Proportional Reasoning - II

Chapter at a Glance

This chapter extends proportional reasoning to complex scenarios, including representative map scaling fractions, multi-term ratios, visual data representations via pie charts, inverse proportions, and collaborative work rates. Students learn to determine geographical distances using map scales, split a single quantity into three or more parts in a given ratio, and calculate central angles to construct pie charts. The chapter details the algebraic difference between direct and inverse proportions, showing how inverse proportions model situations where one variable increases as another decreases at the same rate, and explains how to solve joint work problems by summing individual work rates.

Key Definitions & Terminology

• Representative Fraction (RF): A numerical ratio expressing the relationship between a measured distance on a map and the actual geographical distance on the ground (e.g., $1 : 6,000,000$).
• Multi-Term Ratio: A ratio comparing three or more quantities simultaneously (e.g., $a : b : c : d$).
• Pie Chart: A circular statistical graph divided into slices to illustrate numerical proportions, where the arc length of each slice is proportional to the quantity it represents.
• Direct Proportion: A relationship between two variables where they increase or decrease by the same factor, maintaining a constant ratio ($\frac{x}{y} = k$).
• Inverse Proportion: A relationship between two variables where an increase in one causes a proportional decrease in the other, maintaining a constant product ($x \cdot y = k$).
• Work Rate: The portion of a total job completed by a worker or machine in a unit of time (e.g., $\frac{1}{3}$ tank per hour).

Formulas, Rules & Properties

• Direct Proportion Ratio Rule:
• Two ratios with multiple terms $a : b : c : \dots$ and $p : q : r : \dots$ are in direct proportion if:
  $$\frac{a}{p} = \frac{b}{q} = \frac{c}{r} = \dots = k \quad \text{(where } k \text{ is a constant)}$$
• Multi-Term Sharing Formula:
• When dividing a quantity $x$ in the ratio $a : b : c : \dots$, the sizes of the respective parts are:
  $$\text{Part}_1 = x \cdot \frac{a}{a+b+c+\dots}, \quad \text{Part}_2 = x \cdot \frac{b}{a+b+c+\dots}, \quad \text{Part}_3 = x \cdot \frac{c}{a+b+c+\dots}$$
• Pie Chart Angle Formula:
• To find the central angle $\theta$ for a category representing a quantity $v$ out of a total sum $V$:
  $$\theta = \frac{v}{V} \cdot 360^\circ$$
• Inverse Proportion Product Rule:
• If two quantities $x$ and $y$ are in inverse proportion, then for any two corresponding pairs $(x_1, y_1)$ and $(x_2, y_2)$:
  $$x_1 \cdot y_1 = x_2 \cdot y_2 = k \quad \rightarrow \quad \frac{x_1}{x_2} = \frac{y_2}{y_1}$$
• Combined Work-Rate Formula:
• If agent $A$ completes a task in $t_A$ hours and agent $B$ completes the same task in $t_B$ hours, their combined work rate per hour is:
  $$\text{Combined Rate} = \frac{1}{t_A} + \frac{1}{t_B}$$
  The total time $T$ required to complete the task together is the reciprocal of the combined rate:
  $$T = \frac{1}{\text{Combined Rate}} = \frac{t_A \cdot t_B}{t_A + t_B} \text{ hours}$$

Core Concepts & Topics

• Map Scaling (Representative Fractions):
• An RF scale of $1 : 6,000,000$ indicates that $1\text{ cm}$ measured on the map represents $6,000,000\text{ cm}$ on the ground.
• Converting map centimeters to kilometers:
  $$6,000,000\text{ cm} = \frac{6,000,000}{100}\text{ m} = 60,000\text{ m} = \frac{60,000}{1000}\text{ km} = 60\text{ km}$$
• Slicing the Circle (Constructing Pie Charts):
• Central angles must sum to $360^\circ$.
• Slices are measured sequentially using a protractor, starting from an initial radius and working anticlockwise, with each new slice boundary serving as the baseline for the next angle.
• Inverse Proportional Relationships in Real Life:
• Speed and Time: For a fixed route distance, doubling the speed halves the travel time.
• Workers and Time: For a fixed construction task, doubling the number of workers halves the days needed.
• Pumps and Time: For a fixed tank volume, doubling the pumps halves the hours to fill.
• Triangle Sidelength Feasibility (Triangle Inequality):
• A triangle with sidelengths in a given ratio is only constructible if the sum of the two smaller ratio terms is strictly greater than the largest term. A ratio of $3:4:5$ is feasible ($3+4 > 5$), whereas $1:3:5$ is impossible ($1+3 < 5$).

Worked Examples

• Map Distance Calculation:
• Problem: A map has an RF scale of $1 : 6,000,000$. The measured distance between Bengaluru and Chennai is $5.8\text{ cm}$. Find the geographical distance.
• Solution:
  • $1\text{ cm} \rightarrow 60\text{ km}$.
  • Geographical distance $= 5.8 \times 60 = 348\text{ km}$.
• Multi-Term Concrete Mix (Page 59 Example 3):
• Problem: Concrete is mixed using cement, sand, and gravel in the ratio $1 : 1.5 : 3$. Find the amount of each ingredient needed to produce $110\text{ units}$ of concrete.
• Solution:
  • Sum of parts $= 1 + 1.5 + 3 = 5.5$.
  • Number of batches $= 110 \div 5.5 = 20$.
  • Cement $= 1 \times 20 = 20\text{ units}$.
  • Sand $= 1.5 \times 20 = 30\text{ units}$.
  • Gravel $= 3 \times 20 = 60\text{ units}$.
• Cricket Practice Allocations (Figure it Out Q1, Page 60):
• Problem: A $150$-minute cricket practice is split into warm-up, batting, bowling, and fielding in the ratio $3 : 4 : 3 : 5$. Find the time for each.
• Solution:
  • Sum of parts $= 3 + 4 + 3 + 5 = 15$.
  • Unit block $= 150 \div 15 = 10\text{ minutes}$.
  • Warm-up/cool-down $= 3 \times 10 = 30\text{ minutes}$.
  • Batting $= 4 \times 10 = 40\text{ minutes}$.
  • Bowling $= 3 \times 10 = 30\text{ minutes}$.
  • Fielding $= 5 \times 10 = 50\text{ minutes}$.
• Coin Bag Denominations (Figure it Out Q3, Page 60):
• Problem: A bag containing $100$ coins has denominations of ₹$10$, ₹$5$, ₹$2$, and ₹$1$ in the ratio $4 : 3 : 2 : 1$. Find the total monetary value.
• Solution:
  • Sum of parts $= 4 + 3 + 2 + 1 = 10$.
  • Scaling factor $= 100 \div 10 = 10$.
  • Number of ₹$10$ coins $= 4 \times 10 = 40$ (Value $= 40 \times 10 = $ ₹$400$).
  • Number of ₹$5$ coins $= 3 \times 10 = 30$ (Value $= 30 \times 5 = $ ₹$150$).
  • Number of ₹$2$ coins $= 2 \times 10 = 20$ (Value $= 20 \times 2 = $ ₹$40$).
  • Number of ₹$1$ coins $= 1 \times 10 = 10$ (Value $= 10 \times 1 = $ ₹$10$).
  • Total Value $= 400 + 150 + 40 + 10 = $ ₹$600$.
• Water Supply Depletion (Figure it Out Q3, Page 67):
• Problem: A water tank supplies $20$ families for $6$ days. If $10$ more families move in, how long will the water last?
• Solution: Total families $= 20 + 10 = 30$. Since families and depletion time are in inverse proportion:
  $$x_1 \cdot y_1 = x_2 \cdot y_2 \rightarrow 20 \times 6 = 30 \times y_2 \rightarrow 120 = 30y_2 \rightarrow y_2 = 4\text{ days}$$
  • Assumption: Each family consumes water at the same rate.
• Combined Vegetable Cutting Rate (Page 66 Example 6):
• Problem: Ram takes $1\text{ hour}$ to cut a batch of vegetables, while Shyam takes $1.5\text{ hours}$. How long will they take if they cut them together?
• Solution:
  • Ram's rate $= 1\text{ batch/hour}$.
  • Shyam's rate $= \frac{1}{1.5} = \frac{2}{3}\text{ batch/hour}$.
  • Combined rate $= 1 + \frac{2}{3} = \frac{5}{3}\text{ batch/hour}$.
  • Combined time $T = \frac{1}{\text{Combined Rate}} = \frac{3}{5}\text{ hours} = \frac{3}{5} \times 60 = 36\text{ minutes}$.
• Toy Factory Output (Figure it Out Q11, Page 68):
• Problem: A factory requires $42$ machines to produce toys in $63$ days. How many machines are needed to produce the same quantity in $54$ days?
• Solution: Since machines and days are in inverse proportion:
  $$42 \times 63 = x \times 54 \rightarrow x = \frac{42 \times 63}{54} = \frac{42 \times 7}{6} = 7 \times 7 = 49\text{ machines}$$

Practical Activities & Experiments

• Classroom Scaled Blueprints: Measure the width and length of your classroom, the teacher's desk, and the blackboard in centimeters. Using a Representative Fraction scale of $1 : 50$ (dividing all measurements by $50$), draw a blueprint layout of the classroom in your notebook.
• Pie Chart Survey: Survey $20$ classmates on their favorite subjects. Calculate the proportion of students choosing each subject, convert these proportions into central angles using $\theta = \frac{\text{count}}{20} \cdot 360^\circ$, and construct a color-coded pie chart with a protractor.