7.1 Bar, Line and Pie Charts & Multi-Table Data Interpretation (Tablo ve Grafikler)

7.1 Bar, Line, Pie Charts ($360^\circ$) & Multi-Table Data Interpretation (Tablo ve Grafikler)

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1. Pie Charts (Daire Grafiği — $360^\circ$ Proportional Scaling)

A pie chart partitions a complete circle of $360^\circ$ in direct proportion to each category's share of the total quantity $T$. For any sector with central angle $\theta$, percentage share $P\%$, and raw quantity $Q$:

$$\frac{\theta}{360^\circ} = \frac{P}{100\%} = \frac{Q}{T}$$

From this fundamental ratio, memorize two instant conversion benchmarks:

• $10\% \iff 36^\circ$ and $1\% \iff 3.6^\circ$ (so $5\% \iff 18^\circ$, $15\% \iff 54^\circ$, $20\% \iff 72^\circ$, $25\% \iff 90^\circ$, $30\% \iff 108^\circ$, $35\% \iff 126^\circ$, $40\% \iff 144^\circ$, and $45\% \iff 162^\circ$).
• Whenever all but one sector angle are given, first compute the missing angle via $\theta_{\text{missing}} = 360^\circ - \sum \theta_i$ before setting up any equation.

The GCD Parametric Unit Trick (EBOB ile $k$ Değişkeni Atama)

Writing $\frac{\theta}{360}$ for every sector introduces cumbersome fractions. Instead, take the greatest common divisor $\gcd(\theta_1, \theta_2, \dots, \theta_n)$ of all central angles and assign $k$-units:

$$\text{Unit Scale } d = \gcd(\theta_1, \theta_2, \dots, \theta_n), \qquad Q_i = \left(\frac{\theta_i}{d}\right)k, \qquad T = \left(\frac{360^\circ}{d}\right)k$$

For instance, if four sectors have angles $60^\circ, 90^\circ, 120^\circ,$ and $90^\circ$, their greatest common divisor is $d = 30^\circ$. Dividing each angle by $30^\circ$ immediately assigns sector weights $2k, 3k, 4k,$ and $3k$, with a grand total of $12k$. Any difference, ratio, or percentage condition can now be solved using single-digit integers.

Two-Pie-Chart Chaining (İki Daire Grafiği Arasında Geçiş)

Many TR-YÖS problems display two linked pie charts:

1. Chart 1 (Macro Breakdown): Splits the grand total across main groups $A, B, C, D$ using parameter $k$.
2. Chart 2 (Micro Breakdown): Zooms into a single group—say, Group $A$—and splits only Group $A$'s total across sub-items $A_1, A_2, A_3$ over a full $360^\circ$ circle using a second parameter $m$.

To chain the two charts without fractions, equate Group $A$'s weight in Chart 1 (say $4k$) to the total circle weight in Chart 2 (say $20m$), or scale both parameters to $\operatorname{lcm}(4, 20) = 20$ units so that every sector in both charts is expressed in a single unified variable.

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2. Bar Charts (Sütun Grafiği) & Line Graphs (Çizgi Grafiği)

Percentage Change (Yüzde Değişim)

When reading year-over-year bar charts or time-series line graphs, percentage change is always divided by the original (chronologically earlier or reference) value:

$$\%\Delta = \frac{\text{New Value} - \text{Old Value}}{\text{Old Value}} \times 100\%$$

Linear Growth & Depletion Graphs (Doğrusal Grafikler)

$$y(t) = y_0 + m \cdot t, \qquad m = \frac{y(t_2) - y(t_1)}{t_2 - t_1}$$

• Growth line ($m > 0$): If a plant starts at height $y_0 = 12\text{ cm}$ and reaches $30\text{ cm}$ in $6\text{ months}$, its growth rate is $m = \frac{30 - 12}{6} = +3\text{ cm/month}$.
• Depletion line ($m < 0$): If a candle of initial length $L_0$ burns completely to $0$ in $T_{\text{burn}}$ hours, its burn rate is $m = -\frac{L_0}{T_{\text{burn}}}$, and its remaining height at time $t$ is $L(t) = L_0\left(1 - \frac{t}{T_{\text{burn}}}\right)$.
• Intersection Time ($t^*$): Two linear processes $y_A(t) = y_{A,0} + m_A t$ and $y_B(t) = y_{B,0} + m_B t$ reach the same value at:

$$t^* = \frac{y_{A,0} - y_{B,0}}{m_B - m_A} = \frac{\text{Initial Gap}}{\text{Relative Rate of Convergence}}$$

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3. Histograms & Two-Way Cross-Tabulation Tables (Çapraz Tablolar)

When data are grouped into frequency bars or a two-way matrix with row categories $R_i$ and column categories $C_j$:

1. Weighted Mean (Ağırlıklı Ortalama): If score $x_i$ occurs with frequency (student count or weight) $f_i$, the overall average is:

$$\bar{x} = \frac{\sum_{i=1}^n f_i x_i}{\sum_{i=1}^n f_i}$$

Never take the simple arithmetic mean of the $x_i$ labels when bar heights differ. When all frequencies $f_i$ share a common factor (for example, $f = 15, 25, 20$), divide the frequencies by $\gcd(15, 25, 20) = 5$ to use reduced weights $3, 5, 4$ in the weighted mean formula—the common factor cancels out between numerator and denominator! 2. Marginal Totals in Cross-Tabulation Matrices: In an $R \times C$ table, the sum of all row totals equals the sum of all column totals, which equals the grand total $N$. Reconstruct missing interior cells by scanning for any row or column that has exactly one unknown cell, filling it by subtraction from its marginal total, and propagating that value into the perpendicular row or column.

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Worked Examples

Worked Example 7.1.1: Two-Pie-Chart Chaining with GCD $k$-Units

Problem: Chart 1 displays the distribution of international students at a university across four faculties by central angle: Engineering ($120^\circ$), Medicine ($90^\circ$), Architecture ($60^\circ$), and Economics ($90^\circ$). Chart 2 breaks down only the Engineering students into four class years by central angle: Freshmen ($108^\circ$), Sophomores ($90^\circ$), Juniors ($90^\circ$), and Seniors ($72^\circ$). If there are $36$ more Engineering Freshmen than Engineering Seniors, how many Medicine students attend the university?

Solution:

1. Convert Chart 2 (Engineering sub-chart) into $m$-units: The central angles in Chart 2 are $108^\circ, 90^\circ, 90^\circ,$ and $72^\circ$. Since $\gcd(108, 90, 90, 72) = 18^\circ$, divide each angle by $18^\circ$:

$$\text{Freshmen} = 6m, \quad \text{Sophomores} = 5m, \quad \text{Juniors} = 5m, \quad \text{Seniors} = 4m$$

The total number of Engineering students is $6m + 5m + 5m + 4m = 20m$. 2. Use the difference condition to find $m$ and Total Engineering:

$$6m - 4m = 2m = 36 \implies m = 18$$

Therefore, the total number of Engineering students is $20m = 20 \times 18 = 360$. 3. Convert Chart 1 (University-wide chart) into $k$-units: In Chart 1, $\gcd(120, 90, 60, 90) = 30^\circ$, giving:

$$\text{Engineering} = 4k, \quad \text{Medicine} = 3k, \quad \text{Architecture} = 2k, \quad \text{Economics} = 3k$$

1. Equate the bridge category (Engineering) and solve for Medicine:

$$4k = 360 \implies k = 90 \implies \text{Medicine} = 3k = 3 \times 90 = 270\text{ students.}$$

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Worked Example 7.1.2: Linear Depletion Graph (Two Burning Candles)

Problem: A linear time-height graph shows two homogeneous candles lit at the exact same instant ($t = 0$). Candle A has an initial height of $48\text{ cm}$ and burns completely to $0\text{ cm}$ in $6\text{ hours}$. Candle B has an initial height of $36\text{ cm}$ and burns completely to $0\text{ cm}$ in $9\text{ hours}$. After how many hours $t^*$ is the remaining height of Candle B exactly twice the remaining height of Candle A?

Solution:

1. Determine each candle's hourly burn rate (slope $m$):

$$m_A = -\frac{48\text{ cm}}{6\text{ hr}} = -8\text{ cm/hr} \implies h_A(t) = 48 - 8t$$ $$m_B = -\frac{36\text{ cm}}{9\text{ hr}} = -4\text{ cm/hr} \implies h_B(t) = 36 - 4t$$

1. Apply the target ratio condition $h_B(t^*) = 2 \cdot h_A(t^*)$:

$$36 - 4t^* = 2(48 - 8t^*) = 96 - 16t^*$$

1. Solve for $t^*$:

$$12t^* = 60 \implies t^* = 5\text{ hours}$$

Verification: At $t = 5\text{ hours}$, $h_A(5) = 48 - 40 = 8\text{ cm}$ and $h_B(5) = 36 - 20 = 16\text{ cm}$, confirming $16 = 2 \times 8$.

For an intersection, require unequal slopes and check that the calculated time lies in the model’s time interval. A negative result describes an intersection before the start, not a future crossing. For grouped histogram intervals, using bin midpoints gives an estimated mean; exact weighted means require exact listed values.

Review the Method

Two-way cross-tabulation tables and histograms require tracking marginal row/column totals alongside the weighted mean formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.