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bhanu29
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23 Dec 2025, 04:44
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Bunuel
given rate = 1pkg/s
50 s = 50 /(3600 *8) workday
rate in pkg/wd = 1/(50/(3600*8)) = (3600*8)/50 = 576
rate in pkg/ww = 5 * pkg/wd = 576*5 =2880
pkg/day pkg/week
576 2880
msignatius
23 Dec 2025, 05:00
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I find no method more plausible here than to take out a standardized rate - which would, in this case be, an hourly rate for how quickly the clerk can scan packages.
An hour contains 60 minutes, and each minute contains 60 seconds, so 60*60 = 3,600 seconds, is what we divide by 50, the average per package rate, to get 72, which is how many packages our friend scans every hour.
In a day, the number will just be 8 times this, or 576.
And in a week, the number will just be 5 times the 576, or 2880.
There, management should be satisfied with this much.
An hour contains 60 minutes, and each minute contains 60 seconds, so 60*60 = 3,600 seconds, is what we divide by 50, the average per package rate, to get 72, which is how many packages our friend scans every hour.
In a day, the number will just be 8 times this, or 576.
And in a week, the number will just be 5 times the 576, or 2880.
There, management should be satisfied with this much.
Bunuel
23 Dec 2025, 05:26
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Rate = 1 package per 50s
=> Packages/hour = 3600/50 = 72
Answer:
Packages/day = 72*8 = 576
Packages/week = 576*5 = 2880
=> Packages/hour = 3600/50 = 72
Answer:
Packages/day = 72*8 = 576
Packages/week = 576*5 = 2880
Bunuel
