Question

Productivity for an agency was 350,000 units more for the second year than the first, and productivity for the third year was double the productivity for the second year. If productivity for the third year was 806,000 units, what was productivity for the first year?

Answer 1

Answer: if we call P1 to the productivity of the first year, P2 for the second, and so on, then:

1 ) P2 - P1 = 350,000

2) P3 = 2*P2

and P3 = 806,000

now we can put this value on the equation 2, and obtain the value of P2

then P2 = P3/2 = 806,000/2 = 403,000

now we can put this value on equation 1, and obtain the value of P1.

P2 - P1 = 403,000 - P1 = 350,000

P1 = 403,000 - 350,000 = 53,000

So the productivity in the first year is 53,000 units.

Answer 2

Answer: The productivity during the first year was 53,000 units.

To solve this we can write and solve some equations.

Let the first year = x.

That makes the second year: x + 350000

That mates the third year: 2x + 700000

Since we know the amount of the 3rd year, we can write and solve the following equation.

2x + 700000 = 806000
2x = 106000
x = 53000