SOLUTION: Maria can weed 1,000 m² of cogonal farm surface in 40 hours, while her brother, Bryant, can do the same amount of work in 20 hours. If they work together, how long will it take th

Algebra ->  Rate-of-work-word-problems -> SOLUTION: Maria can weed 1,000 m² of cogonal farm surface in 40 hours, while her brother, Bryant, can do the same amount of work in 20 hours. If they work together, how long will it take th      Log On


   



Question 1188646: Maria can weed 1,000 m² of cogonal farm surface in 40 hours, while her brother, Bryant, can do the same amount of work in 20 hours. If they work together, how long will it take them to weed one hectare (10,000 m2) of cogonal land surface?
Found 3 solutions by Shin123, MathTherapy, ikleyn:
Answer by Shin123(626) About Me  (Show Source):
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In 1 hour, Maria can weed 1000%2F40=25 m2, and Bryant can weed 1000%2F20=50 m2.
Working together, they can weed 25%2B50=75 m2 per hour.
Therefore, it will take them 10000%2F75=133%261%2F3 hours.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Maria can weed 1,000 m² of cogonal farm surface in 40 hours, while her brother, Bryant, can do the same amount of work in 20 hours. If they work together, how long will it take them to weed one hectare (10,000 m2) of cogonal land surface?
As she takes 40 hours to do the entire job (weed 1,000 m²), Maria's per hour rate = 1%2F40
As the brother takes 20 hours to do the entire job (weed 1,000 m²), the brother's per hour rate = 1%2F20
With time that it takes both to do 1 job (1,000 m²) being T, we get: matrix%281%2C3%2C+1%2F40+%2B+1%2F20%2C+%22=%22%2C+1%2FT%29
T + 2T = 40 ------ Multiplying by LCD, 40T 
    3T = 40
Time taken by both to do 1 job (1,000 m²), or matrix%281%2C4%2C+T%2C+%22=%22%2C+40%2F3%2C+hours%29
Time taken by both to do 10 SUCH jobs (weed 10,000 m²) = 

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.

It is, actually, pure Arithmetic problem.

It can be solved without Algebra and without using equations.

It is actually, arithmetic problem for very young students/children of 4-th grade level,
who just know arithmetic, but are not familiar with Algebra and with equations.

Therefore, a preferable method to solve it is using Arithmetic only, without Algebra.


It is how this problem is intended and designed, and how it is expected to be solved.