SOLUTION: With two reapers operating, a field can be harvested in 1 h. If only the newer reaper is used, the crop can be harvested in 1.5 h. How long would it take to harvest the field using

Algebra ->  Rate-of-work-word-problems -> SOLUTION: With two reapers operating, a field can be harvested in 1 h. If only the newer reaper is used, the crop can be harvested in 1.5 h. How long would it take to harvest the field using      Log On


   

Question 1198709: With two reapers operating, a field can be harvested in 1 h. If only the newer reaper is used, the crop can be harvested in 1.5 h. How long would it take to harvest the field using only the older reaper?
Found 4 solutions by ikleyn, josgarithmetic, greenestamps, math_tutor2020:
Answer by ikleyn(52775) About Me  (Show Source):
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.
With two reapers operating, a field can be harvested in 1 h.
If only the newer reaper is used, the crop can be harvested in 1.5 h.
How long would it take to harvest the field using only the older reaper?
~~~~~~~~~~~~~~~ 

The combined rate of work of two reapers is 1_field%2F1_hour = 1 job per hour.


The individual rate of work of the new reaper is  

    1_field/1.5_hours = 1%2F1.5 = 1%2F%28%283%2F2%29%29 = 2%2F3 of the job per hour.


Hence, the rate of work of the old reaper is 1 - 2/3 = 1/3 of the job per jour.


It means that the field can be harvested in 3 hours, if only old reaper is used.


ANSWER.  3 hours.

Solved.



Answer by josgarithmetic(39616) About Me  (Show Source):
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                    WORK RATE

old reaper          1/x

new reaper          1/1.5

both combined       1/1

1%2Fx%2B1%2F1.5=1
-
1%2Fx=1-1%2F1.5
1%2Fx=1-2%2F3=1%2F3
1%2Fx=1%2F3
highlight%28x=3%29

Answer by greenestamps(13198) About Me  (Show Source):
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You have two different responses that show two different ways to solve "working together" problems like this. Here is a third method.

Consider the least common multiple of the two given times, which is 3 hours.

The two reapers together harvest the field in 1 hour; so in 3 hours, the two reapers together could harvest a field that size 3/1 = 3 times.
The newer reaper can harvest the field in 1.5 hours; so in 3 hours, the newer reaper could harvest a field that size 3/1.5 = 2 times.
So in 3 hours, the older reaper could harvest a field that size 3-2 = 1 time.

So it would take the older reaper 3 hours to harvest the field.

Answer by math_tutor2020(3816) About Me  (Show Source):
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Consider a field with 300 plants of potatoes.
There's nothing particularly special about this value. I chose it so that the subsequent division operations result in whole numbers shown below. Feel free to try out other values. The answer will result the same each time.

In this hypothetical numeric example, the goal is to harvest 300 potatoes.
The two machines operate together to get the job done in 1 hour.
Their combined rate is 300/1 = 300 potatoes per hour.
rate = (amount done)/(time)

The newer machine can do the job alone taking 1.5 hours
rate = (amount done)/(time)
rate = (300 plants)/(1.5 hours)
rate = 200 plants per hour

Since the newer machine can do 200 plants per hour on its own, and the combined rate is 300 plants per hour, then it must mean the older piece of equipment handles 300-200 = 100 plants per hour.
This is under the assumption that the two machines work most efficiently together; meaning that neither machine slows the other down.

Then we have one last set of steps to wrap things up
(rate)*(time) = amount done
time = (amount done)/(rate)
time = (300 plants)/(100 plants per hour)
time = 3 hours
This represents how long it takes the older machine to get the job done if working alone.

Answer: 3 hours