14) Work problems Flashcards
Basic work formula
Rate-time-work —> variation of the rate-time-distance formula
Rate * time = work (aka accomplishing a certain job)
e.g., producing units, completing 1 job
Always express work rate as WORK / TIME
> 1 job / time
> units / time
> 0.5 job / time—–> fractional amount of work completed in a specific amount of time can still be turned into a work rate
The quirk with these types of work problems: often involves 2 or more objects working together to complete a job
Work problem variation #1) Single worker problems
Tests basic understanding of rate-time-work equation
Work problem variation #2) Combined worker problems
What is the range of possible total time to complete the job?
Assume Object x takes less than to complete 1 job than Object y (x = x’s time alone, y = y’s time alone)
Multiple objects work together to complete a job
Work done by object 1 + work done by object 2 + … = Total work done
Combined rate of two objects = 1/time for object 1 to complete job + 1/time for object 2 to complete job
(adding individual rates together)
CONCEPTS: ASSUMING w = 1 job
> when two objects complete a job, the job will be completed at a FASTER RATE than if EACH object is working ALONE (“two is better than one”)
RULE 1: TOTAL COMBINED TIME to complete is ALWAYS LESS THAN each individual object’s time to complete
(regardless of faster or slower object)
t < x (faster object’s time) < y (slower object’s time)
Alternatively, TOTAL COMBINED RATE to complete is ALWAYS GREATER THAN each individual object’s rate to complete
1/y < 1/x < 1/t —–> reciprocal of time inequality
RULE 2: HOWEVER, total combined time is GREATER THAN if we had TWO fast objects doing the job, and LESS THAN if we had TWO slow objects doing the job
In other words:
x/2 < t < y/2 ——–> derived by calculating total time to complete 1 job if there were two fast objects and then if there were two slow objects
COMBINED RATE? 1/t …
BE MINDFUL OF TIME:
> objects work on the project for the same amount of time
> objects start the project at the same time, but one stops before completion —-> one object spends ADDITIONAL time
> objects work together, but one object has unknown time to complete the job alone —-> rate = work/t
Other q variations:
> percent of a job done by one object (or fraction of a job done by one object) = work done by object A / work done by object B or work done by object A / total work done
Work problem variation #3) Opposing workers problems
Objects working in opposition
e.g., water filling up a sink and a drain emptying it
e.g., one working digging a hole, and another working putting dirt back into the hole
Individual work must be SUBTRACTED from each other –> total task completed = difference between their work values
Figure out the RATE OF INCREASE = Inflow - outflow
Work problem variation #4) Change in workers problems
e.g., Five workers, each working at the same constant rate, can complete a job working together in six days. If three workers are removed from the group before they begin working, what is the combined rate of the two remaining workers?
What happens to TIME when a certain number of workers added or removed from a group
> Given combined rate of a certain number of workers or machines, then asked to add or more from the group (need to figure out individual rate or use proportion method)
Usually dealing with workers with the SAME (IDENTICAL) CONSTANT RATE
Total combined rate = rate of individual worker * N workers
Method 1) Calculate the rate of ONE worker, then scale up or down
> let r = rate of one worker
Method 2) Proportion Method: Rate of one worker = Rate of one worker
> becomes: Combined rate of X workers / X workers = Combined Rate of Y workers / Y workers
> useful when needed to determine ADDITIONAL number of workers or machines needed to complete a job at a particular rate
e.g., 1/6 / 5 = Ry / 2
Ry = 1 / 15
WATCHOUTS:
> harder questions will express combined rates using a variable to represent an unknown number of workers, then ask you to determine the number of workers required USING the variable
e.g., N workers can complete M jobs in 48 hours —> Individual rate is M/48N
So how many workers are required to produce 1200 jobs in 12 hours?
Let new number of workers = x
x(M/48N)12 = 1200
x = 4800N/M
Robots A, B, and C operate at their respective constant rates. Operating simultaneously, Robots A and B can complete a job in 3 hours, while Robots A, B, and C can complete the job in 2 hours working together. How long would it take Robot C to complete the job working alone at its constant rate?
Combining and subtracting work rates —> turns into a linear equation problem
> set variables equal to individual work rates then solve
> given COMBINED rates of objects —> “combo” questions (instead of solving for each variable separately)
Combined rate = Sum of individual rates in the same units
t = 6 hours = 1 / rate of robot C
Working together at their individual constant rates, Robots A and B can complete a job in 2 hours, Robots A and C can complete the job in 3 hours, and Robots B and C can complete the job in 4 hours. Working together, approximately how long would it take all three robots to complete the job?
Combined rate = sum of individual rates
NOTICE how the three work equations CAN BE SUMMED Together efficiently to yield:
2Ra + 2Rb + 2Rc = 1/2 + 1/3 + 1/4 —–> time saving method (instead of doing elimination) via solving for COMBO
Ra + Rb + Rc = 13/24
Time = 24/13 = ~1.8 hours
Work problem variation #5) Relative work problems
Work rate of one object is expressed relative to the work rate of another worker
Work problem fyis
RATE is constant, but time and work completed is NOT constant
Therefore —> always anchor to using RATES in DS even if the question asks about time
If machines X and Y always work alone at their own constant rates, could Machine Y produce 100 light bulbs in less than 60 percent of the time than it would take machine X to produce the same number of light bulbs?
(1) It takes machine Y 50 percent less time than machine X to produce 4000 light bulbs
(2) Machine Y works twice as fast as Machine X
Work problem:
Machine X: Rx * time to produce 100 bulbs = 100 bulbs
Machine Y: Ry * time to produce 100 bulbs = 100 bulbs
Is time to produce 100 bulbs for X < 3/5 * time to produce 100 bulbs for Y?
ALTERNATIVELY (express in terms of RATES –> fixed across work):
Is 100 / Rx < 3/5 * (100/Ry) —> rearrange into a RATIO to make it easier to compare
Is Rx / Ry < 3/5?
(1) 4000 light bulbs —> express time in terms of RATES
4000 / Rx = 0.5 * 4000 / Ry
Rx / Ry = 0.5 —-> sufficient (always No)
(2) Ry = 2 * Rx
Rx / Ry = 0.5 —> sufficient (always No)
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