SOLUTION: To mow a grass field a team of mowers planned to cover 15 hectares a day. After 4 working days they increased the daily productivity by 33×1/3% , and finished the work 1 day earli

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: To mow a grass field a team of mowers planned to cover 15 hectares a day. After 4 working days they increased the daily productivity by 33×1/3% , and finished the work 1 day earli      Log On


   

Question 1041334: To mow a grass field a team of mowers planned to cover 15 hectares a day. After 4 working days they increased the daily productivity by 33×1/3% , and finished the work 1 day earlier than it was planned.
A) What is the area of the grass field?
B) How many days did it take to mow the whole field?
C) How many days were scheduled initially for this job?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39628) About Me  (Show Source):
You can put this solution on YOUR website!
RT=J rule for work rates. J is for quantity of jobs, here being 1.

Original plan was 15%2At=1 to expect 1%2F15 days.
NO!
Change that to j, unknown number of hectares.

15t=j
t=j%2F15, time in days to do j hectares.

15%2A4%2B%2815%2A%284%2F3%29%29%2Ax=j, ONE unit for j was performed.
x was number of days at the higher rate.
4%2Bx=j%2F15-1-----time taken total was one day less than expected. Solve this for j, and substitute.

I did not finish this, but said how you can.

Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!
To mow a grass field a team of mowers planned to cover 15 hectares a day. After 4 working days they increased the daily productivity by 33×1/3% , and finished the work 1 day earlier than it was planned.
A) What is the area of the grass field?
B) How many days did it take to mow the whole field?
C) How many days were scheduled initially for this job?
Let number of scheduled days be D

Then amount to mow = 15D

After 4 days, 4(15), or 60 hectares were mowed, which leaves 15D – 60 hectares to be mowed

Increasing productivity by matrix%281%2C2%2C33%261%2F3%2C+%22%25%22%29 means that productivity increased to %28%28133%261%2F3%29%2F100%29+%2A+15, or 20 hectares per day

Thus 15D – 60 was left to be completed in D – 4 – 1 days, working at a rate of 20 hectares/day

We then get: 15D – 60 = 20(D – 4 - 1)

15D – 60 = 20(D – 5)

15D – 60 = 20D – 100

15D – 20D = - 100 + 60

– 5D = - 40

D, or scheduled days = highlight_green%28matrix%281%2C3%2C+%28-+40%29%2F%28-+5%29%2C+or%2C+8%29%29



Area of field (amount to mow) = 15(8), or highlight_green%28matrix%281%2C2%2C+120%2C+hectares%29%29

Number of days taken to mow entire field: D – 1, or highlight_green%287%29