10.1 Ratios, Direct/Inverse/Compound Proportions (Oran-Orantı) & Means

10.1 Ratios, Direct/Inverse/Compound Proportions (Oran-Orantı) & Means

On the TR-YÖS exam, Ratios and Proportions (Oran ve Orantı) bridge pure algebraic manipulation and every category of word problem. Whether you are solving a multi-variable fraction system, dividing a quantity among partners, scaling worker output, or computing average speed, a useful modeling route uses a constant of proportionality (orantı sabiti) $k$.

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1. Ratios & Equal-Proportion Properties (Oran ve Orantı Özellikleri)

A ratio (oran) is the quotient $\frac{a}{b}$ of two quantities (with $b \neq 0$). An equal-proportion chain (orantı) equates two or more ratios to a single constant $k$, called the constant of proportionality (orantı sabiti):

$$\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k$$

Parametric Substitution (Parametrik Yerine Koyma)

From $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k$, you can immediately express every numerator in terms of its denominator and $k$:

$$a = bk, \quad c = dk, \quad e = fk$$

Conversely, if a ratio gives concrete numerical denominators such as $\frac{x}{2} = \frac{y}{5} = \frac{z}{7} = k$, setting $x = 2k,\; y = 5k,\; z = 7k$ converts a three-variable system into a single-variable equation in $k$. When two two-variable ratios share a common variable, such as $\frac{x}{y} = \frac{2}{3}$ and $\frac{y}{z} = \frac{4}{5}$, scale both ratios so the shared variable $y$ has the same coefficient ($\text{EKOK}(3, 4) = 12$): $\frac{x}{y} = \frac{8}{12}$ and $\frac{y}{z} = \frac{12}{15}$, allowing the unified substitution $x = 8k,\; y = 12k,\; z = 15k$.

Linear Combination Invariance (Doğrusal Birleşim Özelliği)

For any non-zero scalars $m, n, p$ such that $mb + nd + pf \neq 0$, taking the same linear combination of the numerators and of the denominators preserves the exact same constant $k$:

$$\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k \implies \frac{ma + nc + pe}{mb + nd + pf} = k$$

Critical Exponent Warning: Linear combinations preserve $k^1$ because $k$ factors cleanly out of $m(bk) + n(dk) + p(fk) = k(mb + nd + pf)$. However, multiplying ratios or raising terms to powers changes the exponent of $k$:

• Product of ratios: $\frac{a}{b} \cdot \frac{c}{d} = k^2$ and $\frac{a \cdot c \cdot e}{b \cdot d \cdot f} = k^3$.
• Sum of squared terms: $\frac{a^2 + c^2 + e^2}{b^2 + d^2 + f^2} = \frac{b^2 k^2 + d^2 k^2 + f^2 k^2}{b^2 + d^2 + f^2} = k^2$ (and for any weights $m, n$, $\frac{ma^2 + nc^2}{mb^2 + nd^2} = k^2$).

Reciprocal Sum Pairing (Çarpmaya Göre Ters Toplam Eşleştirmesi)

A classic TR-YÖS question provides $\frac{a}{x} = \frac{b}{y} = \frac{c}{z} = k$ along with the sum $x + y + z$ and the reciprocal sum $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ (or vice versa). Never cross-multiply blindly! Solve for each variable using $k$:

$$\frac{a}{x} = k \implies x = \frac{a}{k} \implies \frac{1}{a} = \frac{1}{kx} \quad \text{and} \quad \frac{1}{x} = \frac{k}{a}$$

Summing across all three terms gives two symmetrical identities:

1. $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{k}\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)$
2. $a + b + c = k(x + y + z)$

Similarly, if a product relation $a \cdot x = b \cdot y = c \cdot z = k$ is given alongside $x + y + z$ and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$, substitute $x = \frac{k}{a},\; y = \frac{k}{b},\; z = \frac{k}{c}$ to obtain:

$$x + y + z = k\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$$

Fourth, Third, and Mean Proportionals (Dördüncü, Üçüncü ve Orta Orantılı)

• Fourth Proportional (Dördüncü Orantılı): The fourth proportional $x$ of three numbers $(a, b, c)$ in that exact order satisfies $\frac{a}{b} = \frac{c}{x} \implies x = \frac{bc}{a}$.
• Mean Proportional (Orta Orantılı): The positive mean proportional $x$ of two positive numbers $(a, b)$ satisfies $\frac{a}{x} = \frac{x}{b} \implies x = \sqrt{ab}$ (which is their geometric mean).
• Third Proportional (Üçüncü Orantılı): The third proportional $x$ of $(a, b)$ satisfies $\frac{a}{b} = \frac{b}{x} \implies x = \frac{b^2}{a}$.

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2. Direct, Inverse, and Compound Proportions (Doğru, Ters ve Bileşik Orantı)

Direct Proportion (Doğru Orantı)

Two quantities $x$ and $y$ are directly proportional (doğru orantılı) if they increase or decrease at the same multiplicative rate, meaning their quotient is constant:

$$(x, y, z) \text{ directly proportional with } (a, b, c) \iff \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k \implies x = ak,\; y = bk,\; z = ck$$

Inverse Proportion (Ters Orantı) & Intermeshed Gears (Dişli Çarklar)

Two quantities $x$ and $y$ are inversely proportional (ters orantılı) if as one increases, the other decreases by the same factor, meaning their product is constant:

$$(x, y, z) \text{ inversely proportional with } (a, b, c) \iff a \cdot x = b \cdot y = c \cdot z = k \implies x = \frac{k}{a},\; y = \frac{k}{b},\; z = \frac{k}{c}$$

The LCM Integer Trick (EKOK Yöntemi): To avoid fractions in inverse proportion problems where $a \cdot x = b \cdot y = c \cdot z$, set the common product equal to $\text{EKOK}(a, b, c) \cdot t$. For example, if $x, y, z$ are inversely proportional to $2, 3,$ and $4$, write $2x = 3y = 4z = 12t$, which immediately yields integer multiples $x = 6t,\; y = 4t,\; z = 3t$.

In intermeshed gears or pulleys (dişli çarklar ve kasnaklar), the number of revolutions (devir sayısı $n_i$) is inversely proportional to the number of teeth (diş sayısı $d_i$) or radius $r_i$: $n_1 \cdot r_1 = n_2 \cdot r_2 = n_3 \cdot r_3$.

Compound Work Proportion (Bileşik Orantı)

When a problem involves multiple groups of workers, machines, or pipes operating at different capacities, hours per day, and number of days to produce different amounts of finished work (yapılan iş), use the universal Compound Work Proportion Formula:

$$\frac{\text{Work}_1}{\text{All Other Factors}_1} = \frac{\text{Work}_2}{\text{All Other Factors}_2} \implies \frac{W_1}{N_1 \cdot t_1 \cdot h_1 \cdot e_1} = \frac{W_2}{N_2 \cdot t_2 \cdot h_2 \cdot e_2}$$

Here $W_i$ is the net work completed (e.g., meters of carpet woven, walls painted, parts manufactured), while $N_i$ (number of workers), $t_i$ (days), $h_i$ (hours per day), and $e_i$ (per-worker efficiency) all sit in the denominator.

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3. Arithmetic, Geometric, and Harmonic Means (Ortalamalar)

For $n$ positive real numbers $x_1, x_2, \dots, x_n$, TR-YÖS tests three classical means:

1. Arithmetic Mean (Aritmetik Ortalama — $AO$):

$$AO = \frac{x_1 + x_2 + \dots + x_n}{n}$$

1. Geometric Mean (Geometrik Ortalama — $GO$):

$$GO = \sqrt[n]{x_1 \cdot x_2 \cdots x_n}$$

1. Harmonic Mean (Harmonik Ortalama — $HO$):

$$HO = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n}} \quad \xrightarrow{\text{for } n = 2} \quad HO(a, b) = \frac{2ab}{a + b}$$

Fundamental Identities for Two Positive Numbers $a$ and $b$

When $n = 2$, $AO = \frac{a+b}{2}$, $GO = \sqrt{ab}$, and $HO = \frac{2ab}{a+b}$. Multiplying $AO$ and $HO$ reveals that the geometric mean is the geometric mean of the arithmetic and harmonic means:

$$AO \cdot HO = \left(\frac{a+b}{2}\right)\left(\frac{2ab}{a+b}\right) = a \cdot b = (GO)^2$$

Furthermore, for any positive real numbers $a$ and $b$, the Mean Inequality Chain holds:

$$HO \le GO \le AO \quad \left(\frac{2ab}{a+b} \le \sqrt{ab} \le \frac{a+b}{2}\right)$$

Equality ($HO = GO = AO$) occurs if and only if $a = b$.

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4. Reference Table of Proportion & Mean Identities

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5. Worked Examples

Worked Example 10.1: Linear Combination Invariance & Reciprocal Pairing

Problem: Given $\frac{a}{x} = \frac{b}{y} = \frac{c}{z} = \frac{2}{5}$, along with $3a - 2b + 4c = 24$ and $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{15}{8}$. Find:

1. The value of $3x - 2y + 4z$.
2. The value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$.

Solution:

1. By linear combination invariance with scalars $m = 3, n = -2, p = 4$:

$$\frac{3a - 2b + 4c}{3x - 2y + 4z} = \frac{2}{5}$$

Substitute $3a - 2b + 4c = 24$:

$$\frac{24}{3x - 2y + 4z} = \frac{2}{5} \implies 3x - 2y + 4z = 24 \cdot \frac{5}{2} = 60$$

1. From $\frac{a}{x} = \frac{2}{5}$, we have $a = \frac{2}{5}x \implies \frac{1}{a} = \frac{5}{2} \cdot \frac{1}{x}$. Applying the same relation to $b$ and $c$ and summing:

$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{5}{2}\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) = \frac{5}{2} \cdot \frac{15}{8} = \frac{75}{16}$$

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Worked Example 10.2: Mixed Direct & Inverse Proportion Division

Problem: A bonus pool of $1{,}200$ TL is distributed among three technicians ($x, y, z$) such that $x$ is directly proportional to $3$, $y$ is inversely proportional to $2$, and $z$ is inversely proportional to $4$. How many TL does technician $x$ receive?

Solution:

1. Write the mixed proportion equation equating all three expressions to the proportionality constant $k$:

$$\frac{x}{3} = 2y = 4z = k$$

1. To avoid fractions, replace $k$ with $4t$ (since $\text{EKOK}(1, 2, 4) = 4$):

$$\frac{x}{3} = 2y = 4z = 4t \implies x = 12t, \quad y = 2t, \quad z = t$$

1. Sum the shares and set equal to $1{,}200$ TL:

$$x + y + z = 12t + 2t + t = 15t = 1{,}200 \implies t = 80$$

1. Technician $x$'s share is $x = 12t = 12(80) = 960$ TL.

Review the Method

For any two positive numbers a and b, the arithmetic, geometric, and harmonic means satisfy HO <= GO <= AO and the identity AO · HO = (GO)^2 = a · b.