EKG Technician Practice Test

EKG EKG Analysis & Interpretation Practice Questions

EKG Analysis & Interpretation makes up 12 of every 50 questions on a Tigertest EKG practice test. Here are 10 of the 48 questions on it, each with the answer and the reason.

Updated 2026-11-10 by the Tigertest team. Weighted to the NHA CET test plan. Rhythm items describe the strip in words; practice with real tracings too.

1. 1. A lead II strip is irregularly irregular, and the technician counts 9 QRS complexes in a 6-second segment. Using the best method for irregular rhythms, the rate is:

   • A.60 beats per minute
   • B.90 beats per minute
   • C.150 beats per minute
   • D.54 beats per minute

   Answer: B. The 6-second method suits irregular rhythms: 9 complexes x 10 = 90 per minute. The 1500 and 300 methods rely on a single R-R interval, so they misstate the rate when R-R intervals vary.

2. 2. A regular rhythm has 18 small boxes between consecutive R waves. Using the 1500 method, the heart rate is about:

   • A.75 per minute
   • B.90 per minute
   • C.100 per minute
   • D.83 per minute

   Answer: D. 1500 / 18 = 83.3, so about 83 per minute. A rate of 75 would require 20 small boxes (1500 / 20 = 75).

3. 3. On a regular strip, consecutive R waves fall exactly 4 large boxes apart. Using the 300-150-100-75-60 sequence, the rate is:

   • A.75 per minute
   • B.60 per minute
   • C.100 per minute
   • D.150 per minute

   Answer: A. Counting large boxes from an R wave on a heavy line gives 300, 150, 100, 75, so 4 boxes = 75 per minute (300 / 4 = 75). A rate of 60 would need 5 large boxes.

4. 4. A 60-year-old patient is scheduled for an exercise stress test. What is the patient's predicted maximum heart rate?

   • A.180 beats per minute
   • B.136 beats per minute
   • C.160 beats per minute
   • D.140 beats per minute

   Answer: C. Predicted maximum heart rate = 220 - age = 220 - 60 = 160. The value 136 is the 85% target (160 x 0.85 = 136), not the maximum.

5. 5. During a treadmill stress test on a 50-year-old, what target heart rate (85% of predicted maximum) should the technician watch for?

   • A.About 170 beats per minute
   • B.About 128 beats per minute
   • C.About 145 beats per minute
   • D.About 153 beats per minute

   Answer: C. Maximum = 220 - 50 = 170, and 170 x 0.85 = 144.5, about 145 per minute. 170 is the full predicted maximum, and 128 is only 75% of it.

6. 6. A very fast regular rhythm is recorded at 50 mm/s paper speed. The R-R interval spans 20 small boxes. The true heart rate is:

   • A.150 per minute
   • B.75 per minute
   • C.300 per minute
   • D.100 per minute

   Answer: A. At 50 mm/s each small box is 0.02 s, so 20 boxes = 0.40 s and 60 / 0.40 = 150 per minute. Applying 1500 / 20 = 75 ignores that doubled paper speed stretches the R-R across twice as many boxes, so the standard-speed result must be doubled.

7. 7. The R-R interval on a regular rhythm strip measures 0.60 seconds. What is the heart rate?

   • A.120 per minute
   • B.90 per minute
   • C.60 per minute
   • D.100 per minute

   Answer: D. Rate = 60 / R-R in seconds = 60 / 0.60 = 100 per minute (the same as 15 small boxes, 1500 / 15 = 100). A rate of 60 would need an R-R of 1.00 s.

8. 8. A technician needs to decide whether a rhythm strip is regular. Which technique gives the most reliable answer?

   • A.Compare the height of each QRS complex across all leads
   • B.Compare R-R intervals across the strip with calipers
   • C.Count the P waves that fall within one large box
   • D.Check whether every T wave on the strip is upright

   Answer: B. Regularity is judged by setting calipers (or marking paper) on one R-R interval and checking that each following R-R matches. QRS height reflects voltage, not timing.

9. 9. At the standard paper speed of 25 mm/s, how much time do one small box and one large box represent?

   • A.0.10 seconds and 0.50 seconds
   • B.0.04 seconds and 0.20 seconds
   • C.0.02 seconds and 0.10 seconds
   • D.0.04 seconds and 0.40 seconds

   Answer: B. At 25 mm/s a 1 mm small box = 0.04 s, and a 5 mm large box = 5 x 0.04 = 0.20 s. A small box equals 0.02 s only when the paper runs at 50 mm/s.

10. 10. Measured from the start of the P wave to the start of the QRS, the PR interval spans 6 small boxes on every beat. This is:

    • A.0.12 seconds, within normal limits
    • B.0.30 seconds, prolonged
    • C.0.24 seconds, within normal limits
    • D.0.24 seconds, prolonged

    Answer: D. 6 x 0.04 s = 0.24 s, longer than the normal 0.12 to 0.20 s, which fits first-degree AV block. The interval runs to the start of the QRS; measuring from the end of the P wave would give only the PR segment.