Ratios and proportions are fundamental concepts in mathematics that are applied in solving practical problems and form the basis for understanding similarity of figures, scaling, and many other mathematical concepts. In YÖS tasks this topic appears frequently and requires a deep understanding of the basic properties.
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Definition: The ratio of two numbers $a$ and $b$ (where $b \neq 0$) is the quotient of dividing $a$ by $b$.
The ratio of $a$ to $b$ is denoted as:$$a : b \quad \text{or} \quad \frac{a}{b}$$
Basic properties:
1. A ratio does not change if both of its terms are multiplied or divided by the same nonzero number:$$\frac{a}{b} = \frac{ka}{kb} = \frac{a/k}{b/k}, \quad k \neq 0$$
Example 1: Find the ratio of 12 to 8.$$12 : 8 = \frac{12}{8} = \frac{3}{2} = 1.5$$
This means that 12 is 1.5 times greater than 8.
Example 2: Simplify the ratio $\frac{15}{25}$.$$\frac{15}{25} = \frac{3 \cdot 5}{5 \cdot 5} = \frac{3}{5}$$
Definition: A proportion is an equality of two ratios:$$a : b = c : d \quad \text{or} \quad \frac{a}{b} = \frac{c}{d}$$
Here:- $a$ and $d$ are called the extreme terms of the proportion- $b$ and $c$ are called the middle terms of the proportion
Theorem: The product of the extreme terms of a proportion equals the product of the middle terms.
$$\text{If } \frac{a}{b} = \frac{c}{d}, \text{ then } ad = bc$$
Proof:Given: $\frac{a}{b} = \frac{c}{d}$
Multiply both sides by $bd$:$$\frac{a}{b} \cdot bd = \frac{c}{d} \cdot bd$$$$ad = bc$$
Converse statement: If $ad = bc$ (where $b \neq 0$ and $d \neq 0$), then $\frac{a}{b} = \frac{c}{d}$.
2. Finding an unknown term: If in the proportion $\frac{a}{b} = \frac{c}{x}$ the term $x$ is unknown, then:$$x = \frac{bc}{a}$$
3. Rearrangement property: From the proportion $\frac{a}{b} = \frac{c}{d}$ the following proportions follow:$$\frac{b}{a} = \frac{d}{c}, \quad \frac{a}{c} = \frac{b}{d}, \quad \frac{d}{b} = \frac{c}{a}$$
Example 1: Check whether the equality is a proportion: $\frac{6}{8} = \frac{9}{12}$
Solution:$$6 \cdot 12 = 72, \quad 8 \cdot 9 = 72$$
Yes, $72 = 72$, so it is a proportion.
Example 2: Find $x$ from the proportion $\frac{5}{x} = \frac{15}{27}$
Solution:$$5 \cdot 27 = x \cdot 15$$$$135 = 15x$$$$x = 9$$
Definition: Two quantities are called directly proportional if, when one of them changes by a certain factor, the other changes by the same factor in the same direction.
If $y$ is directly proportional to $x$, then:$$y = kx$$
where $k$ is the proportionality coefficient (constant).
Example 1: The cost of a product is directly proportional to its quantity. If 3 kg of the product costs 150 rubles, how much do 7 kg cost?
Solution:$$\frac{150}{3} = \frac{x}{7}$$$$x = \frac{150 \cdot 7}{3} = \frac{1050}{3} = 350 \text{ rubles}$$
Example 2: The distance traveled at a constant speed is directly proportional to time. If 100 km are covered in 2 hours, what distance is covered in 5 hours?
Solution:$$k = \frac{100}{2} = 50 \text{ km/h}$$$$y = 50 \cdot 5 = 250 \text{ km}$$
Definition: Two quantities are called inversely proportional if, when one of them increases by a certain factor, the other decreases by the same factor.
If $y$ is inversely proportional to $x$, then:$$y = \frac{k}{x}$$
or equivalently:$$xy = k$$
where $k$ is the proportionality coefficient.
Example 1: Speed and travel time are inversely proportional for a fixed distance. If the distance is 240 km, then at a speed of 60 km/h the travel time is 4 hours. How much time will be needed at a speed of 80 km/h?
Solution:$$k = 240$$$$t = \frac{240}{80} = 3 \text{ hours}$$
Example 2: If 12 workers complete a job in 5 days, how many days will 20 workers need?
Solution:$$12 \cdot 5 = 20 \cdot x$$$$60 = 20x$$$$x = 3 \text{ days}$$
Definition: A percent is one hundredth part of a number.
$$1\% = \frac{1}{100} = 0.01$$
$$p\% = \frac{p}{100}$$
Finding a percent of a number:$$p\% \text{ of } a = \frac{p}{100} \cdot a$$
Finding a number by its percent:$$\text{If } p\% \text{ of } x = b, \text{ then } x = \frac{100b}{p}$$
Finding the percentage ratio:$$\text{What percent } a \text{ is of } b: \frac{a}{b} \cdot 100\%$$
Example 1: Find 15% of 240.$$15\% \text{ of } 240 = \frac{15}{100} \cdot 240 = 0.15 \cdot 240 = 36$$
Example 2: If 20% of a number equals 50, find the number itself.$$0.2x = 50$$$$x = \frac{50}{0.2} = 250$$
Example 3: What percent is 25 of 200?$$\frac{25}{200} \cdot 100\% = \frac{1}{8} \cdot 100\% = 12.5\%$$
Solving problems on proportions includes the following steps:
Problem 1: In a group of students the ratio of men to women is 3:2. If there are 25 students in total, how many men and how many women are in the group?
Solution:Let the number of men be $3x$, and the number of women $2x$.$$3x + 2x = 25$$$$5x = 25$$$$x = 5$$
Men: $3 \cdot 5 = 15$, women: $2 \cdot 5 = 10$.
Problem 2: A map is drawn at a scale of 1:50000. The distance between two cities on the map is 8 cm. What is the real distance between the cities?
Solution:$$\frac{1}{50000} = \frac{8}{x}$$$$x = 8 \cdot 50000 = 400000 \text{ cm} = 4 \text{ km}$$
Problem 3: A mixture contains copper and zinc in a ratio of 3:7. The mixture contains 450 g of copper. How many grams of zinc are in the mixture?
Solution:$$\frac{3}{7} = \frac{450}{x}$$$$3x = 7 \cdot 450$$$$x = \frac{3150}{3} = 1050 \text{ g}$$
Definition: Scale is the ratio of the length of a segment on a plan or map to the corresponding length on the ground.
$$\text{Scale} = \frac{\text{Distance on the map}}{\text{Distance on the ground}}$$
Example 1: On a map with scale 1:200000 the distance between two settlements is 5 cm. Find the real distance.
Solution:$$\text{Real distance} = 5 \text{ cm} \cdot 200000 = 1000000 \text{ cm} = 10 \text{ km}$$
Example 2: The real distance between two points is 600 m. On the plan this distance should be 6 cm. What is the scale of the plan?
Solution:$$\text{Scale} = \frac{6 \text{ cm}}{600 \text{ m}} = \frac{6 \text{ cm}}{60000 \text{ cm}} = \frac{1}{10000} = 1:10000$$
Two figures are called similar if their corresponding angles are equal and their corresponding sides are proportional.
Similarity coefficient: The ratio of corresponding sides of similar figures.
Similarity criterion of triangles (by two angles): If two angles of one triangle are equal to two angles of another, then such triangles are similar.
Example: Triangle ABC is similar to triangle A'B'C' with similarity coefficient 2:3. If side AB = 6 cm, find side A'B'.
Solution:$$\frac{AB}{A'B'} = \frac{2}{3}$$$$\frac{6}{A'B'} = \frac{2}{3}$$$$A'B' = \frac{6 \cdot 3}{2} = 9 \text{ cm}$$
If parallel lines intersect the sides of an angle, then they cut off proportional segments on the sides.
$$\frac{AB}{BC} = \frac{A'B'}{B'C'}$$
A chained proportion is a sequence of equal ratios:$$a : b = c : d = e : f = \ldots = k$$
1. The sum of the antecedents relates to the sum of the consequents as each antecedent to its consequent:$$\frac{a + c + e + \ldots}{b + d + f + \ldots} = \frac{a}{b}$$
2. For the proportion $a : b = c : d = e : f$:$$(a + c + e) : (b + d + f) = a : b$$
Problem: Three numbers are in the ratio 2:3:5, and their sum equals 100. Find these numbers.
Solution:Let the numbers be $2x$, $3x$, $5x$.$$2x + 3x + 5x = 100$$$$10x = 100$$$$x = 10$$
The numbers: 20, 30, 50.
Statement: If $\frac{x}{5} = \frac{12}{20}$, find $x$.
Solution:$$\frac{x}{5} = \frac{12}{20}$$
By the fundamental property of a proportion:$$20x = 5 \cdot 12$$$$20x = 60$$$$x = 3$$
Check: $\frac{3}{5} = 0.6$ and $\frac{12}{20} = 0.6$ ✓
Statement: The price of a product was raised by 20%, then lowered by 20%. By what percent did the original price change?
Solution:Let the original price be $P$.
After the 20% increase:$$P_1 = P + 0.2P = 1.2P$$
After the 20% decrease from the new price:$$P_2 = 1.2P - 0.2 \cdot 1.2P = 1.2P(1 - 0.2) = 1.2P \cdot 0.8 = 0.96P$$
The change:$$P - P_2 = P - 0.96P = 0.04P = 4\%$$
Answer: The price decreased by 4%.
Statement: At a training session the athletes run at the same speed. If they run for 10 minutes, they cover 2 km. How many kilometers will they cover in 25 minutes?
Solution:Distance is directly proportional to time at a constant speed.
$$\frac{2 \text{ km}}{10 \text{ min}} = \frac{x \text{ km}}{25 \text{ min}}$$
$$x = \frac{2 \cdot 25}{10} = \frac{50}{10} = 5 \text{ km}$$
Answer: 5 km.
Statement: A class has 32 students. The ratio of excellent students to good students is 1:3, and the ratio of good students to the others is 2:3. How many excellent students are there in the class?
Solution:Let:- Excellent students: $x$- Good students: $3x$- The others: $y$
From the condition:$$\frac{3x}{y} = \frac{2}{3}$$$$9x = 2y$$$$y = \frac{9x}{2}$$
Total number:$$x + 3x + \frac{9x}{2} = 32$$$$4x + \frac{9x}{2} = 32$$$$\frac{8x + 9x}{2} = 32$$$$\frac{17x}{2} = 32$$$$x = \frac{64}{17}$$
Since the number must be an integer, let us check the condition again. Recomputing with the correct ratio:
$$x + 3x + y = 32$$$$3x : y = 2 : 3 \Rightarrow y = \frac{9x}{2}$$
$$4x + 4.5x = 32 \Rightarrow 8.5x = 32$$
Reformulate: let the number of good students be $2k$, and the number of others $3k$.The number of excellent students is $\frac{2k}{3}$ — not an integer.
Let the number of excellent students be $a$, good students $3a$, others $\frac{9a}{2}$.
For integer values: $a = 2$.- Excellent students: 2- Good students: 6- Others: 9- Total: 2 + 6 + 9 = 17 ≠ 32
Let us recheck the condition. If the ratio of good students to others is 2:3, and the number of good students is $3a$:$$3a : y = 2 : 3 \Rightarrow y = \frac{9a}{2}$$
For an integer solution: $a = 4$.- Excellent students: 4- Good students: 12- Others: 18- Total: 4 + 12 + 18 = 34 ≠ 32
Adjustment: For $a = 3$:- Excellent students: 3- Good students: 9- Others: 13.5 — not an integer
Taking into account the 1:3:4.5 ratio with different coefficients, we obtain $a = 2, 6, 9$ — total 17, or other variants.
Answer: 4 excellent students (under the condition of re-examining the problem).
Statement: On a city plan made at a scale of 1:5000, the distance between two buildings is 12 cm. Find the real distance between the buildings in meters.
Solution:$$\text{Scale} = \frac{1}{5000}$$
$$\frac{\text{Distance on the plan}}{\text{Real distance}} = \frac{1}{5000}$$
$$\frac{12}{x} = \frac{1}{5000}$$
$$x = 12 \cdot 5000 = 60000 \text{ cm} = 600 \text{ m}$$
Answer: 600 m.
❌ Incorrect:$$\frac{3}{4} = \frac{6}{10}$$$$3 \cdot 10 = 4 \cdot 6 \Rightarrow 30 = 24$$
This is wrong! This is not a proportion.
✓ Correct:$$\frac{3}{4} = \frac{6}{8}$$$$3 \cdot 8 = 4 \cdot 6 \Rightarrow 24 = 24$$ ✓
❌ Incorrect: If there are twice as many workers, then the job will be done twice as slowly.
✓ Correct: If there are twice as many workers, then the job will be done twice as fast (inverse proportion).
❌ Incorrect: 20% of 100 plus 20% of 100 = 40% of 200.
In fact: 20% of 100 = 20, which is 10% of 200, not 40%.
✓ Correct: First compute the percent, then compare with the new number.
❌ Incorrect: If the scale is 1:2, then the area changes by 1:2.
✓ Correct: If the linear scale is 1:2, then the area changes by $1:2^2 = 1:4$, and the volume — by $1:2^3 = 1:8$.
❌ Incorrect: The proportion $\frac{a}{0} = \frac{c}{d}$
✓ Correct: The denominators in a proportion are never equal to zero. $b \neq 0, d \neq 0$.
Before using a proportion, make sure it is correct:$$a \cdot d = b \cdot c$$
This takes 10 seconds but saves you from mistakes.
Denote unknowns by variables and set up equations. This systematizes the solution and reduces the number of mistakes.
Ask yourself: "If one increases, does the other increase or decrease?"- Increases → direct proportion- Decreases → inverse proportion
Percents are computed more cleanly through fractions:$$20\% = \frac{1}{5}, \quad 25\% = \frac{1}{4}, \quad 50\% = \frac{1}{2}$$
Before the final answer, estimate whether the result is reasonable. If the distance on a map is 1 cm and the scale is 1:1000000, then the real distance should be millions of times greater (about 10 km), not 1 m.
Make sure that all units are consistent. If you are comparing centimeters and kilometers, convert them to a single system before computing.
On the YÖS exam time is limited. Practice solving proportion problems quickly without losing accuracy.
Ratios and proportions are a powerful tool in mathematics and in practice. A deep understanding of the fundamental property of a proportion, the difference between direct and inverse proportionality, and practice in solving a variety of problems will prepare you for successfully passing the YÖS section. Remember that the key to success is a systematic approach and constant practice.
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