1. **Problem 1: Investment growth with compound and simple interest** Mr. Praveen invests 5000 at 4% annual interest for 5 years. **Formula for compound interest:** $$A = P\left(1 + \frac{r}{n}\right)^{nt}$$ where $A$ is the amount, $P$ is principal, $r$ is annual interest rate (decimal), $n$ is number of times interest applied per year, $t$ is years. Since compounded annually, $n=1$. **Formula for simple interest:** $$A = P(1 + rt)$$ --- 1.1 **Compound interest calculation:** $$A = 5000\left(1 + \frac{0.04}{1}\right)^{1 \times 5} = 5000(1.04)^5$$ Calculate $(1.04)^5$: $$1.04^5 = 1.2166529$$ So, $$A = 5000 \times 1.2166529 = 6083.2645$$ Rounded up: $$A = 6084$$ 1.2 **Simple interest calculation:** $$A = 5000(1 + 0.04 \times 5) = 5000(1 + 0.20) = 5000 \times 1.20 = 6000$$ 2. **Problem 2: Present value and investment for future purchase** Future cost of car = 28000 in 2 years, interest rate 18% compounded annually. **Formula for present value factor:** $$PVF = \frac{1}{(1 + r)^t}$$ 2.1 Calculate present value factor for 2 years: $$PVF = \frac{1}{(1 + 0.18)^2} = \frac{1}{1.18^2} = \frac{1}{1.3924} = 0.7182$$ 2.2 Amount to invest today: $$PV = FV \times PVF = 28000 \times 0.7182 = 20109.6$$ Rounded up: $$PV = 20110$$ 3. **Problem 3: Subscription and linear equation** 3.1 Nancy's subscription cost: Monthly fee = 50, setup fee = 120, total budget = 520. Let $m$ = number of months. $$120 + 50m \leq 520$$ Subtract 120: $$50m \leq 400$$ Divide by 50: $$m \leq 8$$ Nancy can afford up to 8 months. 3.2 Solve linear equation: $$x + \frac{1}{3} + x - \frac{1}{3} = \frac{5}{2}$$ Combine like terms: $$x + x + \frac{1}{3} - \frac{1}{3} = \frac{5}{2}$$ $$2x + 0 = \frac{5}{2}$$ $$2x = \frac{5}{2}$$ Divide both sides by 2: $$x = \frac{5}{4} = 1.25$$ 4. **Problem 4: Intersection of two lines** Given: $$y = 3x - 2$$ $$y = x + 2$$ At intersection, set equal: $$3x - 2 = x + 2$$ Subtract $x$ from both sides: $$2x - 2 = 2$$ Add 2: $$2x = 4$$ Divide by 2: $$x = 2$$ Substitute $x=2$ into $y = x + 2$: $$y = 2 + 2 = 4$$ Intersection point is $(2, 4)$. Final answers: 1.A) 6084 1.B) 6000 2.A) 0.7182 2.B) 20110 3.A) 8 months 3.B) $x=1.25$ 4) Intersection at $(2,4)$