Question:medium
A time study was carried out for a job broken into two elements. For each element, three iterations of task were observed, and observed time was recorded. The observed times (in seconds) and the rating factor for each element are given in the table.
Work element Observed time (seconds) Rating factor A 20; 25; 21 105% B 19; 17; 18 90%
If a further 20% needs to be added for allowances, then the standard output of the worker is ______ units per hour (in integer).
A time study was carried out for a job broken into two elements. For each element, three iterations of task were observed, and observed time was recorded. The observed times (in seconds) and the rating factor for each element are given in the table.
If a further 20% needs to be added for allowances, then the standard output of the worker is ______ units per hour (in integer).
| Work element | Observed time (seconds) | Rating factor |
|---|---|---|
| A | 20; 25; 21 | 105% |
| B | 19; 17; 18 | 90% |
If a further 20% needs to be added for allowances, then the standard output of the worker is ______ units per hour (in integer).
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Find normal time for each element, add allowance, then convert standard time per unit into units produced per hour.
Updated On: Aug 5, 2026
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Correct Answer: 76
Solution and Explanation
Step 1: Understanding the Concept:
We need to calculate the standard time to complete one full job (comprising elements A and B) and then determine how many jobs (units) the worker can produce in exactly one hour.
Step 2: Key Formula or Approach:
Average Observed Time (OT) = Sum of times divided by number of observations.
Normal Time (NT) = OT $ \times $ Rating Factor.
Standard Time (ST) = Total NT $ \times (1 + \text{Allowance Factor}) $.
Output per hour = 3600 seconds divided by ST in seconds.
Step 3: Detailed Explanation:
Element A:
Average OT = $ \frac{20 + 25 + 21}{3} = \frac{66}{3} = 22 $ seconds.
Normal Time ($ NT_A $) = $ 22 \times 1.05 = 23.1 $ seconds.
Element B:
Average OT = $ \frac{19 + 17 + 18}{3} = \frac{54}{3} = 18 $ seconds.
Normal Time ($ NT_B $) = $ 18 \times 0.90 = 16.2 $ seconds.
Total Job:
Total Normal Time = $ 23.1 + 16.2 = 39.3 $ seconds.
Adding 20% allowance to the total Normal Time gives the Standard Time:
\[ ST = 39.3 \times (1 + 0.20) = 39.3 \times 1.2 = 47.16 \text{ seconds/unit} \]
Production Output:
Total time available in an hour = 3600 seconds.
\[ \text{Units per hour} = \frac{3600}{47.16} = 76.335 \text{ units} \]
Final Answer:
Since we require the integer value, the output is 76 units per hour. Apply the rating factor to the average observed time for each individual element before adding them up, applying an average rating factor to the total time would give an incorrect result. \[ \boxed{76 \text{ units per hour}} \]
We need to calculate the standard time to complete one full job (comprising elements A and B) and then determine how many jobs (units) the worker can produce in exactly one hour.
Step 2: Key Formula or Approach:
Average Observed Time (OT) = Sum of times divided by number of observations.
Normal Time (NT) = OT $ \times $ Rating Factor.
Standard Time (ST) = Total NT $ \times (1 + \text{Allowance Factor}) $.
Output per hour = 3600 seconds divided by ST in seconds.
Step 3: Detailed Explanation:
Element A:
Average OT = $ \frac{20 + 25 + 21}{3} = \frac{66}{3} = 22 $ seconds.
Normal Time ($ NT_A $) = $ 22 \times 1.05 = 23.1 $ seconds.
Element B:
Average OT = $ \frac{19 + 17 + 18}{3} = \frac{54}{3} = 18 $ seconds.
Normal Time ($ NT_B $) = $ 18 \times 0.90 = 16.2 $ seconds.
Total Job:
Total Normal Time = $ 23.1 + 16.2 = 39.3 $ seconds.
Adding 20% allowance to the total Normal Time gives the Standard Time:
\[ ST = 39.3 \times (1 + 0.20) = 39.3 \times 1.2 = 47.16 \text{ seconds/unit} \]
Production Output:
Total time available in an hour = 3600 seconds.
\[ \text{Units per hour} = \frac{3600}{47.16} = 76.335 \text{ units} \]
Final Answer:
Since we require the integer value, the output is 76 units per hour. Apply the rating factor to the average observed time for each individual element before adding them up, applying an average rating factor to the total time would give an incorrect result. \[ \boxed{76 \text{ units per hour}} \]
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