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22 January, 18:20

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: 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

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  1. 22 January, 19:33
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    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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