Translating words into mathematics
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Learning objectives
Build a model that preserves the story, then answer the question the story actually asks.
By the end of this section
You should be able to define an unknown with units, translate comparison and grouping language, construct rate and percent relationships, represent constraints, and check a solution against its context. The hard part is often the first equation, not the arithmetic after it.
Define the quantity before choosing the operation
A good variable definition names a quantity and, when relevant, its unit: “Let t be the elapsed time in minutes,” not merely “Let t be time.” For counts, add the restriction that the variable is a nonnegative integer. Distinguish the variable you solve for from the quantity the question ultimately asks you to report.
The modeling sentence
Let _____ represent the __________________, measured in _________. The target is _______________. The information connects these quantities by __________________.
Read relationships, not isolated keywords. The word “more” can describe an addition, a comparison, or a percent increase. The word “per” signals a unit rate, but it does not tell you whether the requested unknown requires multiplication or division. First describe what equals what.
Words | Model |
|---|---|
Five more than x | x + 5 |
Five less than x | x − 5 |
Five minus x | 5 − x |
Three times the sum of x and 2 | 3(x + 2) |
The sum of three times x and 2 | 3x + 2 |
The quotient of x + 2 and 3 | (x + 2)/3 |
A is twice B | A |
A is 6 fewer than B | A |
Test a comparison with an easy number. If “five less than x” is modeled as 5 − x, choose x
Make grouping visible. In “half the sum of n and 10,” the sum happens first, so the model is (n + 10)/2. “The sum of half of n and 10” instead gives n/2 + 10. Parentheses encode which words belong together.
Totals, comparisons, ratios, and percents
A total is a sum of nonoverlapping parts. If 34 tickets are either adult or child tickets and a counts adult tickets, then child tickets number 34 − a. Defining the second count in terms of the first can reduce a two-variable problem to one variable.
Ratios compare quantities, not automatically a part to the whole. A ratio of dogs to cats of 3 : 5 means d
Percent models depend on the base. If p% of A equals B, then (p/100)A
A p% decrease gives B
Description of a new value N | Equation using original O |
|---|---|
N is 80% of O | N |
N is 80% more than O | N |
N is 20% less than O | N |
N is twice O | N |
N is 200% more than O | N |
A fraction of what remains has a new base. After using 3/8 of an amount, 5/8 remains. Using one third of that remainder consumes (1/3)(5/8) of the original, not another 1/3 of the original. Draw a part-whole bar or write the remaining amount before the second operation.
Translate both directions
An equation can be checked by reading it back into words. For A
Rates and units are part of the equation
A rate compares quantities with different units. At a constant rate r, an amount often satisfies
amount
If r is meters per minute and t is minutes, then rt is meters. If t is supplied in seconds, convert it or change the rate before multiplying.
A unit conversion multiplies by 1 in a different form. Because one hour equals 60 minutes, the factors 1 hour/60 minutes and 60 minutes/1 hour each represent equal quantities in numerator and denominator. Choose the orientation that cancels the unwanted unit:
25 min ·
Do not multiply by 60 merely because you recognize a minutes-hours conversion; decide which direction your quantity must move.
Unit rates versus fixed amounts. A $12 initial charge plus $4 per item is C
Quantity | Typical relationship | Unit check |
|---|---|---|
Distance | speed × time | (km/h)(h) |
Cost | price per item × count | (dollars/item)(items) |
Density | mass ÷ volume | g/cm3 |
Mean | total ÷ count | Sum all observations first. |
Area of a rectangle | length × width | (m)(m) |
The domain volumes develop these relationships in depth. Here the goal is to select quantities that can meaningfully be multiplied, divided, or added. Adding meters to square meters does not define a physical total.
A conversion warning for area and volume A linear conversion factor is squared for area and cubed for volume. If 1 m
Constraints, whole items, and multistep models
Words describing q | Constraint |
|---|---|
At least 12; no fewer than 12 | q ≥ 12 |
At most 12; no more than 12 | q ≤ 12 |
More than 12 | q > 12 |
Less than 12 | q < 12 |
Between 12 and 20, inclusive | 12 ≤ q ≤ 20 |
Write the constraint before rounding. To maximize a whole-item count under a budget, solve the budget inequality and choose the greatest permitted integer. To meet a minimum requirement with whole groups or containers, choose the least integer that meets it. Ordinary nearest-integer rounding can fail in both cases. If n ≤ 6.57 and n is a nonnegative integer, the maximum is 6. If n ≥ 5.29 and n is an integer, the minimum is 6. Check the chosen integer and the adjacent one in the original condition. This explains both why your answer works and why a better one is impossible.
Inequality operations. Adding or subtracting the same number on both sides preserves the inequality direction. Multiplying or dividing by a positive number preserves it; multiplying or dividing by a negative number reverses it. Most count-and-cost models here use positive rates, but the sign rule still matters when rearranging.
Build multistep models in the story’s order
Write a line for each dependency: required quantity, quantity already available, additional need, package count, then cost. The final target might be dollars, not packages. Preserve fractions and units until the relevant step is complete. You do not have to combine a clear sequence into one long equation.
Separate useful information from background. A description may identify the setting without supplying a needed numerical relationship. Use a number only when you can say what operation connects it to the target. Conversely, restrictions such as “whole packs only” may be essential even though they contain no numerical value.
The final model audit
Does the equation read back as the original sentence? Does it count every required part exactly once? Do its units agree? Does the solution meet integer, positivity, capacity, and timing restrictions? Does the final response answer the requested quantity rather than an intermediate one?
15 worked examples
5.01. Translate a reversed comparison correctly
Write an expression for five less than twice a number k.
Show worked solutionHide worked solution for example 5.01
Recognize the structure. “Twice a number” is the starting quantity. “Five less than” subtracts 5 from that quantity.
Work it through. First form 2k, then subtract 5: 2k − 5. The phrase does not say “five minus twice the number” or “twice the difference of the number and five.”
Answer: 2k − 5.
Check. For k
Avoid the trap. 5 − 2k reverses the subtraction, while 2(k − 5) subtracts 5 before doubling. These are three different relationships.
Key takeaway. Use a simple input to check a translation, especially when “less than,” “more than,” or “times” appears.
5.02. Let the words determine the parentheses
Write an expression for half the sum of a number n and 10.
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Recognize the structure. The object being halved is the sum. Group that sum before dividing.
Work it through. The sum is n + 10, so half of it is (n + 10)/2. An equivalent expanded form is n/2 + 5.
Answer: .
Check. For n
Avoid the trap. n/2 + 10 means the sum of half the number and 10, which would give 13 for n
Key takeaway. Fractions, products, roots, and powers all act on groups. Mark the full group named by the language before writing its operation.
5.03. Combine a comparison with a total
Alicia has four fewer books than three times the number Ben has. Together they have 36 books. How many books does Alicia have?
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Recognize the structure. There is one comparison and one total. Let b be Ben’s count; then Alicia’s count is 3b − 4.
Work it through. The total gives b + (3b − 4)
Answer: 26 books.
Check. 26 + 10
Avoid the trap. The value 10 solves for the chosen variable but does not answer the final question. Naming the target at the start prevents this common stopping error.
Key takeaway. A carefully chosen variable can represent several related quantities without introducing a separate symbol for each one.
5.04. Convert a part-to-part ratio into a part of the total
The ratio of cats to dogs at an event is 5 : 3. There are 64 cats and dogs altogether. How many dogs are there?
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Recognize the structure. The total contains 5 + 3
Work it through. Let each ratio part represent k animals. Then 5k + 3k
Answer: 24 dogs.
Check. There are 40 cats and 24 dogs. Their ratio is 40 : 24
Avoid the trap. Multiplying 64 by 3/5 treats the dogs-to-cats ratio as dogs-to-total. Those denominators represent different groups.
Key takeaway. For every ratio, state the numerator group and denominator group before converting it to a fraction.
5.05. Use the original amount as the percent-increase base
A fee of $80 increases by 15%. What is the new fee?
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Recognize the structure. The increase is 15% of the original $80, and the new fee includes the original plus the increase.
Work it through. The increase is 0.15(80)
Answer: $92.
Check. The difference is $12, and 12/80
Avoid the trap. The amount of the increase, $12, is not the new total. Adding 15 dollars would treat a percentage as an absolute amount.
Key takeaway. Write “new
5.06. Recover an original amount by dividing by the multiplier
After a 20% discount, an item costs $56. What was its original price?
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Recognize the structure. The discounted price is 80% of the original, not 80% of some new base. Let p be the original price.
Work it through. Write 0.80p
p = = 70.
Answer: $70.
Check. Twenty percent of $70 is $14; 70 − 14
Avoid the trap. Increasing 56 by 20% gives 67.2, which uses 56 as the base. Undoing multiplication by 0.80 requires division by 0.80, not multiplication by 1.20.
Key takeaway. For reverse-percent questions, define the unknown original and translate the stated change before solving.
5.07. Separate a one-time fee from a repeated rate
A bicycle rental costs a one-time $18 charge plus $2.50 per hour, with the hourly charge proportional to time. Write a cost model and find the cost for 6 hours.
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Recognize the structure. The fixed charge occurs once; the hourly charge scales with the number of hours.
Work it through. Let h be rental time in hours and C the cost in dollars. Then C
Answer: C(h)
Check. At 0 hours the formula’s fixed component is $18, and each additional hour increases cost by $2.50. The model should only be used for rental times permitted by the context.
Avoid the trap. 20.50h repeats the one-time charge every hour. The words “one-time” and “per hour” play different roles.
5.08. Make time units match a speed
A cyclist travels at a constant speed of 54 kilometers per hour for 25 minutes. How many kilometers does the cyclist travel?
Show worked solutionHide worked solution for example 5.08
Recognize the structure. Speed is per hour, but time is in minutes. Convert one quantity so the time units cancel.
Work it through. Convert the time: 25/60
d = 54 · h = 22.5 km.
Answer: 22.5 kilometers.
Check. Half an hour at this speed would cover 27 kilometers. Since 25 minutes is slightly less than half an hour, 22.5 kilometers is reasonable.
Avoid the trap. 54(25)
Key takeaway. Write units in the multiplication. They reveal a missing conversion more reliably than a memorized divide-by-60 rule.
5.09. Convert both the numerator and denominator of a rate
Water flows at 1.8 liters per minute. Express the rate in milliliters per second. Use 1 liter = 1,000 milliliters and 1 minute = 60 seconds.
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Recognize the structure. Both the amount unit and the time unit change. Choose conversion-factor orientations that cancel liters and minutes.
Work it through. Write the chain:
1.8 · ·
Answer: 30 milliliters per second.
Check. In 60 seconds, 30(60)
Avoid the trap. Multiplying by 60 would count seconds in the wrong direction for a per-second rate. More time units per minute means less amount per one second.
Key takeaway. A rate is a fraction with units; dimensional cancellation works on its denominator as well as its numerator.
5.10. Maximize a whole-item count under a budget
An order has a $7 delivery fee, and each notebook costs $3.50. With a $30 budget, what is the greatest number of notebooks that can be purchased?
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Recognize the structure. “With a budget” imposes an upper bound, and notebooks are whole items. Let n be a nonnegative integer.
Work it through. Model the total: 7 + 3.50n ≤ 30. Thus 3.50n ≤ 23 and n ≤ 23/3.5 ≈ 6.5714. The greatest permitted integer is 6.
Answer: 6 notebooks.
Check. Six cost 7 + 3.5(6)
Avoid the trap. Nearest-integer rounding gives 7, which violates the constraint. The direction of the inequality, not the first decimal digit, determines the rounding decision.
Key takeaway. Checking the chosen integer and its neighbor proves the maximum rather than merely finding a possible count.
5.11. Meet a minimum with the least whole number
A participant starts with 35 points and earns 17 points per completed task. What is the least number of tasks needed to have at least 125 points?
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Recognize the structure. The total must meet a lower bound. Completed tasks are counted by a nonnegative integer n.
Work it through. Write 35 + 17n ≥ 125. Subtract 35 and divide by positive 17:
n ≥ ≈ 5.2941.
The least integer satisfying this bound is 6.
Answer: 6 tasks.
Check. Five tasks give 35 + 85
Avoid the trap. Rounding 5.2941 to the nearest integer gives 5, which fails the minimum. This is the opposite rounding situation from a maximum count under a budget.
Key takeaway. Translate “at least” or “at most” before deciding which whole number is allowed.
5.12. Turn a geometric description into two connected calculations
A rectangle’s length is 3 meters more than twice its width. Its perimeter is 42 meters. What is its area? Use P = 2l + 2w and A = lw.
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Recognize the structure. Define width w. The comparison gives l
Work it through. Substitute into the perimeter:
2(2w + 3) + 2w = 42 6w + 6 = 42 w = 6.
Then l = 2(6) + 3 = 15, and the area is 15(6)
Answer: 90 square meters.
Check. 2(15) + 2(6)
Avoid the trap. The width 6 and length 15 are intermediate values. Area requires their product and square units, not their sum.
5.13. Represent one category as the total minus the other
An event sells 34 tickets. Adult tickets cost $8 each and child tickets cost $5 each. Total ticket revenue is $218. How many adult tickets were sold?
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Recognize the structure. The categories are exhaustive. If a tickets are adult tickets, then 34 − a are child tickets.
Work it through. The revenue equation is
8a + 5(34 − a)
Expand to get 8a + 170 − 5a
Answer: 16 adult tickets.
Check. The remaining 18 child tickets give 16(8) + 18(5) = 128 + 90 = 218, and 16 + 18
Avoid the trap. Adding the prices 8 + 5 and multiplying by 34 assumes every ticket contributes both prices. Each ticket belongs to exactly one category.
Key takeaway. The alternate “all child tickets” baseline earns $170; each adult ticket adds $3. The extra $48 therefore requires 16 adult tickets.
5.14. Track the changing base in a fraction-of-the-remainder problem
A reader finishes of a book on Monday, then of the remaining pages on Tuesday. After Tuesday, 100 pages remain unread. How many pages are in the book?
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Recognize the structure. Tuesday’s fraction applies to Monday’s remainder, not to the original book. Let N be the original page count.
Work it through. After Monday, (5/8)N remains. Tuesday leaves two thirds of that amount:
· N = N = 100.
Multiply by 12/5 to obtain N
Answer: 240 pages.
Check. Monday’s reading is 3(240)/8
Avoid the trap. Adding 3/8 + 1/3 assumes both fractions share the original total as their base. The word “remaining” changes the reference amount.
Key takeaway. Track each new base explicitly in successive percentages, changing concentrations, and repeated reductions.
5.15. Build a multistep count-to-cost model
A school has 8 classes with 24 students in each class. Every student needs 2 notebooks. The school already has 29 notebooks. Additional notebooks are sold only in packs of 12 for $7.50 per pack, with a one-time $9 delivery fee. What is the minimum total cost of the additional order?
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Recognize the structure. The target is cost, but it depends on the required notebook count, the shortage, and a whole-pack constraint.
Work it through. The school needs 8(24)(2)
Answer: $234.
Check. Twenty-nine packs plus existing stock supply 29(12) + 29
Avoid the trap. Reporting 30 answers the package-count question, not the cost question. Rounding to 29 fails the requirement; charging the delivery fee per pack overcounts it.
Key takeaway. A clear sequence of short equations can be safer than one giant expression. Label every intermediate quantity and finish with the original target.
Section 5 readiness check
You should now be able to explain why your equation matches the story, check the answer in the original conditions, and justify any rounding caused by a count constraint. When the setup is uncertain, test it with a simple number or read the equation back into words before calculating.