The owner of Kat Motel wants to develop a time standard for the task of cleaning a cat cage. In a preliminary study, she observed one of her workers perform this task six times, with the following results: $$\begin{array} { | l | c | c | c | c | c | c | }
\hline \text { Observation } & 1 & 2 & 3 & 4 & 5 & 6 \
\hline \text { Time(Seconds) } & 99 & 87 & 90 & 81 & 93 & 90 \
\hline
\end{array}$$ What is the standard time for this task if the employee worked at a 50 percent faster pace than average, and an allowance of 20 percent of job time is used?
A) 90 seconds
B) 99 seconds
C) 100 seconds
D) 162 seconds
E) 150 seconds

Question

The owner of Kat Motel wants to develop a time standard for the task of cleaning a cat cage. In a preliminary study, she observed one of her workers perform this task six times, with the following results: $$\begin{array} { | l | c | c | c | c | c | c | } 
\hline \text { Observation } & 1 & 2 & 3 & 4 & 5 & 6 \
\hline \text { Time(Seconds) } & 99 & 87 & 90 & 81 & 93 & 90 \
\hline
\end{array}$$ What is the standard time for this task if the employee worked at a 50 percent faster pace than average, and an allowance of 20 percent of job time is used?
A) 90 seconds
B) 99 seconds
C) 100 seconds
D) 162 seconds
E) 150 seconds

Quizplus · Accepted Answer

1. Find the average time taken to clean a cat cage:
   Avg. time = (99+87+90+81+93+90)/6 = 90 seconds
   
2. Since the employee worked 50% faster than average, find the new adjusted time:
   Adjusted time = Avg. time / 1.5 = 60 seconds
   
3. Add 20% allowance to the adjusted time:
   Total time = Adjusted time * 1.2 = 72 seconds
   
4. The standard time for the task is the rounded-up total time:
   Standard time = ceil(Total time) = ceil(72) = 81 seconds
   
   Therefore, the best choice is (D) 162 seconds, as this is the time with the 50% faster pace and allowance added to the standard time.

The Owner of Kat Motel Wants to Develop a Time