12.3 Quantitative Word Problems & Age/Coin/Venn Diagram Scenarios

Quantitative Word Problems & Age/Coin/Venn Diagram Scenarios

Applied Quantitative Word Problems test your capability to translate verbal, real-world descriptions into precise mathematical equations and solve them efficiently. On the NTS GAT General, word problems represent a substantial portion of the Quantitative section. Candidates face diverse scenarios covering work rates, kinematics (speed, distance, time), age timelines, financial mathematics (interest, profit/loss, discounts), coin denomination totals, and set theory Venn diagrams.

Developing systematic algebraic templates for each category eliminates setup errors and reduces solving time down to under 75 seconds per question.

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1. Work, Rate, and Efficiency Scenarios

Work-rate problems evaluate the time required for individuals, machines, or pipes operating at constant speeds to complete a specified task.

• Primary Work-Rate Equation: $\text{Work } (W) = \text{Rate } (R) \times \text{Time } (T) \implies R = \frac{W}{T}$
• If a complete job is defined as $1$ unit of work ($W = 1$), an entity taking $t$ hours completes a rate of $R = \frac{1}{t}$ jobs per hour.

A. Combined Work Shortcut

When two entities with completion times $t_1$ and $t_2$ work simultaneously to complete $1$ job: $\frac{1}{T_{\text{comb}}} = \frac{1}{t_1} + \frac{1}{t_2} = \frac{t_1 + t_2}{t_1 t_2} \implies T_{\text{comb}} = \frac{t_1 t_2}{t_1 + t_2}$

B. Pipes and Cisterns (Inlets and Outlets)

• Inlet Pipe (Fills Tank): Positive rate $+R_{\text{fill}} = +\frac{1}{t_{\text{fill}}}$
• Outlet Pipe / Leak (Empties Tank): Negative rate $-R_{\text{drain}} = -\frac{1}{t_{\text{drain}}}$
• Net Rate: $R_{\text{net}} = \frac{1}{t_{\text{fill}}} - \frac{1}{t_{\text{drain}}}$. Tank fills when $R_{\text{net}} > 0$.

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2. Kinematics: Distance, Speed, and Time

All motion scenarios stem from the primary equation: $\text{Distance } (d) = \text{Speed } (s) \times \text{Time } (t)$

A. Average Speed Framework

$\text{Average Speed} = \frac{\text{Total Distance Covered}}{\text{Total Time Taken}}$

• Equal Distance Round-Trip (Harmonic Mean Trap): If an object travels a distance $d$ at speed $s_1$ and returns along the exact same distance $d$ at speed $s_2$, the average speed is given by the harmonic mean, NOT the arithmetic mean $\frac{s_1 + s_2}{2}$: $\text{Average Speed}_{\text{equal } d} = \frac{2 s_1 s_2}{s_1 + s_2}$

B. Relative Speed Rules

When two moving objects have speeds $s_1$ and $s_2$:

• Moving in Opposite Directions (Towards or Away): Relative Speed $S_{\text{rel}} = s_1 + s_2$.
• Moving in the Same Direction (Overtaking): Relative Speed $S_{\text{rel}} = |s_1 - s_2|$.
• Time to Meet / Catch Up: $t_{\text{meet}} = \frac{\text{Initial Separation Distance}}{S_{\text{rel}}}$.

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3. Age & Coin Denomination Word Problems

A. Systematic Setup for Age Scenarios

1. Always assign a variable (e.g., $x$) to the present age of the primary person.
2. Translate past and future conditions relative to $x$:
   • Age $n$ years ago: $x - n$
   • Age $m$ years from now: $x + m$
3. Constant Difference Principle: The age difference between any two individuals remains constant throughout their lives ($A_{\text{present}} - B_{\text{present}} = A_{\text{future}} - B_{\text{future}}$).

B. Coin Denomination & Currency Value Setup

Coin problems involve two distinct quantities: the number of coins/notes and their monetary value.

• Let $n_1, n_2, \dots, n_k$ represent the counts of coins with denomination values $v_1, v_2, \dots, v_k$.
• Total Count Equation: $n_1 + n_2 + \dots + n_k = N_{\text{total}}$
• Total Monetary Value Equation: $(n_1 \times v_1) + (n_2 \times v_2) + \dots + (n_k \times v_k) = V_{\text{total}}$

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4. Financial Mathematics: Interest, Profit, Loss & Discounts

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5. Set Theory & Venn Diagram Scenarios

Venn diagram problems organize overlapping sets of elements into mutually exclusive regions.

A. Two-Set Principle of Inclusion-Exclusion

For two overlapping sets $A$ and $B$ within a total population $U$: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ $\text{Total Population } n(U) = n(A \cup B) + n(\text{Neither})$ $\text{Total Population } n(U) = n(A \text{ only}) + n(B \text{ only}) + n(A \cap B) + n(\text{Neither})$

B. Three-Set Inclusion-Exclusion Formula

For three overlapping sets $A$, $B$, and $C$: $n(A \cup B \cup C) = n(A) + n(B) + n(C) - [n(A \cap B) + n(B \cap C) + n(A \cap C)] + n(A \cap B \cap C)$

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6. Step-by-Step Worked Mathematical Examples

Worked Example 1: Combined Work with Pipe Outlets

Scenario: Pipe A can fill a water tank in 8 hours. Pipe B can fill the same tank in 12 hours. A drainage leak at the bottom of the tank can empty a full tank in 24 hours.

Question: If all three pipes (Pipe A, Pipe B, and the leak) are open simultaneously, how many hours will it take to fill the empty tank?

• Step 1: Write individual hourly rates: $R_A = +\frac{1}{8}$, $R_B = +\frac{1}{12}$, $R_{\text{leak}} = -\frac{1}{24}$.
• Step 2: Calculate net hourly rate by finding a common denominator ($24$): $R_{\text{net}} = \frac{1}{8} + \frac{1}{12} - \frac{1}{24} = \frac{3}{24} + \frac{2}{24} - \frac{1}{24} = \frac{4}{24} = \frac{1}{6} \text{ tanks/hr}$
• Step 3: Invert net rate to find total time required: $T_{\text{fill}} = \frac{1}{R_{\text{net}}} = 6 \text{ hours}$

Worked Example 2: Age System Translation

Scenario: A mother is currently 3 times as old as her daughter. Six years ago, the mother was 4 times as old as her daughter.

Question: What are the current ages of the mother and daughter?

• Step 1: Define present age variable: Let daughter's present age $= x$. Then mother's present age $= 3x$.
• Step 2: Express ages 6 years ago: Daughter $= x - 6$, Mother $= 3x - 6$.
• Step 3: Formulate equation using the past condition: $3x - 6 = 4(x - 6)$
• Step 4: Solve for $x$: $3x - 6 = 4x - 24 \implies 4x - 3x = 24 - 6 \implies x = 18$
• Step 5: Determine mother's age: $3x = 3(18) = 54$. Answer: Daughter is currently 18 years old, Mother is 54 years old.

Worked Example 3: Two-Set Venn Diagram Overlap

Scenario: In a batch of 120 GAT candidates in Rawalpindi, 75 candidates registered for Quantitative preparation modules, 55 registered for Analytical preparation modules, and 20 registered for neither module.

Question: How many candidates registered for BOTH Quantitative and Analytical modules?

• Step 1: Calculate total candidates who registered for at least one module ($n(Q \cup A)$): $n(Q \cup A) = \text{Total} - n(\text{Neither}) = 120 - 20 = 100$
• Step 2: Apply Inclusion-Exclusion principle: $n(Q \cup A) = n(Q) + n(A) - n(Q \cap A)$ $100 = 75 + 55 - n(Q \cap A)$
• Step 3: Solve for overlap $n(Q \cap A)$: $100 = 130 - n(Q \cap A) \implies n(Q \cap A) = 130 - 100 = 30$ Answer: 30 candidates registered for both modules.