"Work Rate" questions are a common feature of the GRE Quant section. Success depends on quickly choosing the right framework and using logical shortcuts to save time. This guided path will teach you the core methods and advanced strategies you need.
Start with the fundamental principle: converting a total time to complete a job into a rate of work per unit of time (e.g., per hour or day).
Don't add times directly! Convert to Rate (Work per 1 Day).
Takes 10 Days
Takes 15 Days
Rate = 1 / Time
A's Rate: $$ \frac{1}{10} $$ job/day
B's Rate: $$ \frac{1}{15} $$ job/day
Add fractions:
$$ \frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} $$Together, they do 1/6 of the job per day.
Flip the combined rate back to time.
$$ \text{Time} = \frac{1}{\text{Rate}} = \frac{1}{1/6} = \mathbf{6 \text{ Days}} $$Learn how to handle common GRE scenarios with opposing forces, like a draining pipe, by simply using a negative rate.
A drain works against the filling pipes. Treat its rate as Negative.
A: $$ +\frac{1}{12} $$
B: $$ +\frac{1}{18} $$
C: $$ -\frac{1}{24} $$
Flip the net rate:
$$ \text{Time} = \frac{72}{7} \approx \mathbf{10.28 \text{ Hours}} $$Use the concept of "Total Effort" as a powerful shortcut for problems involving groups of workers, a common GRE word problem format.
For groups, calculate Total Effort (Men × Days). This total is constant.
"A project requires 25 people to complete it in 8 days."
If we only have 20 workers?
$$ 20 \times D = 200 $$If we must finish in 5 days?
$$ M \times 5 = 200 $$Master the elegant formula for solving complex problems where both the inputs (workers, time) and the output (work done) change.
When the Output (amount of work) changes, use the ratio:
$$ \frac{\text{Effort}_1}{\text{Output}_1} = \frac{\text{Effort}_2}{\text{Output}_2} $$20 workers, 8 hrs/day, 13 days make 100 toys.
How many days for same workers, 6 hrs/day to make 150 toys?
Case 1: Effort = $$ 20 \times 8 \times 13 $$, Output = 100
Case 2: Effort = $$ 20 \times 6 \times D $$, Output = 150
Simplify LHS: $$ \frac{2080}{100} = 20.8 $$
$$ 20.8 = \frac{120D}{150} $$
$$ D = \frac{20.8 \times 150}{120} = \mathbf{26 \text{ Days}} $$
Learn the three-step process to solve problems that compare the relative efficiency of different workers using ratios.
A more efficient worker takes LESS time. The Time Ratio is the inverse of the Work Ratio.
Worker B is 40% more efficient than Worker A.
Worker A takes 21 days.
How long does Worker B take?
If A does 100 units, B does 140.
$$ A : B = 100 : 140 = 5 : 7 $$Time is the inverse of Work.
Time Ratio A : B = 7 : 5
We know A takes 21 days (which is 7 parts).
$$ 7 \text{ parts} = 21 \text{ days} \implies 1 \text{ part} = 3 \text{ days} $$B takes 5 parts:
$$ 5 \times 3 = \mathbf{15 \text{ days}} $$Learn to solve QC work-rate problems with logical reasoning, which is often much faster than using the on-screen calculator.
Machine A: 6 hours. Machine B: 4 hours.
Put your understanding of GRE Work Rate Problems: A Guide to Efficient Problem Solving to the test with our premium, adaptive practice questions. Build pacing efficiency and eliminate knowledge gaps now.
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