3.2 Ratios, Proportions, and Rates

3.2 Ratios, Proportions, and Rates

Proportional reasoning forms the structural backbone of quantitative aptitude testing. A ratio is a mathematical comparison of two quantities of the same unit, expressed as $a : b$, $a$ to $b$, or $\frac{a}{b}$ (where $b \neq 0$). A proportion is a formal statement asserting that two ratios are equal: $\frac{a}{b} = \frac{c}{d}$ In every true proportion, the product of the extremes ($a$ and $d$) equals the product of the means ($b$ and $c$): $a \times d = b \times c$ Recognizing whether two variables interact through direct, inverse, or partitive relationships allows you to establish the correct mathematical model instantly on your scratch paper.

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The Three Archetypes of Proportion

Understanding the distinct mechanics of direct, inverse, and partitive proportions prevents the common error of inverting fraction setups under time pressure.

1. Direct Proportion

In a direct proportion, as one variable increases or decreases by a factor $m$, the related variable increases or decreases by the exact same factor. The constant of proportionality is $k = \frac{y}{x}$.

• Setup: $\frac{x_1}{y_1} = \frac{x_2}{y_2} \implies x_1 y_2 = x_2 y_1$.
• Example: If 8 meters of uniform fabric cost PHP 1,440, what is the cost of 14 meters of the same fabric? Set up $\frac{8}{1,440} = \frac{14}{C}$. Simplify $\frac{1}{180} = \frac{14}{C}$, yielding $C = 14 \times 180 = \text{PHP }2,520$.

2. Inverse (Indirect) Proportion

In an inverse proportion, as one quantity increases, the other decreases proportionally such that their mathematical product remains constant: $k = x \times y$.

• Setup: $x_1 y_1 = x_2 y_2$.
• Diagnostic Clue: Workforce and timeline problems are almost universally inverse proportions. If 6 construction workers can renovate a classroom in 12 days, how many days will 9 workers take at the same pace? Set up $6 \times 12 = 9 \times D \implies 72 = 9D \implies D = 8\text{ days}$. Doubling the workforce halves the completion time.

3. Partitive Proportion

In a partitive proportion, a whole quantity is divided into unequal parts according to a designated multi-term ratio $a : b : c$. Use the unitary method:

1. Sum the ratio terms to find the total number of parts: $S = a + b + c$.
2. Determine the numerical value of one single part: $\text{Unit Value} = \frac{\text{Total Quantity}}{S}$.
3. Multiply each ratio term by the unit value to obtain individual shares.

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Unit Rates and Scaling Conversions

A rate is a comparison of two quantities measured in different units (such as kilometers per hour, words per minute, or Philippine Pesos per kilogram). A unit rate simplifies this comparison so that the denominator is exactly 1 unit.

Unit Price and Value Comparison

When evaluating purchasing options in word problems, calculate the cost per single unit of measurement:

• Option A: A 750-gram box of cereal costs PHP 210 (Unit rate: $\frac{210}{750} = \frac{21}{75} = \text{PHP }0.28\text{ per gram}$).
• Option B: A 1.2-kilogram (1,200-gram) box of the same cereal costs PHP 312 (Unit rate: $\frac{312}{1200} = \frac{26}{100} = \text{PHP }0.26\text{ per gram}$).
• Option B provides the better value by PHP 0.02 per gram.

Map and Model Scale Factors

A scale of $1 : 25,000$ indicates that 1 centimeter on the map represents 25,000 centimeters in actual terrain. Since $100\text{ cm} = 1\text{ m}$ and $1,000\text{ m} = 1\text{ km}$, $25,000\text{ cm} = 250\text{ m} = 0.25\text{ km}$. Therefore, 4 centimeters on the map corresponds to $4 \times 0.25 = 1\text{ kilometer}$ of physical distance.

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Distance, Rate, and Time Dynamics

Uniform motion problems are governed by the fundamental kinematic formula: $\text{Distance } (d) = \text{Rate } (r) \times \text{Time } (t)$ from which we derive $r = \frac{d}{t}$ and $t = \frac{d}{r}$. Units must always be reconciled prior to calculation (e.g., converting minutes to hours or meters per second to kilometers per hour by multiplying by $\frac{18}{5}$ or $3.6$).

The Round-Trip Average Speed Fallacy

A classic trap on numerical reasoning exams is assuming that average speed equals the simple arithmetic mean of the two speeds. When traveling a distance $d$ at speed $v_1$ and returning the identical distance $d$ at speed $v_2$, more time is spent traveling at the slower speed. Therefore, the slower speed exerts a greater weighting on the total journey.

$\text{Average Speed } (v_{\text{avg}}) = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{d + d}{\frac{d}{v_1} + \frac{d}{v_2}} = \frac{2d}{d\left(\frac{v_1 + v_2}{v_1 v_2}\right)} = \frac{2 v_1 v_2}{v_1 + v_2}$ This formula is the harmonic mean of the two velocities.

Demonstration: A provincial bus travels from Manila to Batangas at $60\text{ km/h}$ and returns along the same route at $40\text{ km/h}$.

• Incorrect simple average: $\frac{60 + 40}{2} = 50\text{ km/h}$.
• Correct harmonic mean: $v_{\text{avg}} = \frac{2(60)(40)}{60 + 40} = \frac{4,800}{100} = 48\text{ km/h}$.

Relative Speed in Motion Scenarios

• Objects Moving in Opposite Directions (Toward or Away from Each Other): The gap between the objects changes at the sum of their speeds: $r_{\text{relative}} = r_1 + r_2$.
• Objects Moving in the Same Direction (Overtaking or Chasing): The gap between the objects closes or expands at the difference of their speeds: $r_{\text{relative}} = r_{\text{faster}} - r_{\text{slower}}$.

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Work and Combined Rate Problems

Work problems follow the identical mathematical architecture as motion problems, replacing distance with the total job completed: $\text{Work } (W) = \text{Rate } (R) \times \text{Time } (t)$

When a single task represents 1 complete unit of work ($W = 1$):

• If Person A completes a job in $t_A$ hours, their hourly work rate is $R_A = \frac{1}{t_A}$.
• If Person B completes the same job in $t_B$ hours, their hourly work rate is $R_B = \frac{1}{t_B}$.

Concurrent Work Formulation

When both individuals work simultaneously without interfering with each other's productivity, their individual work rates are additive: $R_{\text{combined}} = R_A + R_B = \frac{1}{t_A} + \frac{1}{t_B} = \frac{t_A + t_B}{t_A t_B}$ The time required for them to complete the entire job together ($t_{\text{total}}$) is the reciprocal of their combined rate: $t_{\text{total}} = \frac{t_A t_B}{t_A + t_B}$

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Step-by-Step Worked Word Problems

Worked Example 1: Partitive Investment and Profit Allocation

Problem: Three student entrepreneurs—Ben, Carlos, and Dan—establish a campus micro-enterprise selling eco-friendly tote bags. They contribute initial capital in the ratio $3 : 4 : 5$. At the end of the academic year, the enterprise records a net distributable profit of PHP 360,000. How much profit does Carlos receive, and what is the difference in profit between the partner with the largest share and the partner with the smallest share?

Step-by-Step Solution:

1. Determine the total number of ratio parts: $S = 3 + 4 + 5 = 12\text{ parts}$
2. Calculate the monetary value of one unit part: $\text{Unit Part Value} = \frac{\text{PHP }360,000}{12} = \text{PHP }30,000$
3. Calculate individual profit allocations:
   • Ben (3 parts): $3 \times 30,000 = \text{PHP }90,000$
   • Carlos (4 parts): $4 \times 30,000 = \text{PHP }120,000$
   • Dan (5 parts): $5 \times 30,000 = \text{PHP }150,000$
4. Calculate the difference between largest and smallest shares: $\text{Difference} = \text{Dan's share} - \text{Ben's share} = 150,000 - 90,000 = \text{PHP }60,000$ Shortcut: Difference in parts $= 5 - 3 = 2\text{ parts}$. $2 \times 30,000 = \text{PHP }60,000$.

Worked Example 2: Staggered Collaborative Work

Problem: Allan can assemble a complete set of laboratory experimental apparatus in 6 hours. Brenda can assemble the identical apparatus in 12 hours. Allan begins assembling alone and works for 2 hours. At that point, Brenda joins him, and they work together until assembly is finished. What is the total time, from initial start to final completion, required to finish the project?

Step-by-Step Solution:

1. Determine individual hourly rates:
   • Allan's rate: $R_A = \frac{1}{6}\text{ apparatus per hour}$.
   • Brenda's rate: $R_B = \frac{1}{12}\text{ apparatus per hour}$.
2. Calculate work completed by Allan in the first 2 hours: $W_{\text{initial}} = R_A \times 2 = \frac{1}{6} \times 2 = \frac{2}{6} = \frac{1}{3}$
3. Calculate remaining work to be completed: $W_{\text{remaining}} = 1 - \frac{1}{3} = \frac{2}{3}$
4. Determine combined work rate when working together: $R_{\text{combined}} = \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}\text{ apparatus per hour}$
5. Calculate time required for the remaining work: $t_{\text{together}} = \frac{W_{\text{remaining}}}{R_{\text{combined}}} = \frac{\frac{2}{3}}{\frac{1}{4}} = \frac{2}{3} \times 4 = \frac{8}{3} = 2\frac{2}{3}\text{ hours}$ Convert $\frac{2}{3}\text{ hour}$ to minutes: $\frac{2}{3} \times 60 = 40\text{ minutes}$. Thus, $t_{\text{together}} = 2\text{ hours and } 40\text{ minutes}$.
6. Calculate total elapsed time from the start: $t_{\text{total}} = 2\text{ hours (Allan alone)} + 2\frac{2}{3}\text{ hours (together)} = 4\frac{2}{3}\text{ hours (or 4 hours and 40 minutes)}$