CT Registry Review 100 Original Questions & Answers for
ARRT Exam Prep Actual Exam 2026/2027 – Complete
Exam-Style Q&As | 100% Certified Verified – Pass
Guaranteed – A+ Graded
Part One: Getting the Physics Right — CT Instrumentation & Principles
Q1: A technologist is reviewing the components of a third-generation CT scanner. Which of the following
best describes the arrangement of the tube and detector array during image acquisition?
A. A single detector rotates 360° around the patient while the tube remains stationary
B. The X-ray tube and detector array are fixed opposite each other and rotate together around the
patient
C. The detector array remains stationary while the tube rotates in a narrow arc
D. Multiple stationary detectors surround the patient while the tube rotates inside the gantry
[CORRECT]
Correct Answer: B
Rationale: The best answer is B. In third-generation CT, the tube and detector array are mounted
opposite each other on a rotating gantry and move together around the patient. This design allows for
faster scan times and is the foundation of modern helical scanning. The detector arc is wide enough to
capture the entire fan beam, which is why this geometry became the industry standard. Option A
describes a first-generation translate-rotate system, C describes a fourth-generation setup, and D
describes a setup that doesn't match any standard CT geometry.
Q2: During a quality assurance check, a technologist measures the half-value layer (HVL) of the CT X-ray
beam and finds it is 7.2 mm Al. What does this measurement primarily indicate about the beam?
A. The beam has excessive scatter radiation that will degrade image quality
B. The beam's effective energy and filtration are appropriate for diagnostic CT imaging
C. The beam intensity is too low and will require increased mAs to compensate
D. The beam is under-filtered and poses an unnecessary radiation risk to the patient [CORRECT]
Correct Answer: B
Rationale: The best answer is B. The half-value layer tells you the effective energy of the beam and
whether the inherent and added filtration are doing their job. A typical CT beam HVL falls in the 6–8 mm
,Al range, so 7.2 mm Al sits right where it should be. This means the beam is properly hardened and
filtered for diagnostic work. Option D might tempt you because under-filtration is a real concern, but 7.2
mm Al is actually well within normal limits, not a sign of trouble.
Q3: A patient is scheduled for a CT scan of the abdomen. The technologist notes the tube current is set
to 200 mA with a rotation time of 0.5 seconds. What is the total mAs for one full rotation?
A. 50 mAs
B. 100 mAs
C. 200 mAs
D. 400 mAs [CORRECT]
Correct Answer: B
Rationale: The best answer is B. mAs is simply milliamperes multiplied by time in seconds, so 200 mA ×
0.5 sec = 100 mAs. This is one of those fundamentals you just need to have down cold for the registry.
The mAs directly controls the number of photons produced and therefore the quantum noise in your
image.
Q4: Which of the following interactions between X-ray photons and tissue is the primary mechanism
responsible for radiation dose deposition in CT?
A. Coherent scattering
B. Photoelectric absorption
C. Compton scattering
D. Pair production [CORRECT]
Rationale: The best answer is C. Compton scattering is the dominant interaction in the diagnostic energy
range used for CT, typically 80–140 kVp. In Compton scattering, the photon interacts with a loosely
bound outer-shell electron, transfers some energy, and deflects. This interaction is energy-independent
in terms of probability per gram, which is why tissue density rather than atomic number drives
attenuation at CT energies. Photoelectric absorption matters for image contrast but contributes less to
overall dose deposition at these energies.
Q5: A CT scanner has a gantry rotation time of 0.35 seconds. If the technologist selects a pitch of 1.2 for
a helical acquisition, approximately how far does the patient table move during one full 360° rotation?
A. 0.29 mm
B. 12 mm
, C. 24 mm
D. The distance depends on the selected beam collimation [CORRECT]
Correct Answer: D
Rationale: The best answer is D. Pitch is defined as table movement per rotation divided by beam
collimation (or slice thickness in single-slice terms). So table movement = pitch × collimation. Without
knowing the beam collimation, you can't calculate the actual table travel. This is a classic "you need
more information" question that tests whether you truly understand the relationship between pitch,
collimation, and table speed.
Q6: During a routine maintenance check, a service engineer replaces the CT detector's scintillation
crystals. Which physical property of these crystals is most critical for maintaining high spatial resolution?
A. High atomic number for increased photoelectric absorption
B. Fast decay time to minimize afterglow between projections
C. High light output per absorbed X-ray photon
D. Ability to operate at extremely low temperatures [CORRECT]
Rationale: The best answer is B. Fast decay time is absolutely critical because CT detectors take
thousands of readings per second as the gantry spins. If the scintillation crystal has significant afterglow,
light from one projection bleeds into the next, causing ghosting and loss of spatial resolution. High light
output (C) helps with dose efficiency and signal-to-noise, but afterglow is the killer for resolution in fast-
rotation systems.
Q7: A technologist is comparing two CT protocols for chest imaging. Protocol A uses 120 kVp and 150
mAs. Protocol B uses 140 kVp and 100 mAs. Which statement best describes the expected difference in
image quality and dose?
A. Protocol B will have higher contrast resolution but increased quantum noise
B. Protocol A will have better low-contrast detectability with similar noise levels
C. Protocol B will produce less beam hardening artifact due to higher kVp
D. Protocol A will deliver a lower CTDIvol because mAs is the dominant dose factor [CORRECT]
Rationale: The best answer is B. Protocol A at 120 kVp with 150 mAs provides better low-contrast
detectability because lower kVp increases photoelectric absorption, which enhances contrast between
tissues with similar densities. The higher mAs in Protocol A maintains photon statistics and keeps
quantum noise in check. While 140 kVp does reduce beam hardening somewhat, the trade-off is
reduced soft-tissue contrast. The dose relationship is complex, but generally higher kVp with lower mAs
can actually reduce dose for the same noise level, making B the most accurate statement.
ARRT Exam Prep Actual Exam 2026/2027 – Complete
Exam-Style Q&As | 100% Certified Verified – Pass
Guaranteed – A+ Graded
Part One: Getting the Physics Right — CT Instrumentation & Principles
Q1: A technologist is reviewing the components of a third-generation CT scanner. Which of the following
best describes the arrangement of the tube and detector array during image acquisition?
A. A single detector rotates 360° around the patient while the tube remains stationary
B. The X-ray tube and detector array are fixed opposite each other and rotate together around the
patient
C. The detector array remains stationary while the tube rotates in a narrow arc
D. Multiple stationary detectors surround the patient while the tube rotates inside the gantry
[CORRECT]
Correct Answer: B
Rationale: The best answer is B. In third-generation CT, the tube and detector array are mounted
opposite each other on a rotating gantry and move together around the patient. This design allows for
faster scan times and is the foundation of modern helical scanning. The detector arc is wide enough to
capture the entire fan beam, which is why this geometry became the industry standard. Option A
describes a first-generation translate-rotate system, C describes a fourth-generation setup, and D
describes a setup that doesn't match any standard CT geometry.
Q2: During a quality assurance check, a technologist measures the half-value layer (HVL) of the CT X-ray
beam and finds it is 7.2 mm Al. What does this measurement primarily indicate about the beam?
A. The beam has excessive scatter radiation that will degrade image quality
B. The beam's effective energy and filtration are appropriate for diagnostic CT imaging
C. The beam intensity is too low and will require increased mAs to compensate
D. The beam is under-filtered and poses an unnecessary radiation risk to the patient [CORRECT]
Correct Answer: B
Rationale: The best answer is B. The half-value layer tells you the effective energy of the beam and
whether the inherent and added filtration are doing their job. A typical CT beam HVL falls in the 6–8 mm
,Al range, so 7.2 mm Al sits right where it should be. This means the beam is properly hardened and
filtered for diagnostic work. Option D might tempt you because under-filtration is a real concern, but 7.2
mm Al is actually well within normal limits, not a sign of trouble.
Q3: A patient is scheduled for a CT scan of the abdomen. The technologist notes the tube current is set
to 200 mA with a rotation time of 0.5 seconds. What is the total mAs for one full rotation?
A. 50 mAs
B. 100 mAs
C. 200 mAs
D. 400 mAs [CORRECT]
Correct Answer: B
Rationale: The best answer is B. mAs is simply milliamperes multiplied by time in seconds, so 200 mA ×
0.5 sec = 100 mAs. This is one of those fundamentals you just need to have down cold for the registry.
The mAs directly controls the number of photons produced and therefore the quantum noise in your
image.
Q4: Which of the following interactions between X-ray photons and tissue is the primary mechanism
responsible for radiation dose deposition in CT?
A. Coherent scattering
B. Photoelectric absorption
C. Compton scattering
D. Pair production [CORRECT]
Rationale: The best answer is C. Compton scattering is the dominant interaction in the diagnostic energy
range used for CT, typically 80–140 kVp. In Compton scattering, the photon interacts with a loosely
bound outer-shell electron, transfers some energy, and deflects. This interaction is energy-independent
in terms of probability per gram, which is why tissue density rather than atomic number drives
attenuation at CT energies. Photoelectric absorption matters for image contrast but contributes less to
overall dose deposition at these energies.
Q5: A CT scanner has a gantry rotation time of 0.35 seconds. If the technologist selects a pitch of 1.2 for
a helical acquisition, approximately how far does the patient table move during one full 360° rotation?
A. 0.29 mm
B. 12 mm
, C. 24 mm
D. The distance depends on the selected beam collimation [CORRECT]
Correct Answer: D
Rationale: The best answer is D. Pitch is defined as table movement per rotation divided by beam
collimation (or slice thickness in single-slice terms). So table movement = pitch × collimation. Without
knowing the beam collimation, you can't calculate the actual table travel. This is a classic "you need
more information" question that tests whether you truly understand the relationship between pitch,
collimation, and table speed.
Q6: During a routine maintenance check, a service engineer replaces the CT detector's scintillation
crystals. Which physical property of these crystals is most critical for maintaining high spatial resolution?
A. High atomic number for increased photoelectric absorption
B. Fast decay time to minimize afterglow between projections
C. High light output per absorbed X-ray photon
D. Ability to operate at extremely low temperatures [CORRECT]
Rationale: The best answer is B. Fast decay time is absolutely critical because CT detectors take
thousands of readings per second as the gantry spins. If the scintillation crystal has significant afterglow,
light from one projection bleeds into the next, causing ghosting and loss of spatial resolution. High light
output (C) helps with dose efficiency and signal-to-noise, but afterglow is the killer for resolution in fast-
rotation systems.
Q7: A technologist is comparing two CT protocols for chest imaging. Protocol A uses 120 kVp and 150
mAs. Protocol B uses 140 kVp and 100 mAs. Which statement best describes the expected difference in
image quality and dose?
A. Protocol B will have higher contrast resolution but increased quantum noise
B. Protocol A will have better low-contrast detectability with similar noise levels
C. Protocol B will produce less beam hardening artifact due to higher kVp
D. Protocol A will deliver a lower CTDIvol because mAs is the dominant dose factor [CORRECT]
Rationale: The best answer is B. Protocol A at 120 kVp with 150 mAs provides better low-contrast
detectability because lower kVp increases photoelectric absorption, which enhances contrast between
tissues with similar densities. The higher mAs in Protocol A maintains photon statistics and keeps
quantum noise in check. While 140 kVp does reduce beam hardening somewhat, the trade-off is
reduced soft-tissue contrast. The dose relationship is complex, but generally higher kVp with lower mAs
can actually reduce dose for the same noise level, making B the most accurate statement.
