Rate problems cover two families that share one formula: work problems (pipes, printers, machines, people) and distance problems (trains, cyclists, round trips). In both, amount = rate × time. On the GMAT Focus they appear in Quant Problem Solving and in Data Insights, and you solve the Quant ones without a calculator, so clean setups matter.
For work, convert every time into a rate. A printer that finishes a job in 6 hours does 1/6 of the job per hour. Rates of machines working together add; a machine working against another (a drain, a leak) subtracts. Once you have the combined rate, time = 1 ÷ combined rate. For distance, draw a quick timeline and write one equation per traveller. Two objects moving toward each other close the gap at the sum of their speeds; moving in the same direction, at the difference.
Average speed is total distance over total time, never the simple average of the speeds. A car that goes out at 40 and back at 60 averages 48, not 50, because it spends longer at the slower speed.
Most wrong answers are unit slips or intermediate values: minutes mixed with hours, the time one machine works alone instead of the total, or the distance to the meeting point instead of the time. Write the target quantity down before you start.
4 practice questions
Problem SolvingWork/Rate ProblemsEasy
Question 1
Pump A alone can fill a tank in $6$ hours, and pump B alone can fill it in $9$ hours. Both pumps start filling the empty tank together, and after $2$ hours pump A is shut off while pump B continues alone. How many hours after the start does the tank become full?
A$4$
B$4.5$
C$5$
D$5.5$
E$6$
Show answer & explanation
Correct answer: E
Key Concept or Approach
Add rates while pumps work together, then find the time for B to finish the remaining fraction.
Step-by-Step Solution
Step 1: Combined rate: $\frac{1}{6}+\frac{1}{9}=\frac{5}{18}$ tank per hour; in 2 hours they fill $\frac{10}{18}=\frac{5}{9}$.
Step 2: Remaining: $\frac{4}{9}$ of the tank. B needs $\frac{4}{9}\div\frac{1}{9}=4$ hours.
Step 3: Total time from start: $2 + 4 = 6$ hours.
Final Answer
Answer: E
Common trap: 4 - the time B works alone, forgetting to add the first 2 hours.
Data SufficiencyDistance/Rate ProblemsEasy
Question 2
A cyclist rode a $36$-kilometer route. What was the cyclist's average speed, in kilometers per hour, for the whole route?
(1) The cyclist rode the first $18$ kilometers at $10$ kilometers per hour and the remaining $18$ kilometers at $15$ kilometers per hour.
(2) Had the cyclist's average speed been $6$ kilometers per hour greater, the route would have taken $1$ hour less.
AStatement (1) ALONE is sufficient, but statement (2) ALONE is NOT sufficient.
BStatement (2) ALONE is sufficient, but statement (1) ALONE is NOT sufficient.
CBOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
DEACH statement ALONE is sufficient.
EStatements (1) and (2) TOGETHER are NOT sufficient.
Show answer & explanation
Correct answer: D
Key Concept or Approach
Average speed $=$ total distance $\div$ total time. Any statement that fixes the total time, or gives one equation in the unknown speed with a single positive root, is sufficient.
Step-by-Step Solution
Step 1: Statement (1): time $= \frac{18}{10} + \frac{18}{15} = 1.8 + 1.2 = 3$ hours, so average speed $= \frac{36}{3} = 12$ km/h. Sufficient.
Step 2: Statement (2): with average speed $v$, $\frac{36}{v} - \frac{36}{v+6} = 1$. Then $\frac{216}{v(v+6)} = 1$, so $v^2 + 6v - 216 = 0$, i.e. $(v - 12)(v + 18) = 0$. Since $v > 0$, $v = 12$. Sufficient.
Step 3: Each statement alone gives $12$ km/h.
Final Answer
Answer: D
Common trap: A: students see a quadratic in (2) and assume two possible speeds, overlooking that one root (-18) is negative and impossible.
Problem SolvingDistance/Rate ProblemsMedium
Question 3
A courier drives from a depot to a client at an average speed of $30$ miles per hour and returns along the same route at an average speed of $45$ miles per hour. What is the courier's average speed, in miles per hour, for the entire round trip?
A$35$
B$36$
C$37.5$
D$38$
E$40$
Show answer & explanation
Correct answer: B
Key Concept or Approach
Average speed = total distance / total time. Equal distances, so do not average the speeds.
Step-by-Step Solution
Step 1: Let each leg be $90$ miles. Time out: $3$ h; time back: $2$ h.
Step 2: Total: $180$ miles in $5$ hours.
Step 3: Average speed $= 180/5 = 36$ mph.
Final Answer
Answer: B
Common trap: 37.5 - the simple average of 30 and 45, which ignores that more time is spent at the slower speed.
Data SufficiencyWork/Rate ProblemsMedium
Question 4
Machines A and B each pack boxes at their own constant rate. How many hours would it take machine A, working alone, to pack $1{,}200$ boxes?
(1) Working together, machines A and B pack $1{,}200$ boxes in $4$ hours.
(2) Machine A packs $100$ more boxes per hour than machine B.
AStatement (1) ALONE is sufficient, but statement (2) ALONE is NOT sufficient.
BStatement (2) ALONE is sufficient, but statement (1) ALONE is NOT sufficient.
CBOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
DEACH statement ALONE is sufficient.
EStatements (1) and (2) TOGETHER are NOT sufficient.
Show answer & explanation
Correct answer: C
Key Concept or Approach
Let $a$ and $b$ be the hourly rates. The question asks for $\frac{1200}{a}$, so we need $a$.
