10.2 Number, Fraction, Age (Yaş) & Work/Pipe (İşçi-Havuz) Word Problems

10.2 Number, Fraction, Age (Yaş) & Work/Pipe (İşçi-Havuz) Word Problems

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1. Number & Fraction Problems (Sayı ve Kesir Problemleri)

Order of Operations in Verbal Translations

Pay close attention to phrasing order when translating into algebra:

• "$3$ times a number, minus $4$" (Bir sayının 3 katının 4 eksiği): multiply first, then subtract $\implies 3x - 4$.
• "$3$ times the result of a number minus $4$" (Bir sayının 4 eksiğinin 3 katı): subtract first inside parentheses, then multiply $\implies 3(x - 4)$.
• "$\frac{2}{5}$ of $3$ more than half of a number" (Bir sayının yarısının 3 fazlasının 2/5'i): work from the inside out $\implies \frac{2}{5}\left(\frac{x}{2} + 3\right)$.
• Splitting a total $S$ into two parts: If two numbers sum to $S$, never introduce two variables $x$ and $y$! Call the smaller (or requested) number $x$ and the other number $S - x$. Similarly, if two numbers have ratio $\frac{3}{7}$, immediately represent them as $3k$ and $7k$.

The Denominator-Product Variable Choice (Paydaların EKOK'u Kadar $x$ Seçimi)

When a problem repeatedly takes fractions of a quantity—especially using the word "of the remainder" (kalanın)—never name the initial quantity $x$. Instead, set the initial quantity equal to the product of all successive denominators times $x$.

Suppose a problem takes $\frac{1}{3}$ of a quantity and then $\frac{2}{5}$ of the remainder (kalanın):

• Choose initial quantity $= 3 \times 5 \times x = 15x$.
• Step 1: $\frac{1}{3}$ of $15x$ is $5x$, leaving a remainder of $15x - 5x = 10x$.
• Step 2: $\frac{2}{5}$ of the remainder ($10x$) is $\frac{2}{5}(10x) = 4x$, leaving $10x - 4x = 6x$.

Every intermediate quantity is an integer multiple of $x$, eliminating fractional arithmetic errors.

Classic Structural Archetypes

1. Wire / Rod Midpoint Shift (Telin Orta Noktasının Kayması):
   • Cutting length $L$ from one end of a rod of length $x$ shortens the rod to $x - L$ and shifts its midpoint by exactly $\frac{L}{2}$ in the opposite direction of the cut.
   • If $L_{\text{left}}$ is cut from the left end and $L_{\text{right}}$ is cut from the right end, the midpoint shifts by:

$$\Delta\text{Midpoint} = \frac{\lvert L_{\text{left}} - L_{\text{right}}\rvert}{2} \quad (\text{away from the end where more was cut})$$

1. Bench / Hotel Room Sharing (Sıra ve Oda Problemleri): Let $r$ be the number of benches (sıra).
   • "If students sit $2$ per bench, $4$ students remain standing": $N_{\text{students}} = 2r + 4$.
   • "If students sit $3$ per bench, $2$ benches remain completely empty": $r - 2$ benches are used, so $N_{\text{students}} = 3(r - 2)$.
   • Equate the invariant student count: $3(r - 2) = 2r + 4 \implies r = 10,\; N_{\text{students}} = 24$.
2. Staircase Step Problems (Merdiven Basamakları): If a staircase has $s$ steps (basamak) and a person climbs $a$ steps at a time Going Up ($\frac{s}{a}$ strides) and $b$ steps at a time Going Down ($\frac{s}{b}$ strides), taking $k$ fewer strides going down than going up:

$$\frac{s}{a} - \frac{s}{b} = k$$

1. Forward-Backward Step Problems (İleri-Geri Adım Problemleri): A walker who takes $a$ steps forward and $b$ steps backward ($a > b$) completes a cycle of $a + b$ total steps to advance a net displacement of $a - b$ steps. Always divide total steps by $a + b$ using Euclidean division (bölme işlemi), and count any leftover remainder steps strictly as forward steps first.

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2. Age Problems (Yaş Problemleri)

The Three Golden Laws of Age Problems

1. The Age-Difference Invariant (Yaş Farkı Sabiti): The age difference between any two people never changes over time. If Person A is $A$ years old and Person B is $B$ years old today, then $t$ years ago or $t$ years in the future:

$$(A + t) - (B + t) = A - B = \text{constant}$$

1. Birth-Year Inverse Relation (Doğum Yılı İlişkisi): An older person was born in an earlier date; calendar-year differences match the stated age difference only when birthday timing is accounted for: if Person A is $k$ years older than Person B ($\text{Age}_A - \text{Age}_B = k$), then $\text{BirthYear}_B - \text{BirthYear}_A = k$.
2. Group Age Sum & Average Scaling: When a group of $n$ people ages by $t$ years:
   • The sum of their ages increases by $n \cdot t$ (and $t$ years ago, it was $n \cdot t$ less).
   • The arithmetic mean of their ages increases by exactly $+t$ (and $t$ years ago, it was $t$ less).

The Timeline Table Method (Ben Senin Yaşındayken / Sen Benim Yaşıma Geldiğinde)

For problems stating "When I was your age..." or "When you reach my age...", build a 2-column table and equate the age differences between columns:

Since the age difference is invariant across all three rows:

$$x - y = y - p = f - x \implies p = 2y - x \quad \text{and} \quad f = 2x - y$$

Notice that in the timeline $(p,\; y,\; x,\; f)$, the four ages form a 4-term arithmetic progression with common difference $d = x - y$!

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3. Work & Pool/Pipe Problems (İşçi ve Havuz Problemleri)

Unit-Time Fractional Rate Equation (Birim Zamanda Yapılan İş)

Never add completion times directly! Convert every worker or pipe into the fraction of the job completed in $1$ unit of time:

• If Worker A completes a job alone in $a$ hours, Worker A's hourly rate is $\frac{1}{a}$.
• If Worker B completes the job alone in $b$ hours, Worker B's hourly rate is $\frac{1}{b}$.
• If a drain pipe C at the bottom of a pool empties a full pool in $c$ hours, its hourly rate is $-\frac{1}{c}$.

When all three operate simultaneously for $t$ hours to complete the whole job (or fill the pool, represented by $1$):

$$t \cdot \left(\frac{1}{a} + \frac{1}{b} - \frac{1}{c}\right) = 1$$

If Worker A works for $t_A$ hours and Worker B works for $t_B$ hours to complete a fraction $\frac{p}{q}$ of the job:

$$\frac{t_A}{a} + \frac{t_B}{b} = \frac{p}{q}$$

Capacity-Speed Inverse Law (Kapasite-Süre Ters Orantısı)

Completion time $t$ is inversely proportional to worker speed/efficiency $v$:

$$v_A = 3v_B \iff t_B = 3t_A \quad (\text{if Worker A is } 3\times \text{ as fast as B, A finishes in } t \text{ hours while B takes } 3t \text{ hours})$$

Elevated Drain Pipe Caveat: For a constant-cross-section pool, if a drain pipe is located not at the bottom, but at height $\frac{1}{3}$ of the pool's depth, the inlet pipes fill the bottom $\frac{1}{3}$ of the pool without the drain pipe acting; the drain pipe only subtracts rate while the upper $\frac{2}{3}$ of the pool is being filled!

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4. Reference Table of Classic Word-Problem Setups

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5. Worked Examples

Worked Example 10.3: Sequential Remainder Problem with Midpoint Shift

Problem: First $\frac{1}{3}$ of a wire is cut from its left end, and then $\frac{1}{4}$ of the remaining wire is cut from its right end. As a result, the midpoint of the wire shifts $6\text{ cm}$. What was the original length of the wire?

Solution:

1. Since the denominators are $3$ and $4$, let the original length of the wire be $12x$.
2. Left cut: $L_{\text{left}} = \frac{1}{3}(12x) = 4x$. The remaining wire has length $12x - 4x = 8x$.
3. Right cut: $L_{\text{right}} = \frac{1}{4}(8x) = 2x$.
4. Apply the midpoint shift formula:

$$\Delta\text{Midpoint} = \frac{\lvert L_{\text{left}} - L_{\text{right}}\rvert}{2} = \frac{4x - 2x}{2} = x$$

1. Since $\Delta\text{Midpoint} = 6\text{ cm}$, we have $x = 6\text{ cm}$, so the original length of the wire was $12x = 12(6) = 72\text{ cm}$.

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Worked Example 10.4: Staged Work & Capacity Problem

Problem: Master Kerem works $2$ times as fast as Apprentice Mert. Working together, they can finish a project in $8$ days. After they work together for $5$ days, Master Kerem leaves. How many more days will Apprentice Mert need to finish the remaining work alone?

Solution:

1. Since $v_{\text{Kerem}} = 2v_{\text{Mert}}$, their solo completion times are inversely proportional: if Kerem takes $t$ days alone, Mert takes $2t$ days alone.
2. Working together, they finish in $8$ days:

$$8 \cdot \left(\frac{1}{t} + \frac{1}{2t}\right) = 1 \implies 8 \cdot \frac{3}{2t} = 1 \implies t = 12\text{ days (Kerem)}, \quad 2t = 24\text{ days (Mert)}$$

1. In the first $5$ days together, they complete $\frac{5}{8}$ of the job, leaving $1 - \frac{5}{8} = \frac{3}{8}$ of the job unfinished.
2. Mert finishes the remaining $\frac{3}{8}$ of the job at his solo rate of $\frac{1}{24}$ per day:

$$d \cdot \frac{1}{24} = \frac{3}{8} \implies d = 24 \cdot \frac{3}{8} = 9\text{ days}$$

Review the Method

Worker completion time is inversely proportional to worker speed/capacity (v_A = k * v_B implies t_B = k * t_A).