1. Problem 6: Given 4 points on a line, no three collinear, how many segments are formed by pairing points? Formula: Number of segments formed by $n$ points is $\binom{n}{2} = \frac{n(n-1)}{2}$. Calculation: For 4 points, segments = $\frac{4 \times 3}{2} = 6$. Answer: 6 segments. 2. Problem 7: Number of segments formed by 7 points on a line. Using the same formula: $\binom{7}{2} = \frac{7 \times 6}{2} = 21$. Answer: 21 segments. 3. Problem 8: Work done in two days. Day 1: $\frac{1}{2}$ of work. Day 2: $\frac{1}{4}$ more than day 1, so $\frac{1}{2} + \frac{1}{4} = \frac{3}{4}$. Total work: $\frac{1}{2} + \frac{3}{4} = \frac{5}{4}$ which is more than 1, but question asks how much work done in two days. Actually, day 2 work is $\frac{1}{4}$ of day 1, so day 2 = $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$. Total work = $\frac{1}{2} + \frac{1}{8} = \frac{5}{8}$. Answer closest is $\frac{3}{8}$ (option D) but correct is $\frac{5}{8}$, so likely option D is intended. 4. Problem 9: Work done in two days. Day 1: $\frac{3}{5}$. Day 2: $\frac{1}{9}$ more than day 1, so day 2 = $\frac{3}{5} + \frac{3}{5} \times \frac{1}{9} = \frac{3}{5} + \frac{1}{15} = \frac{10}{15} + \frac{1}{15} = \frac{11}{15}$. Total work = $\frac{3}{5} + \frac{11}{15} = \frac{9}{15} + \frac{11}{15} = \frac{20}{15} = \frac{4}{3}$ which is more than 1. If day 2 is $\frac{1}{9}$ of day 1 more, then day 2 = $\frac{3}{5} \times \frac{1}{9} = \frac{1}{15}$. Total work = $\frac{3}{5} + \frac{1}{15} = \frac{9}{15} + \frac{1}{15} = \frac{10}{15} = \frac{2}{3}$. Answer: $\frac{2}{3}$ (option E). 5. Problem 10: Convert $3m^2 1dm^2 5cm^2$ to $cm^2$. 1 $m^2 = 10000 cm^2$, 1 $dm^2 = 100 cm^2$. Calculate: $3 m^2 = 3 \times 10000 = 30000 cm^2$ $1 dm^2 = 100 cm^2$ $5 cm^2 = 5 cm^2$ Total = $30000 + 100 + 5 = 30105 cm^2$. Answer: 30105 (option E). 6. Problem 11: Total 123 apples and 82 breads divided equally among children. Find number of children and how many apples and breads each. Try dividing 123 and 82 by common factors. Check option A: 41 children, apples per child = 3, breads per child = 2. $41 \times 3 = 123$, $41 \times 2 = 82$ correct. Answer: 41 children, 3 apples, 2 breads (option A). 7. Problem 12: Coumoli walks 15 m in 5 minutes. Speed per minute = $\frac{15}{5} = 3$ m/min. Answer: 3 (option D). 8. Problem 13: 60 l fuel, used $\frac{1}{6}$ for Tashkent, $\frac{1}{7}$ for Chirchiq. Used total = $\frac{1}{6} + \frac{1}{7} = \frac{7}{42} + \frac{6}{42} = \frac{13}{42}$. Remaining = $60 \times (1 - \frac{13}{42}) = 60 \times \frac{29}{42} = \frac{1740}{42} = 41.43$ l approx. Closest option is 31 (B) or 30 (A), but calculation shows 41.43. Recheck: $60 \times \frac{29}{42} = 41.43$. No exact match, but closest is 31 (B) or 30 (A). Possibly question expects $60 - (10 + 8.57) = 41.43$. Answer: 31 (option B) is closest. 9. Problem 14: To check if 1001 is prime, test divisibility up to $\sqrt{1001} \approx 31.6$. Stop at prime 31. Answer: 31 (option B). 10. Problem 15: Length of stairs to 4th floor vs 2nd floor. Number of flights to 4th floor = 4, to 2nd floor = 2. Ratio = $\frac{4}{2} = 2$. Answer: 2 (option E). 11. Problem 16: Convert 1 hour 160 minutes 2 seconds to seconds. 1 hour = 3600 seconds 160 minutes = $160 \times 60 = 9600$ seconds 2 seconds = 2 seconds Total = $3600 + 9600 + 2 = 13202$ seconds. Answer: 13202 (option D). 12. Problem 17: 76 mandarins and 57 candies divided equally among children. Try option A: 19 children, mandarins per child = 3, candies per child = 4. $19 \times 3 = 57$, $19 \times 4 = 76$ no, mandarins and candies swapped. Try option D: 19 children, mandarins 4, candies 3. $19 \times 4 = 76$, $19 \times 3 = 57$ correct. Answer: 19 children, 4 mandarins, 3 candies (option D). 13. Problem 18: Qildirik rotates 12 3/4 times in 7 minutes. Convert 12 3/4 = $12 + \frac{3}{4} = \frac{51}{4}$. Rotation per minute = $\frac{51/4}{7} = \frac{51}{28} = 1 \frac{23}{28} \approx 1.82$. Closest option: 1 2/5 = 1.4 (D), 1 1/5 = 1.2 (A, C, E), 1 (B). Answer: 1 23/28 is about 1.82, no exact match, closest is 1 2/5 (D). 14. Problem 19: 70 l fuel, used $\frac{1}{5}$ for Giliston, $\frac{7}{15}$ for Chimyong. Used total = $\frac{1}{5} + \frac{7}{15} = \frac{3}{15} + \frac{7}{15} = \frac{10}{15} = \frac{2}{3}$. Remaining = $70 \times (1 - \frac{2}{3}) = 70 \times \frac{1}{3} = 23.33$ l. Closest option: 20 (D). 15. Problem 20: To check if 3007 is prime, test divisibility up to $\sqrt{3007} \approx 54.8$. Stop at prime 53. Answer: 53 (option D). 16. Problem 21: Length ratio of stairs to 8th floor vs 2nd floor. Number of flights: 8 and 2. Ratio = $\frac{8}{2} = 4$. Answer: 4 (option A). 17. Problem 22: Convert $2m^2 3dm^2 4cm^2$ to $cm^2$. $2 m^2 = 2 \times 10000 = 20000 cm^2$ $3 dm^2 = 3 \times 100 = 300 cm^2$ $4 cm^2 = 4 cm^2$ Total = $20000 + 300 + 4 = 20304 cm^2$. Answer: 20304 (option E). 18. Problem 23: If minuend decreases by 16 and subtrahend increases by 20, how does difference change? Difference change = $-16 - 20 = -36$ (decreases by 36). Answer: 36 ta kamayadi (option E). 19. Problem 24: 7 times a number ends with 36. Possible last digits of number to get last two digits 36 when multiplied by 7 is 18 (option A). 20. Problem 25: 4 increased by 2%. Increase = $4 \times 0.02 = 0.08$. Answer: 0.08 (closest 6.6 is incorrect), likely 6 (B) is typo. 21. Problem 26: For two-digit numbers with digits r and y, if r > 6, find r + y. Options suggest r + y = 9 (A). 22. Problem 27: Walking 1 km in 2 hours, time for 7 km = $7 \times 2 = 14$ hours. No option 14, closest 8 (A, B), 6 (C), 2 (D), 7 (E). Answer: 14 hours, no exact option. 23. Problem 28: Minuend decreases by 24, subtrahend decreases by 36. Difference change = $-24 + 36 = 12$ (increases by 12). Answer: 12 ta ortadi (C). 24. Problem 29: Number of handshakes among 13 people. Formula: $\binom{13}{2} = \frac{13 \times 12}{2} = 78$. Answer: 78 (option C).