ARRT Radiography Practice Questions (2026) — Free Sample Questions with Answers

Answer: A

Estimated glomerular filtration rate (eGFR): Correct. eGFR, calculated from serum creatinine, is the standard measure of renal function used to screen patients before intravenous iodinated contrast.

Hemoglobin and hematocrit (H&H) levels: These assess anemia and blood loss; they do not measure kidney function.

Prothrombin time and international normalized ratio: PT/INR evaluates clotting and is reviewed before invasive procedures such as biopsies, not for contrast nephrotoxicity risk.

Serum glucose concentration: Glucose is relevant to diabetes management (e.g., metformin questions) but is not a measure of renal function.

Renal function before iodinated contrast is judged by creatinine/eGFR (BUN is supporting); low eGFR increases risk.

Answer: B

2 mGy/h: Halving the intensity assumes a linear relationship with distance; intensity actually falls with the square of the distance.

1 mGy/h: Correct. Inverse square law: I₂ = I₁ × (D₁/D₂)² = 4 × (1/2)² = 4 × 1/4 = 1 mGy/h.

8 mGy/h: Intensity decreases, not increases, as distance from the source increases.

16 mGy/h: This multiplies by the square of the distance ratio instead of dividing.

Inverse square law: I₁/I₂ = D₂²/D₁². Doubling distance cuts intensity to one quarter.

Answer: C

15 mAs: 15 mAs uses a conversion factor of 3, which corresponds to a 6:1 grid.

20 mAs: 20 mAs uses a conversion factor of 4, which corresponds to an 8:1 grid.

25 mAs: Correct. The grid conversion factor for a 12:1 grid is 5, so 5 mAs × 5 = 25 mAs.

30 mAs: 30 mAs uses a conversion factor of 6, which corresponds to a 16:1 grid.

GCFs: no grid 1, 5:1 = 2, 6:1 = 3, 8:1 = 4, 12:1 = 5, 16:1 = 6. New mAs = old mAs × (GCF new ÷ GCF old).

Answer: A

10 mAs: Correct. 92 kVp is a 15% increase over 80 kVp (80 × 1.15 = 92), which roughly doubles receptor exposure, so mAs is halved: 20 ÷ 2 = 10 mAs.

20 mAs: Leaving mAs unchanged after a 15% kVp increase would roughly double the exposure to the receptor.

40 mAs: Doubling mAs together with a 15% kVp increase would quadruple receptor exposure.

17 mAs: This reduces mAs by 15% — misapplying the rule, which calls for halving the mAs.

15% rule: +15% kVp ≈ 2× receptor exposure → halve mAs; −15% kVp → double mAs.

Answer: B

AP projection in the right lateral decubitus position: With the right side down, free air rises to the left side, where it can be confused with gas in the stomach and colon.

AP projection in the left lateral decubitus position: Correct. With the left side down, free air rises to the right side and is seen against the homogeneous liver shadow; the patient should stay in position several minutes before exposure.

AP projection with the patient supine: On a supine AP the free air collects anteriorly, so the horizontal beam needed to show air-fluid levels is absent.

PA projection with the patient prone: A prone PA uses a vertical beam, which cannot show air-fluid levels or free air reliably.

Free air: upright or left lateral decubitus with a horizontal beam; air outlines the liver on the right side.

Answer: A

To move the scapulae out of the lung fields: Correct. Rolling the shoulders forward rotates the scapulae laterally so they do not overlap the lung fields.

To raise the diaphragm for a deeper inspiration: Arm position does not raise the diaphragm; full inspiration lowers it.

To bring the clavicles above the lung apices: Shoulders should be depressed so the clavicles move below, not above, the apices.

To rotate the heart into a true lateral position: The PA chest is a frontal projection; the shoulder roll does not rotate the heart.

PA chest: 72 in (180 cm) SID, shoulders rolled forward and depressed, exposure on second full inspiration.