9th Class Math Chapter 10 Notes

Q1 — Graph $S=1000e^{-t}$ for $t=0$ to $9$

A school's strength decays as $S=1000e^{-t}$. Plot the graph from $t=0$ (2016) to $t=9$ (2025).

1. Compute key points: $t=0\to 1000$, $t=1\to 368$, $t=2\to 135$, $t=3\to 50$, $t=4\to 18$, $t=5\to 7$, $t=6\to 2.5$, $t=7\to 0.9$, $t=8\to 0.3$, $t=9\to 0.1$.
2. This traces a steep exponential-decay curve that falls sharply at first then flattens near zero.

Answer: Curve drops from $1000$ toward $0$, roughly $\{1000,368,135,50,18,7,2.5,0.9,0.3,0.1\}$ at $t=0,\dots ,9$

Q2 — Estimate strength in 2019 and 2023

Using the graph, estimate the strength at $t=3$ (2019) and $t=7$ (2023).

1. $S(3)=1000e^{-3}\approx 50$.
2. $S(7)=1000e^{-7}\approx 0.9$ (essentially $0$ — under this model, practically no students remain by 2023).

Answer: 2019: $\approx 50$ students; 2023: $\approx 1$ student (model predicts near-zero)

Q3 — Plot $P_d=400-5Q$ and $P_s=3Q+24$ for $Q=0$ to $300$

Plot each function and note where they meet.

1. $P_d$: starts at $(0,400)$, slope $-5$ (downward line).
2. $P_s$: starts at $(0,24)$, slope $3$ (upward line).
3. Equilibrium: $400-5Q=3Q+24\Rightarrow 376=8Q\Rightarrow Q=47$; $P=3(47)+24=165$.

Answer: Lines cross at the equilibrium point $(Q,P)=(47,165)$

Q4 — Sketch $S(x)=45000+4500x$ for $0\le x\le 5$

Sketch and interpret the salary function.

1. Values: $x=0\to 45000$, $1\to 49500$, $2\to 54000$, $3\to 58500$, $4\to 63000$, $5\to 67500$.
2. A straight line starting at Rs.$45000$ (starting salary) rising by Rs.$4500$ for each year of service.

Answer: Straight line from $(0,45000)$ to $(5,67500)$, slope $4500$/year

Q5 — Break-even for $C(x)=1200+20x$, $R(x)=50x$

Find the break-even point.

1. $1200+20x=50x\Rightarrow 1200=30x$.

Answer: $x=40$ bags

Q6 — Profit/loss at $x=250$ bags

Find the profit or loss when 250 bags are sold.

1. $R(250)-C(250)=50(250)-[1200+20(250)]=12500-6200$.

Answer: Profit of Rs.$6300$

Q7 — Plot both graphs and identify the break-even point

Plot $C(x)$ and $R(x)$ and mark where they cross.

1. $C(x)$ starts at $(0,1200)$ with slope $20$; $R(x)$ starts at the origin with slope $50$.
2. They meet where $x=40$: $R(40)=2000$, $C(40)=1200+800=2000$ ✓

Answer: Break-even point at $(40,2000)$

Q8 — Plot $p(x)=10x-70$; find profit for $500$ copies

Plot the graph and find the profit for $500$ newspapers.

1. A straight line with $y$-intercept $-70$ and slope $10$.
2. $p(500)=10(500)-70=5000-70$.

Answer: Profit of Rs.$4930$

Q9 — Plot $C(x)=1500+10x+0.2x^2$; find cost of $200$ shirts

Plot the graph (stated domain $0\le x\le 150$) and evaluate at $x=200$.

1. An upward-curving parabola-like cost curve over the given domain.
2. $C(200)=1500+10(200)+0.2(200{)}^2=1500+2000+0.2(40000)=1500+2000+8000$ (extending the formula beyond the stated domain of $150$).

Answer: Rs.$11500$ (note: $x=200$ lies outside the stated domain $0\le x\le 150$)

Q10 — Plot $S=200000(1-e^{-0.05t})$ for the first $50$ weeks

Plot the sales-growth curve.

1. Key points: $t=0\to 0$, $t=10\to 78700$, $t=20\to 126420$, $t=30\to 155380$, $t=40\to 172940$, $t=50\to 183580$.
2. The curve rises quickly at first, then levels off, approaching $200000$ asymptotically.

Answer: Growth curve rising from $0$ toward (but never quite reaching) $200000$

Q11 — Sales at $t=5$ and $t=35$ weeks

Calculate the number of magazines sold.

1. $S(5)=200000(1-e^{-0.25})=200000(1-0.7788)=200000(0.2212)$.
2. $S(35)=200000(1-e^{-1.75})=200000(1-0.1738)=200000(0.8262)$.

Answer: $t=5$: $\approx 44,240$; $t=35$: $\approx 165,246$

Q12 — Sketch $P_s=Q^2+5$ and $P_d=Q^2-10Q$ for $Q=-20$ to $20$

Sketch each curve.

1. $P_s=Q^2+5$: an upward parabola with vertex $(0,5)$.
2. $P_d=Q^2-10Q=(Q-5{)}^2-25$: an upward parabola with vertex $(5,-25)$.
3. Setting equal: $Q^2+5=Q^2-10Q\Rightarrow 5=-10Q\Rightarrow Q=-0.5$, giving $P=5.25$ — outside the economically meaningful region $Q\ge 0$.

Answer: Curves cross at $Q=-0.5,P=5.25$ (not economically meaningful; for $Q\ge 0$, supply exceeds demand throughout)

Q13 — Break-even, and profit/loss at $100$ LEDs

$C(x)=60000+250x$, $R(x)=1200x$. Find the break-even point and the profit/loss at $x=100$.

1. Break-even: $60000+250x=1200x\Rightarrow 60000=950x\Rightarrow x=\frac{1200}{19}\approx 63.16$.
2. At $x=100$: $R(100)-C(100)=120000-[60000+25000]=120000-85000$.
3. Since $100>63.16$, the company operates above break-even, hence a profit.

Answer: Break-even $\approx 63.16$ LEDs (i.e. at $64$ units in practice); profit of Rs.$35,000$ at $100$ LEDs