10.3 Percentage, Profit-Loss (Kâr-Zarar), Mixture (Karışım) & Motion (Hareket) Problems

10.3 Percentage, Profit-Loss (Kâr-Zarar), Mixture (Karışım) & Motion (Hareket) Problems

The final pillar of TR-YÖS word problems unites Percentage and Profit-Loss (Yüzde ve Kâr-Zarar), Mixtures (Karışım), and Motion (Hareket / Hız). Each domain is governed by a conservation law: conservation of cost, conservation of pure solute mass, and conservation of relative displacement.

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1. Percentage & Profit-Loss Problems (Yüzde ve Kâr-Zarar Problemleri)

Assigning Cost Price $= 100x$ & Multiplicative Scaling

Whenever a problem applies percentages without stating an initial price, set the Cost Price (Maliyet / Alış Fiyatı) to $C = 100x$:

• Selling Price (Satış Fiyatı) with $p\%$ profit: $SP = C \cdot \left(1 + \frac{p}{100}\right) = 100x + px$.
• Selling Price with $d\%$ loss relative to cost (zarar): $SP = C \cdot \left(1 - \frac{d}{100}\right) = 100x - dx$.
• Profit or Loss Percentage (Kâr / Zarar Yüzdesi): Unless explicitly stated otherwise, profit and loss percentages are always calculated relative to the Cost Price (Maliyet):

$$\text{Profit \%} = \frac{\text{Selling Price} - \text{Cost Price}}{\text{Cost Price}} \times 100\%$$

Successive Percentage Trap: Never add or subtract successive percentages! A $30\%$ markup above cost ($100x \to 130x$) followed by a $20\%$ discount off the label price (etiket fiyatı) multiplies by $0.80$: $130x \times 0.80 = 104x$, resulting in a $4\%$ net profit ($1.30 \times 0.80 = 1.04$), never $10\%$!

Batch-Price Comparison (Adet-Fiyat Eşitleme)

When a merchant buys $a$ items for $b$ TL and sells $c$ items for $d$ TL, equalize the number of items to $N = \text{EKOK}(a, c)$ before computing profit percentage:

• For example, buying $4$ lemons for $10$ TL and selling $5$ lemons for $16$ TL: scale to $\text{EKOK}(4, 5) = 20$ lemons.
• Cost of $20$ lemons $= 5 \times 10 = 50$ TL; revenue from $20$ lemons $= 4 \times 16 = 64$ TL.
• Profit on $50$ TL cost is $14$ TL $\implies \frac{14}{50} \times 100\% = 28\%$ profit.

Spoilage, Shrinkage & Inflation (Fire, Çürüme ve Enflasyon)

1. Spoilage / Shrinkage (Fire / Çürüme): When a merchant buys quantity $Q_{\text{initial}}$ at unit price $P_{\text{initial}}$ and loses a fraction to evaporation, breakage, or rot, the Total Cost (Toplam Maliyet) remains unchanged while the sellable quantity drops to $Q_{\text{remaining}}$:

$$Q_{\text{initial}} \cdot P_{\text{initial}} = Q_{\text{remaining}} \cdot P_{\text{new}}$$

For example, if wet soap loses $20\%$ of its weight upon drying ($Q_{\text{remaining}} = 0.80 Q_{\text{initial}}$), the unit break-even cost becomes $P_{\text{new}} = \frac{P_{\text{initial}}}{0.80} = 1.25 P_{\text{initial}}$ (a $25\%$ increase in unit cost). 2. Inflation & Real Purchasing Power (Enflasyon ve Alım Gücü): If annual inflation is $r_{\text{inflation}}$ and a worker's nominal salary increases by $r_{\text{salary}}$, the real purchasing power multiplier is:

$$1 + r_{\text{real}} = \frac{1 + r_{\text{salary}}}{1 + r_{\text{inflation}}}$$

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2. Mixture Problems (Karışım Problemleri)

The Universal Bowl Balance Equation (Kutu / Kap Yöntemi)

Every homogeneous mixture problem is solved by equating the total mass of pure solute before and after mixing. If a mixture of mass $m_1$ with solute percentage $c_1\%$ is combined with a second mixture of mass $m_2$ with solute percentage $c_2\%$:

$$m_1 \cdot c_1 + m_2 \cdot c_2 = (m_1 + m_2) \cdot c_{\text{final}}$$

Crucial Boundary Concentrations

When applying the Bowl Balance Equation, memorize four boundary rules:

1. Adding pure solute (saf tuz / şeker / alkol ekleme): Treat the added mass as a mixture with concentration $c = 100\%$.
2. Adding pure water (saf su ekleme): Treat the added water as a mixture with concentration $c = 0\%$.
3. Evaporating water (su buharlaştırma): Subtract the evaporated water mass from total mass with concentration $c = 0\%$:

$$m_1 \cdot c_1 - m_{\text{evap}} \cdot 0 = (m_1 - m_{\text{evap}}) \cdot c_{\text{final}} \implies m_1 \cdot c_1 = (m_1 - m_{\text{evap}}) \cdot c_{\text{final}}$$

1. Pouring out part of a homogeneous mixture (karışımdan bir miktar dökme): If $\frac{1}{k}$ of a $30\%$ salt solution is poured out or spilled, the remaining $\left(1 - \frac{1}{k}\right)m_1$ of liquid still has a $30\%$ salt concentration! Only its mass decreases.

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3. Motion / Rate-Time-Distance Problems (Hareket / Hız Problemleri)

Fundamental Law, Relative Speed (Bağıl Hız) & Circular Tracks (Dairesel Pist)

The core equation is $\text{Distance} = \text{Speed} \times \text{Time}$ ($x = v \cdot t$). When two vehicles with speeds $v_1$ and $v_2$ move simultaneously:

• Opposite Directions (Zıt Yönlü — approaching or separating): Their relative speed is the sum of their speeds:

$$\Delta x = (v_1 + v_2) \cdot t$$

• Same Direction (Aynı Yönlü — faster vehicle chasing slower vehicle): Their relative speed is the difference of their speeds:

$$\Delta x = \lvert v_1 - v_2\rvert \cdot t$$

• Circular Track (Dairesel Pist) of Circumference $C$: If two runners start from the same point on a circular track of length $C$, they first meet after $t = \frac{C}{v_1 + v_2}$ when running in opposite directions, and the faster runner first laps (tur bindirme) the slower runner after $t = \frac{C}{\lvert v_1 - v_2\rvert}$ when running in the same direction.

Average Speed (Ortalama Hız)

Average speed over a multi-stage journey is always total distance divided by total time:

$$V_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{x_1 + x_2 + \dots}{t_1 + t_2 + \dots}$$

For a round trip (or two equal-distance halves of a journey) at speeds $v_1$ and $v_2$, $V_{\text{avg}}$ is the Harmonic Mean (Harmonik Ortalama) of $v_1$ and $v_2$:

$$V_{\text{avg}} = \frac{2v_1 v_2}{v_1 + v_2} \quad \left(\text{NEVER } \frac{v_1 + v_2}{2}!\right)$$

Trains & Tunnels (Tren-Tünel) and Boats & Rivers (Akıntı Hızı)

1. Trains & Tunnels: For a train of length $L_{\text{train}}$ to completely pass through a tunnel (or bridge) of length $L_{\text{tunnel}}$, the front of the train must travel the tunnel's length plus the train's own length:

$$L_{\text{train}} + L_{\text{tunnel}} = v_{\text{train}} \cdot t$$

(Unit conversion reminder: $1\text{ m/s} = 3.6\text{ km/h}$, so multiply $\text{km/h}$ by $\frac{5}{18}$ to get $\text{m/s}$.) 2. Boats & Rivers (Nehir / Akıntı Problemleri): For a boat with still-water speed $v_b$ in a river with current speed $v_c$ ($v_b > v_c$):

• Downstream (Akıntıyla aynı yön): $x_{\text{down}} = (v_b + v_c) \cdot t_{\text{down}}$
• Upstream (Akıntıya karşı): $x_{\text{up}} = (v_b - v_c) \cdot t_{\text{up}}$

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4. Reference Table of Profit, Mixture & Motion Equations

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5. Worked Examples

Worked Example 10.5: Multi-Step Mixture with Spillage & Evaporation

Problem: A tank contains $500\text{ g}$ of a $30\%$ salt solution. First, $\frac{1}{5}$ of the solution is poured out. Next, $150\text{ g}$ of a $40\%$ salt solution is added, and $50\text{ g}$ of water is evaporated by heating. What is the final salt percentage of the mixture?

Solution:

1. Pouring out $\frac{1}{5}$ of the $500\text{ g}$ solution leaves $500 - 100 = 400\text{ g}$ of solution, still at $30\%$ concentration.
2. Apply the Bowl Balance Equation with $m_1 = 400\text{ g}$ ($c_1 = 30\%$), $m_2 = 150\text{ g}$ ($c_2 = 40\%$), and evaporated water $m_3 = -50\text{ g}$ ($c_3 = 0\%$):

$$400(30) + 150(40) - 50(0) = (400 + 150 - 50) \cdot c_{\text{final}}$$

1. Compute both sides:

$$12{,}000 + 6{,}000 = 500 \cdot c_{\text{final}} \implies 18{,}000 = 500 \cdot c_{\text{final}} \implies c_{\text{final}} = 36\%$$

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Worked Example 10.6: Circular Track & River Round-Trip Motion

Problem: A motorboat with a still-water speed of $v_b = 20\text{ km/h}$ travels $48\text{ km}$ downstream and immediately returns $48\text{ km}$ upstream to its starting dock in a total of $5$ hours. What is the speed of the river current $v_c$?

Solution:

1. Downstream speed is $20 + v_c$ and upstream speed is $20 - v_c$.
2. Set up the total time equation $t_{\text{down}} + t_{\text{up}} = 5$:

$$\frac{48}{20 + v_c} + \frac{48}{20 - v_c} = 5$$

1. Combine the fractions over the common denominator $(20 + v_c)(20 - v_c) = 400 - v_c^2$:

$$\frac{48(20 - v_c) + 48(20 + v_c)}{400 - v_c^2} = 5 \implies \frac{48 \cdot 40}{400 - v_c^2} = 5$$

1. Divide both sides by $5$:

$$400 - v_c^2 = 48 \cdot 8 = 384 \implies v_c^2 = 16 \implies v_c = 4\text{ km/h}$$

A discount is applied to the stated label or current price, which may differ from cost. The mixture equations assume homogeneous mixtures, a consistent concentration basis, and only the named component lost in evaporation. Motion formulas assume the stated constant rates and units. The harmonic and arithmetic means agree when the two speeds are equal.

Review the Method

Average speed is Total Distance / Total Time; for an equal-distance round trip at speeds v_1 and v_2, average speed equals the harmonic mean 2v_1v_2 / (v_1 + v_2).