NBRC TMC Practice Questions 2026: 8 Exam-Style Items Explained

Answer: D — Check the probe site and signal, and compare the oximeter pulse rate with the palpated or ECG heart rate

The reading does not fit the patient: an SpO2 of 82% would usually come with tachypnea, distress or cyanosis. When a monitor disagrees with the bedside picture, troubleshoot the device first: probe position and fit, motion, a cold or poorly perfused finger, nail polish and ambient light, and whether the oximeter's pulse rate matches the patient's real heart rate. If a good-quality signal still reads low, confirm with an arterial sample.

• A. This treats a number nobody has validated. Oxygen will not fix a misplaced or poorly perfused probe, and it hides the real problem.
• B. An ABG is the confirming test if a verified signal stays low, but an invasive, painful test is not the first step when a quick equipment check may explain the reading.
• C. An SpO2 of 82% cannot simply be filed away; it has to be either explained as artifact or confirmed before it is dismissed.

Answer: B — Pulsus paradoxus

Pulsus paradoxus is an exaggerated inspiratory fall in systolic pressure, defined as more than 10 mm Hg. In severe asthma, large negative pleural pressure swings on inspiration increase left-ventricular afterload and cut stroke volume. A 20 mm Hg fall marks a severe exacerbation (the same sign appears in cardiac tamponade).

• A. Pulsus alternans is a beat-to-beat alternation of strong and weak pulses that does not follow breathing; it signals severe left-ventricular failure.
• C. Orthostatic hypotension is a fall in blood pressure on standing up. It follows posture, not each breath.
• D. Pulsus parvus et tardus is a weak, late-peaking pulse caused by aortic stenosis; it does not vary with the respiratory cycle.

Answer: A — Compensated (chronic) respiratory acidosis

PaCO2 and HCO3- are both high, so one is the primary problem and the other is compensation. A pH of 7.38 sits on the acid side of 7.40, which points to the acid-producing value, the high PaCO2, as primary. A normal pH with a markedly raised HCO3- means the kidneys have had days to compensate: the classic picture of chronic CO2 retention in COPD.

• B. An acute rise of PaCO2 to 68 mm Hg would drop the pH to about 7.18-7.20 and raise HCO3- by only about 3 mEq/L. This pH is normal.
• C. In a primary metabolic alkalosis the pH would sit on the alkaline side of 7.40, and respiratory compensation rarely pushes PaCO2 much above about 55 mm Hg.
• D. An acute metabolic alkalosis would have an alkalemic pH above 7.45; this pH is 7.38.

Answer: B — Release the ulnar artery and watch for the palm to flush

The test asks whether the ulnar artery alone can perfuse the hand if the radial artery is damaged by the puncture. Keep the radial artery compressed and release the ulnar: color returning within about 5-10 seconds (some references allow up to 15) indicates adequate collateral flow. Delayed or absent flushing means choosing another site.

• A. Releasing the radial artery tests the radial supply, the very vessel about to be punctured, so it tells you nothing about collateral circulation.
• C. Releasing both arteries lets either one refill the hand, so the result cannot isolate ulnar flow.
• D. Blanching is the expected effect of occluding both arteries. It is the midpoint of the test, not its result.

Answer: D — Gown and gloves on entry, and hand-washing with soap and water

C. difficile spreads by contact with spores from feces and contaminated surfaces, so Contact Precautions apply: gown and gloves on room entry (IDSA/SHEA 2017, strong recommendation). Alcohol does not kill spores; soap-and-water washing physically removes them and is preferred after contact with fecal contamination and in outbreak or hyperendemic settings, which is why exams key it for C. diff.

• A. Gloves without a gown let clothing carry spores to the next patient, and alcohol rub does not inactivate C. difficile spores.
• B. C. difficile is not airborne. An N95 and a negative-pressure room are for diseases such as tuberculosis, measles and varicella.
• C. A mask adds nothing against a contact-spread organism, the gown is missing, and alcohol rub is the weaker choice against spores.

Answer: C — Increase the PEEP

Ventilation is normal (pH and PaCO2 are normal); the problem is oxygenation. When the FiO2 is already above about 0.50-0.60 and the PaO2 is still under 60 mm Hg, the next step is to raise PEEP. It recruits collapsed alveoli and increases FRC, treating shunt-type hypoxemia without adding to the risk of oxygen toxicity.

• A. A higher rate increases minute ventilation, drives PaCO2 below normal into respiratory alkalosis, and does nothing for oxygenation.
• B. More oxygen may nudge the PaO2 up, but above an FiO2 of 0.5-0.6 the risk of oxygen toxicity and absorption atelectasis climbs, and shunt responds poorly to FiO2 alone. PEEP is the preferred next step.
• D. A PaO2 of 53 mm Hg (SpO2 roughly 87%) is below the usual target of at least 60 mm Hg, so leaving the settings alone is not acceptable.

Answer: A — Endotracheal tube cuff pressure will rise as the aircraft climbs, so it must be checked and adjusted

Boyle's law: as altitude rises, barometric pressure falls and any trapped gas expands. An air-filled ETT cuff can over-inflate during ascent and injure the tracheal mucosa, so cuff pressure is monitored and air removed (some services fill cuffs with saline for flight), then re-checked on descent. The same law explains why each distractor has the physics backwards.

• B. Falling barometric pressure lowers inspired and alveolar PO2, so oxygen needs rise, not fall, as an unpressurized aircraft climbs.
• C. Trapped gas expands on ascent, not descent. An untreated pneumothorax is most dangerous as the aircraft climbs, which is why it is decompressed before flight.
• D. Helicopters are noisier, more cramped and vibrate more, so auscultation and monitoring are harder there than in most fixed-wing aircraft; the statement is reversed.

Answer: C — Increase the set rate to about 16/min

This is an acute respiratory acidosis with adequate oxygenation, so the fix is more alveolar ventilation. The VT is already at the top of the lung-protective range (4-8 mL/kg PBW), so the rate is the lever to pull. New rate = current rate x current PaCO2 / desired PaCO2 = 12 x 54 / 40, or about 16 breaths/min.

• A. A PaO2 of 88 mm Hg is already adequate, and FiO2 changes oxygenation, not CO2 removal.
• B. PEEP is an oxygenation tool; it does not lower PaCO2, and oxygenation is already fine.
• D. 10 mL/kg PBW exceeds the 4-8 mL/kg lung-protective range and would raise plateau pressure and the risk of ventilator-induced lung injury. Rate is the safer lever.