1. **Problem 20:** Given height equation $$h = -16t^2 + 96t$$ , find $$t$$ when $$h = 144$$ feet. 2. Substitute $$h = 144$$: $$144 = -16t^2 + 96t$$ 3. Rearrange: $$-16t^2 + 96t - 144 = 0$$ 4. Divide entire equation by -16 to simplify: $$t^2 - 6t + 9 = 0$$ 5. Recognize this is a perfect square: $$(t - 3)^2 = 0$$ 6. Solve for $$t$$: $$t = 3$$ seconds. --- 1. **Problem 21:** Cost function $$C = 0.5x^2 - 16x + 466$$. Find $$x$$ such that $$C = 338$$. 2. Set up the equation: $$0.5x^2 -16x + 466 = 338$$ 3. Rearrange: $$0.5x^2 -16x + 128 = 0$$ 4. Multiply by 2 to clear decimal: $$x^2 - 32x + 256 = 0$$ 5. Factor: $$(x - 16)^2 = 0$$ 6. Solution: $$x = 16$$ items. --- 1. **Problem 22:** Total investment $$8000$$; two enterprises pay $$5.5\%$$ and $$5\%$$ interest; total income $$425$$. Find amount invested at $$5\%$$. 2. Let $$x$$ = amount invested at $$5.5\%$$; then $$8000 - x$$ at $$5\%$$. 3. Total income equation: $$0.055x + 0.05(8000 - x) = 425$$ 4. Simplify: $$0.055x + 400 - 0.05x = 425$$ 5. Combine like terms: $$0.005x = 25$$ 6. Solve for $$x$$: $$x = 5000$$ invested at $$5.5\%$$. 7. Amount at $$5\%$$: $$8000 - 5000 = 3000$$. --- 1. **Problem 23:** Total $$20000$$ invested: part at 6%, rest at 7%, total interest rate $$6.75\%$$. Find amount invested at 7%. 2. Let $$x$$ = amount at 7%, then $$20000 - x$$ at 6%. 3. Interest total: $$0.06(20000 - x) + 0.07x = 0.0675 \times 20000$$ 4. Calculate right side: $$0.0675 \times 20000 = 1350$$ 5. Expand left: $$1200 - 0.06x + 0.07x = 1350$$ 6. Simplify: $$1200 + 0.01x = 1350$$ 7. Solve for $$x$$: $$0.01x = 150$$ $$x = 15000$$. --- 1. **Problem 24:** Set up system of equations: - Total output: $$x + y + z =45$$ - Twice first line equals sum of other two: $$2x = y + z$$ or $$2x - y - z = 0$$ - Second line 4 more than third: $$y = z + 4$$ or $$y - z = 4$$ 2. Comparing options, correct system is: $$\begin{cases} x + y + z = 45 \\ 2x - y - z = 0 \\ y - z = 4 \end{cases}$$ --- 1. **Problem 25:** Solve $$x^2 + ax = bx + ab$$. 2. Rearrange: $$x^2 + ax - bx - ab = 0$$ 3. Factor: $$x^2 + (a - b)x - ab = 0$$ 4. Factor quadratic: $$(x - b)(x + a) = 0$$ 5. Solutions: $$x = b\text{ or } x = -a$$ --- 1. **Problem 26:** Blend 3 coffees @Rs 65, 70, 75 per gram. Want 100 grams at Rs 71 per gram. Use same amount of two higher priced coffees. 2. Let $$x$$ = grams of lowest priced (Rs 65), then $$y$$ = grams each of Rs 70 and Rs 75 because amounts equal. 3. Total grams: $$x + 2y = 100$$ 4. Total value: $$65x + 70y + 75y = 71 \times 100 = 7100$$ 5. Simplify value equation: $$65x + 145y = 7100$$ 6. From grams: $$2y = 100 - x \Rightarrow y = \frac{100 - x}{2}$$ 7. Substitute into value: $$65x + 145 \times \frac{100 - x}{2} = 7100$$ 8. Multiply both sides: $$65x + 72.5(100 - x) = 7100$$ 9. Expand: $$65x + 7250 - 72.5x = 7100$$ 10. Combine like terms: $$-7.5x + 7250 = 7100$$ 11. Subtract 7250: $$-7.5x = -150$$ 12. Solve for $$x$$: $$x = 20$$ grams. --- 1. **Problem 27:** Total investment $$T$$. $$\frac{3}{10}T + 600$$ invested in venture 1, returns Rs 384. Total return Rs 1120. 2. Let $$T$$ = total invested. 3. Return percentage is same for both ventures; call it $$r$$. 4. Income venture 1: $$r \times \left( \frac{3}{10}T + 600 \right) = 384$$ 5. Income venture 2: $$r \times \left(T - \left( \frac{3}{10}T + 600 \right) \right) = 1120 - 384 = 736$$ 6. Sum of investments: $$T$$, so venture 2 investment: $$T - \frac{3}{10}T - 600 = \frac{7}{10}T - 600$$ 7. From (4), solve for $$r$$: $$r = \frac{384}{\frac{3}{10}T + 600}$$ 8. Substitute into (5): $$\frac{384}{\frac{3}{10}T + 600} \times \left( \frac{7}{10}T - 600 \right) = 736$$ 9. Multiply both sides by denominator: $$384 \times \left( \frac{7}{10}T - 600 \right) = 736 \times \left( \frac{3}{10}T + 600 \right)$$ 10. Expand: $$268.8T - 230400 = 220.8T + 441600$$ 11. Rearrange: $$268.8T - 220.8T = 441600 + 230400$$ 12. Simplify: $$48T = 672000$$ 13. Solve for $$T$$: $$T = 14000$$. 14. Total investment is Rs 14000.