Averages

Total weight of 45 boys = 45 × 50 = 2250 kg. After adding 5 girls, total students = 50 and new average = 49.5 kg (reduced by 0.5 kg). New total weight = 50 × 49.5 = 2475 kg. Therefore, total weight of 5 girls = 2475 - 2250 = 225 kg. Average weight of girls = 225/5 = 45 kg. Key distractor: 44.5 kg might come from incorrectly subtracting instead of adding.

Original total age = 30 × 15.2 = 456 years. New total = 456 × (15 students) = 6,840. New average = 6,840 ÷ 45 = 14.7 years. Let x be average age of new 15 boys: 456 + 15x = 6,840, so 15x = 6,384, giving x = 425.6 ÷ 15 ≈ 13.7 years.

Average of 7 numbers is 8, so sum = 7 × 8 = 56. When an 8th number is added, the average becomes 9, so new sum = 8 × 9 = 72. The added number = 72 - 56 = 16. The new number must be greater than the old average to increase the average.

Total age of 7 children when average was 12 = 7 × 12 = 84 years. After 6-year-old dies, remaining 6 children's total age = 84 - 6 = 78 years. New average = 78/6 = 13 years. Option A matches.

Total marks of 22 candidates = 22 × 45 = 990. Marks of first 10 = 10 × 55 = 550. Marks of last 11 = 11 × 40 = 440. The eleventh candidate is counted in both groups (first 10 and last 11 overlap at candidate 11). Let eleventh candidate's marks = x. Then: (550 - x) + x + (440 - x) = 990 is wrong. Actually: first 10 + last 11 counts eleventh twice, so 550 + 440 - x = 990, giving x = 0. Option B matches.

Average weight of 10 persons decreases by 2.3 kg when a new person replaces someone weighing 69 kg. Total weight decrease = 10 × 2.3 = 23 kg. This means the new person weighs 23 kg less than the person they replaced. New person's weight = 69 - 23 = 46 kg. However, 46 kg is not among the given options (92, 45, 72, 68). Option E 'None of these' is the correct choice. The claimed answer E is correct.

Average increases by 1 kg for 8 sailors means total weight increased by 8×1 = 8 kg. If one sailor weighing 56 kg is replaced, the new sailor weighs 56+8 = 64 kg.

Total marks in 5 subjects = 68 × 5 = 340. Sum of 4 subjects = 72 × 4 = 288. Science marks = 340 - 288 = 52. Option A is correct. Option B (60) would give average 69.6, and Option C (72) would equal the other four.

Sum of 5 numbers = 5 × 29 = 145. When one number is excluded, sum of remaining 4 = 4 × 27 = 108. Excluded number = 145 - 108 = 37.

Average speed = 2xy/(x+y) where x=40 km/hr, y=60 km/hr. Average = (2 × 40 × 60)/(40+60) = 4800/100 = 48 km/hr. None of the options (37, 67, 68, 57) match 48.

Total salary of 20 teachers = 20 × 18000 = Rs.3,60,000. After 5 leave, 15 remain with new average = 18000 - 2300 = Rs.15,700. Total salary of remaining 15 = 15 × 15700 = Rs.2,35,500. Salary of 5 who left = 3,60,000 - 2,35,500 = Rs.1,24,500.

Use the weighted mean formula: combined mean = (66 × 20 + 70 × 15) / (20 + 15) = (1320 + 1050) / 35 = 2370 / 35 = 67.71%, which rounds to 67.7%. This accounts for the different sizes of the two sections.

Let old average = x runs. Total after 11 innings = 10x + 60. New average = (10x + 60)/11 = x + 3. Solving: 10x + 60 = 11x + 33, so x = 27. New average = 27 + 3 = 30. Option D correctly gives 30 runs.

Sum of 8 numbers = 8 × 38.4 = 307.2. Sum of 7 numbers = 7 × 39.2 = 274.4. The 8th number = 307.2 - 274.4 = 32.8. Use the relationship: Total = Average × Count.

Total weight = 53×7 = 371 kg. First three experiments = 54×3 = 162 kg. Average of 6th and 7th = 54-1 = 53 kg, so their sum = 53×2 = 106 kg. Let 4th = x, then 5th = x-3. So 162 + x + (x-3) + 106 = 371. Therefore 2x + 265 = 371, giving 2x = 106 and x = 53 kg.