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Key Facts: GASM REO-1 Exam
STCW IV/2
International Standard
IMO STCW Code Table A-IV/2
70 / 100
Passing Score
GASM Examination Regulation
120 Min
Theory Exam Duration
GASM Examination Guidelines
5 Years
Certificate Validity
STCW Revalidation Standard
At-Sea Maint.
SOLAS Availability Role
SOLAS IV/15 At-Sea Electronic Maintenance
100 MCQs
Study Bank Questions
OpenExamPrep Practice Bank
The Turkish GASM 1. Sınıf Telsiz Elektronik Operatörü (REO-1) examination certifies maritime electronic engineers and radio operators for at-sea maintenance on SOLAS vessels worldwide under STCW Regulation IV/2 and ITU Radio Regulations. The official examination combines a computer-based electronic engineering and radio theory written paper with a practical hardware bench testing and troubleshooting assessment, both requiring a minimum score of 70/100 to pass. OpenExamPrep provides 100 independent English-language practice questions with comprehensive step-by-step calculations and technical rationales.
Sample GASM REO-1 Practice Questions
Try these sample questions to review concepts for the GASM REO-1 exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1A DC power supply delivers 24.0 V to a solid-state driver stage in a marine HF transmitter. When keyed in continuous wave mode, the stage draws a constant current of 4.80 A. What is the equivalent DC input resistance of this amplifier stage?
A.5.00 ohms
B.0.20 ohms
C.115.2 ohms
D.19.2 ohms
Explanation: According to Ohm's Law, DC resistance is calculated by dividing voltage by current: R = V / I. Substituting the given operational parameters yields R = 24.0 V / 4.80 A = 5.00 ohms. This low input resistance is typical for high-power bipolar or power MOSFET RF driver circuits.
2An RF choke installed in the DC bias supply line of a maritime HF transceiver has an inductance of 15.0 microhenries (μH). What is the inductive reactance (XL) of this choke at an operating frequency of 10.0 MHz?
A.150 ohms
B.942 ohms
C.1,500 ohms
D.5,920 ohms
Explanation: Inductive reactance is calculated using the formula XL = 2 * pi * f * L. Substituting 10.0 MHz (10^7 Hz) and 15.0 microhenries (15 * 10^-6 H) yields XL = 2 * 3.14159 * 10^7 * 15 * 10^-6 = 2 * 3.14159 * 150 ≈ 942.5 ohms. This high impedance prevents high-frequency RF signals from leaking into the DC power distribution rail.
3An RF inter-stage coupling capacitor has a capacitance of 100 picofarads (pF). What is its capacitive reactance (XC) at the marine HF band frequency of 8.00 MHz?
A.199 ohms
B.80 ohms
C.1,256 ohms
D.50 ohms
Explanation: Capacitive reactance is calculated as XC = 1 / (2 * pi * f * C). Converting units gives f = 8.0 * 10^6 Hz and C = 100 * 10^-12 F, resulting in XC = 1 / (2 * 3.14159 * 8.0 * 10^6 * 100 * 10^-12) = 1 / (5.0265 * 10^-3) ≈ 198.9 ohms. This provides a low-reactance path for the RF signal while completely blocking DC bias voltages.
4An antenna matching network contains a series circuit comprising a pure resistance R = 30.0 ohms, an inductive reactance XL = 90.0 ohms, and a capacitive reactance XC = 50.0 ohms at the operating frequency. What is the total impedance (Z) of this series circuit?
A.170 ohms
B.70.0 ohms
C.50.0 ohms
D.40.0 ohms
Explanation: In a series AC circuit, total impedance is given by Z = sqrt(R^2 + (XL - XC)^2). First compute the net reactive component: X_net = XL - XC = 90.0 - 50.0 = 40.0 ohms (inductive). Next compute Z = sqrt(30.0^2 + 40.0^2) = sqrt(900 + 1600) = sqrt(2500) = 50.0 ohms.
5An RF preamplifier in a maritime satellite communication terminal accepts an input signal power of 2.50 milliwatts (mW) and delivers an output power of 250 milliwatts (mW). What is the power gain of this amplifier expressed in decibels (dB)?
A.10.0 dB
B.20.0 dB
C.40.0 dB
D.100 dB
Explanation: Power gain in decibels is calculated using the formula Gain (dB) = 10 * log10(P_out / P_in). The power ratio is 250 mW / 2.50 mW = 100. Taking the base-10 logarithm gives log10(100) = 2, and multiplying by 10 yields 10 * 2 = 20.0 dB.
6A 30-meter run of RG-213 coaxial transmission line connects a marine VHF transceiver to the bridge mast antenna. The total cable insertion loss is measured at 3.00 dB at 156.8 MHz. If the transmitter delivers 25.0 W of RF power into the transmission line, how much power reaches the antenna input terminals?
A.8.33 W
B.12.5 W
C.18.5 W
D.22.0 W
Explanation: An attenuation of 3 dB represents a reduction of signal power by half (10^(-3/10) = 10^(-0.3) ≈ 0.5012). Therefore, half of the 25.0 W transmitter output reaches the antenna terminals: 25.0 W * 0.50 = 12.5 W. The remaining 12.5 W is dissipated as heat within the dielectric and conductors of the coaxial cable.
7A coastal station VHF maritime transmitter feeds 50.0 W of RF power to an omnidirectional antenna array that provides a directive gain of +6.00 dBd (decibels relative to a half-wave dipole). Neglecting transmission line losses, what is the Effective Radiated Power (ERP) of this installation?
A.100 W
B.200 W
C.300 W
D.400 W
Explanation: Effective Radiated Power (ERP) is the transmitter power multiplied by the numeric antenna gain relative to a half-wave dipole. A gain of +6 dB corresponds to a power multiplication factor of 10^(6/10) = 10^0.6 ≈ 3.981 (commonly approximated as 4 in RF engineering, since +3 dB doubles power twice: 50 W * 2 * 2 = 200 W). Thus, ERP = 50.0 W * 4 = 200 W.
8What is the physical length of an unshortened quarter-wave (λ/4) whip antenna resonant at the international maritime VHF distress frequency of 156.800 MHz (VHF Channel 16), assuming an end-effect velocity factor of 0.950 and free-space wave velocity c = 3.00 * 10^8 m/s?
A.45.4 cm
B.47.8 cm
C.90.9 cm
D.191.3 cm
Explanation: First calculate the free-space wavelength: λ = c / f = (3.00 * 10^8 m/s) / (156.800 * 10^6 Hz) ≈ 1.9133 m (191.33 cm). The electrical quarter-wavelength is λ / 4 = 1.9133 / 4 ≈ 0.4783 m (47.83 cm). Multiplying by the velocity factor accounts for capacitive end effects: Length = 47.83 cm * 0.950 ≈ 45.4 cm (0.454 m).
9A parallel LC tank circuit in the preselector filter of a marine HF receiver uses an inductor L = 10.0 μH and a tuning capacitor C = 100 pF. What is the resonant frequency (f0) of this circuit?
A.1.59 MHz
B.5.03 MHz
C.10.0 MHz
D.15.9 MHz
Explanation: Resonant frequency is determined by Thomson's formula: f0 = 1 / (2 * pi * sqrt(L * C)). Substituting L = 10.0 * 10^-6 H and C = 100 * 10^-12 F gives L * C = 1.00 * 10^-15 s^2. Taking the square root yields sqrt(L * C) = 3.1623 * 10^-8 s. Thus, f0 = 1 / (2 * 3.14159 * 3.1623 * 10^-8) = 1 / (1.9869 * 10^-7) ≈ 5,033,000 Hz = 5.03 MHz.
10An inline directional RF wattmeter inserted between a marine MF transmitter and an antenna tuning unit indicates 100 W forward power (Pfwd) and 4.00 W reflected power (Prefl). What is the Voltage Standing Wave Ratio (VSWR) on the transmission line?
A.1.22:1
B.1.50:1
C.2.00:1
D.4.00:1
Explanation: The voltage reflection coefficient is given by Γ = sqrt(Prefl / Pfwd) = sqrt(4.00 / 100) = sqrt(0.04) = 0.20. The VSWR is calculated as VSWR = (1 + Γ) / (1 - Γ) = (1 + 0.20) / (1 - 0.20) = 1.20 / 0.80 = 1.50:1. A VSWR of 1.5:1 indicates a well-matched antenna system safely within operating limits.
About the GASM REO-1 Practice Questions
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