Primary & Middle School Mathematics · Grades 1–9

61Proportionality, Speed and Averages

Proportionality (Chapter 49) now meets the physical world: speed, flow rates, unit conversions — quantities that are quotients of two others. The chapter closes with weighted averages, where proportional thinking prevents a classic mistake.

61.1 Average speed

Definition 61.1 (Average speed)

The average speed of a journey is the quotient

v = \frac{d}{t} \qquad \left(\text{distance divided by duration}\right),

in kilometers per hour (km/h), meters per second (m/s), … The formula turns around: $d = v \times t$ and $t = \frac{d}{v}$ .

Example 61.2

A train covers $270$ km in $1$ h $30$ min. First convert the time: $1$ h $30$ min $= 1.5$ h. Then

v = \frac{270}{1.5} = 180 \text{ km/h}.

Distance at $90$ km/h during $2$ h $20$ min $= \frac73$ h: $d = 90 \times \frac73 = 210$ km. Time for $35$ km at $14$ km/h: $t = \frac{35}{14} = 2.5$ h $= 2$ h $30$ min.

Remark 61.3 (Minutes are not decimals)

$1$ h $30$ min is $1.5$ h, but $1$ h $20$ min is not $1.2$ h: it is $1 + \frac{20}{60} = \frac43 \approx 1.33$ h. Always convert minutes to a fraction of an hour ( $60$ min $= 1$ h) before dividing.

At constant speed, distance is proportional to time: a straight line through the origin, whose steepness is the speed.

Method 61.4 (Converting speeds)

To convert km/h into m/s: $1$ km $= 1000$ m and $1$ h $= 3600$ s, so

v \text{ km/h} = \frac{v \times 1000}{3600} \text{ m/s} = \frac{v}{3.6} \text{ m/s}.

To convert m/s into km/h, multiply by $3.6$ .

Example 61.5

$90$ km/h $= \frac{90}{3.6} = 25$ m/s. A sprinter running $100$ m in $10$ s moves at $10$ m/s $= 36$ km/h.

61.2 Quotient quantities

Example 61.6 (Flow, density, price per kilo)

Speed has many cousins, all treated the same way:

a tap fills $48$ L in $4$ min: flow rate $\frac{48}{4} = 12$ L/min, so filling a $150$ L tub takes $\frac{150}{12} = 12.5$ min;
$0.6$ kg of cheese costs $9$ euros: price $\frac{9}{0.6} = 15$ euros per kg;
a car uses $6.3$ L for $90$ km: consumption $\frac{6.3}{90} \times 100 = 7$ L per $100$ km.

Identify the quotient, and the three-way formula ( $q = \frac ab$ , $a = qb$ , $b = \frac aq$ ) does the rest.

61.3 Weighted averages

Definition 61.7 (Weighted average)

When values $v_1, v_2, \dots$ come with counts (or weights) $n_1, n_2, \dots$ , their weighted average is

\bar v = \frac{n_1 v_1 + n_2 v_2 + \dots}{n_1 + n_2 + \dots} :

total of the values, divided by total of the weights.

Example 61.8

A test was taken by two groups: group 1, $12$ students, average $11$ ; group 2, $18$ students, average $14$ . Average of the whole class:

\bar v = \frac{12 \times 11 + 18 \times 14}{12 + 18} = \frac{132 + 252}{30} = \frac{384}{30} = 12.8 .

Not $\frac{11 + 14}{2} = 12.5$ : the larger group pulls the average towards its own — the weights matter.

Example 61.9 (Average speed over two legs)

A cyclist rides $30$ km at $30$ km/h, then $30$ km at $15$ km/h. The average speed over the whole trip is not $\frac{30 + 15}{2} = 22.5$ km/h. Compute with the definition:

times: $\frac{30}{30} = 1$ h, then $\frac{30}{15} = 2$ h; total $3$ h;
total distance $60$ km, so $v = \frac{60}{3} = 20$ km/h.

The slow leg lasts longer, so it weighs more — an average of speeds must be weighted by time.

61.4 Exercises

Exercise 61.1 ★

Compute the average speed: $150$ km in $2$ h; $27$ km in $45$ min; $100$ m in $12.5$ s.

Solution

Solution of Exercise 61.1.

$\frac{150}{2} = 75$ km/h. $45$ min $= 0.75$ h: $\frac{27}{0.75} = 36$ km/h. $\frac{100}{12.5} = 8$ m/s.

Exercise 61.2 ★

Compute the distance: $2$ h at $85$ km/h; $40$ min at $90$ km/h. Compute the time: $315$ km at $90$ km/h (in h and min).

Solution

Solution of Exercise 61.2.

$2 \times 85 = 170$ km. $40$ min $= \frac23$ h: $90 \times \frac23 = 60$ km. $\frac{315}{90} = 3.5$ h $= 3$ h $30$ min.

Exercise 61.3 ★

Convert: $72$ km/h into m/s; $5$ m/s into km/h; $108$ km/h into m/s.

Solution

Solution of Exercise 61.3.

$72 \div 3.6 = 20$ m/s. $5 \times 3.6 = 18$ km/h. $108 \div 3.6 = 30$ m/s.

Exercise 61.4 ★

A tap delivers $15$ L/min. How long to fill a $600$ L tank? How many liters in $2$ h $30$ ?

Solution

Solution of Exercise 61.4.

$600 \div 15 = 40$ min. $2$ h $30 = 150$ min: $15 \times 150 = 2250$ L.

Exercise 61.5 ★

Which is the better buy: $1.2$ kg of apples for $3.30$ euros, or $0.8$ kg for $2.16$ euros? Compare prices per kilogram.

Solution

Solution of Exercise 61.5.

$\frac{3.30}{1.2} = 2.75$ euros/kg against $\frac{2.16}{0.8} = 2.70$ euros/kg: the second offer is (slightly) the better buy.

Exercise 61.6 ★

A class of $25$ students has an average of $12.4$ on a test; another class of $35$ students has an average of $10$ . Compute the average of the two classes together.

Solution

Solution of Exercise 61.6.

\bar v = \frac{25 \times 12.4 + 35 \times 10}{60} = \frac{310 + 350}{60} = \frac{660}{60} = 11 .

Exercise 61.7 ★

Sound travels at about $340$ m/s. You see a lightning flash and hear the thunder $6$ seconds later. How far away did the lightning strike (to the nearest $100$ m)?

Solution

Solution of Exercise 61.7.

$d = 340 \times 6 = 2040$ m $\approx 2$ km.

Exercise 61.8 ★★

A hiker walks $2$ h at $5$ km/h, rests $30$ min, then walks $1$ h $30$ at $4$ km/h. Compute the total distance and the average speed including the rest (total distance over total elapsed time).

Solution

Solution of Exercise 61.8.

Distances: $2 \times 5 = 10$ km, then $1.5 \times 4 = 6$ km: total $16$ km. Elapsed time: $2 + 0.5 + 1.5 = 4$ h. Average speed: $\frac{16}{4} = 4$ km/h.

Exercise 61.9 ★★

Marks with coefficients: a student got $15$ (coefficient $3$ ), $9$ (coefficient $2$ ) and $12$ (coefficient $1$ ). Compute the weighted average. What plain (unweighted) average would she have, and why do they differ?

Solution

Solution of Exercise 61.9.

Weighted: $\frac{3 \times 15 + 2 \times 9 + 1 \times 12}{3 + 2 + 1} = \frac{45 + 18 + 12}{6} = \frac{75}{6} = 12.5$ . Unweighted: $\frac{15 + 9 + 12}{3} = 12$ . They differ because the coefficients give the mark $15$ three times the weight of the mark $12$ .

Exercise 61.10 ★★

A car drives $120$ km at $80$ km/h and then $120$ km at $120$ km/h.

Compute the duration of each leg, then the average speed over the whole trip.
Explain why the answer is less than $100$ km/h, the midpoint of the two speeds.

Solution

Solution of Exercise 61.10.

1. Times: $\frac{120}{80} = 1.5$ h and $\frac{120}{120} = 1$ h. Average speed: $\frac{240}{2.5} = 96$ km/h.

2. The car spends more time at $80$ km/h ( $1.5$ h) than at $120$ km/h ( $1$ h): the slow speed weighs more in the time-weighted average, pulling it below the midpoint $100$ .

Exercise 61.11 ★★★

Two villages are $18$ km apart. Anna leaves the first at $10{:}00$ walking at $5$ km/h towards the second; Boris leaves the second at the same time walking at $4$ km/h towards the first. At what time do they meet, and at what distance from Anna’s village? (Together they close the gap at $5 + 4$ km/h.)

Solution

Solution of Exercise 61.11.

The gap of $18$ km closes at $5 + 4 = 9$ km/h: they meet after $\frac{18}{9} = 2$ h, at $12{:}00$ . Anna has then walked $2 \times 5 = 10$ km: they meet $10$ km from Anna’s village (and $8$ km from Boris’s — check: $10 + 8 = 18$ ).

61.5 Problem: The harmonic mean, or why the return trip ruins the average

Problem 61.1

Weekend problem — the average speed over equal distances is $\frac{2 v_1 v_2}{v_1 + v_2}$ , and it never beats the midpoint of the speeds

Example 61.9 and Exercise 61.10 both ended with the same surprise: ride out at one speed, ride back at another, and the average speed lands below the midpoint of the two. This problem finds the exact formula behind the surprise — a new kind of average, the harmonic mean — proves that the surprise is a theorem, and pushes it to its logical extreme: a catch-up that no speed in the world can achieve.

Part I — The formula for a round trip. A journey covers a distance $d$ at speed $v_1$ , then the same distance $d$ again at speed $v_2$ (all positive numbers).

Express the two durations, then the total duration, as fractions involving $d$ , $v_1$ , $v_2$ .
The average speed of the whole journey is $\bar v = \dfrac{2d}{\ \dfrac{d}{v_1} + \dfrac{d}{v_2}\ }$ — a double-decker fraction (Example 55.8). Simplify it and prove:

$\bar v = \frac{2\, v_1 v_2}{v_1 + v_2} .$
Check the formula against both computations quoted above: $v_1 = 30$ , $v_2 = 15$ km/h (Example 61.9), and $v_1 = 80$ , $v_2 = 120$ km/h (Exercise 61.10).
The distance $d$ has disappeared from the formula. What does that mean, concretely, for the cyclist?
Now suppose the journey instead spends the same time $T$ at each speed. Show that the average speed is then $\frac{v_1 + v_2}{2}$ , the plain midpoint. In the language of Definition 61.7: by what quantity must an average of speeds always be weighted, and what changes between the two scenarios?

Part II — Harmonic against arithmetic. For two positive numbers $v_1$ , $v_2$ , set

H = \frac{2\, v_1 v_2}{v_1 + v_2} \quad\text{(their \emph{harmonic mean}),} \qquad A = \frac{v_1 + v_2}{2} \quad\text{(their \emph{arithmetic mean}).}

Compute $H$ and $A$ for the pairs $(30, 15)$ and $(80, 120)$ . Which of the two means wins each time?
Put $A - H$ over the common denominator $2(v_1 + v_2)$ , expand the numerator with the remarkable identities (Problem 57.1), and prove:

$A - H = \frac{(v_1 - v_2)^2}{2\,(v_1 + v_2)} .$
Deduce the theorem behind all the surprises: for positive speeds, $H \leq A$ always, with equality exactly when $v_1 = v_2$ . Why does a square in the numerator settle the matter?
Prove the elegant identity $H \times A = v_1 \times v_2$ : the two means multiply back to the original product.
A car covers $60$ km at $40$ km/h and $60$ km at $60$ km/h. Predict the average speed with the formula, then confirm it the long way, with times and total distance.

Part III — The impossible catch-up.

A cyclist enters a $30$ km race hoping to average $30$ km/h. What total time may the race take her, at most? She rides the first $15$ km at $15$ km/h: how much of that time budget is left?
Show that the hoped-for average has become literally impossible: compute her average speed if she rides the second half at $90$ km/h, and explain why even an arbitrarily enormous speed leaves her short of $30$ km/h.
She lowers her goal to $20$ km/h. What speed on the second $15$ km achieves it? Check your answer with the harmonic-mean formula.
A cold tap alone fills a tub in $20$ min; a hot tap alone fills it in $30$ min. What fraction of the tub do both taps together fill per minute, and how long do they take? Compare your answer with $H(20, 30)$ — what do you notice?
A plane flies the same route out and back: $900$ km/h with the wind, $700$ km/h against it. Compute the round-trip average speed, and explain in one sentence why a wind, however it blows, always lengthens a round trip.

Solution

Solution of Problem 61.1.

1. By Definition 61.1, $t_1 = \frac{d}{v_1}$ and $t_2 = \frac{d}{v_2}$ ; the whole journey lasts $\frac{d}{v_1} + \frac{d}{v_2}$ for a distance of $2d$ .

2. Factor $d$ out of the denominator and simplify it away:

\bar v = \frac{2d}{d\left(\frac{1}{v_1} + \frac{1}{v_2}\right)} = \frac{2}{\ \frac{1}{v_1} + \frac{1}{v_2}\ } .

Common denominator inside: $\frac{1}{v_1} + \frac{1}{v_2} = \frac{v_2 + v_1}{v_1 v_2}$ , and dividing means multiplying by the inverse (Theorem 55.6):

\bar v = 2 \times \frac{v_1 v_2}{v_1 + v_2} = \frac{2\, v_1 v_2}{v_1 + v_2} .

3. $\dfrac{2 \times 30 \times 15}{30 + 15} = \dfrac{900}{45} = 20$ km/h, as in Example 61.9; and $\dfrac{2 \times 80 \times 120}{80 + 120} = \dfrac{19\,200}{200} = 96$ km/h, as found in Exercise 61.10.

4. The average depends only on the two speeds, not on $d$ : a $1$ km round trip and a $100$ km round trip at the same two speeds have exactly the same average speed. (That is why the formula could be checked on journeys of different lengths.)

5. In time $T$ at each speed, the distances are $v_1 T$ and $v_2 T$ , so

\bar v = \frac{v_1 T + v_2 T}{2T} = \frac{(v_1 + v_2)\,T}{2T} = \frac{v_1 + v_2}{2} .

An average of speeds is always weighted by time (Definition 61.7). With equal times the two speeds carry equal weights: plain midpoint. With equal distances the slower speed occupies more time, hence more weight — and the average slides toward it.

6. For $(30, 15)$ : $H = 20$ and $A = 22.5$ . For $(80, 120)$ : $H = 96$ and $A = 100$ . The arithmetic mean wins both times.

7. Over the common denominator $2(v_1 + v_2)$ :

A - H = \frac{(v_1 + v_2)^2 - 4\, v_1 v_2}{2\,(v_1 + v_2)} ,

and the numerator expands with the identities of Problem 57.1:

(v_1 + v_2)^2 - 4 v_1 v_2 = v_1^2 + 2 v_1 v_2 + v_2^2 - 4 v_1 v_2 = v_1^2 - 2 v_1 v_2 + v_2^2 = (v_1 - v_2)^2 .

8. A square is never negative, and $2(v_1 + v_2)$ is positive: so $A - H \geq 0$ , that is $H \leq A$ , for every pair of positive speeds. Equality requires $(v_1 - v_2)^2 = 0$ , i.e. $v_1 = v_2$ : the moment the two speeds differ, the round trip’s average drops strictly below the midpoint — the surprise is a theorem.

9. Multiply:

H \times A = \frac{2\, v_1 v_2}{v_1 + v_2} \times \frac{v_1 + v_2}{2} = v_1 v_2 ,

after cancelling the factor $2$ and the factor $v_1 + v_2$ .

10. Formula: $\bar v = \dfrac{2 \times 40 \times 60} {40 + 60} = \dfrac{4800}{100} = 48$ km/h. The long way: $\frac{60}{40} = 1.5$ h and $\frac{60}{60} = 1$ h, so $120$ km in $2.5$ h: $\frac{120}{2.5} = 48$ km/h. They agree.

11. Averaging $30$ km/h over $30$ km means a total time of at most $\frac{30}{30} = 1$ h. The first $15$ km at $15$ km/h already take $\frac{15}{15} = 1$ h: the entire time budget is spent, with half the course to go.

12. At $90$ km/h the second half takes $\frac{15}{90} = \frac16$ h, so the race takes $\frac76$ h and

\bar v = \frac{30}{\ 7/6\ } = \frac{180}{7} \approx 25.7 \text{ km/h} < 30 .

Whatever the second-half speed, its duration is a positive number, so the total time exceeds $1$ h and the average stays strictly below $30$ km/h. (With the formula: $\frac{2 \times 15 \times v_2}{15 + v_2} = 30$ would force $30 v_2 = 450 + 30 v_2$ , i.e. $0 = 450$ — no speed $v_2$ works.)

13. A $20$ km/h average allows $\frac{30}{20} = 1.5$ h; after the first hour, $15$ km remain for $0.5$ h: she needs $\frac{15}{0.5} = 30$ km/h. Check: $H(15, 30) = \frac{2 \times 15 \times 30}{45} = \frac{900}{45} = 20$ km/h.

14. Per minute, the taps fill $\frac{1}{20}$ and $\frac{1}{30}$ of the tub; together

\frac{1}{20} + \frac{1}{30} = \frac{3}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12}

of the tub per minute: the tub fills in $12$ min. And $H(20, 30) = \frac{2 \times 600}{50} = 24 = 2 \times 12$ : the two taps together take exactly half the harmonic mean of their solo times — rates add, so it is again harmonic-mean territory.

15. Same distance each way, so

\bar v = \frac{2 \times 900 \times 700}{900 + 700} = \frac{1\,260\,000}{1\,600} = 787.5 \text{ km/h},

below the still-air $800$ km/h. In one sentence: the plane spends longer flying against the wind than with it, so the slow leg gets the larger weight in the time-weighted average — a wind can only lengthen a round trip.