PROPORTIONAL REASONING-1

Detailed Chapter Roadmap

• Observation, Scaling, and Distortion (Pages 1–2):
  • Core Idea: Building intuitive foundations through visual representation. Using digital images (such as tigers or geometrical figures) to explore how resizing an image works. If an image is stretched horizontally without a proportional vertical stretch, it appears distorted. True scaling preserves the aspect ratio, which serves as our first physical encounter with proportions.
• Formal Ratios and Simplest Form (Pages 3–4):
  • Core Idea: Transitioning from visual perception to numeric quantification. Defining a ratio as a comparison of two quantities of the same kind using division ($a:b$ or $\frac{a}{b}$). Introducing the Highest Common Factor (HCF) to reduce ratios to their simplest form, establishing that $2:3$, $4:6$, and $10:15$ represent the exact same multiplicative relationship.
• The Language of Proportion and Cross-Multiplication (Pages 4–6):
  • Core Idea: Establishing the equality of two ratios using the proportion symbol :: (e.g., $a:b :: c:d$). Proving that for any proportion, the product of the extremes equals the product of the means ($ad = bc$), providing a universal algebraic check for proportional equivalence.
• The Rule of Three (Trairasika) (Pages 6–9):
  • Core Idea: A historical and practical algorithmic method. Given three known quantities in a proportional relationship, find the fourth unknown quantity. This method forms the backbone of unitary and proportional calculations worldwide.
• Sharing Quantities in a Given Ratio (Pages 14–17):
  • Core Idea: Distributing a total amount among multiple recipients according to a specific ratio (e.g., dividing resources, money, or chemical mixtures into parts like $m:n$).
• Non-Proportional Relationships and Population Density (Pages 10–13):
  • Core Idea: Recognizing when quantities do not scale proportionally. Analyzing real-world data like population density (people per square kilometer) to evaluate crowding and resource distribution.
• Comprehensive Units and Conversions (Pages 17–19):
  • Core Idea: Mastering dimensional consistency across metric and imperial systems for length, area, volume, mass, time, and temperature before applying proportional formulas.

Deep-Dive Case Studies and Real-Life Applications

Case Study 1: Digital Image Scaling and Aspect Ratios

• The Problem: Graphic designers and software engineers must resize images for mobile phones, tablets, and billboards without distorting the picture. If a photograph of a historical monument has a dimension of $1200\text{ pixels wide} \times 800\text{ pixels high}$, what must its height be if it is scaled down to a width of $300\text{ pixels}$ for a mobile screen?
• The Analysis:
  1. The original ratio of width to height is $1200 : 800$.
  2. Simplify this ratio by finding the HCF of 1200 and 800, which is 400.
  3. Dividing both terms by 400 yields the simplest form: $3 : 2$. This means for every 3 units of width, there must be 2 units of height.
  4. Set up the proportion where the new width is 300: $\text{Width} : \text{Height} = 3 : 2 = 300 : h$
  5. Using cross-multiplication ($ad = bc$): $3 \times h = 2 \times 300$ $3h = 600 \implies h = 200\text{ pixels}$
• Real-World Takeaway: Preserving the $3:2$ aspect ratio prevents stretching or squishing the image, maintaining visual integrity.

Case Study 2: Earth's Orbital Mechanics and Proportional Travel

• The Problem: The Earth travels approximately $940,000,000\text{ km}$ in a complete orbit around the Sun, taking 52 weeks. Using proportional reasoning, calculate the exact distance traveled by Earth in exactly 1 week.
• The Analysis:
  1. Identify the known relationship: $940,000,000\text{ km}$ corresponds to $52\text{ weeks}$.
  2. Set up the proportion with $x$ representing the distance in 1 week: $940,000,000 : 52 :: x : 1$
  3. Convert the proportion to an equation using the cross-multiplication property: $52 \times x = 940,000,000 \times 1$ $x = \frac{940,000,000}{52} \approx 18,076,923.08\text{ km}$
• Real-World Takeaway: Astronomical distances that are hard to comprehend are easily broken down into manageable weekly intervals using the Rule of Three.

Case Study 3: Masonry and Construction Efficiency

• The Problem: A masonry crew lays 1,450 bricks to build a section of a boundary wall that is $10\text{ feet}$ long. Assuming uniform thickness and height, how many bricks are required to complete a total wall length of $108\text{ feet}$?
• The Analysis:
  1. The relationship between length and brick count is direct and proportional.
  2. Set up the proportion: $10\text{ feet} : 1,450\text{ bricks} :: 108\text{ feet} : x\text{ bricks}$
  3. Apply the cross-multiplication rule: $10 \times x = 1,450 \times 108$ $10x = 156,600$ $x = \frac{156,600}{10} = 15,660\text{ bricks}$
• Real-World Takeaway: Civil engineers and contractors rely on baseline samples (such as bricks per 10 feet) to order materials accurately, minimizing waste and financial loss.

Step-by-Step Problem Solving Strategies & Detailed Proofs

The Cross-Multiplication Theorem for Proportions

• Statement: If $\frac{a}{b} = \frac{c}{d}$ (or written as $a:b :: c:d$), where $b \neq 0$ and $d \neq 0$, then $ad = bc$.
• Proof:
  1. Start with the given proportional equality in fractional form: $\frac{a}{b} = \frac{c}{d}$
  2. Multiply both sides of the equation by the common denominator $bd$: $bd \cdot \left(\frac{a}{b}\right) = bd \cdot \left(\frac{c}{d}\right)$
  3. Cancel out common terms on both sides: $d \cdot a = b \cdot c$
  4. Rearrange terms to get the final property: $ad = bc \quad \text{(Hence Proved)}$

Algorithm for Sharing Quantities in a Ratio ($m:n$)

To divide a total quantity $Q$ into two parts based on the ratio $m:n$:

1. Find the Total Number of Parts: Add the components of the ratio: $\text{Total Parts} = m + n$
2. Determine the Value of One Part: Divide the total quantity $Q$ by the total number of parts: $\text{Value of One Part} = \frac{Q}{m + n}$
3. Calculate Individual Shares: Multiply each ratio component by the value of one part: $\text{First Part} = m \times \left(\frac{Q}{m+n}\right)$ $\text{Second Part} = n \times \left(\frac{Q}{m+n}\right)$
4. Verification Step: Ensure that the sum of the calculated parts equals the original total quantity: $\text{First Part} + \text{Second Part} = Q$

Higher-Order Thinking Skills (HOTS) Questions

Question 1 (Altering Ratios via Addition)

A chemical solution consists of acid and water mixed in the ratio $3 : 5$. If $16\text{ mL}$ of water is added to the mixture, the new ratio of acid to water becomes $3 : 13$. Find the original volume of the acid and the water in the mixture.

• Solution Strategy:
  1. Let the original quantity of acid be $3x$ and water be $5x$.
  2. According to the problem, $16\text{ mL}$ of water is added. The new quantity of water is $5x + 16$, while the acid remains $3x$.
  3. Write down the new ratio equation: $\frac{3x}{5x + 16} = \frac{3}{13}$
  4. Cross-multiply to solve for $x$: $13(3x) = 3(5x + 16)$ $39x = 15x + 48$
  5. Isolate $x$: $39x - 15x = 48$ $24x = 48 \implies x = 2$
  6. Substitute $x$ back into the original quantities:
     • $\text{Acid} = 3(2) = 6\text{ mL}$
     • $\text{Water} = 5(2) = 10\text{ mL}$
  7. Verification: New water = $10 + 16 = 26\text{ mL}$. New ratio = $6 : 26 = 3 : 13$. Matches perfectly!

Question 2 (Comparative Population Density and Crowding)

City A has a population of $2,500,000$ people spread across an area of $500\text{ sq. km}$. City B has a population of $1,800,000$ people spread across an area of $300\text{ sq. km}$. Which city is more densely populated, and by what factor?

• Solution Strategy:
  1. Calculate population density for City A ($\text{People} / \text{Area}$): $\text{Density}_A = \frac{2,500,000}{500} = 5,000\text{ people per sq. km}$
  2. Calculate population density for City B: $\text{Density}_B = \frac{1,800,000}{300} = 6,000\text{ people per sq. km}$
  3. Compare values: $6,000 > 5,000$, meaning City B is more crowded.
  4. Find the ratio/factor of difference: $\text{Ratio} = \frac{6,000}{5,000} = 1.2$
  5. Conclusion: City B is more densely populated than City A by a factor of 1.2 (or 20% denser).

Previous Year Questions (PYQs) with Solutions

Question 1 (Direct Proportion & Unitary Method)

• Question: If 15 meters of cloth cost ₹980, how many meters of the same cloth can be purchased for ₹3,920?
• Solution:
  1. Let the unknown length of cloth be $x$ meters.
  2. Set up the direct proportion: $15\text{ m} : ₹980 :: x\text{ m} : ₹3,920$
  3. Apply the cross-multiplication property ($ad = bc$): $980 \times x = 15 \times 3,920$
  4. Calculate the right-hand side: $15 \times 3,920 = 58,800$
  5. Solve for $x$: $x = \frac{58,800}{980} = 60\text{ meters}$
  6. Answer: $60\text{ meters}$ of cloth can be purchased for ₹3,920.

Question 2 (Sharing Quantities)

• Question: Divide a total sum of ₹2,400 among three individuals—A, B, and C—in the ratio $2 : 3 : 5$. Find the exact share received by each person.
• Solution:
  1. Calculate the total number of parts in the ratio: $\text{Total Parts} = 2 + 3 + 5 = 10\text{ parts}$
  2. Find the monetary value of one single part: $\text{Value of 1 Part} = \frac{₹2,400}{10} = ₹240$
  3. Calculate individual shares by multiplying each ratio term by the value of one part:
     • $\text{A's Share} = 2 \times ₹240 = ₹480$
     • $\text{B's Share} = 3 \times ₹240 = ₹720$
     • $\text{C's Share} = 5 \times ₹240 = ₹1,200$
  4. Verification: $480 + 720 + 1,200 = ₹2,400$. Matches the total amount.

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NCERT Textbook Questions & Detailed Answers

Question 1 (Simplifying Ratios)

• Question: Express the ratio $45 : 75$ in its simplest form.
• Detailed Answer:
  1. Write the ratio as a fraction: $\frac{45}{75}$.
  2. Find the Highest Common Factor (HCF) of 45 and 75:
     • Factors of 45: $1, 3, 5, 9, 15, 45$
     • Factors of 75: $1, 3, 5, 15, 25, 75$
     • The HCF is $15$.
  3. Divide both the numerator and denominator by the HCF: $\frac{45 \div 15}{75 \div 15} = \frac{3}{5}$
  4. Final Answer: The simplest form of $45 : 75$ is $3 : 5$.

Question 2 (Verifying Proportions)

• Question: Check whether the ratios $12 : 18$ and $20 : 30$ form a proportion.
• Detailed Answer:
  1. Simplify the first ratio ($12 : 18$): HCF of 12 and 18 is 6. $\frac{12 \div 6}{18 \div 6} = \frac{2}{3} \quad (\text{or } 2 : 3)$
  2. Simplify the second ratio ($20 : 30$): HCF of 20 and 30 is 10. $\frac{20 \div 10}{30 \div 10} = \frac{2}{3} \quad (\text{or } 2 : 3)$
  3. Alternatively, use the cross-multiplication method ($ad = bc$) for $12 : 18 :: 20 : 30$:
     • $\text{Extremes product} = 12 \times 30 = 360$
     • $\text{Means product} = 18 \times 20 = 360$
  4. Since both products are equal ($360 = 360$), the ratios are equal.
  5. Final Answer: Yes, $12 : 18$ and $20 : 30$ form a valid proportion ($12 : 18 :: 20 : 30$).

Question 3 (Acid-Water Mixture Ratio Application)

• Question: A chemical laboratory has $240\text{ mL}$ of a specialized cleaning solution containing acid and water mixed in the ratio $1 : 5$. Calculate the exact volumes of acid and water present in the mixture.
• Detailed Answer:
  1. Identify the ratio of components: $\text{Acid} : \text{Water} = 1 : 5$.
  2. Find the total number of parts in the ratio: $\text{Total Parts} = 1 + 5 = 6\text{ parts}$
  3. Determine the volume represented by a single part from the total volume of $240\text{ mL}$: $\text{Value of 1 Part} = \frac{240\text{ mL}}{6} = 40\text{ mL}$
  4. Calculate the individual volumes:
     • $\text{Volume of Acid} = 1 \times 40\text{ mL} = 40\text{ mL}$
     • $\text{Volume of Water} = 5 \times 40\text{ mL} = 200\text{ mL}$
  5. Final Answer: The mixture contains $40\text{ mL}$ of acid and $200\text{ mL}$ of water.