Primary & Middle School Mathematics · Grades 1–9

49Proportionality

Grade 6 met proportional tables (Chapter 44); this chapter turns proportionality into a fully-fledged tool: the cross rule for finding a fourth value, percentages, and scales of maps and models. Speed, the most famous proportionality of all, is studied in Chapter 61.

49.1 Recognizing and completing tables

Definition 49.1 (Proportionality table)

A table of two rows is a proportionality table when the numbers of the second row are those of the first multiplied by one fixed number, the coefficient. Equivalently: all column quotients $\frac{\text{bottom}}{\text{top}}$ are equal.

Example 49.2

top	$4$	$10$	$14$
bottom	$6$	$15$	$21$

Quotients: $\frac64 = 1.5$ , $\frac{15}{10} = 1.5$ , $\frac{21}{14} = 1.5$ : proportional, coefficient $1.5$ .

Theorem 49.3 (Cross rule)

In a proportionality table with columns $\begin{pmatrix} a \\ b \end{pmatrix}$ and $\begin{pmatrix} c \\ d \end{pmatrix}$ , the cross products are equal:

a \times d = b \times c .

Consequently the fourth value can be computed from the three others: $d = \dfrac{b \times c}{a}$ .

Proof. Let $k$ be the coefficient: $b = ka$ and $d = kc$ . Then

a \times d = a \times kc = k \times ac, \qquad b \times c = ka \times c = k \times ac :

both cross products equal $k \times ac$ . Dividing $ad = bc$ by $a$ gives the formula for $d$ . ∎

Example 49.4 (Fourth proportional)

If $7$ identical books weigh $2.8$ kg, how much do $12$ such books weigh? Table and cross rule:

books	$7$	$12$
kg	$2.8$	$m$

7 \times m = 2.8 \times 12 \quad\Longrightarrow\quad m = \frac{2.8 \times 12}{7} = \frac{33.6}{7} = 4.8 \text{ kg}.

Check by the unit: one book weighs $0.4$ kg, twelve weigh $4.8$ kg.

49.2 Percentages

Method 49.5 (Applying and finding percentages)

Apply $t\,\%$ : multiply by $\frac{t}{100}$ ( $12\,\%$ of $350$ is $0.12 \times 350 = 42$ );
find the percentage that a part represents: compute $\frac{\text{part}}{\text{whole}}$ and rewrite over $100$ ( $21$ out of $60$ : $\frac{21}{60} = \frac{35}{100} = 35\,\%$ );
always ask: a percentage of what? The reference whole matters.

Example 49.6

In a school, $180$ of the $400$ students are girls. Percentage:

\frac{180}{400} = \frac{180 \div 4}{400 \div 4} = \frac{45}{100} = 45\,\% .

The other way: how many are the $55\,\%$ boys? $0.55 \times 400 = 220$ . Check: $180 + 220 = 400$ .

49.3 Scales

Definition 49.7 (Scale)

On a map or model at scale $\frac{1}{n}$ (written $1 : n$ ), the distances on the map are proportional to the real distances, with coefficient $\frac1n$ :

\text{map distance} = \frac{1}{n} \times \text{real distance},

both measured in the same unit.

Example 49.8

On a $1 : 50\,000$ map, two towns are $7$ cm apart. Real distance:

7 \times 50\,000 = 350\,000 \text{ cm} = 3\,500 \text{ m} = 3.5 \text{ km}.

Conversely, a $12$ km hiking trail measures on the map $12$ km $= 1\,200\,000$ cm, so $1\,200\,000 \div 50\,000 = 24$ cm.

A scale is a proportionality: map distances against real distances line up on a straight line through the origin (here 1 cm 4 km; the dashes read 7 cm 28 km). — A scale is a proportionality: map distances against real distances line up on a straight line through the origin (here $1$ cm $\leftrightarrow$ $4$ km; the dashes read $7$ cm $\leftrightarrow$ $28$ km).

Remark 49.9

Graphs are the quickest test of proportionality (Chapter 44): points on a straight line through the origin mean proportional quantities; a line missing the origin (like a taxi fare with a fixed charge) means not proportional, even though it is a straight line. Affine functions, in Chapter 67, will make this precise.

49.4 Exercises

Exercise 49.1 ★

Which tables are proportionality tables? Give the coefficient when it exists.

	$3$	$7$	$11$
	$7.5$	$17.5$	$27.5$

	$2$	$5$	$9$
	$6$	$15$	$28$

Solution

Solution of Exercise 49.1.

First table: quotients $\frac{7.5}{3} = 2.5$ , $\frac{17.5}{7} = 2.5$ , $\frac{27.5}{11} = 2.5$ : proportional, coefficient $2.5$ .

Second table: $\frac62 = 3$ and $\frac{15}{5} = 3$ , but $\frac{28}{9} \approx 3.1$ : not proportional.

Exercise 49.2 ★

Complete using the cross rule, writing the computation: $5$ kg of potatoes cost $6$ euros; how much do $8$ kg cost?

Solution

Solution of Exercise 49.2.

With the cross rule: $5 \times p = 6 \times 8$ , so $p = \frac{48}{5} = 9.60$ euros.

Exercise 49.3 ★

A printer prints $36$ pages in $3$ minutes. At the same rate, how many pages in $5$ minutes? How long for $96$ pages?

Solution

Solution of Exercise 49.3.

Rate: $36 \div 3 = 12$ pages per minute. In $5$ minutes: $60$ pages. For $96$ pages: $96 \div 12 = 8$ minutes.

Exercise 49.4 ★

Compute: $30\,\%$ of $250$ ; $15\,\%$ of $60$ ; $t\,\%$ such that $t\,\%$ of $80$ is $20$ .

Solution

Solution of Exercise 49.4.

$30\,\%$ of $250$ : $0.3 \times 250 = 75$ . $15\,\%$ of $60$ : $0.15 \times 60 = 9$ . $t\,\%$ of $80$ is $20$ : $\frac{20}{80} = \frac{25}{100}$ , so $t = 25$ .

Exercise 49.5 ★

Out of $25$ shots, a basketball player scored $18$ . What is her success percentage? (Rewrite the fraction over $100$ .)

Solution

Solution of Exercise 49.5.

$\frac{18}{25} = \frac{18 \times 4}{25 \times 4} = \frac{72}{100} = 72\,\%$ .

Exercise 49.6 ★

On a $1 : 25\,000$ map, a path measures $9$ cm. What is its real length in km? A lake is $2$ km long: how long is it on the map?

Solution

Solution of Exercise 49.6.

Path: $9 \times 25\,000 = 225\,000$ cm $= 2.25$ km. Lake: $2$ km $= 200\,000$ cm on the ground, so $200\,000 \div 25\,000 = 8$ cm on the map.

Exercise 49.7 ★

A model car is built at scale $1 : 43$ . The real car is $4.3$ m long. How long is the model, in cm?

Solution

Solution of Exercise 49.7.

$4.3$ m $= 430$ cm, and $430 \div 43 = 10$ cm.

Exercise 49.8 ★★

A recipe uses $240$ g of chocolate for $8$ servings. Aline has $300$ g of chocolate. For how many servings is that enough (whole number of servings)?

Solution

Solution of Exercise 49.8.

Chocolate per serving: $240 \div 8 = 30$ g. With $300$ g: $300 \div 30 = 10$ servings exactly.

Exercise 49.9 ★★

The price of a jacket drops from $80$ to $60$ euros.

What is the discount in euros? In percent of the original price?
Later the price goes back up from $60$ to $80$ euros. Explain why this rise is not $25\,\%$ — compute the correct percentage of increase.

Solution

Solution of Exercise 49.9.

1. Discount: $20$ euros, i.e. $\frac{20}{80} = 25\,\%$ of the original price.

2. The rise of $20$ euros is now compared with the new reference $60$ : $\frac{20}{60} = \frac13 \approx 33\,\%$ . A percentage always refers to a whole; the whole changed, so the percentage does too.

Exercise 49.10 ★★

Two rows of a table are proportional:

	$4$	$x$	$10$
	$y$	$10.5$	$17.5$

Find the coefficient, then $x$ and $y$ .

Solution

Solution of Exercise 49.10.

Coefficient from the complete column: $\frac{17.5}{10} = 1.75$ . Then $y = 4 \times 1.75 = 7$ and $x = \frac{10.5}{1.75} = 6$ .

Exercise 49.11 ★★★

A photocopier reduces documents to $80\,\%$ of their size. A segment of $10$ cm is copied, then the copy is copied again with the same setting.

How long is the segment after one copy? After two?
Why is the answer after two copies not $60\,\%$ of the original? What single percentage corresponds to two copies?

Solution

Solution of Exercise 49.11.

1. After one copy: $10 \times 0.8 = 8$ cm. After two: $8 \times 0.8 = 6.4$ cm.

2. The second reduction applies to the already-reduced copy, not to the original: the factors multiply, $0.8 \times 0.8 = 0.64$ , so two copies reduce to $64\,\%$ of the original — not $60\,\%$ .

49.5 Problem: Measuring the Earth with a stick

Problem 49.1

Weekend problem — shadows are a proportionality, and how Eratosthenes computed the circumference of the Earth in 240 BC

Twenty-two centuries ago, with no telescope, no satellite and no calculator, the librarian of Alexandria measured the Earth — using a stick, a well, a camel caravan and this chapter’s mathematics. This problem retraces his steps: first the proportionality of shadows, then the famous computation itself, and finally what his answer looks like at human scale.

Part I — The shadow of a stick. The sun is so far away that its rays reach us parallel to one another. At a given moment, all vertical objects and their shadows are therefore proportional.

At noon, a vertical stick of $1$ m casts a shadow of $0.4$ m, and a tree’s shadow measures $3.2$ m. How tall is the tree (Theorem 49.3)?
At the same moment, a tower casts a $26$ m shadow: how tall is the tower? And how long is the shadow of a child $1.5$ m tall?
Explain in one sentence why all these computations are valid only at the same moment of the day.
Legend says Thales measured the Great Pyramid by waiting for the moment when his own shadow was exactly as long as he was tall. What is the proportionality coefficient at that moment, and how tall is the pyramid if the tip of its shadow lies $147$ m from the point of the ground directly below its apex?
Summarize Part I: at one given moment, which quantity is the same for the stick, the tree, the tower and the pyramid (Definition 49.1)?

Part II — Eratosthenes’ computation. Eratosthenes knew two facts. In the town of Syene, at noon on the summer solstice, the sun stood exactly overhead: sunlight reached the bottom of the deepest wells, and vertical sticks cast no shadow. In Alexandria, $800$ km due north, at the same moment, a vertical stick did cast a shadow, and the sun’s rays made an angle of $7.2^\circ$ with the vertical.

Explain why these two observations together prove that the ground of Syene and the ground of Alexandria are not parallel — that is, the Earth’s surface is curved. (What would parallel rays do to two sticks on a flat Earth?)
On a drawing of the round Earth with parallel sun rays, the $7.2^\circ$ in Alexandria reappears at the center of the Earth, as the angle between the directions of the two cities. Make the drawing and convince yourself (the clean justification uses the equal-angle pairs of Chapter 51).
What fraction of a full turn is $7.2^\circ$ ?
The arc from Syene to Alexandria ( $800$ km) corresponds to $7.2^\circ$ ; the whole circumference corresponds to $360^\circ$ . Set up the proportionality and compute the circumference of the Earth.
Deduce the diameter of the Earth, using $\pi \approx 3.14$ (Proposition 43.3) and rounding to the nearest hundred kilometers. (The modern value is $12\,742$ km — how close was a man with a stick in 240 BC?)

Part III — The Earth at human scale.

A globe is built at scale $1 : 40\,000\,000$ . Using the circumference found in question 9, show that the globe’s circumference is exactly $1$ m. What is its diameter, to the nearest centimeter?
Mont Blanc rises about $4.8$ km. Convert $4.8$ km to centimeters, divide by $40\,000\,000$ , and express the mountain’s height on the globe in millimeters. What does the answer say about how “bumpy” the Earth really is?
A walker covers $40$ km per day. At that pace, how many days for Eratosthenes’ full circumference — and roughly how many years is that?
The breathable atmosphere is concentrated in roughly the first $10$ km above the ground. What percentage of the Earth’s radius (about $6\,370$ km) is that (Method 49.5)? Round to the nearest tenth of a percent.
Historians estimate that Eratosthenes’ announced value, converted to modern units, may have been about $42\,000$ km. Taking $40\,000$ km as the true value, compute his percentage of error. Conclude in one sentence about sticks, wells and proportionality.

Solution

Solution of Problem 49.1.

1. Heights and shadows are proportional, so by the cross rule (Theorem 49.3), with the stick’s column $(1,\ 0.4)$ and the tree’s $(h,\ 3.2)$ : $0.4 \times h = 1 \times 3.2$ , hence $h = \frac{3.2}{0.4} = 8$ m.

2. Tower: $\frac{26}{0.4} = 65$ m. Child’s shadow: $1.5 \times 0.4 = 0.6$ m (shadow $=$ height $\times$ the moment’s coefficient $0.4$ ).

3. As the sun moves across the sky, the coefficient linking heights to shadows changes; only measurements taken at the same moment share the same coefficient.

4. At that moment the coefficient is $1$ : every height equals its shadow. The pyramid’s apex is therefore $147$ m high — the height of the Great Pyramid (its shadow’s tip, measured from the point below the apex, is all one needs).

5. The column quotient $\frac{\text{shadow}}{\text{height}}$ — the proportionality coefficient of the moment — is common to every vertical object: that is exactly what makes the table of heights and shadows a proportionality table (Definition 49.1).

6. Sun rays arrive parallel. On a flat Earth, two parallel rays would strike two vertical sticks at the same angle: both would cast proportional shadows — either both no shadow, or both a shadow. One stick with no shadow (Syene) and one with a shadow (Alexandria), at the same instant, is impossible on a flat Earth: the two verticals must point in different directions — the surface curves.

7. On the drawing, the Syene vertical points straight at the sun; the Alexandria vertical is tilted by the angle between the two city directions, seen from the center. Since the rays are parallel, that tilt is exactly the $7.2^\circ$ measured between ray and stick in Alexandria.

8. $\frac{7.2}{360} = \frac{72}{3600} = \frac{1}{50}$ : one fiftieth of a full turn.

9. Arc lengths are proportional to angles: if $7.2^\circ$ — one fiftieth of the turn — corresponds to $800$ km, the full turn corresponds to

50 \times 800 = 40\,000 \text{ km}.

10. Diameter $= \frac{\text{circumference}}{\pi} \approx \frac{40\,000}{3.14} \approx 12\,700$ km — against the modern $12\,742$ km: correct to well within one percent, with a stick.

11. $40\,000$ km $= 4\,000\,000\,000$ cm, and $4\,000\,000\,000 \div 40\,000\,000 = 100$ cm $= 1$ m of circumference. Diameter: $100 \div 3.14 \approx 32$ cm — a handsome desk globe.

12. $4.8$ km $= 480\,000$ cm, and $480\,000 \div 40\,000\,000 = 0.012$ cm $= 0.12$ mm. The highest mountain of the Alps is a tenth of a millimeter on a meter-round globe: a speck of dust. At this scale the Earth is smoother than most polished balls.

13. $40\,000 \div 40 = 1\,000$ days, and $1\,000 \div 365 \approx 2.7$ : nearly three years of walking.

14. $\frac{10}{6\,370} \approx 0.0016$ , that is about $0.2\,\%$ (more precisely $0.16\,\%$ ): the atmosphere is a whisper-thin skin on the planet.

15. Error: $42\,000 - 40\,000 = 2\,000$ km, and $\frac{2\,000}{40\,000} = \frac{5}{100} = 5\,\%$ . One sentence: with a stick, a well, a measured road and one proportionality, Eratosthenes measured a planet to within a few percent — mathematics travels far on very little.