The 5-Step Framework
Every math word problem, no matter how complex, can be approached with the same five steps.
- Step 1 — Read carefully.
Read the entire problem at least twice before writing anything. Identify what is being asked. Underline or highlight key information: quantities, relationships, and units.
- Step 2 — Define variables.
Assign a letter to each unknown quantity. Be specific: write "let x = Maria's current age" rather than just "let x = age." This prevents confusion later.
- Step 3 — Translate to equations.
Convert the English sentences into algebraic equations. Words like "is" mean equals, "more than" means addition, "less than" means subtraction, "of" often means multiplication, and "per" indicates division.
- Step 4 — Solve the equations.
Use the appropriate algebraic method: solving linear equations, systems of equations, or quadratic equations.
- Step 5 — Check and interpret.
Plug your answer back into the original problem (not just the equation) to verify it makes sense. State the answer in a complete sentence with units.
This framework works because it breaks the intimidating task of translating English into math into manageable, repeatable steps.
Age Problems
Age problems are among the most common word problem types. They typically involve the ages of two or more people at different points in time.
"Sarah is 5 years older than Tom. In 3 years, Sarah's age will be twice Tom's age. How old are they now?"
Step 1: We need to find Sarah's and Tom's current ages.
Step 2: Let t = Tom's current age. Then Sarah's current age = t + 5.
Step 3: In 3 years, Sarah will be (t + 5) + 3 = t + 8, and Tom will be t + 3. The condition says Sarah's future age is twice Tom's: t + 8 = 2(t + 3).
Step 4: t + 8 = 2t + 6 → 8 − 6 = 2t − t → t = 2. So Tom is 2 and Sarah is 7.
Step 5: Check: In 3 years, Sarah is 10 and Tom is 5. Is 10 twice 5? Yes.
Tom = 2, Sarah = 7The key to age problems is setting up a clear table with "now" and "future" (or "past") columns. This organizes the information visually and makes writing equations straightforward.
Solver AI handles age problems beautifully — snap a photo and see the variable setup, equation, and solution.
Distance, Rate, and Time Problems
distance = rate × time (or d = rt). They appear in many forms: two objects traveling toward or away from each other, round trips, or problems involving current or wind speed."A car travels from City A to City B at 60 mph and returns at 40 mph. If the round trip takes 5 hours, what is the distance between the cities?"
Step 2: Let d = distance between cities.
Step 3: Time going = d/60, time returning = d/40. Total time = 5 hours: d/60 + d/40 = 5.
Step 4: Find the LCD (120): 2d/120 + 3d/120 = 5 → 5d/120 = 5 → d = 120 miles.
Step 5: Check: time going = 120/60 = 2 hours, time returning = 120/40 = 3 hours, total = 5 hours.
d = 120 milesFor problems with two objects, draw a diagram showing starting positions and directions of travel. For current/wind problems, remember that the effective rate is r + c downstream and r − c upstream. Always use a table with columns for distance, rate, and time to keep the information organized.
Mixture Problems
"How many liters of a 30% acid solution must be mixed with 10 liters of a 60% acid solution to produce a 50% acid solution?"
Step 2: Let x = liters of 30% solution.
Step 3: Acid from 30% solution: 0.30x. Acid from 60% solution: 0.60(10) = 6. Total mixture: (x + 10) liters at 50% acid. Equation: 0.30x + 6 = 0.50(x + 10).
Step 4: 0.30x + 6 = 0.50x + 5 → 1 = 0.20x → x = 5 liters.
Step 5: Check: 0.30(5) + 0.60(10) = 1.5 + 6 = 7.5. Total mixture: 15 liters. 7.5/15 = 0.50 = 50%. Correct.
x = 5 litersMixture problems also appear with prices: mixing two types of coffee at different prices per pound to achieve a target blend price. The same framework applies — just replace concentration with price per unit. A table with columns for amount, concentration, and total ingredient makes these problems manageable.
Work-Rate Problems
1/A + 1/B jobs per hour."Pipe A fills a tank in 6 hours. Pipe B fills the same tank in 4 hours. How long will it take both pipes working together?"
Step 2: Let t = time for both pipes together.
Step 3: Pipe A's rate = 1/6 tank per hour, Pipe B's rate = 1/4 tank per hour. Together: 1/6 + 1/4 = 1/t.
Step 4: Find LCD: 2/12 + 3/12 = 5/12 = 1/t → t = 12/5 = 2.4 hours (or 2 hours 24 minutes).
Step 5: Check: In 2.4 hours, Pipe A fills 2.4/6 = 0.4 of the tank, Pipe B fills 2.4/4 = 0.6 of the tank, total = 1.0. Correct.
t = 2.4 hoursVariations include problems where one person starts before the other, or where one person is draining while the other fills. For draining, use a negative rate.
Work-rate problems have a satisfying structure — once you recognize the 1/A + 1/B = 1/T pattern, they become quite straightforward.
Percent Problems
part = percent × whole, or equivalently percent = part / whole."A shirt originally costs $40. It is marked up by 25%. What is the new price?"
New price = 40 + 0.25(40) = 40 + 10 = $50.
Shortcut: 40 × 1.25 = $50.
$50"After a 20% discount, a laptop costs $640. What was the original price?"
If the discount is 20%, the customer pays 80%. So 0.80 × original = 640, giving original = 640/0.80 = $800.
$800"A stock goes up 10% one year and down 10% the next. Is it back to the original price?"
If the original is $100, after +10% it is $110. After −10% of $110, it is $110 − $11 = $99.
So it is not back to the original — a common trap on exams.
$99 (not $100)A +10% then −10% sequence does not return to the original value — successive percent changes apply to different bases.
Percent problems become much easier when you remember to express the percent as a decimal (25% = 0.25) and set up a clear equation. Solver AI can walk through percent problems step by step, which is especially useful for multi-step problems involving tax plus discount or compound interest.