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20 fully solved word problems across 6 categories — each with a complete step-by-step solution you can reveal. Filter by type or difficulty to focus on what you’re working on.

20
Solved examples
6
Problem types
3
Difficulty levels
All examples
Rate & Distance Easy
Grade 6–7

“Maria drives from her house to the airport at 55 mph. The trip takes 2 hours. How far is the airport?”

Givenr = 55 mph, t = 2 hours
Formulad = r × t
Solved = 55 × 2 = 110
110 miles — verify: 110 ÷ 55 = 2 hrs ✓
Rate & Distance Medium
Grade 7–8

“Two cities are 360 miles apart. Car A leaves city 1 at 60 mph. Car B leaves city 2 at the same time at 90 mph toward Car A. How long until they meet?”

SetupBoth leave at same time → same t. Combined speed = 60 + 90 = 150 mph
Equation150t = 360
Solvet = 360 ÷ 150 = 2.4 hours = 2h 24min
2 hours 24 minutes — verify: 60×2.4 + 90×2.4 = 144 + 216 = 360 ✓
Rate & Distance Medium
Grade 8–9

“A cyclist leaves town at 12 mph. A car leaves the same place 30 minutes later at 48 mph in the same direction. How long before the car catches the cyclist?”

Let t= hours the car travels. Cyclist travels (t + 0.5) hours.
Equation48t = 12(t + 0.5) → 48t = 12t + 6 → 36t = 6
Solvet = 6 ÷ 36 = 1/6 hour = 10 minutes
10 minutes after the car departs — verify: 48×(1/6) = 8 mi = 12×(2/3) ✓
Rate & Distance Hard
SAT / Grade 10+

“A boat travels 24 miles downstream in 2 hours and 24 miles upstream in 3 hours. What is the speed of the current?”

Let b, c= boat speed, current speed. Downstream: b+c. Upstream: b−c.
Equations(b+c) = 24/2 = 12  |  (b−c) = 24/3 = 8
Add them2b = 20 → b = 10 mph
Solve cc = 12 − 10 = 2 mph
Current = 2 mph, boat speed = 10 mph ✓
Percentage Easy
Grade 6

“A store has 240 items. 15% are on sale. How many items are on sale?”

FormulaPart = (% ÷ 100) × Whole
SolvePart = (15 ÷ 100) × 240 = 0.15 × 240 = 36
36 items are on sale ✓
Percentage Easy
Grade 7

“James scored 42 out of 60 on a test. What percentage did he score?”

Formula% = (Part ÷ Whole) × 100
Solve% = (42 ÷ 60) × 100 = 0.7 × 100 = 70%
70% — verify: 70% of 60 = 42 ✓
Percentage Medium
Grade 8

“A jacket costs $180 and is discounted by 30%. Tax of 9% is applied to the discounted price. What is the final price?”

Discount180 × (1 − 0.30) = 180 × 0.70 = $126
Tax126 × (1 + 0.09) = 126 × 1.09 = $137.34
$137.34 final price ✓
Percentage Hard
SAT / Grade 9+

“A population of 12,000 increased by 15% one year and then decreased by 10% the following year. What is the population after two years?”

Year 112,000 × 1.15 = 13,800
Year 213,800 × 0.90 = 12,420
NoteNet change ≠ +5% — sequential % changes don’t simply add up.
12,420 — net increase is only +3.5% ✓
Ratio Easy
Grade 6

“A bag contains red and blue marbles in ratio 3:5. There are 40 marbles total. How many are red?”

Find kk = 40 ÷ (3+5) = 40 ÷ 8 = 5
Red3 × 5 = 15
15 red marbles — verify: 15 + 25 = 40 ✓
Ratio Medium
Grade 7

“A recipe for 6 servings needs 2.5 cups of flour. How much flour is needed for 15 servings?”

Proportion6/2.5 = 15/x → 6x = 37.5
Solvex = 37.5 ÷ 6 = 6.25 cups
6.25 cups — verify: 6.25 ÷ 2.5 = 2.5 = 15 ÷ 6 ✓
Ratio Hard
Grade 8–9

“Three partners share profits in ratio 2:3:5. If the largest share is $4,500, what is the total profit?”

LargestLargest share = 5k. So 5k = 4,500 → k = 900
Parts2k = 1,800  |  3k = 2,700  |  5k = 4,500
Total1,800 + 2,700 + 4,500 = 9,000
$9,000 total profit
Mixture Medium
Grade 9

“How many liters of 20% salt solution must be added to 8 liters of 50% solution to get a 30% solution?”

Let x= liters of 20% solution to add
Equation0.20x + 0.50(8) = 0.30(x + 8)
Expand0.20x + 4 = 0.30x + 2.4 → 1.6 = 0.10x
Solvex = 16 liters
16 liters of 20% solution ✓
Mixture Medium
Grade 9

“A store mixes $3/lb nuts with $7/lb nuts to make 20 lbs of mix worth $5/lb. How many pounds of each type?”

Let x= lbs of $3 nuts → (20−x) lbs of $7 nuts
Equation3x + 7(20−x) = 5×20 → 3x + 140 − 7x = 100
Solve−4x = −40 → x = 10
10 lbs at $3 and 10 lbs at $7
Mixture Hard
SAT / Grade 10+

“Pure acid is added to 40 liters of a 25% acid solution to produce a 40% solution. How much pure acid is added?”

Let x= liters of pure acid (100%) to add
Equation1.00x + 0.25(40) = 0.40(x + 40)
Expandx + 10 = 0.40x + 16 → 0.60x = 6
Solvex = 10 liters
10 liters of pure acid ✓
Work Rate Easy
Grade 8

“Worker A paints a fence in 6 hours. Worker B does it in 3 hours. How long does it take them working together?”

FormulaT = (A×B)÷(A+B) = (6×3)÷(6+3) = 18÷9 = 2
2 hours together — verify: 1/6+1/3 = 1/6+2/6 = 3/6 = 1/2 job/hr → 2 hrs ✓
Work Rate Medium
Grade 9

“Pipe A fills a tank in 10 hours. Pipes A and B together fill it in 6 hours. How long does Pipe B take alone?”

RatesA+B together = 1/6 per hr. A alone = 1/10 per hr.
B alone1/B = 1/6 − 1/10 = 5/30 − 3/30 = 2/30 = 1/15
SolveB = 15 hours
15 hours for Pipe B alone ✓
Work Rate Hard
SAT / Grade 10+

“A fill pipe fills a pool in 8 hours. A drain empties it in 12 hours. The pool is half full. Both are opened. How long to fill it completely?”

Net rate1/8 − 1/12 = 3/24 − 2/24 = 1/24 pool/hr
RemainingPool is half full → 1/2 pool left to fill
Timet = (1/2) ÷ (1/24) = (1/2) × 24 = 12 hours
12 hours to fill the remaining half ✓
Geometry Easy
Grade 6

“A rectangular garden has a length of 14 m and a width of 9 m. What is its perimeter and area?”

PerimeterP = 2(14) + 2(9) = 28 + 18 = 46 m
AreaA = 14 × 9 = 126 m²
P = 46 m, A = 126 m²
Geometry Medium
Grade 7–8

“A rectangle has a perimeter of 90 cm. Its length is 3 times its width. Find the dimensions.”

Let w= width → length = 3w
Equation2(3w) + 2(w) = 90 → 6w + 2w = 90 → 8w = 90
Solvew = 11.25 cm  |  l = 3×11.25 = 33.75 cm
Width = 11.25 cm, Length = 33.75 cm — verify: 2(33.75)+2(11.25) = 90 ✓
Geometry Hard
Grade 9 / SAT

“A circular fountain has a diameter of 10 m. A path 2 m wide surrounds it. What is the area of the path only?”

Outer rr_outer = 10/2 + 2 = 7 m  |  r_inner = 5 m
Outer Aπ × 7² = 49π ≈ 153.94 m²
Inner Aπ × 5² = 25π ≈ 78.54 m²
Path A49π − 25π = 24π ≈ 75.40 m²
≈ 75.4 m² (exactly 24π m²) ✓
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How to use these word problem examples

Each example on this page follows the same structure: a real-world word problem, a complete step-by-step solution, and a reference to the formula used. The goal is to show the full translation process — from sentence to equation to verified answer — so you can recognize the same pattern when it appears in a slightly different form on a test or homework assignment.

For the best results, try solving each problem yourself before revealing the solution. Cover the answer, work through the steps on paper, then compare your method to the worked solution. Differences in method aren’t necessarily wrong — there are often multiple valid approaches — but the underlying formula and verification step should match.

If you have a specific word problem that isn’t covered here, the word problem solver will work through any problem with the same step-by-step format. For the formulas behind each category, see the complete word problem formulas reference.

What makes a good word problem solution?

A complete solution to a math word problem has four components, not just a final number:

  • Variable definition — explicitly stating what x (or any variable) represents, with units
  • Equation setup — showing the relationship between knowns and unknowns before solving
  • Step-by-step algebra — each transition shown and justified
  • Verification — substituting the answer back into the original problem to confirm it satisfies all stated conditions

The verification step is what separates a complete solution from a guess that happens to be correct. It’s also the step most students skip — and the one that catches the most errors. Every example on this page includes a verification line for that reason.

Difficulty levels explained

Examples are labeled Easy, Medium, or Hard based on the number of steps required and the algebraic complexity of the setup — not the arithmetic difficulty.

  • Easy — one or two steps, single unknown, direct formula application. Typical Grade 6–7.
  • Medium — two to four steps, may require expressing one variable in terms of another, or applying a formula in a non-obvious direction. Typical Grade 7–9.
  • Hard — multiple steps, system of equations, sequential operations, or a conceptual twist (like sequential percentage changes not being additive). Typical Grade 9–12 and SAT.

Which problem types appear most on standardized tests?

On the SAT Math section, rate & distance problems and percentage problems appear in almost every test. Ratio and proportion appear in both calculator and no-calculator sections. The ACT includes more geometry word problems proportionally. State assessments vary by grade, but mixture and work rate problems typically begin in Grade 9 algebra courses and continue through precalculus. For a complete breakdown of formulas by type, see the word problem formulas page. For a method that works across all types, see the guide on how to solve math word problems.

Frequently asked questions

The same formula can be applied in multiple directions. d = r × t can be solved for distance, rate, or time depending on which value is unknown. Similarly, the mixture formula can be solved for the volume to add, the final concentration, or the initial concentration — the algebraic manipulation changes but the core equation stays the same. Recognizing this flexibility is what allows you to apply a single formula to many different word problem variations.
Yes — there are often multiple valid approaches to a word problem. A rate problem can be solved with d = rt applied separately or with a combined-speed equation. A ratio problem can use the k multiplier method or cross-multiplication. What matters is that your variable is clearly defined, your equation correctly represents the problem, and your answer is verified. The solutions shown here represent the most systematic approach, but they aren’t the only valid one.
Use a system of equations when a problem has two unknowns that can’t be expressed in terms of each other using only the information given in a single equation. The boat speed / current problem (example 4) is a classic case: you have two unknowns (boat speed and current speed) and two relationships (downstream rate and upstream rate), which gives two equations. If you can write one unknown as a direct function of the other — like “length is 3 times width” — a single equation is sufficient.
Research on math skill acquisition consistently shows that 8–10 problems per category is sufficient to internalize the pattern, as long as the problems vary in structure (not just the numbers). The goal is to reach the point where you can identify the problem type in the first sentence and immediately know which formula applies — that pattern recognition is what makes tests feel fast. Working through this entire page (20 examples across 6 types) covers that threshold for each category.
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