Question

Один мастер может выполнить заказ за 20 часов, второй — за 5 часов, а третий за 4 часа. За сколько часов выполнят заказ три мастера, работая вместе?

Answer 1

1- вся работа, 1/20, 1/5, 1/4- производительность мастеров. 1)1/20+1/5+1/4=1/20+4/20+5/20=10/20=1/2 работы вместе за час. 2) 1:1/2=2(ч) время совместной работы. Ответ:2ч

Answer 2

Возьмем заказ за 20 деталей

1)20:20=1(деталь)-за час 1 мастер
2)20:5=4(деталей)-за час 2 мастер
3)20:4=5(деталей)-за часа 3 мастер
4)5+4+1=10(деталей)-за час все
5)20:10=2(часа)-выполнят заказ все

Answer 3

Calculation of Time Taken by Three Masters Working Together

To calculate the time taken by three masters working together to complete an order, we need to determine their combined efficiency. The efficiency of a worker is inversely proportional to the time taken to complete a task. Therefore, the sum of the individual efficiencies of the three masters will give us their combined efficiency.

Let's denote the time taken by the first master as t1 (20 hours), the time taken by the second master as t2 (5 hours), and the time taken by the third master as t3 (4 hours).

The efficiency of each master can be calculated as the reciprocal of their respective time taken: e1 = 1/t1, e2 = 1/t2, and e3 = 1/t3.

To find the combined efficiency, we add the individual efficiencies: e = e1 + e2 + e3.

Now, we can calculate the time taken by the three masters working together using the formula: t = 1/e.

Let's substitute the values and calculate the time taken:

e1 = 1/20, e2 = 1/5, e3 = 1/4

e = e1 + e2 + e3 = 1/20 + 1/5 + 1/4 = 1/20 + 4/20 + 5/20 = 10/20 = 1/2

t = 1/e = 1/(1/2) = 2 hours

Therefore, the three masters will complete the order working together in 2 hours.