE Code III.4 and the NCEES Model Rules: engineers respect the confidential information of others and do not use it without consent; the proper response to a misdirected transmission is to notify the sender and not use the material.

Respecting others’ confidential information means not using or spreading it and telling the owner of the error.

Spreading the information inside the firm multiplies the breach.

Retaining it for later use is still using another firm’s confidential information.

Using information that is plainly proprietary to someone else is misappropriation, however it arrived.

Question 63 of 70 · Mathematics

Which expression is the indefinite integral ∫ 4xe^2x dx, evaluated by integration by parts?

Given	Integrand 4xe^2x
Find	Antiderivative

Solution

Integration by parts (FE Reference Handbook, Mathematics — Integral Calculus): u = 4x, du = 4 dx, dv = e^2x dx, v = ½e^2x.

∫u dv = uv − ∫v du = 2xe^2x − ∫2e^2x dx = 2xe^2x − e^2x + C = e^2x(2x − 1) + C

u = 4x, v = ½e^2x, so 2xe^2x − ∫2e^2x dx = 2xe^2x − e^2x + C.

This keeps only the uv term and forgets to subtract ∫v du.

This differentiates dv instead of integrating it (v = 2e^2x instead of ½e^2x).

This drops the ½ in the second integration: ∫2e^2x dx is e^2x, not 2e^2x.

This adds ∫v du instead of subtracting it; differentiating gives 4xe^2x + 4e^2x, not the integrand.

Question 64 of 70 · Mathematics

An equilibrium analysis gives three unknown forces T₁, T₂ and T₃ (in newtons) that satisfy: T₁ + T₂ + T₃ = 100 2T₁ − T₂ + T₃ = 50 −T₁ + 2T₂ − T₃ = 0 What is T₃?

Given	T₁ + T₂ + T₃ = 100 2T₁ − T₂ + T₃ = 50 −T₁ + 2T₂ − T₃ = 0
Find	T₃ (N)

Solution

Add rows 1 and 3, then subtract row 2 from row 1:

3T₂ = 100 → T₂ = 33.3 N

−T₁ + 2T₂ = 50 → T₁ = 16.7 N

T₃ = 100 − 16.7 − 33.3 = 50.0 N

16.7 N is T₁; the elimination stopped one back-substitution early.

33.3 N is T₂, which drops out first (rows 1 + 3 give 3T₂ = 100).

T₂ = 33.3 N, T₁ = 16.7 N, and row 1 then gives T₃ = 100 − 50 = 50.0 N.

66.7 N is T₁ + T₃, i.e. 100 − T₂; the remaining unknown was not separated.

83.3 N is 100 − T₁: T₂ was left inside the result instead of being subtracted too.

Question 65 of 70 · Mathematics

A steady-state mass balance gives this linear system for three stream flow rates x, y and z (kg/h): 2x + 3y − z = 10 4x − y + 2z = 15 6x + 2y + z = 25 Using the determinant of the coefficient matrix and a consistency check, what is the nature of the solution?

Given	Coefficient rows [2, 3, −1], [4, −1, 2], [6, 2, 1] Right side [10, 15, 25]
Find	Nature of the solution

Solution

Expand the determinant along row 1 (FE Reference Handbook, Mathematics — Matrices):

det A = 2(−1 − 4) − 3(4 − 12) − 1(8 + 6) = −10 + 24 − 14 = 0

Row 3 = row 1 + row 2 on both sides (6x + 2y + z = 10 + 15 = 25), so the system is consistent with one redundant equation → infinitely many solutions.

That would need det = 0 with an inconsistent right side; here 10 + 15 matches 25, so the equations agree.

A linear system has zero, one or infinitely many solutions; three unknowns do not mean three solutions.

The trivial solution belongs to a homogeneous system; with right sides 10, 15 and 25 zero flows do not satisfy any equation.

Det = 0 and row 3 equals row 1 + row 2 including the right side (10 + 15 = 25), so one equation is redundant.

A unique solution needs a nonzero determinant; here det = 0, so the coefficient matrix is singular.

Question 66 of 70 · Mathematics

What is the general solution of the differential equation dy/dx + 2xy = 4x?

Given	dy/dx + 2xy = 4x
Find	General solution y(x)

Solution

Linear first-order equation with P(x) = 2x (FE Reference Handbook, Mathematics — Differential Equations):

μ = e^{∫2x dx} = e^x²

(ye^x²)′ = 4xe^x² → ye^x² = 2e^x² + C

y = 2 + Ce^−x²

μ = e^x², so (ye^x²)′ = 4xe^x², ye^x² = 2e^x² + C, and y = 2 + Ce^−x².

This flips the sign of the exponent and forgets to divide 4x by the 2x that comes with μ.

The homogeneous part is right, but 2x does not satisfy the equation: substituting gives 2 + 4x² ≠ 4x.

The particular part 2 is right, but the homogeneous solution of y′ + 2xy = 0 is Ce^−x², not Cx².

Question 67 of 70 · Probability and Statistics

Confidence level	Lower bound (%)	Upper bound (%)
90%	0.15	2.45
95%	−0.07	2.67
99%	−0.50	3.10

An engineer compares the mean yield difference μ_d = μ_A − μ_B of two catalysts, with H₀: μ_d = 0 and H_a: μ_d ≠ 0. The two-sided confidence intervals for the mean difference are shown in the table. Based on the duality between confidence intervals and two-sided tests, which decision is correct?

Solution

A two-sided test at level α rejects H₀ exactly when the (1 − α) interval excludes the null value 0.

90%: [0.15, 2.45] excludes 0 → reject at α = 0.10

95%: [−0.07, 2.67] contains 0 → fail to reject at α = 0.05

Failing to reject never proves H₀; it only means the data are consistent with zero difference at that level.

At α = 0.10 the 90% interval [0.15, 2.45] excludes zero, so H₀ is rejected at that level.

If the wider 95% interval excluded 0, the narrower 90% one would too, so rejecting at 0.05 but not at 0.10 is impossible.

Rejection at α = 0.05 needs the 95% interval to exclude 0, but its lower bound is −0.07.

The 90% interval excludes 0 and the 95% interval contains it, so the p-value lies between 0.05 and 0.10.

Question 68 of 70 · Probability and Statistics

A plant alarm system monitors a workspace for hydrogen sulfide leaks. Its null hypothesis is that no leak is present. If the monitoring algorithm commits a Type II error, what happens in the plant?

Solution

H₀: no leak. Type I error = reject a true H₀ (false alarm). Type II error = fail to reject a false H₀ (missed detection): the leak is real and the alarm does not sound.

This is a correct rejection of a false H₀ (the test’s power), not an error at all.

This is a Type I error, rejecting a true H₀: a false alarm with no leak behind it.

Recalibration is a maintenance action, not a test outcome; raising the threshold would in fact make Type II errors more likely.

Costs are a business consequence of any alarm program, not the outcome of a single wrong decision about H₀.

A Type II error fails to reject a false H₀: a leak exists, the alarm stays silent, and people keep working in the gas.

Question 69 of 70 · Ethics and Professional Practice

An environmental engineer is asked to seal a plan set that includes the structural design of a heavy-equipment platform. The engineer is not competent in structural design but discussed it with a licensed structural colleague, who said verbally that it is adequate. What should the engineer do?

Solution

NSPE Code II.2.a–b and the NCEES Model Rules: engineers perform services and affix their seal only in areas of their competence, on work prepared under their direction and control.

A co-signature does not give the sealing engineer competence or responsible charge for the structural design.

A handbook check by a non-specialist is not the competence needed to take responsibility for a structural design.

A seal says the work is the sealer’s responsibility; a disclaimer on the same sheet contradicts it.

A verbal assurance from someone else does not give the sealing engineer competence or responsible charge.

Licensees seal only work within their competence and responsible charge; the structural sheets should be sealed by the qualified engineer who takes charge of them.

Question 70 of 70 · Ethics and Professional Practice

A process engineer finds a toxic gas leak that threatens a nearby neighborhood. Management refuses to shut down the plant, citing production quotas, and internal appeals have been rejected. What does the engineer’s code of ethics require next?

Solution

NSPE Code II.1.a: if engineers’ judgment is overruled under circumstances that endanger life or property, they shall notify their employer or client and such other authority as may be appropriate. Public safety is paramount (I.1).

The danger is already confirmed and imminent; more review only delays action.

Sabotage is unlawful and can create new hazards; the proper channel is the regulator.

Leaving the job does not protect the neighborhood; the hazard and the duty remain.

Documentation is useful but does nothing to stop the exposure; the public still is not protected.

When an employer overrules a judgment that endangers the public, engineers must notify the appropriate authority.

Answer every question to see your score

Mathematics

All five specifications on this site test analytic geometry and calculus. Beyond that they diverge:

Differential equations: Mechanical, Electrical and Computer, Environmental and Chemical; Chemical lists partial differential equations and Laplace transforms too.
Linear algebra and vectors: Mechanical and Electrical and Computer by name, Civil as vector operations.
Numerical methods (approximation, error propagation, Taylor series): Mechanical, Environmental and Chemical.
Complex numbers and discrete mathematics: Electrical and Computer, whose math range is also the widest of the five.

On test day the formulas are in the Handbook’s Mathematics chapter. Significant-figure rules are not: NCEES errata place them in Units and Conversion Factors (June 2023 notice).

Common traps: a calculator left in degree mode while you differentiate or integrate a trig function; log and ln swapped in a rearranged equation; an answer read to more figures than the data support. Applied math also turns up inside other sections, such as Laplace transforms in linear systems and vector moments in statics.

Probability and statistics

Weights vary widely here: FE Civil folds statistics into its math section, the four other disciplines on this site give it 4–6 questions, and FE Industrial and Systems gives it 10–15. What the specifications list:

All five: distributions (normal, binomial, discrete, continuous) and measures of central tendency and dispersion.
Confidence intervals and regression: Civil, Mechanical, Environmental and Chemical.
Hypothesis testing (t-tests, outlier tests, analysis of variance): Environmental and Chemical.
Expected value in decisions: Mechanical, Electrical and Computer, and Chemical.
Control limits: Chemical, under statistical control.

The Handbook chapter is Engineering Probability and Statistics, and it has been corrected twice in recent printings, so drill from the copy in your MyNCEES account. Common traps: reading a two-tail area from the unit normal table when the question asks for one tail, and pooling data that the problem treats as two separate samples. Reliability questions on the electrical side lean on these rules too (computer systems).

Ethics and professional practice

Ethics questions are short scenarios with one defensible action. The core, codes of ethics, is on all five specifications here; each discipline adds its own angle:

NCEES Model Law: named in the Mechanical and Electrical and Computer specifications, the Model Rules in Electrical and Computer. Both are free to read (Model Law, Model Rules).
Intellectual property (copyright, trade secrets, patents, trademarks): Mechanical, Electrical and Computer, and Chemical.
Contracts: Civil and Chemical. Professional liability and licensure: Civil, Environmental and Chemical.
Regulations: Environmental names federal statutes, among them the Clean Water and Clean Air Acts, RCRA and OSHA.
Safety: Electrical and Computer lists grounding, PPE and radiation protection.

Read each scenario for three things before the options: whom the engineer works for, who carries the risk, and what has already been tried. Options that are each reasonable in isolation are the usual way these items discriminate. Construction contract scenarios with more numbers in them live on FE Civil construction.

The fourth shared section

Engineering economics is on all seven FE specifications too, but it has its own page with the interest-factor formulas and a worked example: FE engineering economics.

Discipline practice tests

FE Civil practice test60 FE Civil problems across all 14 sections, with worked solutions
FE Mechanical practice testA mixed set across all 14 areas, sized by the specification's question ranges
FE Electrical and Computer practice test60 problems spread by the July 2020 specification, every option explained
FE Environmental practice testProblems in the proportions of the specification, every step worked
FE Chemical practice testProblems weighted like the July 2020 Chemical outline, every option explained
PE Mechanical HVAC & Refrigeration practice testA 60-problem mixed set in USCS units, weighted like the October 2025 outline

Discipline problems on your phone

FE Civil app

FE Civil questions, answers and worked solutions in a study app.

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FE Mechanical app

FE Mechanical questions, answers and worked solutions in a study app.

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FE Electrical and Computer app

FE Electrical and Computer questions, answers and worked solutions in a study app.

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FE Environmental app

FE Environmental questions, answers and worked solutions in a study app.

Download on theApp Store Android: practice in the browser Get the FE Environmental app

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FE Chemical app

FE Chemical questions, answers and worked solutions in a study app.

Download on theApp Store Android: practice in the browser Get the FE Chemical app

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FE practice test, quick answers

Does NCEES have an FE practice exam?

Yes. For each of the five disciplines here NCEES sells two 50-question practice exams at $50 each (store price as of October 2026), built on the July 2020 specifications. This page is not one of them and uses none of their items.

Are these real FE exam questions?

No. Each one was set for this site against a section of the July 2020 specification and solved again by a second reviewer without the key; NCEES keeps its exam items confidential.

How much of the FE is math?

It depends on the discipline: 6–9 questions on FE Mechanical, 11–17 on FE Electrical and Computer, and 8–12 on FE Civil, where statistics is included.