Question

Определить ЗП рабочего за месяц при сдельно - прогрессивной системе оплаты труда, если дневная выработка рабочего 130 деталей. За месяц он должен был обработать 4000 деталей, а фактически сделал 5500 штук по 4 разряду. По действующей на данном предприятии шкале при выполнении плана свыше 100% сдельная расценка увеличится в 2, 5 раза. Тс4 – 60 руб.

Answer 1

Ответ:

Рсд. = Тс*Тсм /Нвыр.дн= 60*8/130=3,69руб  

ЗП сд. прог. = Рсд. * Nпл + Рсд. * Кув * (Nф – Nпл)= 3,69*4000+3,69*2,5*(5500-4000)= 28597,5 руб

Объяснение:

Answer 2

Calculation of Monthly Salary for a Worker under Piece-Rate Progressive Payment System

To determine the monthly salary of a worker under a piece-rate progressive payment system, we need to consider the worker's daily output, the target output for the month, the actual output, and the applicable rate scale for exceeding the target.

Given the following information: - Daily output of the worker: 130 pieces - Target output for the month: 4000 pieces - Actual output of the worker: 5500 pieces - Rate scale for exceeding the target: 2.5 times the regular rate (for output above 100% of the target)

We can calculate the worker's monthly salary as follows:

1. Calculate the regular salary for the target output: - Regular rate per piece (Tс4): 60 rubles - Regular salary for the target output: 4000 pieces * 60 rubles/piece = 240,000 rubles

2. Calculate the additional salary for the excess output: - Excess output: 5500 pieces - 4000 pieces = 1500 pieces - Additional rate for excess output: 2.5 times the regular rate - Additional salary for the excess output: 1500 pieces * 60 rubles/piece * 2.5 = 225,000 rubles

3. Calculate the total monthly salary: - Total monthly salary = Regular salary for the target output + Additional salary for the excess output - Total monthly salary = 240,000 rubles + 225,000 rubles = 465,000 rubles

Therefore, the worker's monthly salary under the given conditions would be 465,000 rubles.

Please note that the calculation assumes that the rate scale applies only to the excess output and not to the entire output.