Grade IX Mathematics: Linear Equations, Rational Numbers, and Applications

Linear Equations and Problem Solving: Taxi Fare Modeling

• Fare Structure Definition:

  • The initial cost (fixed charge) for the first kilometre traveled is set at $\text{₹}\,15$.
  • The rate for any subsequent distance after the first kilometre is $\text{₹}\,10$ per kilometre.

• Derivation of the Linear Equation:

  • Let $x$ represent the total distance traveled in kilometres ($\text{km}$).
  • Let $y$ represent the total fare in rupees ($\text{₹}$).
  • The first kilometre costs $\text{₹}\,15$. The remaining distance is $(x - 1)\,\text{km}$.
  • The cost of the remaining distance is $(x - 1) \times 10$.
  • The total fare equation is: $y = 15 + 10(x - 1)$.
  • Simplifying the equation: $y = 15 + 10x - 10$, which yields $y = 10x + 5$.

• Case Study: Deepak's Taxi Ride:

  • Distance traveled ($x$): $10\,\text{km}$.
  • Amount paid to the driver: $\text{₹}\,200$.
  • Calculation of Actual Fare:
  • Substituting $x = 10$ into the linear equation: $y = 10(10) + 5$.
  • $y = 100 + 5 = 105$.
  • The actual fare is $\text{₹}\,105$.
  • Donation Calculation:
  • Deepak asks the driver to donate the balance amount to an orphanage.
  • Balance = $\text{Amount Paid} - \text{Actual Fare}$.
  • $\text{Balance} = 200 - 105 = 95$.
  • The taxi-driver will donate $\text{₹}\,95$ to the orphanage.

Linear Modeling in Sports: Cricket Match Scoring and Donations

• Scenario and Ratio Analysis:

  • The donation to an old-age home is equal to the total runs scored by the opening pair, Ratan and Naval.
  • Let $x$ be the runs scored by Ratan and $y$ be the runs scored by Naval.
  • Ratio Given: For every $1$ run scored by Ratan, Naval scores $3$ runs.
  • This establishes the relationship: $y = 3x$.

• Mathematical Model:

  • Linear Equation: $y - 3x = 0$ or $y = 3x$.
  • Total runs ($N$) to be donated: $N = x + y$. Using the relationship, $N = x + 3x = 4x$.

• Calculations for Specific Scoring Scenarios:

  • Scenario A (Naval scored 180 runs):
  • $y = 180$.
  • Since $y = 3x$, then $180 = 3x \Rightarrow x = \frac{180}{3} = 60$.
  • Ratan scored $60$ runs.
  • Scenario B (Ratan scored 99 runs):
  • $x = 99$.
  • Naval's score ($y$) = $3 \times 99 = 297$.
  • Total runs/Donation amount ($N$) = $99 + 297 = 396$.
  • The amount donated to the old-age home is $\text{₹}\,396$.
  • Scenario C (Naval scored 126 runs):
  • $y = 126$.
  • $126 = 3x \Rightarrow x = \frac{126}{3} = 42$.
  • Ratan scored $42$ runs.

Linear Decay and Asset Depreciation: DSLR Camera Valuation

• Initial Parameters:

  • Cost of high-end DSLR camera at purchase ($t = 0$): $\text{₹}\,45,000$.
  • Annual depreciation (wear and tear/obsolescence): $\text{₹}\,3,500$ per year.

• Linear Expression Formulation:

  • Let $v$ be the market value and $t$ be the time in years.
  • The value decreases by a constant amount each year, indicating a linear function with a negative slope.
  • Equation: $v = 45000 - 3500t$.

• Valuation Schedule (Table of Values):

  • Time $t=0$: $v = 45000 - 3500(0) = 45,000$
  • Time $t=1$: $v = 45000 - 3500(1) = 41,500$
  • Time $t=2$: $v = 45000 - 3500(2) = 38,000$
  • Time $t=3$: $v = 45000 - 3500(3) = 34,500$
  • Time $t=4$: $v = 45000 - 3500(4) = 31,000$
  • Time $t=5$: $v = 45000 - 3500(5) = 27,500$
  • Time $t=6$: $v = 45000 - 3500(6) = 24,000$

• Linear Decay Explanation:

  • This specific relationship represents linear decay because the value of the camera decreases by a fixed, constant amount ($\text{₹}\,3,500$) for every unit increase in time (per year). The rate of change ($dv/dt$) is constant and negative.

• Projection for Value Drop Below Limit:

  • Criterion: $v < 15,000$.
  • $45000 - 3500t < 15000$.
  • $45000 - 15000 < 3500t$.
  • $30000 < 3500t$.
  • $t > \frac{30000}{3500} \approx 8.57$.
  • Therefore, after $9$ years, the camera's value will drop below $\text{₹}\,15,000$.

Number Systems: Rational Numbers and Repeating Decimals

• Proof Concept for Non-Terminating Repeating Decimals:

  • Every non-terminating repeating decimal can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$, making it a rational number.
  • General Method:
  1. Let $x$ equal the repeating decimal.
  2. Identify the number of repeating digits (period length, $n$).
  3. Multiply $x$ by $10^n$.
  4. Subtract the original equation from the new equation to eliminate the repeating block.
  5. Solve for $x$.
  • Illustration with Example: $x = 0.\bar{3}$.
  • $x = 0.333...$ (Equation 1).
  • Multiply by $10^1$ ($n=1$): $10x = 3.333...$ (Equation 2).
  • Subtract (Eq 2 - Eq 1): $9x = 3$.
  • $x = \frac{3}{9} = \frac{1}{3}$.
  • Since $1/3$ is a ratio of two integers, the number is rational.

• Converse Statement:

  • The converse state that if a number is a rational number, its decimal expansion is either terminating or non-terminating repeating.

Advanced Number Properties: Cyclic Numbers and Decimal Expansions

• Cyclic Numbers through 1/7:

  • The decimal expansion of $\frac{1}{7}$ is $0.\overline{142857}$.
  • The repeating block is $142857$.
  • It is called a cyclic number because the same digits (in the same circular order) appear in the decimal expansions of other fractions with the same denominator ($n/7$):
  • $\frac{1}{7} = 0.\overline{142857}$
  • $\frac{2}{7} = 0.\overline{285714}$
  • $\frac{3}{7} = 0.\overline{428571}$
  • $\frac{4}{7} = 0.\overline{571428}$
  • $\frac{5}{7} = 0.\overline{714285}$
  • $\frac{6}{7} = 0.\overline{857142}$
  • Each product is a cyclic permutation of the original digits.

• Decimal Expansion of 13/11:

  • Calculation: $\frac{13}{11} = 1.181818... = 1.\overline{18}$.
  • Characterization: This expansion is non-terminating and repeating, making it a rational number with a period of length $2$.