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ANSWER ASAP: New machine has productivity (measured in the number of produced details) 12% greater than the last model. After replacing the machines during renovation, a plant has 5% fewer new models than it had old models before. By what percent did the number of produced details on this plant change?

User Jimkont
by
6.4k points

2 Answers

3 votes
the total number of produced details increased by 6.4%

suppose the starting number was x.
suppose x=100
100 +12%= 112
112-5%=106.4
106.4-100=6.4
6.4%

User Uwe Hafner
by
6.0k points
1 vote

Answer:

6.4%

Explanation:

Given:

Productivity of new machine = 12% greater than last model

Number of new machines = -5%

Let's take number of machines = n

Take p as productivity of each machine.

Total productivity = np.

Productivity of new machine will be:

(100% + 12%) * p

= 112%.p

= 1.12p

Number of new machines will be:

(100% - 5%) * n

= 0.95n

From the calculations, total productivity will now be

1.12p * 0.95n

= 1.064np

Percentage Change in total productivity will be:


= (new productivity - old productivity)/(old productivity) * 100


= (1.064np - np)/(np) * 100

Converting np to 1 (or any number of your choice), we have:


= ((1064*1)-1)/(1) * 100

= 6.4%

Therefore, the percentage change = 6.4%

User Jeffmayn
by
6.9k points
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