Professional + Subprofessional

Civil Service Exam word problems: recognize the type, apply the formula.

Last reviewed May 2026

Word problems are one of the three official topics in Numerical Ability (alongside basic operations and number sequence), and they look intimidating because each one is wrapped in a paragraph of narrative. The trick is that most of them fall into five recurring types (work, distance, age, mixture, percentage), and each type has a standard formula. Once you can recognize the type from the first sentence, the rest is mechanical. The strongest candidates on word problems don't compute faster. They recognize faster. Note: the CSC does not publish a per-subtest or per-topic breakdown, so any item-count or percentage figures you see online are estimates, not official.

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Quick facts

Primary subtest: Numerical Ability
Calculator: Not allowed
Official scope: Word problems (Numerical Ability)
Difficulty to improve: Medium. Five formulas.

The five types, and how to recognize each

Every word problem on the exam is one of five types. Learn the recognition cues in the first column and the formula in the last, and you can name the type from the opening sentence.

The five word-problem types and their formulas
Type	Recognition cues	Standard formula
Work	Two or more workers, pipes, or machines doing the same task; "alone," "together," or a rate (per hour, per day).	If A takes a hours alone and B takes b hours alone, together they take ab/(a+b) hours.
Distance	A vehicle, person, or object moving; "speed," "rate," "km/h," "miles per hour."	distance = rate × time. Three structures: meeting, chase, round trip.
Age	"How old will she be," "twice as old as," "x years ago," "in x years."	Set up one equation using current ages, then translate the time shift into +x or −x.
Mixture	"Solution," "concentration," "percent alcohol/salt/acid," "liters of X% mixed with liters of Y%."	Pure substance = volume × concentration. Conserve total pure substance before and after.
Percentage	"Discount," "markup," "interest," "tax," "commission," "increased by," "decreased by."	Identify the original (the base) first, then apply the factor.

Percentage word problems The mechanics match basic-operations percentages, but the narrative often hides which value is the original. Find the base before you apply any factor.

Work problems: the most common type

Standard formula for two workers. If A takes a hours alone and B takes b hours alone, combined time is: combined = ab / (a + b) Example. A takes 6 hours, B takes 12 hours. combined = (6 × 12) / (6 + 12) = 72 / 18 = 4 hours

For three workers, work with rates instead of times: 1/a + 1/b + 1/c = 1/combined For pipes where one fills and another drains, subtract the drain rate. Example. Pipe A fills in 4 hours; drain B empties in 12 hours. Net rate = 1/4 − 1/12 = 3/12 − 1/12 = 2/12 = 1/6 Combined fill time = 6 hours

Reverse work problems give you the combined time and ask for one worker's solo time. "Together in 6 hours; A alone in 10 hours; how long does B alone take?" 1/10 + 1/b = 1/6 1/b = 1/6 − 1/10 = 5/30 − 3/30 = 2/30 = 1/15 b = 15 hours

Distance problems: the three sub-structures

Meeting. Two objects start apart and move toward each other. Add the speeds. time to meet = distance apart / combined speed Example. Car A at 60 km/h, car B at 80 km/h, 350 km apart. combined speed = 60 + 80 = 140 km/h time to meet = 350 / 140 = 2.5 hours

Chase. One object catches up to another. Subtract the speeds (relative speed). time to catch = head start / relative speed Example. A leaves at 8 AM at 60 km/h. B leaves at 9 AM at 80 km/h. A's head start at 9 AM = 60 km relative speed = 80 − 60 = 20 km/h time to catch = 60 / 20 = 3 hours after B leaves → 12 noon

Round trip. Same distance both ways, different speeds. Use the harmonic-mean formula: avg speed = 2ab / (a + b) This is NOT the arithmetic mean. Example. Out at 40 km/h, back at 60 km/h. avg = 2(40)(60) / (40 + 60) = 4800 / 100 = 48 km/h ← not 50

Age problems: the equation setup

Always work with current ages as variables. Translate the future or past statement into an equation. Example. "Maria is twice as old as Pedro. In 5 years, she'll be 1.5 times as old." Let Pedro's current age = p Maria's current age = 2p In 5 years: 2p + 5 = 1.5(p + 5) 2p + 5 = 1.5p + 7.5 0.5p = 2.5 p = 5 So Pedro is 5 now, Maria is 10.

For "x years ago" problems, subtract from current ages. Example. "Five years ago, the sum of their ages was 30." If current ages are M and P: (M − 5) + (P − 5) = 30 M + P − 10 = 30 M + P = 40

Common trap: forgetting to apply the time shift to BOTH ages. "In 10 years, Maria will be twice as old as Pedro IS NOW" → M + 10 = 2P (Pedro stays at his current age) "In 10 years, Maria will be twice as old as Pedro" → M + 10 = 2(P + 10) (both ages shift) The difference is one word. Read the sentence twice before writing the equation.

Mixture problems: the conservation principle

The amount of pure substance is conserved across the mix. If you have V1 liters at concentration c1 and add V2 liters at c2, the result is: final concentration = (V1·c1 + V2·c2) / (V1 + V2)

Example. 10 liters of 20% acid + 30 liters of 50% acid. Pure acid from first: 10 × 0.20 = 2 Pure acid from second: 30 × 0.50 = 15 Total pure acid: 2 + 15 = 17 liters Total volume: 10 + 30 = 40 liters Final concentration: 17 / 40 = 42.5%

Reverse mixture. "How many liters of 70% solution must be added to 20 liters of 30% to get 50%?" Let x = liters added. Conservation: (20 × 0.30) + (x × 0.70) = 0.50 × (20 + x) 6 + 0.70x = 10 + 0.50x 0.20x = 4 x = 20 liters

How to drill

Three weeks is enough to turn word problems from a weakness into a strength. Each week has a single focus, and they build on each other.

1
Week one: identify the type before solving
Take twenty word problems a day, untimed. For each, write down the type before doing any computation. "This is a work problem. Standard formula ab/(a+b)." By day five, type-identification should take under five seconds.
2
Week two: drill one type per day
Monday work, Tuesday distance, Wednesday age, Thursday mixture, Friday percentage. Twenty items per day, focused on one type. Look up every formula you miss and keep a single-page formula sheet.
3
Week three: mixed timed sets
Twenty-five word problems in twenty-five minutes. This is exam pace. Track which type slows you down. That's the type to drill again the following day.

Worked examples

These items are written specifically for this guide. The actual practice bank pulls from a separate pool of original CSE-style items reviewed by passers.

Item 01
A pipe can fill a tank in 4 hours. Another pipe can fill it in 6 hours. A drain can empty the tank in 12 hours. If all three are open at once, how long does it take to fill the tank?
- A2 hours
- B3 hoursCorrect
- C4 hours
- D5 hours
Solution
1. 1
  Write each rate as fraction of tank per hour
  Pipe 1 fills: +1/4 Pipe 2 fills: +1/6 Drain empties: −1/12
2. 2
  Find LCD (12) and rewrite
  +1/4 = 3/12 +1/6 = 2/12 −1/12 = −1/12
3. 3
  Combine the rates
  3/12 + 2/12 − 1/12 = 4/12 = 1/3 tank per hour
4. 4
  Invert to get total time
  If 1/3 of the tank fills per hour, total time = 3 hours
Answer
3 hours
Trap to watch. Forgetting to subtract the drain rate gives 1/4 + 1/6 = 5/12, fill time 12/5 = 2.4 hours, close to option A. Always identify which rates add and which subtract BEFORE computing.
Item 02
A car travels from city A to city B at 60 km/h and returns at 40 km/h. What is the average speed for the entire trip?
- A48 km/hCorrect
- B50 km/h
- C52 km/h
- D55 km/h
Solution
1. 1
  Recognize the round-trip structure
  Same distance both ways, different speeds. Formula: avg = 2ab / (a + b)
2. 2
  Plug in the values
  a = 60, b = 40 avg = 2(60)(40) / (60 + 40)
3. 3
  Compute
  = 4800 / 100 = 48 km/h
Answer
48 km/h
Trap to watch. Option B (50 km/h) is the arithmetic mean of the two speeds. That's wrong because more time is spent at the slower speed (the return), so the average pulls toward 40, not toward the midpoint.
Item 03
Maria is 3 times as old as her son. In 12 years, she will be only twice as old. How old is Maria now?
- A30
- B36Correct
- C40
- D45
Solution
1. 1
  Define variables
  Let son's current age = s Maria's current age = 3s
2. 2
  Write the future condition (in 12 years)
  Maria's future age = 3s + 12 Son's future age = s + 12 Maria = 2 × son → 3s + 12 = 2(s + 12)
3. 3
  Expand and solve
  3s + 12 = 2s + 24 3s − 2s = 24 − 12 s = 12
4. 4
  Compute Maria's age
  Maria = 3s = 3 × 12 = 36
5. 5
  Verify
  In 12 years: Maria = 48, son = 24, and 48 = 2 × 24 ✓
Answer
Maria is 36
Trap to watch. The most common error is to apply the +12 only to Maria's side: 3s + 12 = 2s (treating son's age as unchanged in the comparison). Both ages shift when the time shifts.
Item 04
How many liters of a 30% salt solution must be added to 20 liters of a 60% salt solution to produce a 40% solution?
- A10 liters
- B20 liters
- C30 liters
- D40 litersCorrect
Solution
1. 1
  Define the unknown
  Let x = liters of 30% solution to add.
2. 2
  Write the salt-conservation equation
  salt from 30% sol + salt from 60% sol = salt in final mix 0.30x + 0.60(20) = 0.40(x + 20)
3. 3
  Expand
  0.30x + 12 = 0.40x + 8
4. 4
  Solve for x
  12 − 8 = 0.40x − 0.30x 4 = 0.10x x = 40 liters
5. 5
  Verify
  Salt: 40 × 0.30 + 20 × 0.60 = 12 + 12 = 24 L Volume: 40 + 20 = 60 L Concentration: 24/60 = 40% ✓
Answer
40 liters of the 30% solution
Trap to watch. The trap is to assume the answer is some simple proportion of 20 liters (10 or 20). Always set up the conservation equation; intuition fails on mixture items.
Item 05
A merchant marks up an item by 40% and then offers a 20% discount on the marked price. What is the merchant's profit percentage relative to the original cost?
- A12%Correct
- B15%
- C20%
- D30%
Solution
1. 1
  Let the cost be 100 for easy arithmetic
  Cost = 100
2. 2
  Apply 40% markup
  Marked price = 100 × 1.40 = 140
3. 3
  Apply 20% discount to marked price
  Sale price = 140 × 0.80 = 112
4. 4
  Compute profit
  Profit = 112 − 100 = 12 Profit % = 12/100 = 12%
Answer
12% profit
Trap to watch. Option C (20%) subtracts percentages directly (40% − 20%). That's wrong because the 20% discount applies to the higher marked price (140), not the cost (100). Always multiply factors: 1.40 × 0.80 = 1.12 → 12% profit.
Item 06
Two trains start from stations 480 km apart and travel toward each other. Train A travels at 70 km/h, train B at 50 km/h. How long until they meet?
- A3 hours
- B4 hoursCorrect
- C5 hours
- D6 hours
Solution
1. 1
  Identify the structure
  Two objects moving TOWARD each other → add the speeds.
2. 2
  Compute combined closing speed
  70 + 50 = 120 km/h
3. 3
  Divide the gap by the combined speed
  480 ÷ 120 = 4 hours
4. 4
  Verify
  In 4 hours: A travels 70 × 4 = 280 km B travels 50 × 4 = 200 km 280 + 200 = 480 km ✓
Answer
4 hours
Trap to watch. Add speeds for meeting (toward each other). Subtract speeds for chase (one catches up). Knowing which to do is half the battle.
Item 07
Worker A can finish a job in 8 days. Worker B can finish the same job in 12 days. They work together for 3 days, then worker A leaves. How many more days does worker B need to finish alone?
- A3 days
- B4 days
- C4.5 daysCorrect
- D5 days
Solution
1. 1
  Phase 1: both workers together
  Combined rate: 1/8 + 1/12 LCD 24: 3/24 + 2/24 = 5/24 of job per day
2. 2
  Compute work done in 3 days together
  3 × 5/24 = 15/24 = 5/8 of job done
3. 3
  Compute remaining work
  1 − 5/8 = 3/8 still to do
4. 4
  Phase 2: B alone
  B's rate alone: 1/12 of job per day Time to do 3/8 at rate 1/12: (3/8) ÷ (1/12) = (3/8) × 12 = 36/8 = 4.5 days
Answer
4.5 more days
Trap to watch. For multi-step problems, break the timeline into phases. Compute work done in each phase, then move to the next. The single-phase formula ab/(a+b) doesn't apply when one worker leaves partway through.

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Study tactics that actually move the score

01
Identify the type before doing any math. "This is a work problem" or "This is a mixture problem" should be your first thought, not the variables. Type-recognition saves more time than computation speed.
02
Memorize the five standard formulas: ab/(a+b) for work, distance = rate × time, 2ab/(a+b) for round-trip average speed, conservation of pure substance for mixture, and factor multiplication for percentage.
03
Always check your answer by plugging back into the problem conditions. Word problems are easy to mis-set-up, but they're hard to mis-verify. The check catches setup errors quickly.
04
For multi-step problems, write out each phase before computing. "Phase 1: both workers, 3 days, 5/8 done. Phase 2: B alone, remaining 3/8." The visual breakdown prevents you from skipping the conservation step.
05
Test the answer options when the algebra is messy. Plugging in is often faster than solving, especially for items with integer answers and four reasonable options.

Frequently asked questions

How many word problems appear on the exam?

The CSC does not publish a per-topic item count, so no exact number is official. What is official: the Professional paper has 170 items total and the Subprofessional paper has 165, and word problems are one of the three named topics under Numerical Ability. Treat any specific count you see online as an estimate. Plan to be solid on all five types rather than betting on a particular tally.

Which type is most common?

No official source breaks word problems down by type, so claims that one type appears more than another are not verifiable. Practically, the safe move is to be comfortable with all five types (work, distance, age, mixture, percentage) so the exam's mix doesn't matter.

Are there problems that combine multiple types?

Occasionally yes. A percentage discount applied to a distance-based pricing scheme, for example. Identify the dominant type first (usually the one driving the question being asked), then handle the secondary type as a sub-step.

Should I memorize formulas or derive them?

Memorize them. The exam doesn't reward derivation under time pressure. The five formulas listed in this guide cover the great majority of word problems you are likely to see; learn them cold.

What's the most common setup error?

Forgetting to apply the time shift to both sides of an age equation. "In 5 years, A will be twice as old as B" almost always means (A + 5) = 2(B + 5), not (A + 5) = 2B. When in doubt, write the variable expression for both people's ages at the relevant time, then write the relationship.

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The five word-problem types and their formulas

Type

Recognition cues

Standard formula

Work

Two or more workers, pipes, or machines doing the same task; "alone," "together," or a rate (per hour, per day).

If A takes a hours alone and B takes b hours alone, together they take ab/(a+b) hours.

Distance

A vehicle, person, or object moving; "speed," "rate," "km/h," "miles per hour."

distance = rate × time. Three structures: meeting, chase, round trip.

Age

"How old will she be," "twice as old as," "x years ago," "in x years."

Set up one equation using current ages, then translate the time shift into +x or −x.

Mixture

"Solution," "concentration," "percent alcohol/salt/acid," "liters of X% mixed with liters of Y%."

Pure substance = volume × concentration. Conserve total pure substance before and after.

Percentage

"Discount," "markup," "interest," "tax," "commission," "increased by," "decreased by."

Identify the original (the base) first, then apply the factor.