Free FE Electrical and Computer Practice Test

Correct answer: B. V_Th = 8 V, R_Th = 2 kΩ

V_Th is the open-circuit voltage at a–b. With nothing connected, R1 and R2 form a divider: V_Th = 24 × 3/(6 + 3) = 8 V. For R_Th, turn the voltage source off (replace it with a short). Looking back from a–b, R1 and R2 are now in parallel: (6 × 3)/(6 + 3) = 2 kΩ.

A adds R1 and R2 in series, which forgets that the shorted source puts them in parallel. C uses the divider backwards (R1 on top instead of R2), giving the voltage across R1. D skips the divider and reports the source voltage.

Source: MIT OCW 6.002 Lecture 3, Thévenin equivalent, PDF p. 16; OpenStax University Physics Vol. 2 §10.2, Resistors in Series and Parallel

Correct answer: D. 5∠126.9°

Magnitude: |z| = √((−3)² + 4²) = 5. The point (−3, 4) sits in the second quadrant, so the angle must be between 90° and 180°. Your calculator's arctan(4/−3) returns −53.1°, which points into the fourth quadrant. Add 180° to land in the right quadrant: −53.1° + 180° = 126.9°.

A is the raw calculator answer with the quadrant ignored. B drops the sign of the real part. C adds the components (3 + 4) instead of using the Pythagorean magnitude.

Source: OpenStax Precalculus 2e §8.5, Polar Form of Complex Numbers

Correct answer: A. x = 2, y = 3

From the first equation, y = 7 − 2x. Substitute into the second: x − 3(7 − 2x) = −7, so 7x − 21 = −7 and x = 2. Then y = 7 − 2(2) = 3. Check both: 2(2) + 3 = 7 and 2 − 9 = −7.

B, C and D each satisfy the first equation but fail the second (3 − 3 = 0, 1 − 15 = −14, −2 − 33 = −35). Plugging an answer back into both equations is the fastest check on exam day.

Source: OpenStax College Algebra 2e §7.1, Systems of Linear Equations: Two Variables

Correct answer: C. 12

An antiderivative is t³ + 2t. Evaluate at the limits: (2³ + 2·2) − (0 + 0) = 8 + 4 = 12.

A evaluates the integrand at t = 2 (3·4 + 2) instead of integrating. B keeps only the t³ term and drops the 2t contribution. D forgets to divide by 3 when integrating 3t², using 3t³ + 2t. Integrate every term, then subtract the lower-limit value.

Source: OpenStax Calculus Vol. 1 §5.3, Fundamental Theorem of Calculus, Part 2

Correct answer: C. 63.6°

A · B = (2)(1) + (−1)(2) + (2)(2) = 4. Both vectors have magnitude 3 (√(4 + 1 + 4) and √(1 + 4 + 4)). So cos θ = 4/(3 × 3) = 0.444 and θ = 63.6°.

A is 90° − 63.6°, a complement mix-up. B would require a zero dot product. D is 180° − 63.6°, which would need a negative dot product.

Source: OpenStax Calculus Vol. 3 §2.3, The Dot Product (Eq. 2.4)

Correct answer: A. 56

Order of the chosen positions doesn't matter (a string with 1s in bits 2, 5 and 7 is one string), so this is a combination: C(8, 3) = 8!/(3! · 5!) = 56.

B is the permutation count P(8, 3) = 8 × 7 × 6, which counts the same set of positions six times. D is every possible 8-bit string (2⁸). C is 8 × 3, which isn't a counting rule.

Source: OpenStax College Algebra 2e §9.5, Counting Principles (combinations formula)

Correct answer: D. y = C₁e^(−2t) + C₂e^(−3t)

Write the characteristic equation r² + 5r + 6 = 0, which factors as (r + 2)(r + 3) = 0. The roots are r = −2 and r = −3: real and distinct. That gives y = C₁e^(−2t) + C₂e^(−3t), an overdamped response that decays without oscillating.

A flips the signs of the roots. B is the form for a repeated root, which happens only when the discriminant is zero (here 25 − 24 = 1). C is the form for complex roots, which would need a negative discriminant.

Source: OpenStax Calculus Vol. 3 §7.1, Second-Order Linear Equations (characteristic equation)

Correct answer: B. 0.315

This is binomial with n = 10 and p = 0.05. P(X = 1) = C(10, 1)(0.05)¹(0.95)⁹ = 10 × 0.05 × 0.6302 = 0.315.

C is P(X = 0) = 0.95¹⁰, the chance of no defects. D is 1 − 0.599, the chance of at least one defect, which is a different question from exactly one. A is just p for a single part.

Source: NIST/SEMATECH e-Handbook §1.3.6.6.18, Binomial Distribution

Correct answer: D. 1.58 Ω

Mean = 500/5 = 100 Ω. Squared deviations: 4, 0, 1, 1, 4, summing to 10. The sample standard deviation divides by n − 1 = 4: s = √(10/4) = 1.58 Ω.

A divides by n = 5, which is the population formula. B is the sample variance (10/4) before taking the square root, and its units would be Ω². C is the mean.

Source: NIST/SEMATECH e-Handbook §1.3.5.6, Measures of Scale (standard deviation)

Correct answer: B. Notify the employer or client and any other appropriate authority

Model Rules §240.15 A.3 says licensees shall notify their employer or client, and such other authority as may be appropriate, when their professional judgment is overruled and public health, safety or welfare is endangered. Protecting the public comes first (§240.15 A.1).

A treats the duty as transferable to a manager; it isn't. C removes the engineer but leaves the hazard unreported. D conflicts with A.5, which says public opinions must rest on adequate knowledge of the facts. Your state board's adopted rules control your actual license.

Source: NCEES Model Rules (2026), §240.15 A.1, A.3 and A.5, printed p. 11 (PDF p. 15)

Correct answers: A, D

A is correct. Copyright covers original works of authorship, including literary works, and computer programs are protected as written expression (17 U.S.C. §102(a)).

B is wrong. Section 102(b) says copyright never extends to an idea, procedure, process, system or method of operation. Algorithms fall in that excluded group.

C is wrong. The federal definition of a trade secret (18 U.S.C. §1839(3)) requires reasonable secrecy measures and independent economic value from not being generally known or readily ascertainable through proper means by others who could benefit from it. It doesn't require registration.

D is correct because it meets both parts of that definition.

Source: U.S. Copyright Office, 17 U.S.C. §102(a)–(b); 18 U.S.C. §1839(3), definition of trade secret; USPTO, Trade secret policy: definition and elements

Correct answer: A. $7,473

P = F/(1 + i)ⁿ = 10,000/1.06⁵ = 10,000/1.3382 = $7,473.

B uses simple interest (10,000/1.30). C is the future worth of $10,000 invested today, not the present worth of a future payment. D discounts for only one year.

Source: OpenStax Contemporary Mathematics §6.4, Compound Interest (present value formula)

Correct answer: 3,480 $ per year

Use the capital recovery relationship, A = P × i(1 + i)ⁿ/[(1 + i)ⁿ − 1]. With i = 0.08 and n = 8, (1.08)⁸ = 1.8509, so the factor is 0.08 × 1.8509/0.8509 = 0.17401. A = 20,000 × 0.17401 = $3,480 per year.

A common slip is dividing 20,000 by 8 ($2,500), which ignores interest. Another is multiplying by the interest rate alone ($1,600), which pays the interest but never repays the principal.

Source: OpenStax Principles of Finance §8.2, Annuities (present value of an annuity)

Correct answer: C. 5,000 units

Each unit contributes $25 − $15 = $10 toward fixed costs. Break-even units = 50,000/10 = 5,000.

A divides by the full price ($25), ignoring variable cost. B divides by the variable cost ($15). D is the result if the margin were $5.

Source: OpenStax Principles of Managerial Accounting §3.2, Break-Even Point in Units

Correct answer: 1.15 Ω

R = ρL/A. Convert the area first: 1.5 mm² = 1.5 × 10⁻⁶ m². Then R = (1.72 × 10⁻⁸)(100)/(1.5 × 10⁻⁶) = 1.15 Ω.

The usual mistake is converting mm² as if it were mm (using 1.5 × 10⁻³ m²), which makes the answer a thousand times too small. Another is inverting the ratio (A/L), which gives nonsense units.

Source: OpenStax University Physics Vol. 2 §9.3, Resistivity and Resistance (Eq. 9.9)

Correct answers: A, B

A donor atom has one more valence electron than silicon needs for bonding. That extra electron is easily freed, so donor doping makes an n-type semiconductor (A), and free electrons become the majority carriers (B).

C is wrong: acceptors have one fewer valence electron (for example, boron) and create holes. D describes p-type material, which acceptor doping produces.

Source: OpenStax University Physics Vol. 3 §9.6, Semiconductors and Doping; Royal Society of Chemistry, Phosphorus: Fact box

Correct answer: 9.6 V

Write KCL at the node, with currents leaving as positive: (V − 12)/4 + V/6 − 1 = 0. Multiply by 12: 3(V − 12) + 2V − 12 = 0, so 5V = 48 and V = 9.6 V.

If you get 7.2 V, you dropped the current source and solved only the divider. If you get 4.8 V, you reversed the 1 A source: it pushes current into the node, so it appears with the opposite sign of the currents leaving.

Source: OpenStax University Physics Vol. 2 §10.3, Kirchhoff's Rules

Correct answer: B. 17.3 V

RMS means the square root of the average of v² over one period. Squaring gives 100 + 400 sin(ωt) + 400 sin²(ωt). Over a full period, the sin(ωt) term averages to zero and sin²(ωt) averages to ½. So the mean of v² is 100 + 200 = 300 V², and V_rms = √300 = 17.3 V. In short, V_rms = √(V_dc² + V_ac,rms²).

A is only the AC part (20/√2). C adds the DC value and the AC RMS value directly, but RMS values don't add linearly. D uses the AC peak instead of its RMS value.

Source: OpenStax University Physics Vol. 2 §15.4, Power in an AC Circuit

Correct answer: D. 2.4∠−53.1° A

X_L = ωL = 2π(60)(0.1061) ≈ 40 Ω. The impedance is Z = 30 + j40 Ω = 50∠53.1° Ω. Current I = V/Z = 120∠0°/50∠53.1° = 2.4∠−53.1° A. Current lags voltage in an inductive circuit, so the angle is negative.

A has the right magnitude but shows current leading, which would mean a capacitive load. B assumes R and X_L are equal. C uses only the resistor and ignores the inductor.

Source: OpenStax University Physics Vol. 2 §15.2, Simple AC Circuits (Eq. 15.8); OpenStax University Physics Vol. 2 §15.3, RLC Series Circuits with AC (Eqs. 15.9, 15.11)

Correct answer: C. 22 Ω

Reduce the parallel pair first: 20 × 30/(20 + 30) = 12 Ω. Then add the series resistor: 10 + 12 = 22 Ω.

A puts all three in series. B puts all three in parallel. D adds 10 and 20 and drops the 30 Ω branch entirely.

Source: OpenStax University Physics Vol. 2 §10.2, Resistors in Series and Parallel

Correct answer: A. 0.125 m

λ = c/f = (3.0 × 10⁸)/(2.4 × 10⁹) = 0.125 m, or 12.5 cm.

The other choices are the same digits off by powers of ten. That usually means a prefix slip (GHz is 10⁹ Hz, not 10⁸ or 10¹⁰). Sanity check: Wi-Fi at 2.4 GHz has a wavelength of a few inches, not meters.

Source: OpenStax University Physics Vol. 2 §16.2, Plane Electromagnetic Waves

Correct answer: D. 8.65 V

τ = RC = (10 × 10³)(100 × 10⁻⁶) = 1 s. For charging, v_C(t) = 10(1 − e^(−t/τ)), so v_C(2) = 10(1 − e⁻²) = 8.65 V.

A is the value after one time constant (63.2%). B uses e⁻² alone, which is the discharge form. C is the value after four time constants.

Source: OpenStax University Physics Vol. 2 §10.5, RC Circuits (Eqs. 10.8–10.9)

Correct answer: A. 1.59 kHz and Q = 10

ω₀ = 1/√(LC) = 1/√(0.01 × 10⁻⁶) = 10,000 rad/s, so f₀ = ω₀/2π = 1.59 kHz. For a series RLC, Q = ω₀L/R = (10,000)(0.01)/10 = 10.

B reports ω₀ in rad/s as if it were Hz. C inverts Q (R/ω₀L). D is off by a factor of ten in f₀ and in Q; neither value follows from the formulas.

Source: OpenStax University Physics Vol. 2 §15.5, Resonance in an AC Circuit (Eqs. 15.17–15.19)

Correct answer: C. 6e^(−3t)u(t)

The impulse response is the inverse Laplace transform of H(s). The causal pair e^(−at)u(t) ↔ 1/(s + a), scaled by 6, gives h(t) = 6e^(−3t)u(t).

A is the step response, which is the inverse transform of H(s)/s. B has the wrong exponent sign; that would need a pole at s = +3, and it would grow without bound. D swaps the gain and the pole location.

Source: MIT OCW 6.003 Lecture 6, Laplace Transform, PDF p. 8

Correct answer: B. 3 kHz

The sampling rate must exceed twice the signal frequency (14 kHz here) to avoid aliasing. At 10 kHz, anything above the 5 kHz folding frequency wraps back. The alias appears at |f − f_s| = |7 − 10| = 3 kHz.

A is the original tone frequency and C is f + f_s. Both are outside the requested 0–5 kHz baseband, so neither is the principal alias. D is the sampling rate itself.

Source: MIT OCW 6.003 Lecture 21, Sampling, PDF p. 60

Correct answer: D. 9.95 kHz

f_c = 1/(2πRC) = 1/(2π × 1.6 × 10³ × 10 × 10⁻⁹) = 1/(1.005 × 10⁻⁴) ≈ 9.95 kHz.

A is 1/(RC) = 62,500, which is the cutoff in rad/s, not Hz. B and C are factor-of-ten slips: using 100 nF instead of 10 nF gives about 995 Hz, while using 1 nF gives about 99.5 kHz.

The capacitor impedance is 1/(jωC), so the divider gives H(jω) = 1/(1 + jωRC). Its magnitude is 1/√(1 + (ωRC)²); the −3 dB point occurs at ωRC = 1.

Source: MIT OCW 6.061, AC Power Flow in Linear Networks, §§4 and 7

Correct answers: A, B, E

Take the z-transform: Y(z) = X(z) + 0.5z⁻¹Y(z), so H(z) = 1/(1 − 0.5z⁻¹), confirming A. The output feeds back on itself, so an impulse produces 1, 0.5, 0.25, … forever, which is the definition of IIR (B). DC gain is H(1) = 1/(1 − 0.5) = 2 (E).

C is wrong: the pole is at z = 0.5, inside the unit circle, so the causal filter is stable. D is wrong because the impulse response never ends.

Source: MIT OCW 6.003 Lecture 5, Z Transform, PDF p. 11; MIT OCW RES.6-007, Lecture 23 transcript, pp. 3–5

Correct answer: B. −1.5 V

Inverting gain is −R_f/R_in = −10/2 = −5. v_out = −5 × 0.3 = −1.5 V. That's well inside the ±15 V rails, so the amplifier stays linear.

A drops the minus sign. C uses the non-inverting gain, 1 + R_f/R_in = 6. D assumes saturation, which happens only if the computed output exceeds the supply.

Source: MIT OCW 6.002 Lecture 20, Operational Amplifier Circuits, p. 3

Correct answer: C. 12.7 V

For a full-wave rectified sine, V_avg = 2V_m/π = 2(20)/π = 12.7 V.

A is the half-wave result, V_m/π, where every other half-cycle is missing. B is the RMS value, V_m/√2, which isn't the average. D is the peak.

Source: ANU ENGN2211 notes, Full-Wave Rectifiers, Eq. (79)

Correct answer: A. 8 V

I_C = βI_B = 100 × 20 µA = 2 mA. The collector loop gives V_CE = V_CC − I_C R_C = 12 − (2 mA)(2 kΩ) = 8 V. This is not a saturated collector-emitter voltage and is consistent with the stated active-region model.

B is the drop across R_C, not V_CE. C uses the base current in place of the collector current. D assumes saturation without checking.

Source: MIT OCW 6.071J, Bipolar Junction Transistor Circuits, PDF p. 1

Correct answer: D. 1 mV

Twelve bits give 2¹² = 4,096 levels. For the equal-width bins specified here, step size = span/2ⁿ = 4.096 V/4,096 = 1.000 mV.

A is the 8-bit result (4.096/256). B is the 10-bit result (4.096/1,024). C is the 13-bit result. Counting bits correctly is the whole problem.

Source: MIT OCW 6.003 Lecture 22, Sampling and Quantization, PDF p. 13

Correct answer: C. 42.1 kvar

Use the power triangle, Q = P tan θ. At 0.80 PF, θ = 36.9° and Q₁ = 100 × 0.75 = 75.0 kvar. At 0.95 PF, θ = 18.2° and Q₂ = 100 × 0.3287 = 32.9 kvar. The capacitor must supply the difference: 75.0 − 32.9 = 42.1 kvar.

A is the full correction to unity power factor. B is the reactive power left after correction, not the amount added. D adds the two values instead of subtracting.

Source: MIT OCW 6.061, AC Power Flow in Linear Networks, §§4 and 7

Correct answer: A. 23.0 kW

In a wye connection, line voltage is √3 times phase voltage, so each phase sees 480/√3 = 277 V. Each phase draws about 27.7 A and dissipates (480/√3)²/10 = 7.68 kW. Total P = 3 × 7.68 = 23.0 kW.

B applies the full 480 V across each phase resistor, which is the delta answer. C is only one phase. D computes the current as 480/10 = 48 A and then uses √3 × V_L × I_L, mixing line and phase quantities.

Source: MIT OCW 6.061, Polyphase Networks, §4, PDF p. 7; OpenStax University Physics Vol. 2 §15.4, Power in an AC Circuit

Correct answer: B. 800 Ω

With a = N₁/N₂ = 10, the primary voltage is 10 times the secondary voltage and the primary current is one-tenth the secondary current. So Z_primary = V₁/I₁ = (10V₂)/(I₂/10) = 100 × (V₂/I₂) = a²Z_L = 100 × 8 = 800 Ω.

A scales by a instead of a². C divides by a², which reflects a primary impedance to the secondary side. D ignores the transformer.

Source: OpenStax University Physics Vol. 2 §15.6, Transformers (Eqs. 15.21–15.22)

Correct answer: D. 3.0%

Synchronous speed n_s = 120f/P = 120 × 60/4 = 1,800 rpm. Slip s = (n_s − n)/n_s = (1,800 − 1,746)/1,800 = 0.030, or 3.0%.

A divides by the rotor speed instead of synchronous speed. B uses the 2-pole synchronous speed of 3,600 rpm. C is the ratio n/n_s, which is 1 − s.

Source: University of Utah ECE 3600, Induction Motors notes, PDF p. 1

Correct answer: B. 90.9%

Feeder loss = I²R = 60² × 0.50 = 1,800 W. Input power = delivered + loss = 18,000 + 1,800 = 19,800 W. Efficiency = 18,000/19,800 = 90.9%.

A computes 1 − loss/output, which uses the wrong denominator. C uses only one conductor's resistance (0.25 Ω), but the problem already gives the round-trip value. D treats the 30 V drop (IR) as if it were the power loss in watts.

Source: OpenStax University Physics Vol. 2 §9.5, Electrical Energy and Power (Eq. 9.13)

Correct answer: A. 354 pF

Capacitance: C = ε_rε₀A/d = 4 × 8.854 × 10⁻¹² × 0.01/(1.0 × 10⁻³) = 3.54 × 10⁻¹⁰ F = 354 pF.

B leaves out the dielectric constant (vacuum value). C and D are factor-of-ten slips, usually from converting 1 mm to 10⁻² or 10⁻⁴ m.

Source: OpenStax University Physics Vol. 2 §8.1, Capacitors and Capacitance (Eq. 8.3); OpenStax University Physics Vol. 2 §8.4, Capacitor with a Dielectric (Eq. 8.11)

Correct answer: C. Γ = 0.2, VSWR = 1.5

Γ = (Z_L − Z₀)/(Z_L + Z₀) = (75 − 50)/(75 + 50) = 0.2. VSWR = (1 + |Γ|)/(1 − |Γ|) = 1.2/0.8 = 1.5. Quick check for this resistive load above Z₀: VSWR = 75/50 = 1.5.

A flips the subtraction; for a resistive load, Γ is negative only when Z_L < Z₀. B computes 1/(1 − |Γ|), dropping the 1 + |Γ| numerator. D uses 25/50 for Γ, dividing by Z₀ instead of the sum.

Source: Steer, Fundamentals of Microwave and RF Design (LibreTexts), Ch. 3 summary, Eqs. 3.3.6 and 3.3.20

Correct answer: D. 16.3%

Closed loop: T(s) = G/(1 + G) = 16/(s² + 4s + 16). Match it to ω_n²/(s² + 2ζω_n s + ω_n²): ω_n = 4 rad/s and 2ζω_n = 4, so ζ = 0.5. Percent overshoot = 100 e^(−ζπ/√(1−ζ²)) = 100 e^(−1.814) = 16.3%.

A is the overshoot for ζ ≈ 0.707. B reports ζ as a percentage. C has no basis in the formula.

Source: University of Illinois ECE 486, Lecture 6: Overshoot and Peak Time; University of Illinois ECE 486, Lecture 7: Routh–Hurwitz criterion

Correct answer: B. 0.25

The closed-loop characteristic polynomial is s² + 5s + 20, whose roots have negative real parts, so the steady-state error limit is applicable.

G(s) has one pole at the origin, so the system is type 1. For a ramp, e_ss = 1/K_v with K_v = lim s→0 of sG(s) = 20/5 = 4. So e_ss = 1/4 = 0.25.

A is the type-1 error for a step input, not a ramp. C is the type-0 error for a ramp. D substitutes K_v = 4 into 1/(1 + K_v), but the ramp-error relationship here is 1/K_v.

Source: University of Illinois ECE 486, Lecture 9: System Type; University of Illinois ECE 486, Lecture 7: Routh–Hurwitz criterion

Correct answers: A, C

A factors as (s + 1)(s + 2): roots −1 and −2, stable.

C passes the Routh test. The first column is 1, 2, (2·3 − 1·1)/2 = 2.5, 1, with no sign changes, so all three roots are in the left half-plane.

B has a negative coefficient, which guarantees at least one right-half-plane root. D fails Routh: the s¹ entry is (1·2 − 1·8)/1 = −6, giving two sign changes and two unstable roots. E has roots ±j2 on the imaginary axis, so its roots are not strictly in the left half-plane.

Source: University of Illinois ECE 486, Lecture 7: Routh–Hurwitz criterion

Correct answer: C. 98 kHz and 102 kHz

Expand the product. The term 0.5 cos(2π·2,000t) cos(2π·100,000t) equals 0.25[cos(2π·102,000t) + cos(2π·98,000t)]. Multiplying by the carrier shifts the 2 kHz message up and down around the carrier, so the sidebands sit at 100 ± 2 kHz.

A includes the carrier, which isn't a sideband. B lists the message and carrier frequencies. D doubles the message frequency.

Source: MIT OCW 6.003 Lecture 23, Modulation Part 1, PDF p. 14

Correct answer: A. 180 kHz

Carson's rule: B ≈ 2(Δf + f_m) = 2(75 + 15) = 180 kHz.

B counts only the deviation (2Δf). C forgets the factor of 2. D counts only the message (2f_m), which is the AM-style bandwidth.

Source: Stanford EE179 Lecture 9, Carson's rule, PDF p. 4

Correct answer: D. 1.536 Mb/s

The stated sample rate is 8,000 samples/s per channel, twice the 4 kHz band limit. Each channel carries 8,000 × 8 = 64 kb/s. Twenty-four channels give 24 × 64 kb/s = 1.536 Mb/s.

A samples at 4 kHz instead of the stated 8 kHz. B is a single channel. C multiplies 24 × 8 kHz but forgets the 8 bits per sample.

Source: MIT OCW 6.003 Lecture 21, Sampling, PDF p. 60

Correct order: Physical → Data Link → Network → Transport → Session → Presentation → Application

From the bottom up: Physical (1), Data Link (2), Network (3), Transport (4), Session (5), Presentation (6), Application (7).

A useful anchor for networking questions: Ethernet switches forward frames at Layer 2, routers forward IP packets at Layer 3, and TCP and UDP sit at Layer 4.

Source: ITU-T X.200 (07/1994), §6.1.2, printed p. 28; Cisco, EtherSwitch Network Module: Layer 2 Ethernet Switching; RFC 1122, §3.2.1.3 (special addresses)

Correct answer: 62 addresses

A /26 leaves 32 − 26 = 6 host bits, which gives 2⁶ = 64 total addresses. On an ordinary subnet, the all-zeros host address identifies the network and the all-ones host address is the directed broadcast. Neither can be assigned to a host, so 64 − 2 = 62 are usable.

If you entered 64, you counted the total block. Point-to-point /31 links are the special case where both addresses can be used (RFC 3021), but that exception doesn't apply to a /26.

Source: RFC 4632 (CIDR), §3.1; RFC 1122, §3.2.1.3 (special addresses); RFC 3021, 31-bit prefixes on point-to-point links

Correct answer: B. −76

The leading 1 means the number is negative. Read it as unsigned (180), then subtract 2⁸: 180 − 256 = −76. You can also invert and add 1: 0100 1011 + 1 = 0100 1100 = 76, so the value is −76.

A is the unsigned reading. C is the magnitude without the sign. D is a sign-magnitude style misreading that treats the low seven bits (52) as the magnitude.

Source: Ward, Computation Structures notes §3.2.1.2, Two's Complement

Correct answer: A. F = A′ + BC

Minterms 0–3 are every combination with A = 0, which groups into A′. Minterm 7 (ABC) pairs with minterm 3 (A′BC) to give BC. So F = A′ + BC: two terms, three literals.

B also covers minterm 6 (ABC′), which isn't in the function. C misses minterm 2 (A′BC′). D gives the right function, but ABC can shrink to BC, so it isn't minimal.

Source: Ward, Computation Structures notes §7.6, Sum-of-Products and Simplification

Correct answer: C. 6

n flip-flops hold at most 2ⁿ states, so you need the smallest n with 2ⁿ ≥ 50. 2⁵ = 32 is too few and 2⁶ = 64 is enough, so n = 6.

A can count only 32 states. B works but isn't the minimum. D describes a one-hot design, which is valid but uses far more flip-flops than necessary.

Source: Ward, Computation Structures notes §9.2.4, FSM state count

Correct answer: D. 66.7 MHz

The clock period must cover launch delay, logic delay and setup: T_min = 3 + 10 + 2 = 15 ns. f_max = 1/15 ns = 66.7 MHz.

A ignores setup time (13 ns). B ignores clock-to-Q delay (12 ns). C counts only the logic (10 ns).

Source: Ward, Computation Structures notes §8.3.2, Adding Logic to Clocked Devices

Correct answer: 16 address lines

64 KiB = 64 × 1,024 = 65,536 bytes = 2¹⁶ locations, one per byte. N address lines select 2ᴺ locations, so N = 16.

If you got 19, you counted bits (65,536 × 8 = 2¹⁹) instead of byte locations. If you got 6, you used log₂(64) without converting KiB to bytes.

Source: University of Washington CSE 390B, Storing Data: Memory, 'Size Limitations'

Correct answer: C. 7 ns

AMAT = hit time + miss rate × miss penalty = 2 + 0.05 × 100 = 7 ns. The question defines the penalty as added time, so every access pays the 2 ns hit time and 5% of accesses pay 100 ns more.

A drops the hit time. B applies the miss rate to the hit time instead of the penalty. D is the time of a single miss, not the average.

Source: MIT OCW 6.004, Memory Hierarchy & Caches Worksheet, p. 1

Correct answer: A. 960

Each character takes 1 + 8 + 1 = 10 bit times. At 9,600 bits/s, that's 9,600/10 = 960 characters/s.

B counts only the 8 data bits. C counts 9 bits, dropping one framing bit. D confuses bits with characters.

Source: Texas Instruments SLAA083, §2, PDF p. 2 and §2.3/Fig. 3, PDF p. 5

Correct order: Θ(1) → Θ(log n) → Θ(n) → Θ(n log n) → Θ(n²) → Θ(2ⁿ)

Constant time doesn't grow at all, then logarithmic (binary search on sorted data), linear (one pass over the data), n log n (efficient comparison sorts), quadratic (nested loops over the data) and exponential (trying every subset).

Θ (big-theta) describes a tight growth bound up to constant factors; Big-O alone gives an upper bound. For small inputs a 'worse' algorithm can still run faster because constant factors matter there.

Source: NIST Dictionary of Algorithms and Data Structures: Θ (big-theta); NIST Dictionary of Algorithms and Data Structures: big-O notation; NIST DADS: binary search

Correct answers: A, B, D

A and B are the defining behaviors of a stack and a queue. D is correct because an array's index maps directly to a memory location.

C is wrong: inserting at the head only rewires the new node's link and the head pointer, which is O(1). E is wrong: binary search discards half the data at each step, which is valid only if the data are sorted.

Source: NIST DADS: stack; NIST DADS: queue; NIST DADS: linked list; NIST DADS: binary search