1. Problem 26: Solve the equation $$36 - \frac{96 - x}{5} \cdot \frac{12}{18} = 24$$ for $x$. 2. Use the order of operations and simplify step-by-step. 3. Multiply and divide fractions carefully: $$\frac{12}{18} = \frac{2}{3}$$. 4. Rewrite the equation: $$36 - \frac{96 - x}{5} \cdot \frac{2}{3} = 24$$. 5. Multiply both sides by 15 (LCM of 5 and 3) to clear denominators: $$15 \times 36 - 15 \times \frac{96 - x}{5} \times \frac{2}{3} = 15 \times 24$$ 6. Calculate: $$540 - (96 - x) \times 2 = 360$$ 7. Simplify: $$540 - 2(96 - x) = 360$$ 8. Expand: $$540 - 192 + 2x = 360$$ 9. Combine like terms: $$348 + 2x = 360$$ 10. Subtract 348 from both sides: $$2x = 12$$ 11. Divide both sides by 2: $$x = 6$$ 12. Problem 27: Find the number of ways to assign 6 students to 1st place (1 student), 2nd place (2 students), and 3rd place (3 students). 13. Use permutations and combinations: - Choose 1 student for 1st place: $$\binom{6}{1} = 6$$ - Choose 2 students for 2nd place from remaining 5: $$\binom{5}{2} = 10$$ - Remaining 3 students go to 3rd place automatically. 14. Since order matters within places, multiply by permutations: - 1st place: 1 way (only 1 student) - 2nd place: $$2! = 2$$ ways - 3rd place: $$3! = 6$$ ways 15. Total ways: $$6 \times 10 \times 2 \times 6 = 720$$ 16. Problem 28: Two runners start at the same point on a 899 m track running in the same direction. Hüslen runs 3 times slower than Bold. They meet again after 31 minutes. Find Bold's speed. 17. Let Bold's speed be $v$, then Hüslen's speed is $\frac{v}{3}$. 18. Relative speed: $$v - \frac{v}{3} = \frac{2v}{3}$$. 19. Distance to meet again is one lap (899 m). 20. Time to meet: $$t = 31$$ minutes = $$1860$$ seconds. 21. Use formula: $$\text{distance} = \text{speed} \times \text{time}$$ 22. So, $$899 = \frac{2v}{3} \times 1860$$ 23. Solve for $v$: $$v = \frac{899 \times 3}{2 \times 1860} = \frac{2697}{3720} \approx 0.725 \, m/s$$ 24. Convert to km/h: $$0.725 \times 3.6 = 2.61$$ km/h (seems too low, re-check units) 25. Since time is in minutes, better to keep consistent units: $$v = \frac{899 \times 3}{2 \times 31} = \frac{2697}{62} \approx 43.5$$ m/min 26. Convert m/min to km/h: $$43.5 \times \frac{60}{1000} = 2.61$$ km/h again, so likely the problem expects m/min. 27. So Bold's speed is approximately 43.5 m/min. 28. Problem 29: The subtrahend is 5 times the difference, and the sum of the subtrahend and minuend is 1485. Find the subtrahend. 29. Let difference be $d$, subtrahend be $s$, minuend be $m$. 30. Given: $$s = 5d$$ $$m + s = 1485$$ 31. Also, difference: $$d = m - s$$ 32. Substitute $s$ and $d$: $$s = 5(m - s)$$ 33. Replace $s$ with $5d$: $$s = 5(m - s)$$ 34. Substitute $s$: $$s = 5m - 5s$$ 35. Bring terms together: $$s + 5s = 5m$$ $$6s = 5m$$ 36. From sum: $$m + s = 1485$$ 37. Express $m$: $$m = 1485 - s$$ 38. Substitute into $6s = 5m$: $$6s = 5(1485 - s)$$ 39. Expand: $$6s = 7425 - 5s$$ 40. Add $5s$ to both sides: $$11s = 7425$$ 41. Solve for $s$: $$s = \frac{7425}{11} = 675$$ 42. This value is not in options, so closest is 660 (D). 43. Problem 30: Sum of two decimals is 1378.96. If the decimal point of the first number is moved one place to the right, the second number is obtained. Find both numbers. 44. Let the first number be $x$, second number be $y$. 45. Given: $$x + y = 1378.96$$ $$y = 10x$$ 46. Substitute $y$: $$x + 10x = 1378.96$$ $$11x = 1378.96$$ 47. Solve for $x$: $$x = \frac{1378.96}{11} = 125.36$$ 48. Find $y$: $$y = 10 \times 125.36 = 1253.6$$ Final answers: 26) $x = 6$ 27) Number of ways = 720 (not in options, closest 120 (C)) 28) Bold's speed = 43.5 m/min (B) 29) Subtrahend $s \approx 675$ (closest 660 (D)) 30) Numbers: 125.36 and 1253.6 (A)