CSSGB Full 100 Randomly Drawn Questions Mock Exam (240 Minutes)

Free sample questions

5 questions from this set, shown in full with the correct answer and the reasoning behind it. Read them as a study resource.

1. Q1.During a walk of the shop floor a team records four observations.
   
   $$\begin{array}{c|l} \hline \textbf{No.} & \textbf{Observation} \\ \hline 1 & \text{Finished units are trucked to a second building for packing} \\ 2 & \text{Six weeks of resin is held in the warehouse} \\ 3 & \text{The operator turns and reaches behind for a tool every cycle} \\ 4 & \text{Assemblers stand idle waiting for the first tote of the day} \\ \hline \end{array}$$
   
   Which observation is an example of the waste of motion?

   • A.Observation 1
   • B.Observation 2
   • C.Observation 3✓ Correct answer
   • D.Observation 4

   Explanation

   Motion is unnecessary movement of the worker's body within the workstation, so turning and reaching on every cycle is motion. Observation 1 is the primary foil because it is also movement, but it moves the product rather than the person, which makes it the waste of transportation.

   Reference: CSSGB BoK I

2. Q2.A failure mode currently carries severity 7, occurrence 6 and detection 5. Three corrective actions have been proposed, and the ratings each one would leave behind are shown below.
   
   $$\begin{array}{l|c|c|c} \hline \textbf{Action} & \textbf{Sev} & \textbf{Occ} & \textbf{Det} \\ \hline \text{A: add an end of line test} & 7 & 6 & 2 \\ \text{B: error proof the loading fixture} & 7 & 2 & 5 \\ \text{C: redesign so the effect is milder} & 4 & 6 & 5 \\ \hline \end{array}$$
   
   Which ranking orders the three actions from lowest to highest revised risk priority number?

   • A.C, A, B
   • B.A, B, C
   • C.B, A, C✓ Correct answer
   • D.B, C, A

   Explanation

   The risk priority number is the product of the three ratings:
   
   $$RPN = S \times O \times D$$
   
   Applying it to each proposal:
   
   $$RPN_A = 7 \times 6 \times 2 = 84, \quad RPN_B = 7 \times 2 \times 5 = 70, \quad RPN_C = 4 \times 6 \times 5 = 120$$
   
   So the order is B, then A, then C, against a current value of $7 \times 6 \times 5 = 210$. Action C is the trap: cutting severity feels like the strongest possible move, but leaving occurrence and detection untouched leaves it with the highest revised RPN of the three.

   Reference: CSSGB BoK I

3. Q3.An interrelationship digraph drawn for chronic schedule slippage is shown below.
   
   Which issue should the team attack as the root driver?

   

   • A.Frequent design changes
   • B.Rework at assembly
   • C.Unclear specifications✓ Correct answer
   • D.Late shipments

   Explanation

   In an interrelationship digraph the driver is the node with the most arrows leaving it and none arriving, because it causes other issues without being caused by any of them. Rework at assembly is the primary foil: it has the most arrows arriving, which makes it look central, but a node dominated by incoming arrows is an outcome, and attacking it treats a symptom.

   Reference: CSSGB BoK II

4. Q4.A green belt reviews the flowchart below during a documentation review.
   
   Which defect in the flowchart should be corrected?

   

   • A.A process step is drawn as a parallelogram rather than a rectangle
   • B.A terminator symbol is missing from the end of the chart
   • C.A decision symbol has only one exit path✓ Correct answer
   • D.An inspection step is drawn as a rectangle rather than a diamond

   Explanation

   A decision symbol asks a yes or no question, so it must have at least two exit paths; a diamond with one exit means the chart has no route for the nonconforming part and the rework loop has been left off the map. Reading the measurement step as a missing diamond is the tempting alternative, but measuring is an activity and belongs in a rectangle. The decision that follows it is what needs both branches.

   Reference: CSSGB BoK III

5. Q5.An $\bar{X}$ chart has a centre line at $100.0$, an upper control limit at $106.0$ and a lower control limit at $94.0$, so one sigma on the chart equals $2.0$. The eight most recent subgroup means plotted as follows.
   
   $$\begin{array}{c|c|c|c|c|c|c|c|c} \hline \textbf{Subgroup} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \bar{X} & 100.5 & 99.2 & 101.0 & 104.6 & 102.1 & 105.2 & 100.4 & 99.8 \\ \hline \end{array}$$
   
   Which run rule is violated?

   • A.Four of five consecutive points beyond one sigma on the same side of the centre line
   • B.Two of three consecutive points beyond two sigma on the same side of the centre line✓ Correct answer
   • C.A single point falling beyond a three sigma control limit
   • D.Eight consecutive points falling on the same side of the centre line

   Explanation

   The zone boundaries follow from the sigma value given:
   
   $$\text{two sigma lines} = 100.0 \pm 2(2.0) = 96.0 \text{ and } 104.0$$
   
   Subgroups 4, 5 and 6 are consecutive, and two of them, $104.6$ and $105.2$, sit beyond the upper two sigma line, which fires the zone A rule. The one sigma line is at $102.0$, and no window of five consecutive points holds four beyond it, so the four of five rule is the primary foil rather than the answer. No point reaches $106.0$, and subgroups 2 and 8 fall below the centre line, so neither of the remaining rules applies.

   Reference: CSSGB BoK VI

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