What is the standard time (ST) for this task if the employee worked at a twenty-percent faster pace than is average, and an allowance of twenty percent of job time is used?
A)4.5 seconds
B)5 seconds
C)20 seconds
D)28.8 seconds
E)100 seconds

Question

Quizplus · Accepted Answer

We need to first find the original standard time (OST) by dividing the given time by (1 + allowance percentage):
OST = 6 / (1 + 0.2) = 5 seconds 
The employee works at a 20% faster pace, so their actual time is 80% of the original standard time:
Actual time = 0.8 x OST = 0.8 x 5 = 4 seconds 
Thus, the total time including allowance is:
Total time = Actual time + (allowance percentage x actual time) 
Total time = 4 + (0.2 x 4) = 4.8 seconds 
Therefore, the answer is D, 28.8 seconds, which is found by multiplying the total time by the factor of 6 (to match the given units of "per 6 parts"). 
28.8 = 4.8 x 6

What Is the Standard Time (ST) for This Task If