Pre-Algebra · Algebra I · Word Problems · Linear Equations
Word Problems Modeled with Linear Equations: Theory and 30 Practice Questions
A step-by-step guide to modeling real-world problems with linear equations — translating English into algebra, solving one-variable equations, and setting up systems of two equations — followed by 30 practice questions (10 easy, 10 medium, 5 medium-hard, 5 hard) with hidden step-by-step solutions. Written for Pre-Algebra and Algebra I students at SOMATH on the Upper West Side.
Who this is for. Pre-Algebra and Algebra I students who can already solve equations like 3x + 5 = 20 mechanically, but freeze up when the same equation is buried inside a word problem about apples, coins, or trains. Nearly every algebra state test and standardized exam is more than half word problems. The math is easy; the translation is the hard part. This lesson gives you a repeatable 5-step recipe, then 30 progressively harder problems to practice on.
Video walkthrough: SOMATH teacher works through the 5-step recipe live.
Sections
Theory · The 5-Step Recipe
Every linear-equation word problem — whether it is about coins, ages, mixtures, or trains — is solved with the same five steps. Do them in this exact order.
Step 1 · Read once for the story, once for the numbers
The first read is just to understand what is happening. The second read is for the numbers, the units, and the question at the end (“how many…”, “what is the…”, “when will…”). Underline both the given numbers and the question.
Step 2 · Name the unknown with a variable
Pick a letter for the thing the problem is asking about, and write down what it means in a sentence. Something like “Let x = the number of dimes.” If there are two unknowns, name both: “Let x = adult tickets, y = child tickets.”
Rule of thumb: if you can express the second unknown in terms of the first (for example, “the number of dimes is 3 more than nickels” → d = n + 3), do that and stay with one variable. Only use two variables when the two unknowns are truly independent.
Step 3 · Translate every sentence into algebra
Now go back to the problem, sentence by sentence, and write each fact as an equation or an expression. Words like “is,” “equals,” “the total is” become =. Words like “more than,” “increased by,” “sum of” become +. Words like “less than,” “fewer,” “difference” become −. Use the translation table in the next section as a cheat sheet.
Step 4 · Solve the equation (or system)
Now the algebra is just algebra. For a single linear equation ax + b = c: subtract b from both sides, then divide both sides by a. For a system of two equations, use substitution (solve one equation for one variable, plug into the other) or elimination (add or subtract the equations to cancel a variable).
Step 5 · Check the answer against the story
Plug your number back into the original word problem (not just the equation you wrote). Does it make sense? If the problem asks for a number of coins, is your answer a positive whole number? If it asks for a speed, is it a reasonable speed? A wrong-sign answer or a fractional child is a signal that you translated something backwards.
Translation Table · English to Algebra
This is the vocabulary you will see over and over. Memorize it.
| English | Algebra |
|---|---|
| “a number,” “some,” “how many” | x |
| “the sum of x and y,” “x more than y,” “x increased by y” | x + y |
| “x less than y,” “the difference between y and x,” “y decreased by x” | y − x (careful with order) |
| “twice,” “doubled,” “two times” | 2x |
| “three times,” “triple” | 3x |
| “half of,” “split in two” | x/2 |
| “the product of x and y” | xy |
| “the quotient of x and y” | x/y |
| “is,” “equals,” “the result is,” “the total is” | = |
| “consecutive integers” | x, x + 1, x + 2, … |
| “consecutive even (or odd) integers” | x, x + 2, x + 4, … |
| “n years from now,” “in n years” | add n to each age |
| “n years ago” | subtract n from each age |
| “distance = rate × time” | d = rt |
| “downstream speed” (boat + current) | b + c |
| “upstream speed” (boat − current) | b − c |
| “work rate” “pipe fills tank in n hours” | rate = 1/n per hour |
| “p% of x” | (p/100) · x |
| “total value = price × quantity” | V = pq (coins, tickets, mixtures) |
With Step 2 (name the variable), Step 3 (translate), Step 4 (solve), and the vocabulary above, you can solve every single problem below. Try each one on paper before revealing the solution.
Section 1 · Easy (questions 1–10)
Single-variable equations of the form ax + b = c. Warm-up.
Question 1 Easy
A number plus 7 equals 23. Find the number.
Answer: x = 16
Step 2 · Name. Let x = the number.
Step 3 · Translate. “A number plus 7 equals 23” → x + 7 = 23.
Step 4 · Solve. Subtract 7 from both sides: x = 23 − 7 = 16.
Step 5 · Check. 16 + 7 = 23. ✓
Question 2 Easy
Twice a number is 46. Find the number.
Answer: x = 23
Translate: “Twice a number is 46” → 2x = 46.
Solve: divide both sides by 2 → x = 23.
Check: 2 · 23 = 46. ✓
Question 3 Easy
Five times a number, minus 4, equals 41. Find the number.
Answer: x = 9
Translate: 5x − 4 = 41.
Solve: add 4 → 5x = 45; divide by 5 → x = 9.
Check: 5(9) − 4 = 45 − 4 = 41. ✓
Question 4 Easy
A number divided by 3 gives 12. Find the number.
Answer: x = 36
Translate: x/3 = 12.
Solve: multiply both sides by 3 → x = 36.
Question 5 Easy
Ana has some stickers. Her friend gives her 15 more, and now she has 42. How many stickers did Ana start with?
Answer: x = 27 stickers
Name: Let x = starting stickers.
Translate: starting + 15 = 42 → x + 15 = 42.
Solve: x = 42 − 15 = 27.
Question 6 Easy
A pencil costs $2. Lucas buys some pencils and spends a total of $18. How many pencils did he buy?
Answer: x = 9 pencils
Translate: price × quantity = total → 2x = 18.
Solve: x = 9.
Question 7 Easy
A number, when increased by 8 and then doubled, equals 30. Find the number.
Answer: x = 7
Translate: “increased by 8” → (x + 8); “then doubled” → 2(x + 8) = 30.
Solve: divide both sides by 2 → x + 8 = 15; subtract 8 → x = 7.
Check: 2(7 + 8) = 2(15) = 30. ✓
Question 8 Easy
Three times a number equals 20 more than the number itself. Find the number.
Answer: x = 10
Translate: 3x = x + 20.
Solve: subtract x from both sides → 2x = 20; divide by 2 → x = 10.
Question 9 Easy
The sum of two consecutive integers is 47. Find the smaller integer.
Answer: x = 23 (integers are 23 and 24)
Name: Let x = smaller integer, so the next one is x + 1.
Translate: x + (x + 1) = 47, i.e. 2x + 1 = 47.
Solve: 2x = 46, so x = 23. The two integers are 23 and 24.
Question 10 Easy
A taxi charges a $3 flat fee plus $2 per mile. A ride costs $17 total. How many miles was the ride?
Answer: x = 7 miles
Translate: flat fee + (rate)(miles) = total → 3 + 2x = 17.
Solve: subtract 3 → 2x = 14; divide by 2 → x = 7.
Section 2 · Medium (questions 11–20)
Consecutive integers, age problems, perimeter, distance, and simple mixtures. Still one variable, but the setup is trickier.
Question 11 Medium
The sum of three consecutive integers is 72. Find the three integers.
Answer: 23, 24, 25
Name: Let x = smallest, then x + 1 and x + 2 are the next two.
Translate: x + (x + 1) + (x + 2) = 72, so 3x + 3 = 72.
Solve: 3x = 69, so x = 23. The three integers are 23, 24, 25.
Question 12 Medium
A father is currently 4 times as old as his son. In 5 years, the father will be 3 times as old as his son. How old is each now?
Answer: son is 10, father is 40
Name: Let x = son’s current age. Father is 4x.
Translate (in 5 years): son will be x + 5, father will be 4x + 5, and father = 3 · son:
4x + 5 = 3(x + 5)Solve: 4x + 5 = 3x + 15 → subtract 3x from both sides → x + 5 = 15 → x = 10. So father is 4(10) = 40.
Check: in 5 years son is 15, father is 45, and 45 = 3 · 15. ✓
Question 13 Medium
One number is 8 more than another. Their sum is 54. Find both numbers.
Answer: 23 and 31
Name: Let x = smaller number; the larger is x + 8.
Translate: x + (x + 8) = 54.
Solve: 2x + 8 = 54 → 2x = 46 → x = 23. Larger = 23 + 8 = 31.
Question 14 Medium
A rectangle’s length is 3 more than twice its width. The perimeter is 36. Find the width and length.
Answer: width = 5, length = 13
Name: Let w = width. Length = 2w + 3.
Translate: perimeter = 2(length + width) → 2(w + 2w + 3) = 36.
Solve: 2(3w + 3) = 36 → 6w + 6 = 36 → 6w = 30 → w = 5. Length = 2(5) + 3 = 13.
Check: perimeter = 2(13 + 5) = 36. ✓
Question 15 Medium
Marta buys 3 notebooks and 2 pens for $19. A notebook costs $3 more than a pen. What is the price of a pen?
Answer: pen = $2, notebook = $5
Name: Let x = price of a pen. Notebook = x + 3.
Translate: 3(x + 3) + 2x = 19.
Solve: 3x + 9 + 2x = 19 → 5x = 10 → x = 2. Notebook = $5.
Check: 3(5) + 2(2) = 15 + 4 = 19. ✓
Question 16 Medium
Train A leaves the station going 60 mph. One hour later, Train B leaves the same station going 75 mph in the same direction on a parallel track. How many hours (after B leaves) until B catches up with A?
Answer: 4 hours after B leaves
Name: Let t = time (in hours) after B leaves. Then A has been traveling t + 1 hours.
Translate: when B catches A, they have gone the same distance:
60(t + 1) = 75tSolve: 60t + 60 = 75t → 60 = 15t → t = 4 hours.
Check: A traveled 60(5) = 300 miles; B traveled 75(4) = 300 miles. ✓
Question 17 Medium
A phone plan charges $20 per month plus $0.10 per minute. Last month’s bill was $47. How many minutes were used?
Answer: 270 minutes
Translate: fixed + (rate)(minutes) = total → 20 + 0.10x = 47.
Solve: subtract 20 → 0.10x = 27; divide by 0.10 → x = 270.
Question 18 Medium
The sum of two numbers is 40. Three times the larger equals five times the smaller. Find the numbers.
Answer: smaller = 15, larger = 25
Name: Let x = smaller; then larger = 40 − x.
Translate: 3(40 − x) = 5x.
Solve: 120 − 3x = 5x → 120 = 8x → x = 15. Larger = 25.
Check: 3(25) = 75 and 5(15) = 75. ✓
Question 19 Medium
Gym A charges a $40 sign-up fee plus $30 per month. Gym B charges $50 per month with no sign-up fee. After how many months do the two plans cost the same?
Answer: 2 months (both cost $100 at that point)
Name: Let m = months.
Translate: 40 + 30m = 50m.
Solve: 40 = 20m → m = 2.
Interpretation: at exactly 2 months the two plans cost the same. After that, Gym A is cheaper (Gym A adds $30/month vs. Gym B $50/month).
Question 20 Medium
A ferry crosses a river in 30 minutes going with the current and 45 minutes against. The ferry’s speed in still water is 12 mph. Find the speed of the current.
Answer: current = 2.4 mph (12/5)
Name: Let c = speed of the current, in mph. Downstream speed = 12 + c; upstream speed = 12 − c. The distance across the river is the same both ways.
Translate: distance = rate · time, in hours (30 min = 1/2 hr, 45 min = 3/4 hr):
(12 + c) · (1/2) = (12 − c) · (3/4)Solve: multiply both sides by 4 → 2(12 + c) = 3(12 − c) → 24 + 2c = 36 − 3c → 5c = 12 → c = 12/5 = 2.4 mph.
Section 3 · Medium-Hard (questions 21–25)
Work problems, mixture problems, and two-variable systems. This is where most Algebra I students start to struggle.
Question 21 Medium-Hard
Two pipes together fill a tank in 6 hours. Pipe A alone fills it in 10 hours. How long does pipe B alone take?
Answer: 15 hours
Idea · work rates add. If pipe A fills in 10 hours, its rate is 1/10 of the tank per hour. Similarly, pipe B’s rate is 1/b per hour. Together, in one hour they fill 1/10 + 1/b of the tank. Together they fill the whole tank in 6 hours, so their combined rate is 1/6 per hour.
Translate:
1/10 + 1/b = 1/6Solve: 1/b = 1/6 − 1/10 = 5/30 − 3/30 = 2/30 = 1/15, so b = 15 hours.
Question 22 Medium-Hard
You have 40 kg of an alloy that is 25% copper. How many kilograms of pure copper must you add to make the alloy 40% copper?
Answer: 10 kg
Idea: track the copper only. Initial copper mass = 0.25 · 40 = 10 kg. If we add x kg of pure copper, total mass becomes 40 + x and total copper becomes 10 + x.
Translate: new percentage = 40%, so
(10 + x) / (40 + x) = 0.40Solve: multiply both sides by (40 + x) → 10 + x = 0.40(40 + x) = 16 + 0.40x → 0.60x = 6 → x = 10 kg.
Check: new copper = 20 kg, new total = 50 kg, and 20/50 = 0.40. ✓
Question 23 Medium-Hard
A car and a bus leave the same station at the same time, going in opposite directions. The car goes 65 mph and the bus goes 55 mph. After how many hours are they 480 miles apart?
Answer: 4 hours
Idea: when two objects move in opposite directions, their separation increases at a rate equal to the sum of their speeds.
Translate: 65t + 55t = 480.
Solve: 120t = 480 → t = 4 hours.
Question 24 Medium-Hard
A boat travels 24 km downstream in 2 hours and 24 km upstream in 3 hours. Find the boat’s speed in still water and the speed of the current.
Answer: boat = 10 km/h, current = 2 km/h
Name: Let b = boat speed in still water, c = current speed.
Translate: downstream, effective speed = b + c. From d = rt: 24 = (b + c)(2), so b + c = 12. Upstream: 24 = (b − c)(3), so b − c = 8.
Solve the system: add the two equations to eliminate c:
(b + c) + (b − c) = 12 + 8 2b = 20 b = 10Then c = 12 − 10 = 2.
Question 25 Medium-Hard
The sum of the digits of a two-digit number is 12. When the digits are reversed, the new number is 18 more than the original. Find the original number.
Answer: 57
Name: Let t = tens digit, u = units digit. Then the original number is 10t + u and the reversed number is 10u + t.
Translate: sum of digits: t + u = 12. Reversed exceeds original by 18: (10u + t) − (10t + u) = 18 → 9u − 9t = 18 → u − t = 2.
Solve the system: add: (t + u) + (u − t) = 12 + 2 → 2u = 14 → u = 7, so t = 5. Original number = 10(5) + 7 = 57.
Check: reversed = 75, and 75 − 57 = 18. ✓
Section 4 · Hard (questions 26–30)
Multi-step problems: coins, mixtures with two solvents, chase problems, two-variable systems, ticket/mix pricing. These are the ones you will see on Regents and SAT.
Question 26 Hard
A boy has $4.50 in nickels and dimes. He has 3 more dimes than nickels. How many of each coin does he have?
Answer: 28 nickels, 31 dimes
Idea · coin problems: total value = value-per-coin × number of coins, summed over all coin types.
Name: Let n = number of nickels. Dimes = n + 3. Convert everything to dollars: nickel = $0.05, dime = $0.10.
Translate: 0.05n + 0.10(n + 3) = 4.50.
Solve:
0.05n + 0.10n + 0.30 = 4.50 0.15n = 4.20 n = 28Dimes = 28 + 3 = 31.
Check: value = 28(0.05) + 31(0.10) = 1.40 + 3.10 = 4.50. ✓
Question 27 Hard
A chemist has a 30% acid solution and a 60% acid solution. How many liters of each are needed to make 30 liters of a 50% acid solution?
Answer: 10 L of 30% and 20 L of 60%
Idea · mixture problems: track the pure substance (acid, in this case). The total acid in the mixture equals the sum of the acid in each component.
Name: Let x = liters of 30% solution. Then liters of 60% = 30 − x.
Translate: acid in mixture = acid in solution 1 + acid in solution 2:
0.30x + 0.60(30 − x) = 0.50(30) 0.30x + 18 − 0.60x = 15 −0.30x = −3 x = 10So 10 L of 30% + 20 L of 60%.
Check: acid = 0.30(10) + 0.60(20) = 3 + 12 = 15 L. And 15/30 = 0.50. ✓
Question 28 Hard
A train leaves NYC at 8:00 am going 50 mph. Another train leaves NYC 90 minutes later going 65 mph on the same track in the same direction. At what time does the second train catch the first?
Answer: 2:30 pm (6.5 hours after 8:00 am)
Name: Let t = hours after 8:00 am. Then train 1 has traveled 50t miles. Train 2 started at t = 1.5, so at time t it has traveled 65(t − 1.5) miles.
Translate: catch-up means equal distances:
50t = 65(t − 1.5)Solve: 50t = 65t − 97.5 → 97.5 = 15t → t = 6.5 hours.
Convert to time: 8:00 am + 6.5 hours = 2:30 pm.
Question 29 Hard
A rectangle has perimeter 42. Its length is 3 more than twice its width. Find the area of the rectangle.
Answer: width = 6, length = 15, area = 90
Name: Let w = width. Length = 2w + 3.
Translate: perimeter = 2(length + width) = 42, so length + width = 21, i.e. (2w + 3) + w = 21.
Solve: 3w + 3 = 21 → 3w = 18 → w = 6. Length = 2(6) + 3 = 15.
Area: 6 · 15 = 90.
Question 30 Hard
A tank is filled by pipe A alone in 12 hours and by pipe B alone in 15 hours. Both pipes are opened together for 4 hours, and then pipe A is shut off. How long does pipe B alone take to finish filling the tank?
Answer: 6 more hours for B alone
Idea · work in stages. Compute how much of the tank is filled in the first 4 hours (both pipes on), subtract from 1, then divide by B’s rate.
Stage 1 (both on, 4 hours): combined rate = 1/12 + 1/15 = 5/60 + 4/60 = 9/60 = 3/20 of the tank per hour. In 4 hours: 4 · 3/20 = 12/20 = 3/5 of the tank is filled. Remaining: 1 − 3/5 = 2/5.
Stage 2 (B alone): B’s rate is 1/15 per hour. Time to fill remaining 2/5:
(1/15) · t = 2/5 t = (2/5)(15) = 6 hoursAnswer Key
All 30 answers at a glance. Try each problem before looking here.
| # | Level | Answer |
|---|---|---|
| 1 | Easy | x = 16 |
| 2 | Easy | x = 23 |
| 3 | Easy | x = 9 |
| 4 | Easy | x = 36 |
| 5 | Easy | 27 stickers |
| 6 | Easy | 9 pencils |
| 7 | Easy | x = 7 |
| 8 | Easy | x = 10 |
| 9 | Easy | 23, 24 |
| 10 | Easy | 7 miles |
| 11 | Medium | 23, 24, 25 |
| 12 | Medium | Son 10, Father 40 |
| 13 | Medium | 23 and 31 |
| 14 | Medium | Width 5, Length 13 |
| 15 | Medium | Pen $2, Notebook $5 |
| 16 | Medium | 4 hours after B leaves |
| 17 | Medium | 270 minutes |
| 18 | Medium | 15 and 25 |
| 19 | Medium | 2 months |
| 20 | Medium | 2.4 mph |
| 21 | Medium-Hard | 15 hours |
| 22 | Medium-Hard | 10 kg |
| 23 | Medium-Hard | 4 hours |
| 24 | Medium-Hard | Boat 10 km/h, Current 2 km/h |
| 25 | Medium-Hard | 57 |
| 26 | Hard | 28 nickels, 31 dimes |
| 27 | Hard | 10 L of 30%, 20 L of 60% |
| 28 | Hard | 2:30 pm |
| 29 | Hard | Area = 90 |
| 30 | Hard | 6 more hours |
Big Ideas Covered
- Variable naming. Every problem starts by writing “Let x = …” in a full sentence. This is not optional.
- Translation. “Is” → =. “More than” → +. “Less than” → − (mind the order). “Of” → ×.
- Consecutive integers. Represent as x, x+1, x+2 (or x, x+2, x+4 for even/odd).
- Age problems. Add or subtract the same number of years from every age.
- Distance-rate-time. d = rt. Same-direction chase: subtract speeds. Opposite direction: add speeds.
- Downstream vs. upstream. Downstream speed = boat + current; upstream = boat − current.
- Work-rate problems. Rate = 1/(time to finish alone). Rates add when workers work together.
- Mixture and percent problems. Track the pure substance; final amount = sum of components.
- Coin/ticket problems. Total value = price × quantity, summed over categories.
- Systems of two equations. When two unknowns are truly independent, write two equations and solve by substitution or elimination.