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Math word problem
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Every formula you need for math word problems — rate & distance, percentages, ratios, mixtures, work rate, and geometry — with variable definitions and a worked example for each.

Math word problems become significantly easier once you recognize which formula applies to the scenario being described. The challenge isn’t the algebra itself — it’s knowing which equation to set up before you start solving. This reference page covers every major formula category you’ll encounter in middle school, high school, and introductory college math: rate and distance problems, percentage calculations, ratio and proportion, mixture problems, combined work rate, and geometry formulas for perimeter, area, and volume.

Each formula includes a definition of every variable, notes on when to apply it, and a worked example you can expand. Use the quick calculator in the sidebar to test values instantly. When you have a specific word problem to solve, the word problem solver applies these formulas automatically and walks through every step.

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Rate & Distance Formulas

3 formulas

Rate and distance problems — sometimes called speed, distance, time problems — are among the most common word problem types on standardized tests including the SAT, ACT, and state math assessments. The core relationship is always the same: distance equals rate multiplied by time (d = r × t). The variation comes from how many objects are moving, whether they travel in the same or opposite directions, and whether one has a head start.

Basic distance formula
Use when one object travels at a constant speed for a given time.
d = r × t
d = distance  |  r = rate (speed)  |  t = time
Rearranged: r = d/t  |  t = d/r
“A car travels at 65 mph for 3 hours. How far does it go?”
Givenr = 65 mph, t = 3 hours
Applyd = 65 × 3 = 195
195 miles
Two objects — same direction (catching up)
One object starts ahead or earlier. Find when the faster one catches up.
r₁ × t₁ = r₂ × t₂
Both travel the same distance when they meet.
If one has a head start: t₂ = t₁ − head_start_time
“Train A leaves at 8am at 50 mph. Train B follows at 9am at 75 mph on the same track. When does B catch A?”
Let t= hours Train B travels. Then Train A travels (t+1) hours.
Eq.75t = 50(t+1) → 75t = 50t + 50 → 25t = 50 → t = 2
2 hours after 9am = 11:00am
Two objects — opposite directions (moving apart or toward each other)
Add the speeds to get the combined closing or separating rate.
d_total = (r₁ + r₂) × t
Moving apart: total distance = sum of distances
Moving toward: same formula — they meet when d₁ + d₂ = total gap
“Two cities are 480 miles apart. Car A leaves city 1 at 60 mph, Car B leaves city 2 at 80 mph toward each other. How long until they meet?”
Combinedspeed = 60 + 80 = 140 mph
Time= 480 ÷ 140 = 3.43 hrs ≈ 3h 26min
≈ 3 hours 26 minutes
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Percentage Formulas

4 formulas

Percentage word problems cover four main scenarios: finding the part (what is X% of Y?), finding the percentage (X is what percent of Y?), finding the whole (X is Y% of what number?), and calculating percent change (increase or decrease). Discount and tax problems combine two of these in sequence — apply the discount first, then apply tax to the discounted price. Converting percentages to decimals before calculating eliminates most arithmetic errors.

Problem type
Formula
Solve for
Find the part
Part = (% ÷ 100) × Whole
“What is 35% of 80?”
Find the percent
% = (Part ÷ Whole) × 100
“18 is what % of 72?”
Find the whole
Whole = Part ÷ (% ÷ 100)
“30 is 40% of what?”
Percent change
% change = (New − Old) ÷ Old × 100
Increase or decrease
Price after discount + tax (two-step)
Apply discount first, then tax — order matters.
Final = Original × (1 − discount%) × (1 + tax%)
Convert percentages to decimals first.
Example: 20% off = × 0.80  |  8% tax = × 1.08
“A $120 jacket is 25% off. Then 8% tax is applied. What is the final price?”
After disc.120 × 0.75 = $90
After tax90 × 1.08 = $97.20
$97.20
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Unit check: The answer to a percentage problem should always be in the same units as the “Whole”. If the whole is dollars, the part is dollars — not a bare number.
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Ratio & Proportion Formulas

3 formulas

Ratio problems ask you to divide a total into parts according to a given ratio, or to scale quantities proportionally. The key technique for ratio division is the multiplier method: assign the ratio parts as a·k and b·k, then use the known total to solve for k. Proportion problems — where two equivalent ratios are set equal — are solved by cross-multiplication. Both techniques appear frequently in recipe scaling, map distance, and unit conversion problems.

Ratio parts — finding actual quantities
When a total is divided in a given ratio a:b, use the multiplier k.
Part₁ = a·k    Part₂ = b·k    a·k + b·k = Total
Solve for k first: k = Total ÷ (a + b)
Then: Part₁ = a × k  |  Part₂ = b × k
“A class of 35 has boys to girls in ratio 3:4. How many of each?”
k= 35 ÷ (3+4) = 35 ÷ 7 = 5
Boys= 3 × 5 = 15  |  Girls = 4 × 5 = 20
15 boys, 20 girls
Cross-multiplication (solving proportions)
When two ratios are equal, cross-multiply to find the missing value.
a/b = c/d  →  a × d = b × c
Use when: “if 3 items cost $12, how much do 7 cost?”
3/12 = 7/x → 3x = 84 → x = 28
“A recipe for 4 servings uses 3 cups of flour. How much flour for 10 servings?”
Setup4/3 = 10/x → 4x = 30
Solvex = 30 ÷ 4 = 7.5
7.5 cups of flour
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Mixture Problem Formulas

2 formulas

Mixture problems involve combining two substances — solutions, coffee blends, coin types, ticket prices — to produce a result with a target concentration or average value. The core principle is that the total amount of the active ingredient (acid, caffeine, dollar value) before mixing must equal the total after mixing. Always track the substance, not the solution: multiply volume by concentration, or quantity by price, on both sides of the equation.

Mixing two solutions of different concentrations
Track the amount of pure substance (not the solution volume) on each side.
V₁·C₁ + V₂·C₂ = (V₁+V₂)·C_final
V = volume of solution  |  C = concentration (as decimal)
If one is pure water: C = 0  |  If pure substance: C = 1
“How many liters of 30% acid solution must be added to 10 liters of 60% solution to get a 45% solution?”
Let x= liters of 30% solution to add
Eq.0.30x + 0.60(10) = 0.45(x + 10)
Solve0.30x + 6 = 0.45x + 4.5 → 1.5 = 0.15x → x = 10
10 liters of 30% solution
Mixing two items at different prices (cost mixture)
Coffee blends, coin problems, ticket prices — same structure as concentration.
Q₁·P₁ + Q₂·P₂ = Q_total · P_avg
Q = quantity  |  P = price per unit
Often Q₁ + Q₂ = Q_total is a second equation to form a system
“Blend $4/lb and $6/lb coffee to make 20 lbs worth $4.70/lb. How much of each?”
Let x= lbs of $4 coffee → (20−x) lbs of $6
Eq.4x + 6(20−x) = 4.70×20 → 4x + 120 − 6x = 94
Solve−2x = −26 → x = 13
13 lbs at $4 and 7 lbs at $6
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Work Rate Formulas

2 formulas

Work rate problems — also called combined work problems or pipe and cistern problems — ask how long it takes multiple workers (or pipes) to complete a job together. The key insight is that rates add: if worker A completes 1/6 of a job per hour and worker B completes 1/4, together they complete 1/6 + 1/4 = 5/12 per hour, so the job takes 12/5 = 2.4 hours. When one agent fills and another drains, subtract their rates instead.

Two workers completing a job together
Each worker’s rate = fraction of job done per unit time. Add rates together.
1/A + 1/B = 1/T
A = hours for worker A alone  |  B = hours for worker B alone
T = hours together → T = AB ÷ (A+B)
“Pipe A fills a tank in 6 hours. Pipe B fills it in 4 hours. How long do both together take?”
RatesA does 1/6 per hr, B does 1/4 per hr
TogetherT = (6×4)÷(6+4) = 24÷10 = 2.4 hrs
2 hours 24 minutes
One worker filling, one worker draining
Subtract the draining rate from the filling rate.
1/T = 1/fill − 1/drain
If filling rate > draining rate: tank eventually fills
If draining rate > filling rate: tank eventually empties
“A pipe fills a pool in 8 hours. A drain empties it in 12 hours. If both are open, how long to fill?”
Net rate1/8 − 1/12 = 3/24 − 2/24 = 1/24
TimeT = 24 hours
24 hours to fill
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Geometry Word Problem Formulas

8 formulas

Geometry word problems give you one or more measurements of a shape and ask you to find another — a missing side, perimeter, area, or volume. The most common pattern is a perimeter problem where a relationship between dimensions is described in words (“the length is 5 meters more than twice the width”) and must be substituted into the perimeter formula to create a solvable equation. Always sketch and label the shape before writing any equation.

Shape
Perimeter / Circumference
Area / Volume
Rectangle
P = 2l + 2w
A = l × w
Square
P = 4s
A = s²
Triangle
P = a + b + c
A = ½ × b × h
Circle
C = 2πr = πd
A = πr²
Rectangular box
SA = 2(lw+lh+wh)
V = l × w × h
Cylinder
SA = 2πr² + 2πrh
V = πr²h
Pythagorean theorem
a² + b² = c²  (c = hypotenuse)
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Word problem pattern: When a problem gives perimeter and a relationship between dimensions (e.g. “length is 3 more than twice the width”), substitute the relationship into the perimeter formula and solve. Always draw a labeled diagram first.
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How to choose the right formula for a word problem

The most common difficulty students face with word problems isn’t applying the formula — it’s identifying which formula to use. A reliable approach is to read the problem twice and ask two questions: what type of quantities are involved, and what relationship connects them?

If the problem mentions speed, distance, or travel time, reach for d = r × t. If it mentions “percent of”, “discount”, or “increase by”, it’s a percentage problem. If it describes dividing something into parts according to a given ratio, use the multiplier method. If it combines two substances or products at different values, it’s a mixture problem. If two agents work on the same task simultaneously, it’s a combined work rate problem. And if it asks about measurements of a physical shape, identify the shape and apply the corresponding geometry formula.

For a detailed walkthrough of how to translate word problem language into equations — including what to do when the problem type isn’t immediately obvious — see the complete guide on how to solve math word problems step by step.

Which formulas appear most on standardized tests?

On the SAT Math section, percentage problems and rate-distance problems are the two most frequently tested word problem types. Ratio and proportion appear regularly in both the calculator and no-calculator sections. The ACT tends to include more geometry word problems — particularly perimeter and area problems where a relationship between dimensions must be set up as an equation. State math assessments vary by grade level, but mixture and work-rate problems typically appear in 8th grade and above.

For practice problems with full solutions organized by problem type, see the full word problem examples with solutions.

Tips for applying formulas correctly

Three habits prevent most formula-application errors. First, always write out your variable definitions before setting up the equation — “let x = the number of liters of 30% solution” is far less error-prone than keeping it implicit. Second, carry units through every calculation step. Units act as a self-check: if you’re multiplying mph by hours, the result should be miles; if it isn’t, something is wrong with the setup. Third, substitute your answer back into the original problem and verify that all stated conditions hold — this catches arithmetic errors and wrong-variable answers before they cost points.

— common questions

Frequently asked questions

d = r × t (distance equals rate times time) is the single most tested formula across all grade levels and standardized tests. It underpins every rate and distance problem, and its structure — two known values multiplied to find a third — is the same pattern used in percentage problems (Part = % × Whole) and work rate problems. Mastering d = r × t and understanding why it works makes the other formula types significantly easier to learn.
Add rates when both agents work toward the same goal (two pipes filling a tank, two workers building a fence). Subtract rates when one agent works against the other (a fill pipe and a drain pipe, a worker building and a separate worker removing). The net rate tells you how quickly the overall job progresses — positive means it will eventually complete, negative means it will never complete.
A ratio problem divides a known total into parts according to a given ratio — you use the multiplier k method. A proportion problem sets two equivalent ratios equal to each other and asks you to find a missing value — you use cross-multiplication. Many word problems combine both: a recipe uses a ratio of ingredients (ratio problem) and asks how much you’d need to scale it up (proportion problem).
Because the percentages apply to different volumes. Adding 30% + 60% would give 90%, which is meaningless without accounting for how much of each solution you’re combining. By multiplying volume × concentration, you calculate the actual amount of the active substance in each component. When you add those amounts and divide by the total volume, you get the true final concentration — regardless of the volumes used.
For tests that allow a reference sheet, no. For tests that don’t (like most state math assessments and the SAT without the formula sheet), the most important ones to memorize are: d = r × t, the percentage formula, cross-multiplication for proportions, and the basic geometry formulas for rectangles, triangles, and circles. Work rate and mixture formulas can usually be re-derived quickly if you understand the underlying logic — rates add, and you’re always tracking the substance rather than the solution.