Respiratory Therapist (TMC – CRT/RRT)
Respiratory Therapist (TMC – CRT/RRT) is a trademark of its owner, which is not affiliated with and does not endorse Test Treasure Publication or TTP Prep.
Respiratory Therapist (TMC – CRT/RRT) is a trademark of its owner, which is not affiliated with and does not endorse Test Treasure Publication or TTP Prep.
10 sample questions for the Respiratory Therapist (TMC – CRT/RRT), with the correct answer and an explanation for each. No account needed. Try each one before you open the answer.
Sections covered on the exam: Patient Data Evaluation and Recommendations, Troubleshooting and Quality Control of Devices, and Infection Control, Initiation and Modification of Interventions.
Question 1 · Patient Data Evaluation and Recommendations
| Parameter | Value |
|---|---|
| pH | 7.25 |
| PaCO2 | 62 mm Hg |
| HCO3- | 26 mEq/L |
| PaO2 | 58 mm Hg |
| SaO2 | 88% |
Answer: B. Acute (uncompensated) respiratory acidosis with hypoxemia
The pH is low and the PaCO2 is high, so the primary disorder is respiratory acidosis. In acute hypercapnia the HCO3 rises only about 1 mEq/L per 10 mm Hg rise in PaCO2, so 26 mEq/L is the expected acute value, not renal compensation. The hypoxemia is explained by hypoventilation alone: PAO2 = 0.21(713) - 62/0.8 = about 72 mm Hg, giving a normal A-a gradient of about 14 mm Hg. Partial or full compensation would require a clearly elevated HCO3, and a metabolic acidosis would show a low HCO3 and low PaCO2.
Question 2 · Patient Data Evaluation and Recommendations
Answer: A. 10.8 mL O2/dL
CaO2 = (Hb x 1.34 x SaO2) + (PaO2 x 0.003) = (8 x 1.34 x 0.98) + (95 x 0.003) = 10.51 + 0.29 = 10.8 mL O2/dL, about half the normal value of roughly 20 mL/dL despite a normal PaO2 and saturation, which shows why anemia impairs oxygen delivery. 10.5 mL/dL omits the dissolved oxygen. 20.0 mL/dL assumes a normal hemoglobin of 15 g/dL. 8.1 mL/dL omits the 1.34 mL O2/g hemoglobin binding factor. Adding oxygen beyond normal saturation will not substantially correct the deficit; restoring hemoglobin will.
Question 3 · Troubleshooting and Quality Control of Devices, and Infection Control
Answer: C. Remove the HME and provide active humidification with a heated humidifier
Thick, copious, or bloody secretions are a contraindication to an HME because the device can become occluded, raising resistance and work of breathing, and its humidity output may be inadequate. A heated humidifier provides more effective humidification without adding resistance at the airway. Frequent HME changes do not address the inadequate humidity or the occlusion risk. A nebulizer placed between the HME and the airway saturates the HME and increases resistance. Routine saline instillation is not recommended and can dislodge bacteria into the lower airways.
Question 4 · Initiation and Modification of Interventions
| Parameter | Value |
|---|---|
| Peak cough flow | 140 L/min |
| Vital capacity | 1.1 L |
| MIP | -30 cm H2O |
| SpO2 (2 L/min NC) | 93% |
Answer: C. Mechanical insufflation-exsufflation
A peak cough flow below about 160 L/min means the cough cannot clear secretions, and a VC of 1.1 L with MIP -30 cm H2O confirms neuromuscular weakness. Mechanical insufflation-exsufflation delivers a deep positive-pressure breath and then a rapid shift to negative pressure, simulating an effective cough. Incentive spirometry is for lung expansion and does not move secretions. Oscillating PEP needs adequate expiratory effort. Huff coughing relies on the same expiratory muscles that are already too weak.
Question 5 · Initiation and Modification of Interventions
Answer: B. Increase the peak inspiratory flow
A scooped (concave) pressure waveform during a volume-controlled breath shows flow starvation: the patient is pulling harder than the set flow can deliver, causing dyssynchrony and increased work of breathing. Raising peak flow (for example to 60 to 80 L/min), or switching to a pressure-targeted mode with variable flow, matches the patient's demand. A less sensitive trigger adds to work of breathing. A 10 mL/kg PBW tidal volume abandons lung protection. A higher set rate does not change flow during each breath.
Question 6 · Patient Data Evaluation and Recommendations
| Sign | Finding |
|---|---|
| Heart rate | 88/min |
| Respiratory effort | Slow, irregular |
| Muscle tone | Some flexion of extremities |
| Reflex irritability (nasal catheter) | Grimace |
| Color | Pink body, blue hands and feet |
Answer: C. 5
Each sign scores 0 to 2. Heart rate below 100/min = 1; slow, irregular respirations = 1; some flexion of the extremities = 1; grimace to the nasal catheter = 1; pink body with blue extremities (acrocyanosis) = 1. Total = 5. A score of 4 results from scoring the heart rate of 88/min as 0, which applies only when no heart rate is present. Scores of 6 or 7 require at least one sign to be scored 2, such as a heart rate of 100/min or more or a vigorous cry.
Question 7 · Patient Data Evaluation and Recommendations
Answer: D. Potassium 7.2 mEq/L
Missed dialysis allows potassium to accumulate. Hyperkalemia first produces tall, peaked T waves, then flattened or absent P waves and a widening QRS complex that can progress to a sine-wave pattern and ventricular fibrillation. Hypokalemia produces flattened T waves and prominent U waves. Hypercalcemia shortens the QT interval. Hypomagnesemia is associated with QT prolongation and torsades de pointes, not peaked T waves.
Question 8 · Troubleshooting and Quality Control of Devices, and Infection Control
Answer: C. Connect two nebulizers set at 50% in tandem
At 50% the air-to-oxygen ratio is 1.7:1, so total flow = 12 x (1.7 + 1) = 32.4 L/min. Peak inspiratory flow is roughly 3-4 times minute ventilation (48-64 L/min), so room air is entrained around the mask, as shown by the mist vanishing on inspiration, and the delivered FiO2 falls. Two nebulizers in tandem roughly double total flow while keeping the concentration at 50%. Raising the dial to 70% reduces entrained air and total flow. Lowering the oxygen flow reduces total flow. A cannula at 6 L/min cannot deliver 50%.
Question 9 · Initiation and Modification of Interventions
Answer: D. Increase the set PEEP to partially offset the auto-PEEP
Auto-PEEP is 10 - 4 = 6 cm H2O. The patient must first generate enough effort to overcome this trapped pressure before the trigger can be reached, so many efforts fail (ineffective triggering). Raising the set PEEP to a level below the auto-PEEP reduces the pressure gap the patient must overcome. A less sensitive trigger makes it harder to trigger. A higher rate or larger VT shortens expiratory time or adds volume, which increases air trapping and worsens the problem.
Question 10 · Initiation and Modification of Interventions
| Parameter | Value |
|---|---|
| PaO2 | 58 mm Hg |
| SpO2 | 89% |
| Hematocrit | 58% |
| ECG | Normal sinus rhythm |
Answer: C. He qualifies for continuous home oxygen because of secondary erythrocytosis
Home oxygen is covered when resting PaO2 is 55 mm Hg or less or SpO2 is 88% or less. It is also covered when PaO2 is 56-59 mm Hg or SpO2 is 89% and there is evidence of chronic hypoxemia, such as a hematocrit above 56%, cor pulmonale, or dependent edema. A PaO2 of 58 mm Hg with a hematocrit of 58% qualifies. A PaO2 above 55 mm Hg does not disqualify him when secondary criteria are met. Neither the sleep-only nor the 85% claim is a criterion.
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