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Free CT practice questions.
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Try 15 questions on CT acquisition, image quality, dose, and patient care. Choose your answer, then use the explanation to check your reasoning.

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15 questions.
Clear explanations.

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  • Explanations for all four choices
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A quick note before you start

Take your time and choose one answer per question. You can change your choices before checking the set. For calculations, use the conditions stated in the question; results can differ with the scanner and reconstruction method.

This is a short review, not a full ARRT CT practice exam or a prediction of your score. The questions are independently written, not actual or recalled ARRT items. MRITechStudio is not affiliated with ARRT. For patient care, follow approved protocols and your clinical team.

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Your practice set

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Question 1 / 15 · Acquisition & displayOn the conventional Hounsfield scale, what CT number is assigned to water?
Answer and explanation

Answer B: 0 HU

Water defines 0 HU; air is approximately −1,000 HU. A real water-phantom measurement may vary slightly and is checked against the applicable QC tolerance. A single HU value does not identify a tissue by itself.

How the choices compare

  • A. Approximately −1,000 HU is the air reference.
  • B. Water is the reference at 0 HU.
  • C. +100 HU is above the water reference.
  • D. +1,000 HU is far above the water reference; it is not the assigned value for water.
Question 2 / 15 · Image quality & artifactsIn a quantum-noise-dominated CT acquisition reconstructed with filtered back projection, mAs increases from 100 to 400. Patient size, kVp, pitch, section thickness and reconstruction settings stay fixed. What is the approximate image-noise ratio, new/original?
Answer and explanation

Answer D: 0.50

Under these conditions, noise is approximately proportional to 1/√mAs. The ratio is √(100/400) = 0.50. This comparison concerns noise magnitude, not a guaranteed change in diagnostic performance; nonlinear reconstruction may behave differently.

How the choices compare

  • A. 0.25 treats the relationship as inverse-linear instead of inverse-square-root.
  • B. About 0.71 would follow a doubling of mAs.
  • C. 2.00 reverses the direction: more photons reduce quantum noise.
  • D. A fourfold increase in mAs halves the modeled noise.
Question 3 / 15 · Dose & patient careWhich statement best describes the CTDIvol displayed for a CT acquisition?
Answer and explanation

Answer A: A standardized phantom-based scanner-output index

CTDIvol is a standardized scanner-output index reported in mGy, based on a specified reference phantom. It does not directly measure an individual patient’s organ dose. Record the reference phantom when comparing values.

How the choices compare

  • A. This describes the role and limitation of CTDIvol.
  • B. Organ dose also depends on patient anatomy, size and irradiation geometry.
  • C. CTDIvol is reported in mGy, not effective-dose units of mSv.
  • D. Patient-size correction belongs to SSDE, not unadjusted CTDIvol.
Question 4 / 15 · Acquisition & displayWith standard linear CT windowing, you narrow the window width while keeping the level fixed. For CT numbers that remain inside the window, what changes?
Answer and explanation

Answer C: The same HU difference produces a larger displayed brightness difference.

A narrower window maps a smaller HU range across the available grayscale, increasing displayed contrast within that range. Values outside it saturate at the display limits. Windowing changes presentation, not the acquired data or spatial resolution.

How the choices compare

  • A. Window level sets the center and was held fixed.
  • B. Windowing changes the display mapping, not the stored CT numbers.
  • C. A narrower window makes HU differences within the window more visually distinct.
  • D. Narrowing the window reduces the HU range between the display limits.
Question 5 / 15 · Image quality & artifactsThe acquired CT data support 1 mm sections. You reconstruct them at 1 mm instead of 5 mm, using filtered back projection with the same display FOV, matrix and kernel. Which trade-off is most likely?
Answer and explanation

Answer C: Less through-plane averaging and more image noise

Thinner reconstructed sections reduce averaging along the section direction but generally contain more noise at fixed acquisition and reconstruction conditions. Producing a thinner reconstruction from existing data does not itself add an x-ray exposure. Actual results depend on the available acquired data and reconstruction method.

How the choices compare

  • A. The averaging direction is right, but thinner sections generally increase noise in this comparison.
  • B. This better describes increasing, rather than decreasing, section thickness.
  • C. This captures the through-plane detail versus noise trade-off.
  • D. The noise direction is right, but through-plane averaging decreases.
Question 6 / 15 · Dose & patient careFor one uniform acquisition, CTDIvol is 10 mGy and the scan length used in the DLP calculation is 30 cm. What is the DLP?
Answer and explanation

Answer B: 300 mGy·cm

DLP = CTDIvol × scan length = 10 mGy × 30 cm = 300 mGy·cm. This is a scanner-output metric, not an organ-dose measurement. For a multiphase examination, the reported total can include multiple acquisitions; this question describes only one.

How the choices compare

  • A. Dividing 10 by 30 does not calculate DLP.
  • B. Multiplying the index by length gives 300 mGy·cm.
  • C. Adding the two numbers is not dimensionally valid.
  • D. Dividing length by CTDIvol uses the wrong relationship and units.
Question 7 / 15 · Acquisition & displayA reconstructed CT image has a 256 mm display FOV and a 512 × 512 matrix. What is its nominal in-plane pixel width?
Answer and explanation

Answer D: 0.50 mm

Pixel width = display FOV ÷ matrix width = 256/512 = 0.50 mm. This is the sampling dimension in the reconstructed image. Actual spatial resolution also depends on the acquisition, focal spot, kernel and other system factors.

How the choices compare

  • A. 0.25 mm would require 1,024 pixels across this FOV.
  • B. 1.00 mm would require 256 pixels across this FOV.
  • C. 2.00 mm reverses the numerical ratio and does not give the pixel width.
  • D. 256 mm divided by 512 pixels gives 0.50 mm per pixel.
Question 8 / 15 · Image quality & artifactsUsing filtered back projection, you reconstruct the same acquired data with a sharper kernel instead of a smoother one. Section thickness, FOV and matrix stay fixed. What is the usual trade-off?
Answer and explanation

Answer A: More visible noise, with stronger edge definition

A sharper kernel emphasizes higher spatial frequencies, which can improve edge definition while increasing visible noise. A smoother kernel suppresses noise but softens edges. Kernel choice should match the diagnostic task; sharper is not automatically better.

How the choices compare

  • A. This is the usual sharp-kernel trade-off under the stated conditions.
  • B. This assumes a free gain in both noise and edge detail.
  • C. Noise generally increases rather than decreases with a sharper kernel.
  • D. A sharper kernel emphasizes, rather than suppresses, edges.
Question 9 / 15 · Dose & patient careCompared with CTDIvol alone, what additional information is incorporated when calculating a size-specific dose estimate (SSDE)?
Answer and explanation

Answer C: A patient-size metric and the appropriate conversion factor

SSDE combines CTDIvol with a size-dependent conversion factor appropriate to the reference phantom and method. Water-equivalent diameter can account for attenuation as well as physical size. SSDE remains an estimate; it is not an exact organ dose, effective dose or individual cancer-risk prediction.

How the choices compare

  • A. Length and phase count affect examination output, but do not supply the patient-size correction for SSDE.
  • B. Image count and reconstruction interval do not supply the patient-size correction.
  • C. Patient size and a matching conversion method distinguish SSDE from CTDIvol alone.
  • D. Effective dose and SSDE are different quantities; organ-weighting factors do not calculate SSDE.
Question 10 / 15 · Acquisition & displayIn helical CT, the table travels 60 mm during one rotation and the total nominal collimated beam width is 40 mm. What is the pitch?
Answer and explanation

Answer A: 1.50

Pitch is table travel per rotation divided by total nominal collimated beam width: 60/40 = 1.50. The denominator is the total beam width, not one detector row or a reconstructed section. Pitch alone does not predict patient dose when automatic exposure controls change tube output.

How the choices compare

  • A. 60 mm divided by 40 mm gives a dimensionless pitch of 1.50.
  • B. 0.67 is the inverted ratio, 40/60.
  • C. 1.00 would mean table travel equals the total beam width.
  • D. 0.50 is not the ratio of the stated travel to beam width.
Question 11 / 15 · Image quality & artifactsWhat primarily causes a polychromatic x-ray beam to harden as it passes through the patient?
Answer and explanation

Answer B: Lower-energy photons are preferentially removed from the transmitted beam.

Lower-energy photons are attenuated more strongly, so the transmitted spectrum has a higher mean energy. Reconstruction errors from this spectral change can produce cupping or dark bands. Metal streaks can also involve photon starvation and other mechanisms; not every streak is explained by beam hardening alone.

How the choices compare

  • A. Scatter can degrade CT data, but adding lower-energy scatter does not explain beam hardening.
  • B. Preferential loss of lower-energy photons raises the transmitted beam’s mean energy.
  • C. Equal proportional losses would preserve the relative energy distribution.
  • D. The patient changes the transmitted spectrum; beam hardening does not require a rise in tube voltage or maximum photon energy.
Question 12 / 15 · Dose & patient careBefore IV iodinated contrast, which history is the strongest specific predictor of another allergic-like reaction to that same class of contrast?
Answer and explanation

Answer D: A previous allergic-like reaction to iodinated contrast

A previous allergic-like or unknown-type reaction to the same contrast class is the most important specific risk factor for a future allergic-like reaction. Nausea alone is generally physiologic, and shellfish allergy has no unique relationship to iodinated contrast. Clarify and document the earlier event and follow the facility’s assessment process; this answer does not determine whether contrast should be given.

How the choices compare

  • A. An unrelated allergy may increase risk modestly but is not the strongest specific predictor listed.
  • B. Shellfish allergy does not have a special iodine-based cross-reactivity with contrast.
  • C. Nausea alone is usually a physiologic reaction rather than an allergic-like one.
  • D. A prior same-class allergic-like reaction is the key history among these choices.
Question 13 / 15 · Acquisition & displayWhich approach describes bolus tracking for a contrast-enhanced CT acquisition?
Answer and explanation

Answer C: Monitoring attenuation in a vessel ROI to time the scan

Bolus tracking monitors contrast-related attenuation change in a selected ROI, commonly in a vessel. A protocol-specific threshold and additional delay help time the diagnostic acquisition. ROI placement, trigger value and delay depend on the examination; no single threshold works for every study or patient.

How the choices compare

  • A. A fixed delay is a different scan-timing approach; it does not monitor bolus arrival in an ROI.
  • B. Test-bolus timing uses a separate small injection and is a different approach from bolus tracking.
  • C. Attenuation in the monitoring ROI is the relevant measurement.
  • D. ECG gating coordinates acquisition with cardiac motion rather than monitoring contrast arrival.
Question 14 / 15 · Image quality & artifactsA thin circular band centered on the scanner’s rotation axis persists on repeated scans of a uniform water phantom. Which equipment check best matches this pattern?
Answer and explanation

Answer A: Check detector-channel response and calibration.

A ring that persists in a uniform phantom points toward a scanner-related detector response or calibration issue. Follow the manufacturer’s QC and service process and the facility’s escalation policy. Repeating a patient scan before investigating the equipment problem may add exposure without solving it.

How the choices compare

  • A. The persistent scanner-centered pattern supports investigating detector response and calibration.
  • B. Table-drive errors can affect helical sampling but are not the characteristic first explanation for this circular band.
  • C. An enlarged focal spot mainly degrades edge sharpness rather than producing a stable circular band.
  • D. Tube arcing more typically produces streaks from inconsistent projection data, rather than a stable detector-centered ring.
Question 15 / 15 · Dose & patient careDuring a power injection of IV iodinated contrast, the patient reports pain and new swelling appears around the IV site. What is the best immediate response?
Answer and explanation

Answer B: Stop the injection and promptly arrange assessment of the site and symptoms.

Stop the injection when extravasation is suspected and obtain prompt evaluation according to the facility’s protocol. Assessment includes pain, swelling and neurovascular findings. Further care and escalation depend on symptoms and examination, not a single universal volume threshold. This short question is not a complete management protocol.

How the choices compare

  • A. Slowing the injection can still deliver more contrast into the tissues.
  • B. Stopping limits further delivery into the tissues and allows prompt assessment.
  • C. Symptoms and examination guide care; a small estimated volume does not remove the need for assessment.
  • D. Flushing a potentially infiltrated IV can deliver more fluid into the tissues and delay assessment.
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Turn a missed answer into a useful next step

Acquisition and display

Separate what the scanner acquires from how you reconstruct and display it. Pixel size is not the same as measured resolution; windowing changes the displayed grayscale, not the stored CT numbers. Pitch uses the total collimated beam width.

Dose and patient care

CTDIvol describes standardized scanner output. DLP also accounts for length, while SSDE adds a patient-size correction. None is an exact organ-dose or personal risk measurement. For contrast questions, review the patient’s actual reaction history and the facility’s assessment and escalation procedures.

For image-quality questions, revisit the CT artifact guide. To plan a focused review session, try turning missed questions into a study plan. Studying MRI too? Try the free MRI practice questions.