Setting up proportions from word problems
A crew of 6 filled 15 bags of litter, and a crew of 14 is coming. Four numbers, one unknown. Which goes on top, which goes underneath, and does it matter?
Where you are 3 in Proportion
When you reach for it
A litter-pick crew of 6 volunteers fills 15 bags in a morning. A bigger crew of 14 is coming next week to work the same way, and the organizer wants to know how many bags they will fill. Four numbers are in play and three are known: 15 bags, 6 volunteers, 14 volunteers, and the unknown number of bags, b.
Solving a proportion finds a missing number once the proportion is written down. This lesson is about writing it down: which number goes on top, which goes underneath, and on which side.
Getting that wrong produces no error message. The equation \displaystyle \frac{15}{6} = \frac{14}{b} solves perfectly well, to b = 5.6, and claims that more than twice as many volunteers fill far fewer bags. The arithmetic is fine; the four numbers are in the wrong places. Reach for this lesson whenever a word problem describes two amounts that grow together at a fixed rate (bags per volunteer, pages per minute, goals per game), gives one complete pair, and asks about another.
The procedure
This lesson is taught over two days. Part 1 is placing the numbers: the rule about units, and why it works. Part 2 is the word problem around the proportion: whether a proportion fits at all, units that differ, a total in place of a part, and reading the answer back into the story.
Part 1: putting each number in its place
Name the unknown, with its unit. Here b is a number of bags. The unit is what tells you where the letter belongs, so a letter without one is a setup waiting to go wrong.
Lay the four amounts out in a small table, one column for each situation and one row for each kind of quantity:
| first crew | bigger crew | |
|---|---|---|
| bags | 15 | b |
| volunteers | 6 | 14 |
Each column is one complete situation. Each row holds one kind of quantity, in one unit. Every number gets exactly one cell, and the story decides the cell, not the order in which the numbers happen to appear in the sentence. With practice the table stays in your head, but it is the whole idea.
Read the proportion down the columns. The first crew’s ratio of bags to volunteers is \displaystyle \frac{15}{6}, and the bigger crew’s is \displaystyle \frac{b}{14}. The story says each volunteer fills bags at the same rate whatever the size of the crew, so the two ratios name the same comparison, which is exactly what a proportion states:
Bags on top on both sides, volunteers underneath on both sides. This is the rule the lesson rests on: whatever unit is on top of one ratio is on top of the other, and the same underneath. Each side is then a number of bags per volunteer, so the equation says one rate equals another, which is the claim the story makes. Cross products give 6b = 15 \cdot 14 = 210, so b = 35 bags. Solving a proportion sets out every way of finishing it.
Or read it along the rows. The bigger crew has \displaystyle \frac{14}{6} times as many volunteers as the first, so at the same rate it fills \displaystyle \frac{14}{6} times as many bags. Comparing the two crews row by row gives
Now each ratio compares like with like, bags with bags and volunteers with volunteers, and both put the first crew on top. This is just as correct, and its cross products are again 6b = 15 \cdot 14.
Four right ways, and why they agree. Write the table with letters:
| situation 1 | situation 2 | |
|---|---|---|
| first kind | a | c |
| second kind | b | d |
Reading down the columns gives \displaystyle \frac{a}{b} = \frac{c}{d}, or with both ratios turned over, \displaystyle \frac{b}{a} = \frac{d}{c}. Reading along the rows gives \displaystyle \frac{a}{c} = \frac{b}{d}, or \displaystyle \frac{c}{a} = \frac{d}{b}. Cross multiply any of the four (every letter standing for a number that is not zero) and the result is the same equation,
the two numbers on one diagonal of the table multiplied together equal the two on the other diagonal. So all four proportions have exactly the same solution, and you are free to choose the one that puts the unknown on top, where it takes one step to solve.
What makes a setup wrong. A mixed reading such as \displaystyle \frac{a}{b} = \frac{d}{c} cross multiplies to ac = bd: it multiplies two numbers from the same row instead of from the diagonals. That is a different equation with a different answer. In the crew story it is \displaystyle \frac{15}{6} = \frac{14}{b}, bags over volunteers on the left and volunteers over bags on the right, and it leads to 15b = 84. Its answer, b = 5.6, says that more volunteers fill fewer bags.
Check the size. Once the proportion is solved, ask whether the answer moved the right way and by roughly the right factor. 14 volunteers is a little more than twice 6, so the bags should be a little more than twice 15, and 35 is. A size check cannot prove a setup right, but an upside-down setup usually fails it, because in a proportional story more of one quantity means more of the other, in the same proportion.
Part 2: from the story to the table
Is it a proportion at all? A proportion fits only when the two quantities keep a fixed ratio: double one and the other doubles, and when one is zero so is the other. Bags and volunteers, pages and minutes at a steady pace, goals and games at a steady scoring rate all behave this way, at least as a model. Some stories that look alike do not. A ticket order with a booking fee charged once per order does not double when the number of tickets doubles, because the fee does not double. The time a crew needs to clear a park goes down as the crew grows, so time and crew size keep no fixed ratio at all. Example 7 shows what that does to a proportion.
The same unit, not just the same kind. The rule says the unit on top of one ratio is the unit on top of the other. Minutes and hours are both time, but pages over minutes is a rate in pages per minute and pages over hours is a rate in pages per hour, and at one steady pace the second is 60 times the first. Setting them equal claims something false. Convert one amount first so the whole row is in one unit; Example 3 shows what happens otherwise.
The row the question needs. A story may give a ratio of two parts (5 student tickets for every 3 adult tickets) and then a total. The table must have a row for the total before the proportion can be read from it: every group of 5 + 3 = 8 tickets holds 5 student tickets. Example 4 builds that row.
Answer the question that was asked. The proportion returns a number; the story decides what to do with it. A number of sketchbooks to buy is rounded up, since rounding down leaves someone without one. A number of complete pages read is rounded down. “How many more” asks for the rest, not the total. A sentence with the unit in it finishes the problem.
Worked examples
Examples 1 to 3 go with Part 1. Examples 4 to 7 go with Part 2.
Choosing where the unknown sits
Odile reads 48 pages of a novel in 40 minutes. At the same pace, how long will the remaining 210 pages take her?
SolutionThe unknown is a time, t minutes. In the table, one column is what she has read and the other is what is left:
| read so far | still to read | |
|---|---|---|
| pages | 48 | 210 |
| minutes | 40 | t |
Reading down the columns with pages on top gives \displaystyle \frac{48}{40} = \frac{210}{t}. That is correct, but it leaves t underneath. Turning both ratios over is just as correct, minutes on top on both sides, and it brings t to the top:
Multiplying both sides by 210 keeps the equation true:
The first version agrees, as it must: its cross products give 48t = 40 \cdot 210 = 8400, so t = 175 again.
Check the size. 210 pages is 210 \div 48 = 4.375 times 48 pages, so the time should be 4.375 times 40 minutes, and 4.375 \times 40 = 175. The rest of the book takes her 175 minutes, which is 2 hours and 55 minutes.
Sorting right setups from wrong ones
A tree-planting crew of 4 plants 30 saplings an hour. A crew of 10, every planter working at the same pace, plants s saplings an hour. Which of these proportions describe the situation?
(a) \displaystyle \frac{4}{30} = \frac{10}{s} \qquad (b) \displaystyle \frac{30}{4} = \frac{10}{s} \qquad (c) \displaystyle \frac{4}{10} = \frac{30}{s} \qquad (d) \displaystyle \frac{s}{30} = \frac{10}{4} \qquad (e) \displaystyle \frac{30}{10} = \frac{s}{4}
SolutionThe table has the crew of 4 in one column and the crew of 10 in the other:
| crew of 4 | crew of 10 | |
|---|---|---|
| planters | 4 | 10 |
| saplings | 30 | s |
(a) Planters over saplings on both sides: a column reading. Correct.
(b) Saplings over planters on the left, planters over saplings on the right. The units are not in the same places. Wrong.
(c) Planters over planters and saplings over saplings, the crew of 4 on top both times: a row reading. Correct.
(d) Saplings over saplings and planters over planters, the crew of 10 on top both times: a row reading with both ratios turned over. Correct.
(e) The left side takes the saplings of one crew over the planters of the other, so it is neither a column nor a row. Wrong.
The three correct setups all cross multiply to 4s = 300, so s = 75 saplings an hour. The two wrong ones give different equations: (b) gives 30s = 40, so \displaystyle s = \frac{4}{3}, and (e) gives 10s = 120, so s = 12. Both claim that the larger crew plants fewer saplings than the smaller one, which the size check rules out at once: 10 planters is 2.5 times 4, so the saplings should be 2.5 \times 30 = 75.
Hours and minutes in one problem
Jun reads 18 pages in 25 minutes. At that pace, how many pages does he read in 2 hours?
SolutionWriting \displaystyle \frac{18}{25} = \frac{p}{2} puts minutes underneath on the left and hours underneath on the right. It gives 25p = 36, so p = 1.44 pages in two hours, which is absurd. The row for time has two different units in it.
Convert first: 2 hours is 2 \times 60 = 120 minutes. Now the time row is all in minutes:
| the sample | two hours | |
|---|---|---|
| pages | 18 | p |
| minutes | 25 | 120 |
Multiplying both sides by 120 gives \displaystyle p = \frac{120 \cdot 18}{25} = \frac{2160}{25} = 86.4.
Converting the other way works just as well. 25 minutes is \displaystyle \frac{25}{60} = \frac{5}{12} of an hour, and \displaystyle \frac{18}{5/12} = \frac{p}{2} gives \displaystyle p = 2 \cdot 18 \cdot \frac{12}{5} = 86.4 again. Either unit is fine, so long as one unit fills the whole row.
At that pace Jun reads 86.4 pages in two hours: 86 complete pages and part of the next.
A ratio of parts and a total
At a school play, student tickets and adult tickets were sold in the ratio 5 to 3. There were 344 tickets sold in all. How many of each kind were sold?
SolutionThe ratio compares two parts, student tickets with adult tickets, but the number given is the total. So the table needs a row for all tickets. In every group of 5 + 3 = 8 tickets, 5 are student tickets:
| one group | the whole sale | |
|---|---|---|
| student tickets | 5 | s |
| all tickets | 8 | 344 |
Student tickets on top on both sides, all tickets underneath on both sides:
Multiplying both sides by 344 gives \displaystyle s = \frac{344 \cdot 5}{8} = \frac{1720}{8} = 215. The adult tickets are the rest, 344 - 215 = 129.
Check. 215 and 129 are both 43 times a whole number, 215 = 5 \times 43 and 129 = 3 \times 43, so their ratio is 5 to 3, and 215 + 129 = 344. The strip shows the same thing: 344 tickets in 8 equal parts is 43 tickets a part.
The proportion \displaystyle \frac{5}{3} = \frac{s}{344} would put adult tickets underneath on the left and all tickets underneath on the right. It gives 3s = 1720, so \displaystyle s = 573\frac{1}{3}, more student tickets than tickets sold. 215 student tickets and 129 adult tickets were sold.
Rounding up for supplies
An art teacher plans for at least 2 sketchbooks for every 3 students, and 40 students are taking the class. Sketchbooks come in packs of 5. How many sketchbooks does the teacher need, and how many packs?
SolutionThe unknown is a number of sketchbooks, k. Sketchbooks on top, students underneath, on both sides:
Cross products give 3k = 80, so \displaystyle k = \frac{80}{3} = 26\frac{2}{3}.
A teacher cannot buy two thirds of a sketchbook, and the story decides which way to round. 26 sketchbooks would fall short of the plan: comparing \displaystyle \frac{26}{40} with \displaystyle \frac{2}{3} by cross products, 26 \cdot 3 = 78 is less than 40 \cdot 2 = 80. So round up to 27.
The packs are a second proportion, with sketchbooks on top and packs underneath:
which gives 5p = 27, so p = 5.4. Five packs hold only 25 sketchbooks, so again round up. The teacher needs 27 sketchbooks, which means buying 6 packs, 30 sketchbooks, with 3 to spare.
How many more
A soccer team scored 26 goals in the first 8 games of a 20-game season. If it keeps scoring at the same rate, how many more goals will it score in the rest of the season?
SolutionThe question asks about the games still to come, 20 - 8 = 12 of them. Goals on top, games underneath:
Multiplying both sides by 12 gives \displaystyle r = \frac{12 \cdot 26}{8} = \frac{312}{8} = 39.
A second route answers a different question first and then subtracts. The whole season’s goals, g, satisfy \displaystyle \frac{26}{8} = \frac{g}{20}, so \displaystyle g = \frac{520}{8} = 65, and the goals still to come are 65 - 26 = 39. The two routes agree. What does not work is \displaystyle \frac{26}{8} = \frac{r}{20} with r called “more”: the 20 is the whole season, so that proportion gives the season’s total.
At its current rate the team scores 39 more goals. That is what the rate predicts, not a guarantee: a team’s scoring is only roughly steady, and the proportion is a model of it.
When the ratio is not fixed
Four volunteers take 3 hours to clear the litter from a park. Fen wants to know how long 6 volunteers would take, each working at the same pace, and writes \displaystyle \frac{4}{3} = \frac{6}{t}. Is that a correct setup?
SolutionThe units are in place, volunteers over hours on both sides, and the proportion solves without complaint: 4t = 18, so t = 4.5 hours. But more people taking longer to do the same job cannot be right, and the size check catches it.
The fault comes before the placing of numbers. Test the story against Part 2: doubling the crew to 8 volunteers does not double the time, it halves it. Volunteers and hours keep no fixed ratio, so no proportion between them describes this story, however carefully its units are lined up. What stays fixed here is the product, 4 \times 3 = 12 volunteer-hours of work, so 6 volunteers need 12 \div 6 = 2 hours. Amounts that behave this way are inversely proportional, a different relationship with its own lesson.
Boundaries
The two quantities must keep a fixed ratio. Double one and the other doubles; when one is zero, so is the other. A fixed charge added once (a booking fee), a starting amount (pages already read before the timing began) or a quantity that falls as the other rises (time against crew size) breaks this, and a proportion then gives a wrong answer with no warning. In the booking-fee case the part that does scale, the cost of the tickets alone, can still be handled by a proportion.
Each ratio takes its two numbers from one column or one row. Column readings compare two kinds of quantity within one situation; row readings compare two situations within one kind. A ratio with one number from each, such as saplings of one crew over planters of the other, belongs to neither and produces a different equation.
Both ratios keep the same order. Bags over volunteers on one side means bags over volunteers on the other. In a row reading, the same situation goes on top of both ratios.
One unit per row. Minutes and hours in the same row make two different rates look equal. Convert before writing the proportion.
No zero underneath. A ratio with 0 underneath is not a number, so a story with 0 games played or 0 minutes read gives no rate to scale from.
The proportion’s answer is the rate’s answer. It is exact when the rate really is fixed (bags per volunteer when every volunteer fills the same amount) and a prediction when the rate is only roughly steady (goals per game). Either way the story decides whether to round up, round down, or keep a fraction.
Where this goes next
Once the proportion is on paper, solving it is mechanical, and a column reading in particular is a statement about a unit rate: \displaystyle \frac{15}{6} bags per volunteer is 2.5 bags for one volunteer. That fixed number is the constant of proportionality, and writing the column reading as b = 2.5v turns the proportion into an equation of a proportional relationship, which answers every crew size at once instead of one at a time.
The row reading, “the same situation on top of both ratios”, is the habit that scale drawings and proportions in similar triangles depend on, where the rows become matching lengths in two figures. And when the amounts in a story are expressions rather than numbers, or one ratio is a percent out of 100, the setup is the same and the solving moves to equations from proportions and percents.
Exercises
Part 1: right way up or not
- Mira reads 35 pages in 50 minutes. At that pace she needs t minutes for 84 pages. Which of these proportions are correct setups? Solve one of them. (a) \displaystyle \frac{35}{50} = \frac{84}{t} \quad (b) \displaystyle \frac{50}{35} = \frac{84}{t} \quad (c) \displaystyle \frac{35}{84} = \frac{50}{t} \quad (d) \displaystyle \frac{t}{50} = \frac{35}{84}
Part 1: write a proportion and solve it
- A tree-planting crew plants 45 trees in 3 hours. At the same rate, how many trees does it plant in 7 hours?
- A litter-pick crew collects 12 bags of litter from 2 km of riverbank. At the same rate, how many kilometers of riverbank give 30 bags?
- Wren has made 9 of her 15 free throws. At the same rate, how many free throws would she need to attempt to make 24?
- Three packs of markers hold 36 markers. How many packs hold 84 markers? Write two different correct proportions, one with packs on top and one with markers on top, and show they give the same answer.
Part 1: matching the units
- Tomas reads 15 pages in 20 minutes. At that pace, how many pages does he read in 1.5 hours?
- A crew plants 40 trees in 2.5 hours. At the same rate, how many minutes does it take to plant 56 trees?
Part 2: parts and totals
- Tickets for a school play sell in the ratio of 3 adult tickets for every 7 student tickets. If 210 student tickets are sold, how many adult tickets are sold?
- Of every 9 tickets sold for a play, 4 are for the Friday show. If 315 tickets are sold, how many are for Friday, and how many for the other nights?
- A team’s goals from headers and goals from kicks are in the ratio 2 to 7. The team scored 63 goals. How many were headers?
Part 2: answers in context
- A teacher plans for at least 4 glue sticks for every 5 students, for a class of 32. Glue sticks come in packs of 6. How many glue sticks are needed, and how many packs?
- Dagny reads 11 pages in 15 minutes. At that pace, how many complete pages does she read in 40 minutes?
- A striker scored 14 goals in the first 10 games of a 25-game season. At the same rate, how many more goals will she score this season?
- A crew planted 56 trees in its first 4 hours. It has to plant 140 trees in all. At the same rate, how many more hours does it need?
- A litter-picker fills \displaystyle \frac{2}{3} of a bag every 5 minutes. At that rate, how many minutes does it take to fill 4 bags?
Part 2: is it a proportion?
- Play tickets cost 8 dollars each, plus a booking fee of 3 dollars charged once per order. An order of 2 tickets costs 19 dollars. Halvor writes \displaystyle \frac{19}{2} = \frac{c}{5} to find the cost c of an order of 5 tickets. What does his proportion give, what does the order really cost, and why do they differ?
- Three volunteers clear a field of litter in 4 hours. Yusra writes \displaystyle \frac{3}{4} = \frac{6}{t} to find how long 6 volunteers would take. What does her proportion give, what is the right answer, and why?
Reasoning
- Show that \displaystyle \frac{a}{b} = \frac{c}{d} and \displaystyle \frac{a}{c} = \frac{b}{d} always have the same cross products, for numbers a, b, c, d that are not zero. In the crew story at the start of the lesson, say what each of the two proportions compares.
- A crew plants 12 trees in 30 minutes. Sam wants the minutes x it takes to plant 45 trees and writes \displaystyle \frac{12}{30} = \frac{x}{45}. What is wrong with his setup, what answer does it give, and what is the right answer?
- A crew plants 30 trees in 45 minutes. Lio wants the number of trees n it plants in 3 hours and writes \displaystyle \frac{30}{45} = \frac{n}{3}. Both ratios have trees on top and time underneath. Explain why the setup is still wrong and, before solving, whether his answer will be too big or too small, and by what factor. Then find both answers.
- For the crew story at the start of the lesson, suppose 40 volunteers come instead of 14. Solve the wrong setup \displaystyle \frac{15}{6} = \frac{40}{b} and the right setup \displaystyle \frac{15}{6} = \frac{b}{40}. Does the wrong answer pass a check that only asks whether the answer went up? What check catches it?
Answers
- (a) and (c) are correct, and t = 120 minutes. (a) has pages over minutes on both sides, and (c) has 35 pages and 50 minutes, the first situation, on top of both ratios. (b) mixes minutes over pages with pages over minutes, and (d) puts the second situation on top on the left and the first on top on the right. From (a), 35t = 50 \cdot 84 = 4200, so t = 120.
- 105 trees. \displaystyle \frac{45}{3} = \frac{n}{7}, so 3n = 315.
- 5 km. \displaystyle \frac{12}{2} = \frac{30}{d}, so 12d = 60.
- 40 attempts. \displaystyle \frac{9}{15} = \frac{24}{a}, so 9a = 360.
- 7 packs. \displaystyle \frac{3}{36} = \frac{p}{84} and \displaystyle \frac{36}{3} = \frac{84}{p} both cross multiply to 36p = 252.
- 67.5 pages at that pace, so 67 complete pages. 1.5 hours is 90 minutes, and \displaystyle \frac{15}{20} = \frac{p}{90} gives 20p = 1350.
- 210 minutes. In hours, \displaystyle \frac{40}{2.5} = \frac{56}{h} gives 40h = 140, so h = 3.5 hours, and 3.5 \times 60 = 210.
- 90 adult tickets. Adult tickets over student tickets on both sides: \displaystyle \frac{3}{7} = \frac{a}{210}, so 7a = 630.
- 140 tickets for Friday and 175 for the other nights. \displaystyle \frac{4}{9} = \frac{f}{315}, so 9f = 1260, and 315 - 140 = 175.
- 14 headers. Every 2 + 7 = 9 goals include 2 headers, so \displaystyle \frac{2}{9} = \frac{h}{63} and 9h = 126.
- 26 glue sticks in 5 packs. \displaystyle \frac{4}{5} = \frac{g}{32} gives g = 25.6, rounded up to 26 since 25 would fall short; then \displaystyle 26 \div 6 = 4\frac{1}{3}, rounded up to 5 packs.
- 29 complete pages. \displaystyle \frac{11}{15} = \frac{p}{40} gives 15p = 440, so \displaystyle p = 29\frac{1}{3}.
- 21 more goals. There are 25 - 10 = 15 games left, and \displaystyle \frac{14}{10} = \frac{r}{15} gives 10r = 210.
- 6 more hours. 140 - 56 = 84 trees are left, and \displaystyle \frac{56}{4} = \frac{84}{h} gives 56h = 336.
- 30 minutes. \displaystyle \frac{2/3}{5} = \frac{4}{t} gives \displaystyle \frac{2}{3}t = 20, so t = 30. Check: 4 bags is 6 times \displaystyle \frac{2}{3} of a bag, and 6 \times 5 = 30.
- His proportion gives c = 47.5 dollars, but the order costs 43 dollars. 2c = 95 from his proportion; the real cost is 5 \times 8 + 3 = 43. The fee is charged once, so the total cost does not double when the tickets double, and cost and tickets keep no fixed ratio. The ticket part alone is proportional: \displaystyle \frac{16}{2} = \frac{x}{5} gives x = 40, and 40 + 3 = 43.
- Her proportion gives t = 8 hours; the right answer is 2 hours. 3t = 24 from her setup, which says that twice the crew takes twice as long. More volunteers take less time, so the ratio is not fixed. The work is 3 \times 4 = 12 volunteer-hours, and 12 \div 6 = 2.
- Multiplying both sides of \displaystyle \frac{a}{b} = \frac{c}{d} by bd gives ad = bc. Multiplying both sides of \displaystyle \frac{a}{c} = \frac{b}{d} by cd gives ad = bc as well. Since each step can be undone by dividing by a number that is not zero, both proportions say exactly what ad = bc says. In the crew story, \displaystyle \frac{15}{6} = \frac{b}{14} compares bags per volunteer in the two crews, and \displaystyle \frac{15}{b} = \frac{6}{14} compares the first crew with the bigger one, once in bags and once in volunteers.
- His setup puts trees over minutes on the left and minutes over trees on the right. It gives 30x = 540, so x = 18 minutes, less time for more trees. The right setup is \displaystyle \frac{12}{30} = \frac{45}{x}, which gives 12x = 1350, so x = 112.5 minutes.
- His answer is 60 times too small: n = 2 trees, against the right n = 120. A ratio is a rate in the units it is written in, so \displaystyle \frac{30}{45} is trees per minute and \displaystyle \frac{n}{3} is trees per hour. At one steady pace a crew plants 60 times as many trees in an hour as in a minute, so setting the two rates equal makes the hourly rate, and with it n, 60 times too small. His setup gives 45n = 90. Writing 3 hours as 180 minutes, \displaystyle \frac{30}{45} = \frac{n}{180} gives 45n = 5400, so n = 120.
- The wrong setup gives 15b = 240, so b = 16; the right one gives 6b = 600, so b = 100. 16 is more than 15, so the wrong answer passes a check that asks only whether the answer went up. A check on the factor catches it: 40 volunteers is about 6.7 times 6, so the bags should be about 6.7 times 15, close to 100, and 16 is nowhere near.