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3 years ago

7

The owner of Kat Motel wants to develop a time standard for the task of cleaning a cat cage. In a preliminary study, she observe

d one of her workers perform this task six times, with the following results: Observation 1 2 3 4 5 6 Time(Seconds) 99 87 90 81 93 90 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? 162 seconds 99 seconds 90 seconds

1 answer:

babunello [35]3 years ago

3 0

Answer:

The standard time for this task is 162 seconds ⇒ 1st answer

Step-by-step explanation:

* Lets explain how to solve the problem

- Kat observed one of her workers perform this task six times, with the

following results

- Observation : 1 2 3 4 5 6

Time (sec.) : 99 87 90 81 93 90

- Lets calculate the average of the time

∵ The average time = Total time ÷ number of times

∵ Total = 99 + 87 + 90 + 81 + 93 + 90 = 540

∵ The number of times is 6

∴ The average time = 540 ÷ 6 = 90 seconds

- The standard time = normal time + allowances

∵ Normal time = average time × rating

∵ The average time is 90 seconds

∵ The rate is 50 percent faster pace than average

∴ The rate is (1 + 50/100) = 1.5

∴ The normal time = 90 × 1.5 = 135 seconds

- The allowance of 20 percent of job time is used

∵ The allowance = average time × allowance rate

∵ The allowance is 20%

∴ The allowance = 135 × 20/100 = 135 × 0.2 = 27 seconds

* Lets calculate the standard time

∵ The standard time = normal time + allowances

∵ the normal time is 135 seconds

∵ The allowance time is 27 seconds

∴ The standard time = 135 + 27 = 162 seconds

* The standard time for this task is 162 seconds

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