1. Problem 48: Given $$S=\frac{2x+t}{r}$$, solve for $$x$$. Multiply both sides by $$r$$ to clear the denominator: $$rS=2x+t$$ Subtract $$t$$ from both sides: $$rS - t = 2x$$ Divide both sides by 2: $$x=\frac{rS - t}{2}$$ Answer: (b) $\frac{rS - t}{2}$. 2. Problem 49: Noman and Murtuza earn 8 per hour each. Noman worked 3 hours more than Murtuza. Total earnings: 120. Let Murtuza's hours be $$h$$, then Noman's hours are $$h + 3$$. Total money: $$8(h) + 8(h + 3) = 120$$ Simplify: $$8h + 8h + 24 = 120$$ $$16h + 24 = 120$$ Subtract 24: $$16h = 96$$ Divide by 16: $$h = 6$$ Noman's hours: $$6 + 3 = 9$$ Answer: (b) 9. 3. Problem 50: Width = $$2x - 3$$, Area = 43. Area = length × width: $$x(2x - 3) = 43$$ Answer: (d) $$x(2x - 3) = 43$$. 4. Problem 51: Equation: $$x^2 + 6x -7 = 0$$ Complete the square: $$x^2 + 6x = 7$$ Add $$9$$ (since $$(6/2)^2 = 9$$) to both sides: $$x^2 + 6x + 9 = 7 + 9$$ $$(x + 3)^2 = 16$$ Answer: (d) $$(x + 3)^2 = 16$$. 5. Problem 52: $$f(x) = \frac{\sqrt{2x + 3}}{6x - 5}$$, find $$f(\frac{1}{2})$$. Calculate numerator inside square root: $$2(\frac{1}{2}) + 3 = 1 + 3 = 4$$ $$\sqrt{4} = 2$$ Calculate denominator: $$6(\frac{1}{2}) - 5 = 3 - 5 = -2$$ So: $$f(\frac{1}{2}) = \frac{2}{-2} = -1$$ Answer: (c) -1. 6. Problem 53: Find $$x$$ such that $$f(x) = g(x)$$, Given $$f(x) = x^2 - 2x -8$$ and $$g(x) = \frac{1}{4}x - 1$$. Set equal: $$x^2 - 2x - 8 = \frac{1}{4}x - 1$$ Multiply both sides by 4: $$4x^2 - 8x - 32 = x - 4$$ Bring all terms to one side: $$4x^2 - 8x - 32 - x + 4 = 0$$ $$4x^2 - 9x - 28 = 0$$ Use quadratic formula: $$x = \frac{9 \pm \sqrt{(-9)^2 - 4 \times 4 \times (-28)}}{2 \times 4} = \frac{9 \pm \sqrt{81 + 448}}{8} = \frac{9 \pm \sqrt{529}}{8}$$ $$\sqrt{529} = 23$$ So, $$x = \frac{9 \pm 23}{8}$$ Two roots: $$x = \frac{9 + 23}{8} = 4$$ $$x = \frac{9 - 23}{8} = -\frac{14}{8} = -1.75$$ Answer: (b) -1.75 and 4. 7. Problem 54: $$f(n) = (n - 1)^2 + 3n$$. Check each choice: (a) $$f(-2) = (-2 -1)^2 + 3(-2) = (-3)^2 - 6 = 9 -6 = 3 \neq -15$$ (b) $$f(-15) = (-15 -1)^2 + 3(-15) = (-16)^2 -45 = 256 - 45 = 211 \neq -2$$ (c) $$f(-2) = 3$$ (already found in (a)) which is true. (d) $$f(3) = (3 -1)^2 + 3(3) = (2)^2 + 9 = 4 + 9 = 13 \neq -2$$ Answer: (c) $$f(-2) = 3$$. 8. Problem 55: Cost price = 3.40, profit = 20% on selling price. Let selling price be $$x$$. Profit = Selling Price - Cost Price = $$x - 3.40$$ Profit is 20% of selling price: $$x - 3.40 = 0.20 x$$ $$x - 0.20 x = 3.40$$ $$0.80 x = 3.40$$ $$x = \frac{3.40}{0.80} = 4.25$$ Answer: (b) Rs 4.25. 9. Problem 56: Nadia's sister age = $$2N - 7$$, sum of ages = 41. Equation: $$N + (2N - 7) = 41$$ Answer: (b) $$N + (2N - 7) = 41$$.