Rate-based questions are a cornerstone of the CAT Quant section. Whether dealing with cars on a highway or workers on a project, the underlying logic is the same: Output = Rate × Time. This guided path will teach you the core frameworks and advanced shortcuts for mastering both Time, Speed, & Distance and Time & Work problems with speed and confidence.
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 CAT 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" (Man-Days) as a powerful shortcut for problems involving groups of workers, a common CAT 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 work 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:
$$ 5 \times 3 = \mathbf{15 \text{ days}} $$Master the fundamental Distance = Speed × Time relationship, explore the 3 algebraic rearrangements, and practice instant unit conversions with our interactive tool.
All motion questions test the relationship between three variables: Distance, Speed (Rate), and Time.
Distance = Speed × Time
Speed = Distance / Time
Time = Distance / Speed
Learn the crucial difference between simple average of speeds and the correct "Total Distance / Total Time" formula for complex journeys.
Car travels A to B at 40 km/hr.
Returns B to A at 60 km/hr.
What is the average speed?
Master the two core scenarios for relative speed (objects moving in opposite vs. same directions) using a visual "race" simulator.
A (2 m/s) and B (3 m/s) are 100m apart.
Apply relative speed concepts to the classic, tricky CAT problem types involving trains crossing platforms, boats in streams, and escalators.
A dedicated question set for Rate Problems is currently being authored. In the meantime, test your baseline score on the full Diagnostic:
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