6.3 Ratios, Proportions, Mixtures & Work-Time Rates

Ratios, Proportions, Mixtures & Work-Time Rates

Rate calculations, proportion dynamics, mixture compositions, and work-time scenarios represent some of the most frequent word problem types on the NTS GAT General Quantitative Reasoning section. Although these problems present diverse real-world contexts—such as combining chemical solutions, predicting team construction timelines, or calculating train arrival times—they all rely on unified rate equations ($y = k x$, $W = R \cdot T$, $D = S \cdot T$). Developing structural mastery over these rate frameworks allows rapid problem translation and error-free execution.

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1. Ratio Principles, Proportion Rules & Multi-Ratio Combination

A ratio is a mathematical comparison of two quantities of the same unit, expressed as $A : B$ or $\frac{A}{B}$.

Combining Ratios (The LCM Method)

When given separate ratios sharing a common variable (e.g., $A : B = a : b$ and $B : C = c : d$), combine them into a single continuous ratio $A : B : C$ by finding the LCM of the common term ($B$):

$$\text{Multiplier for first ratio} = \frac{\text{LCM}(b, c)}{b}$$ $$\text{Multiplier for second ratio} = \frac{\text{LCM}(b, c)}{c}$$

Example: Given $A : B = 2 : 3$ and $B : C = 4 : 5$. Common element $B$ has values $3$ and $4$. $\text{LCM}(3,4) = 12$.

• Scale first ratio by $4$: $A : B = 8 : 12$.
• Scale second ratio by $3$: $B : C = 12 : 15$.
• Combined Ratio: $A : B : C = 8 : 12 : 15$.

Proportion Fundamentals

A proportion asserts that two ratios are equal ($A : B :: C : D \iff \frac{A}{B} = \frac{C}{D}$).

• Extremes and Means Rule: The product of extremes equals the product of means: $$A \times D = B \times C$$
• Mean Proportion: The mean proportion between two numbers $a$ and $b$ is $\sqrt{a \cdot b}$.
• Third Proportion: If $a : b :: b : c$, then $c = \frac{b^2}{a}$.

Direct vs. Inverse Variation

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2. Weighted Averages, Mixtures & The Rule of Alligation

When combining two solutions or ingredients of different concentrations/prices, compute the final concentration using weighted averages or the Rule of Alligation.

The Rule of Alligation (Cross Matrix)

To find the ratio in which two ingredients of prices/concentrations $C$ (cheaper) and $D$ (dearer) must be mixed to produce a mixture of mean price/concentration $M$:

$$\frac{\text{Quantity of Cheaper } (Q_c)}{\text{Quantity of Dearer } (Q_d)} = \frac{D - M}{M - C}$$ $$\begin{matrix} \text{Cheaper Price } (C) & & \text{Dearer Price } (D) \\ & \searrow \quad \swarrow & \\ & \text{Mean Price } (M) & \\ & \swarrow \quad \searrow & \\ (D - M) & & (M - C) \end{matrix}$$

Repeated Liquid Replacement Formula

If a vessel initially contains $x$ units of pure liquid, and $y$ units are repeatedly drawn out and replaced with water $n$ times:

$$\text{Final Quantity of Pure Liquid Remaining} = x \left(1 - \frac{y}{x}\right)^n$$

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3. Combined Work-Time Rates & Pipe/Cistern Principles

Work problems follow the core rate relationship:

$$\text{Work Done } (W) = \text{Rate } (R) \times \text{Time } (T)$$

Unit Work Assumption

Set total work to $W = 1$ complete job. If a person completes a job in $D$ days, their daily work rate is $R = \frac{1}{D}$.

Combined Work Rates

When two individuals with completion times $A$ and $B$ work together, their rates add linearly:

$$R_{\text{together}} = R_A + R_B = \frac{1}{A} + \frac{1}{B} = \frac{A + B}{A \cdot B}$$ $$T_{\text{together}} = \frac{A \cdot B}{A + B}$$

Pipes and Cisterns Variant

• Inlet Pipe (Fills): Positive work rate $+ \frac{1}{A}$.
• Outlet Pipe / Leak (Empties): Negative work rate $- \frac{1}{B}$.
• Net Rate (Both Open): $R_{\text{net}} = \frac{1}{A} - \frac{1}{B} = \frac{B - A}{A \cdot B}$.

4. Distance, Speed & Time Rates

Motion problems operate on the motion equation:

$$\text{Distance } (D) = \text{Speed } (S) \times \text{Time } (T)$$

Unit Conversions

• $\text{Kilometers per hour (km/h)} \rightarrow \text{Meters per second (m/s)}$: Multiply by $\frac{5}{18}$.
• $\text{Meters per second (m/s)} \rightarrow \text{Kilometers per hour (km/h)}$: Multiply by $\frac{18}{5}$.

Average Speed Principles

• General Formula: $$\text{Average Speed} = \frac{\text{Total Distance Covered}}{\text{Total Time Taken}}$$
• Equal Distance Rule: If an object covers distance $D$ at speed $S_1$ and returns over the same distance $D$ at speed $S_2$, average speed is the harmonic mean: $$S_{\text{avg}} = \frac{2 \cdot S_1 \cdot S_2}{S_1 + S_2}$$

Relative Speed Mechanics

When two objects move at speeds $S_1$ and $S_2$:

• Moving in Opposite Directions (Towards or Away): Relative Speed $= S_1 + S_2$.
• Moving in Same Direction: Relative Speed $= |S_1 - S_2|$.

Train Crossing Distances

• Train crossing a stationary point object (pole, standing person): Distance $= L_{\text{train}}$.
• Train crossing a platform, bridge, or tunnel of length $L_{\text{platform}}$: Distance $= L_{\text{train}} + L_{\text{platform}}$.

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5. Comprehensive GAT Worked Examples

Worked Example 1: Combined Work with Worker Leaving

Question: Worker A can build a wall in $10$ days, while Worker B can build it in $15$ days. Both work together for $4$ days, after which Worker A leaves. How many additional days will Worker B take to complete the remaining wall?

Solution:

1. Individual daily work rates: $R_A = \frac{1}{10}$, $R_B = \frac{1}{15}$.
2. Combined daily rate: $R_{\text{combo}} = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6}$.
3. Work completed in $4$ days together: $W_{\text{done}} = 4 \times \frac{1}{6} = \frac{2}{3}$.
4. Remaining work: $W_{\text{rem}} = 1 - \frac{2}{3} = \frac{1}{3}$.
5. Time needed for B to complete $W_{\text{rem}}$: $T_B = \frac{W_{\text{rem}}}{R_B} = \frac{1/3}{1/15} = \frac{15}{3} = 5$ days.
6. Final Answer: $5$ days.

Worked Example 2: Rule of Alligation Mixture Problem

Question: In what ratio must a rice merchant mix rice costing PKR $80$/kg with rice costing PKR $110$/kg to obtain a blend worth PKR $92$/kg?

Solution:

1. Cheaper price $C = 80$, Dearer price $D = 110$, Mean price $M = 92$.
2. Apply Alligation ratio formula: $\frac{Q_c}{Q_d} = \frac{D - M}{M - C}$.
3. Compute numerator and denominator:
   • $D - M = 110 - 92 = 18$.
   • $M - C = 92 - 80 = 12$.
4. Ratio $= \frac{18}{12} = \frac{3}{2} \rightarrow 3 : 2$.
5. Final Answer: $3 : 2$.

Worked Example 3: Train Crossing Platform

Question: A train $200$ meters long traveling at $72$ km/h crosses a railway platform in $25$ seconds. What is the length of the platform?

Solution:

1. Convert train speed to m/s: $72 \times \frac{5}{18} = 4 \times 5 = 20$ m/s.
2. Total distance covered in $25$ seconds: $D_{\text{total}} = \text{Speed} \times \text{Time} = 20 \times 25 = 500$ meters.
3. Total distance equals train length plus platform length: $L_{\text{train}} + L_{\text{platform}} = 500$.
4. $200 + L_{\text{platform}} = 500 \implies L_{\text{platform}} = 300$ meters.
5. Final Answer: $300$ meters.