Two workers finished a job in 12 days. How long would it take each worker to do the job by himself if one of the workers needs 10 more days to finish the job than the other worker

Two workers finished a job in 7.5 days.

How long would it take each worker to do the job by himself if one of the workers needs 8 more days to finish the job than the other worker?

let t = time required by one worker to complete the job alone

then

(t+8) = time required by the other worker (shirker)

let the completed job = 1

A typical shared work equation

7.5%2Ft + 7.5%2F%28%28t%2B8%29%29 = 1

multiply by t(t+8), cancel the denominators, and you have

7.5(t+8) + 7.5t = t(t+8)

7.5t + 60 + 7.5t = t^2 + 8t

15t + 60 = t^2 + 8t

form a quadratic equation on the right

0 = t^2 + 8t - 15t - 60

t^2 - 7t - 60 = 0

Factor easily to

(t-12) (t+5) = 0

the positive solution is all we want here

t = 12 days, the first guy working alone

then

the shirker would struggle thru the job in 20 days.

7 + 17 = 24÷2 (since there are 2 workers) =12. Also, ½(7) + ½17 = 3.5 + 8.5 = 12. So, we know that the faster worker will take 7 days and the slower worker will take 17 days. Hope this helps! jul15

Step-by-step explanation: