Ratios and proportions
A "16:9" screen, a map with a scale of "1:300,000", a recipe that calls for "2 of flour for every 1 of water"… These are all ratios: ways of comparing two quantities by dividing. If on the map of the Autopista Duarte 1 cm equals 3 km, how do you figure out the distance between Santo Domingo and Santiago? We'll solve it at the end of the page.
Ratio between two quantities
In one class at the high school there are 12 girls and 18 boys. The girls-to-boys ratio is $\frac{12}{18} = \frac{2}{3}$: for every 2 girls there are 3 boys.
Ratios can also compare different quantities, such as a speed: $95\text{ km}$ in $4\text{ h}$ gives $\frac{95}{4} = 23.75\text{ km/h}$. Here the units matter: a ratio is always interpreted together with them ("23.75 km for every hour").
Scales
A scale is the ratio between a measurement in the drawing and the real measurement. A scale of $1 : 300{,}000$ means that 1 cm on the map is $300{,}000$ cm = 3 km in real life.
Calculate the value of the ratio of each pair of quantities and interpret the result: a) $3\ \text{kg}$ of cherries cost $\text{RD\$}\ 630…
Calculate the value of the ratio of each pair of quantities and interpret the result:
- a) $3\ \text{kg}$ of cherries cost $\text{RD\$}\ 630$.
- b) A gravel cyclist covers $95\ \text{km}$ in $4\ \text{h}$.
- c) A bottle contains $2\ \text{L}$ and a drum, $25\ \text{L}$.
Two metal rods measure $8\ \text{dm}$ and $12\ \text{dm}$. a) Calculate the ratio of the length of the first to the second, and simplify it. b) Calculate the ratio of…
Two metal rods measure $8\ \text{dm}$ and $12\ \text{dm}$.
- a) Calculate the ratio of the length of the first to the second, and simplify it.
- b) Calculate the ratio of the second to the first.
- c) How are the two ratios obtained related to each other? Does this always happen?
A road map is drawn to a scale of $1 : 300,000$. a) Explain the meaning of this ratio. b) What real-life distance is represented by $5\ \text{cm}$ on the map? c) Two…
A road map is drawn to a scale of $1 : 300,000$.
- a) Explain the meaning of this ratio.
- b) What real-life distance is represented by $5\ \text{cm}$ on the map?
- c) Two towns are $21\ \text{km}$ apart in reality. How far apart are they on the map?
A javelin thrower trains with a mark set at $60\ \text{m}$ and takes $4.8\ \text{s}$ to cover that distance in her run-up. a) Calculate the ratio of distance to time, in…
A javelin thrower trains with a mark set at $60\ \text{m}$ and takes $4.8\ \text{s}$ to cover that distance in her run-up.
- a) Calculate the ratio of distance to time, in $\text{m/s}$.
- b) Express it in $\text{km/h}$.
- c) Is it correct to calculate the ratio between $60\ \text{m}$ and $4.8\ \text{s}$ without considering the units? Why must units always be specified when interpreting a ratio?
Proportions
Check, by calculating the product of the extremes and the product of the means, whether the following pairs of ratios form a proportion: a) $\frac{6}{4}$ and…
Check, by calculating the product of the extremes and the product of the means, whether the following pairs of ratios form a proportion:
- a) $\frac{6}{4}$ and $\frac{15}{10}$
- b) $\frac{5}{8}$ and $\frac{7}{12}$
- c) $\frac{2.5}{4}$ and $\frac{5}{8}$
Using the numbers $3$, $6$, $4$ and $8$, form: a) A proportion in which $3$ and $8$ are the extremes. b) A proportion in which $3$ and $8$ are the means. c) Check in…
Using the numbers $3$, $6$, $4$ and $8$, form:
- a) A proportion in which $3$ and $8$ are the extremes.
- b) A proportion in which $3$ and $8$ are the means.
- c) Check in both cases that the product of the extremes is equal to the product of the means.
Finding the missing term
Thanks to the fundamental property, if you know three terms you can find the fourth (the fourth proportional):
$$\frac{x}{4} = \frac{9}{12} \ \Rightarrow\ 12x = 36 \ \Rightarrow\ x = 3$$And if the means are equal, $\frac{a}{x} = \frac{x}{b}$ (a continued proportion), then $x^2 = ab$ and $x = \sqrt{ab}$ is the mean proportional. For example, between 4 and 9: $x = \sqrt{36} = 6$.
Swapping the means or the extremes
If $\frac{a}{b} = \frac{c}{d}$, the proportions that keep the same cross product $a \times d = b \times c$ are still true:
-
swapping the means: $\dfrac{a}{c} = \dfrac{b}{d}$
-
swapping the extremes: $\dfrac{d}{b} = \dfrac{c}{a}$
For example, from $\frac{6}{9} = \frac{10}{15}$ we get $\frac{6}{10} = \frac{9}{15}$. On the other hand, moving the numbers around "by eye" doesn't work: $\frac{6}{10} = \frac{15}{9}$ is false, because $6 \times 9 = 54$ and $10 \times 15 = 150$.
Find the value of $x$ in these proportions: a) $\frac{x}{2.4} = \frac{7}{4.8}$ b) $\frac{3}{x} = \frac{\frac{1}{2}}{\frac{3}{4}}$ c) $\frac{1.5}{4.5} = \frac{x}{13.5}$
Find the value of $x$ in these proportions:
- a) $\frac{x}{2.4} = \frac{7}{4.8}$
- b) $\frac{3}{x} = \frac{\frac{1}{2}}{\frac{3}{4}}$
- c) $\frac{1.5}{4.5} = \frac{x}{13.5}$
Find the mean proportional in each case, that is, the positive value of $x$ that satisfies the continuous proportion: a) $\frac{x}{4} = \frac{25}{x}$ b)…
Find the mean proportional in each case, that is, the positive value of $x$ that satisfies the continuous proportion:
- a) $\frac{x}{4} = \frac{25}{x}$
- b) $\frac{3}{x} = \frac{x}{27}$
- c) $\frac{x}{2} = \frac{50}{x}$
A carpenter makes picture frames whose central decorative section is a square. So that the frame maintains the same proportion in all sizes, the side $x$ of the square…
A carpenter makes picture frames whose central decorative section is a square. So that the frame maintains the same proportion in all sizes, the side $x$ of the square must be the mean proportional between the inner width and the outer width of the frame.
- a) For a frame with an inner width of $9\ \text{cm}$ and an outer width of $25\ \text{cm}$, set up the continuous proportion $\frac{9}{x} = \frac{x}{25}$ and find the side length of the square.
- b) Check that the two ratios have the same value.
- c) If the inner width is $8\ \text{cm}$ and the decorative square has side length $12\ \text{cm}$, what is the outer width of the frame?
A student states the following: > "Since $\frac{6}{9} = \frac{10}{15}$, then $\frac{6}{10} = \frac{15}{9}$ also holds, because I have only changed the order of the…
A student states the following:
"Since $\frac{6}{9} = \frac{10}{15}$, then $\frac{6}{10} = \frac{15}{9}$ also holds, because I have only changed the order of the numbers."
- a) Check whether the first equality is a true proportion.
- b) Check whether the second equality is also true. Is she correct?
- c) The correct rule states that in a proportion you can swap the means or swap the extremes. Apply this to $\frac{6}{9} = \frac{10}{15}$ and write down two equivalent true proportions.
State whether each statement is true or false, assuming that $\frac{a}{b} = \frac{c}{d}$ is a true proportion. Justify your answer using the product of the extremes and…
State whether each statement is true or false, assuming that $\frac{a}{b} = \frac{c}{d}$ is a true proportion. Justify your answer using the product of the extremes and means:
- a) "$\frac{a}{c} = \frac{b}{d}$"
- b) "$\frac{d}{b} = \frac{c}{a}$"
- c) "$\frac{a + b}{b} = \frac{c + d}{d}$"
- d) "$\frac{a}{b} = \frac{a + c}{b + d}$"
Summary
Back to the beginning: if 1 cm on the map is 3 km, the proportion $\frac{1}{3} = \frac{x}{\text{real km}}$ lets you find any distance. With $x = 60$ cm on the map, it would be $180$ km.