Question

3 Реши задачу.Один мастер изготовил 760 деталей, а другой, при той женорме выработки в день 520 деталей. Сколько дней работалкаждый мастер в отдельности, если второй мастер работална 6 дней меньше?можно УСЛОВИЕ пожалуйста ​

Answer 1

Ответ:

760-520=240:6=40

Пошаговое объяснение:

Дано

I-760Шт

II-520шт

II=х-6 дней

I=x дней

Решение

760÷40=19(д.)1 мастер

520÷40=13(д.)2 мастер

Ответ:19 дней 1 мастер, а 2 мастер 13 дней.

Answer 2

Problem Statement:

Два мастера изготавливают детали. Первый мастер изготовил 760 деталей, а второй мастер, при той же норме выработки в день, изготовил 520 деталей. Второй мастер работал на 6 дней меньше, чем первый мастер. Необходимо определить, сколько дней каждый мастер работал в отдельности.

Solution:

Let's assume that the first master worked for x days. Since the second master worked 6 days less, the second master worked for x - 6 days.

We know that the first master produced 760 parts, and the second master produced 520 parts. We can set up the following equation based on their productivity:

x * P1 = 760 (where P1 is the productivity of the first master) (x - 6) * P2 = 520 (where P2 is the productivity of the second master)

To find the number of days each master worked, we need to solve these equations simultaneously.

Let's solve the first equation for P1: P1 = 760 / x

Substituting this value into the second equation: (x - 6) * (760 / x) = 520

Simplifying the equation: 760(x - 6) = 520x

Expanding and rearranging the equation: 760x - 4560 = 520x

Bringing like terms to one side: 760x - 520x = 4560

Simplifying: 240x = 4560

Dividing both sides by 240: x = 19

Therefore, the first master worked for 19 days, and the second master worked for 19 - 6 = 13 days.

So, the first master worked for 19 days, and the second master worked for 13 days.

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