NBRC TMC Practice Questions (2026) — Free Sample Questions with Answers

6 free NBRC TMC questions from our 25-question bank, grouped by content domain. Each answer explains why the right option is right — and why every wrong option is wrong.

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Patient Data

Question 1

Shortly after placement of a left subclavian central line, a patient becomes acutely short of breath. The left chest has diminished breath sounds and is hyperresonant to percussion, and the trachea is shifted to the right. What is the most likely cause?

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Answer: A

Left-sided pneumothorax: Correct. Hyperresonance, absent or diminished breath sounds and tracheal shift away from the affected side after subclavian puncture indicate a (possibly tension) pneumothorax.

Left-sided pleural effusion: An effusion is dull, not hyperresonant, to percussion.

Left lower lobe consolidation: Consolidation is dull to percussion and usually produces bronchial breath sounds without tracheal shift.

Right upper lobe atelectasis: Atelectasis pulls the trachea toward the affected side and is dull to percussion.

Percussion: hyperresonant = air (pneumothorax, hyperinflation); dull = fluid or solid tissue (effusion, consolidation, atelectasis). Tension pneumothorax pushes the trachea away.

Question 2

A drowsy patient who received a large opioid dose has these arterial values on room air: pH 7.25, PaCO2 60 mm Hg, HCO3− 26 mEq/L, PaO2 58 mm Hg. Which interpretation is correct?

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Answer: B

Partially compensated respiratory acidosis: Partial compensation would show a bicarbonate clearly above the expected acute rise (about 30 mEq/L or more for this PaCO2); 26 mEq/L reflects only the acute buffer effect.

Acute respiratory acidosis with hypoxemia: Correct. pH is low, PaCO2 is high, and HCO3− rose only about 1 mEq/L per 10 mm Hg PaCO2, as expected acutely; PaO2 58 indicates hypoxemia from hypoventilation.

Metabolic acidosis with respiratory compensation: Metabolic acidosis requires a low HCO3−, and compensation would lower, not raise, the PaCO2.

Combined respiratory and metabolic acidosis: A combined disorder would show a bicarbonate below normal (below about 22 mEq/L); here HCO3− is normal.

Acute respiratory acidosis: HCO3− rises about 1 mEq/L for each 10 mm Hg rise in PaCO2; chronic compensation raises it about 3.5–4 mEq/L per 10 mm Hg.

Troubleshooting and Quality Control of Devices, and Infection Control

Question 3

A therapist must give a nebulizer treatment to a patient with suspected active pulmonary tuberculosis. Which precautions are required?

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Answer: A

N95 respirator and airborne isolation room: Correct. TB is spread by droplet nuclei; airborne precautions require a negative-pressure airborne infection isolation room and an N95 or higher respirator.

Surgical mask within three feet, private room: Droplet precautions do not protect against airborne droplet nuclei.

Gown and gloves, standard private room: Contact precautions address organisms spread by touch, not airborne transmission.

Standard precautions only, door left open: Standard precautions alone are inadequate, and an open door defeats isolation.

Airborne diseases (TB, measles, varicella) = N95 + negative-pressure room; droplet = surgical mask; contact = gown and gloves.

Question 4

An E cylinder of oxygen (factor 0.28 L/psi) reads 1,100 psi. A patient will use it at 4 L/min. Approximately how long will the cylinder last until empty?

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Answer: C

308 minutes: 308 is the volume remaining in liters (1,100 × 0.28); it still must be divided by the flow.

275 minutes: This divides psi by flow and leaves out the cylinder factor.

77 minutes: Correct. Duration = (psi × factor) / flow = (1,100 × 0.28) / 4 = 308 L ÷ 4 L/min = 77 minutes.

863 minutes: This uses the H/K cylinder factor (3.14) instead of the E cylinder factor.

Cylinder duration (min) = gauge pressure (psi) × cylinder factor ÷ flow (L/min). Factors: E = 0.28, H/K = 3.14.

Initiation and Modification of Interventions

Question 5

A patient on volume control ventilation (VT 500 mL, rate 12/min, no spontaneous breaths) has a PaCO2 of 60 mm Hg. The physician wants a PaCO2 of 45 mm Hg without changing tidal volume. What rate should be set?

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Answer: A

16/min: Correct. New rate = (current rate × current PaCO2) / desired PaCO2 = (12 × 60) / 45 = 16/min.

9/min: This inverts the ratio (12 × 45 / 60); lowering the rate would raise PaCO2 further.

20/min: 20/min overshoots, predicting a PaCO2 of about 36 mm Hg.

14/min: 14/min is too small a change and would leave PaCO2 near 51 mm Hg.

PaCO2 is inversely proportional to alveolar ventilation: new rate (or VE) = current rate × current PaCO2 ÷ desired PaCO2.

Question 6

A 70-inch (178 cm) tall man weighing 100 kg is intubated for ARDS. Using the ARDSNet low tidal volume strategy, what initial tidal volume is appropriate?

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Answer: B

About 600 mL: This uses actual body weight (100 kg × 6 mL/kg); lung size relates to height, not weight.

About 440 mL: Correct. Male PBW = 50 + 2.3 × (70 − 60) = 73 kg; 6 mL/kg × 73 kg ≈ 440 mL.

About 585 mL: This is 8 mL/kg PBW, the upper starting value rather than the 6 mL/kg target.

About 730 mL: 10 mL/kg PBW is a traditional volume that increases ventilator-induced lung injury in ARDS.

ARDSNet: VT 6 mL/kg predicted body weight (range 4–8), plateau pressure ≤ 30 cm H2O. Male PBW = 50 + 2.3(height in inches − 60); female = 45.5 + 2.3(height in inches − 60).

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