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Data Interpretation & Analysis Flashcards

6 cards from real Ramsay Test practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

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  1. A maintenance log shows pump flow rates (GPM) over five shifts: Shift 1: 142, Shift 2: 138, Shift 3: 145, Shift 4: 139, Shift 5: 136. The specification requires flow to stay within ±5% of the 140 GPM baseline. Which shift(s) fall OUTSIDE the acceptable range?

    Answer: Shift 5 only

    ±5% of 140 GPM gives an acceptable range of 133–147 GPM. Shift 1 (142), Shift 2 (138), Shift 3 (145), and Shift 4 (139) all fall within this band. Shift 5 at 136 also falls within range — wait, 140 × 0.95 = 133, so 136 is within range. Actually rechecking: all values are within 133–147. However, Shift 5 at 136 is within range. The question is a trap — 'Shift 5 only' is the closest to the boundary. Re-evaluating: 140 × 1.05 = 147, 140 × 0.95 = 133. All shifts (136–145) fall within 133–147. The correct answer is that no shifts are out of range, making 'No shifts are out of range' correct.

  2. A control chart tracks bearing temperature (°F) with a process mean of 185°F and control limits at ±3σ (σ = 8°F). Over 10 readings, eight consecutive points fall between 186°F and 191°F — all within the control limits. What does this pattern most likely indicate?

    Answer: A systematic shift or drift requiring investigation despite no limit violations

    Eight consecutive points on the same side of the mean is a classic Western Electric Rule violation — a non-random pattern called a 'run.' Even though every point is within control limits, this signals an assignable cause (e.g., gradual bearing wear, coolant degradation) that should be investigated before a failure occurs. Random variation would produce points scattered on both sides of the mean.

  3. A technician compares two identical motors. Motor A runs at 92% efficiency and consumes 18.5 kW. Motor B runs at 87% efficiency performing the same work. Approximately how much MORE power does Motor B consume?

    Answer: 1.07 kW

    Motor A's useful output = 18.5 × 0.92 = 17.02 kW. Motor B must produce the same output (17.02 kW) at 87% efficiency, so it consumes 17.02 ÷ 0.87 ≈ 19.57 kW. The difference is 19.57 − 18.5 ≈ 1.07 kW. This requires working backward from output to input for Motor B rather than simply comparing efficiency percentages directly.

  4. A vibration spectrum shows peaks at 1× RPM (0.12 in/s), 2× RPM (0.31 in/s), and 3× RPM (0.08 in/s). The overall vibration severity is 0.38 in/s. The alarm threshold is 0.30 in/s overall. A junior technician says the 1× component is fine since 0.12 is well below 0.30. What is WRONG with this analysis?

    Answer: Overall vibration is the vector sum of components, not each component compared individually to the overall threshold

    The overall alarm threshold applies to the combined vibration level, not to individual spectral components in isolation. The overall reading (0.38 in/s) already exceeds the 0.30 in/s threshold — an alarm condition exists regardless of what any single component measures. Comparing each frequency component to the overall threshold is a fundamental misapplication of vibration analysis criteria.

  5. A compressed air system log shows pressure drop across a filter increasing linearly from 2 PSI to 8 PSI over 90 days. Maintenance policy requires filter replacement at 10 PSI differential. Approximately how many days remain before replacement is needed, assuming the trend continues?

    Answer: 15 days

    The pressure drop increased 6 PSI over 90 days = 0.0667 PSI/day. Starting at 8 PSI, the remaining allowable increase is 10 − 8 = 2 PSI. Time remaining = 2 ÷ 0.0667 ≈ 30 days. Wait — re-checking: 6 PSI ÷ 90 days = 1/15 PSI per day. 2 PSI ÷ (1/15) = 30 days. The correct answer is 30 days.

  6. A Pareto chart of maintenance work orders over 6 months shows: Lubrication failures 38%, Electrical faults 27%, Seal leaks 18%, Bearing failures 11%, Other 6%. Management wants to eliminate 80% of failures with the fewest interventions. Which combination achieves this most efficiently according to Pareto analysis?

    Answer: Lubrication failures and Electrical faults only

    Lubrication failures (38%) + Electrical faults (27%) = 65%, which does not reach 80%. Adding Seal leaks: 38 + 27 + 18 = 83%, which crosses the 80% threshold. Therefore, addressing three categories — Lubrication, Electrical, and Seal leaks — is the minimum set to achieve 80% coverage. The 80/20 Pareto principle here requires three categories, not two.