---
title: "Proportionality, Speed and Averages"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 61
exercises: 11
source: https://one-course.com/books/math/1/en/chapter/61-proportionality-speed-and-averages
---

# Chapter 61 — Proportionality, Speed and Averages

[Proportionality](https://one-course.com/books/math/1/en/chapter/49-proportionality#def-g7-prop-table) ([Chapter 49](https://one-course.com/books/math/1/en/chapter/49-proportionality#ch-g7-prop)) now meets the physical world: [speed](#def-g8-speed-def), flow rates, unit conversions — quantities that are *[quotients](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder)* of two others. The chapter closes with [weighted averages](#def-g8-speed-average), where [proportional](https://one-course.com/books/math/1/en/chapter/44-proportionality-and-data#def-g6-propdata-def) thinking prevents a classic mistake.

## 61.1 Average speed

**Definition 61.1 (Average speed).**

The *average speed* of a journey is the [quotient](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder)

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

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](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) 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.](https://one-course.com/images/onecourse/chapters/math-1/g8-speed/fig-ef26fdf66004.svg)

*At constant [speed](#def-g8-speed-def), distance is [proportional](https://one-course.com/books/math/1/en/chapter/44-proportionality-and-data#def-g6-propdata-def) to time: a straight line through the origin, whose steepness *is* the [speed](#def-g8-speed-def).*

**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](#def-g8-speed-def) 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](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder), 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](#def-g8-speed-def) over the whole trip is *not* $\frac{30 + 15}{2} = 22.5$ km/h. Compute with the definition:

1. times: $\frac{30}{30} = 1$ h, then $\frac{30}{15} = 2$ h; total $3$ h;
2. 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](#def-g8-speed-def) must be weighted by *time*.

## 61.4 Exercises

**Exercise 61.1 ★.**

Compute the [average speed](#def-g8-speed-def): $150$ km in $2$ h; $27$ km in $45$ min; $100$ m in $12.5$ s.

**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 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 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 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 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 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 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](#def-g8-speed-def) *including the rest* (total distance over total elapsed time).

**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](#def-g8-speed-def): $\frac{16}{4} = 4$ 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](#def-g8-speed-average). What plain (unweighted) average would she have, and why do they differ?

**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.

1. Compute the duration of each leg, then the [average speed](#def-g8-speed-def) over the whole trip.
2. Explain why the answer is less than $100$ km/h, the midpoint of the two [speeds](#def-g8-speed-def) .

**Solution of Exercise 61.10.**

*1.* Times: $\frac{120}{80} = 1.5$ h and $\frac{120}{120} = 1$ h. [Average speed](#def-g8-speed-def): $\frac{240}{2.5} = 96$ 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](#def-g8-speed-def) 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 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](#def-g8-speed-def) over equal distances is $\frac{2 v_1 v_2}{v_1 + v_2}$, and it never beats the midpoint of the [speeds](#def-g8-speed-def)

[Example 61.9](#ex-g8-speed-twolegs) and [Exercise 61.10](#exo-g8-speed-10) both ended with the same surprise: ride out at one [speed](#def-g8-speed-def), ride back at another, and the [average speed](#def-g8-speed-def) 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](#def-g8-speed-def) in the world can achieve.

**Part I — The formula for a round trip.** A journey covers a distance $d$ at [speed](#def-g8-speed-def) $v_1$, then the same distance $d$ again at [speed](#def-g8-speed-def) $v_2$ (all positive numbers).

1. Express the two durations, then the total duration, as [fractions](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) involving $d$ , $v_1$ , $v_2$ .
2. The [average speed](#def-g8-speed-def) of the whole journey is $\bar v = \dfrac{2d}{\ \dfrac{d}{v_1} + \dfrac{d}{v_2}\ }$ — a double-decker [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) ([Example 55.8](https://one-course.com/books/math/1/en/chapter/55-fractions-all-four-operations#ex-g8-fractions-complex)). Simplify it and prove: $$\bar v = \frac{2\, v_1 v_2}{v_1 + v_2} .$$
3. Check the formula against both computations quoted above: $v_1 = 30$ , $v_2 = 15$ km/h ( [Example 61.9](#ex-g8-speed-twolegs) ), and $v_1 = 80$ , $v_2 = 120$ km/h ( [Exercise 61.10](#exo-g8-speed-10) ).
4. The distance $d$ has disappeared from the formula. What does that mean, concretely, for the cyclist?
5. Now suppose the journey instead spends the same *time* $T$ at each [speed](#def-g8-speed-def) . Show that the [average speed](#def-g8-speed-def) is then $\frac{v_1 + v_2}{2}$ , the plain midpoint. In the language of [Definition 61.7](#def-g8-speed-average) : by what quantity must an average of [speeds](#def-g8-speed-def) always be weighted, and what changes between the two scenarios?

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

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

6. Compute $H$ and $A$ for the pairs $(30, 15)$ and $(80, 120)$ . Which of the two means wins each time?
7. Put $A - H$ over the common [denominator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) $2(v_1 + v_2)$, expand the [numerator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) with the remarkable identities ([Problem 57.1](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#pb-g8-equations-1)), and prove: $$A - H = \frac{(v_1 - v_2)^2}{2\,(v_1 + v_2)} .$$
8. Deduce the theorem behind all the surprises: for positive [speeds](#def-g8-speed-def) , $H \leq A$ always, with equality exactly when $v_1 = v_2$ . Why does a square in the [numerator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) settle the matter?
9. Prove the elegant identity $H \times A = v_1 \times v_2$ : the two means multiply back to the original [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) .
10. A car covers $60$ km at $40$ km/h and $60$ km at $60$ km/h. Predict the [average speed](#def-g8-speed-def) with the formula, then confirm it the long way, with times and total distance.

**Part III — The impossible catch-up.**

11. 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?
12. Show that the hoped-for average has become literally impossible: compute her [average speed](#def-g8-speed-def) if she rides the second [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) at $90$ km/h, and explain why [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) an arbitrarily enormous [speed](#def-g8-speed-def) leaves her short of $30$ km/h.
13. She lowers her goal to $20$ km/h. What [speed](#def-g8-speed-def) on the second $15$ km achieves it? Check your answer with the harmonic-mean formula.
14. A cold tap alone fills a tub in $20$ min; a hot tap alone fills it in $30$ min. What [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) 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?
15. 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](#def-g8-speed-def) , and explain in one sentence why a wind, however it blows, always *lengthens* a round trip.

**Solution of Problem 61.1.**

**1.** By [Definition 61.1](#def-g8-speed-def), $t_1 = \frac{d}{v_1}$ 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](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) 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](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) 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](https://one-course.com/books/math/1/en/chapter/55-fractions-all-four-operations#def-g8-fractions-inverse) ([Theorem 55.6](https://one-course.com/books/math/1/en/chapter/55-fractions-all-four-operations#thm-g8-fractions-div)):

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

**3.** $\dfrac{2 \times 30 \times 15}{30 + 15}
= \dfrac{900}{45} = 20$ km/h, as in [Example 61.9](#ex-g8-speed-twolegs); and $\dfrac{2 \times 80 \times 120}{80 + 120} = \dfrac{19\,200}{200}
= 96$ km/h, as found in [Exercise 61.10](#exo-g8-speed-10).

**4.** The average depends only on the two [speeds](#def-g8-speed-def), not on $d$: a $1$ km round trip and a $100$ km round trip at the same two [speeds](#def-g8-speed-def) have exactly the same [average speed](#def-g8-speed-def). (That is why the formula could be checked on journeys of different lengths.)

**5.** In time $T$ at each [speed](#def-g8-speed-def), 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](#def-g8-speed-def) is always weighted by *time* ([Definition 61.7](#def-g8-speed-average)). With equal times the two [speeds](#def-g8-speed-def) carry equal weights: plain midpoint. With equal distances the slower [speed](#def-g8-speed-def) 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](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) $2(v_1 + v_2)$:

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

and the [numerator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) expands with the identities of [Problem 57.1](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#pb-g8-equations-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](#def-g8-speed-def). Equality requires $(v_1 - v_2)^2 = 0$, i.e. $v_1 = v_2$: the moment the two [speeds](#def-g8-speed-def) 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](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) the course to go.

**12.** At $90$ km/h the second [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-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](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) [speed](#def-g8-speed-def), 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](#def-g8-speed-def) $v_2$ 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](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) the [harmonic mean](#pb-g8-speed-1)* 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.