The total of the 10 numbers is 10 * 50 = 500. If one number is increased by 10, the new total becomes 500 + 10 = 510. The new average is 510 / 10 = 51.

The sum of the 10 numbers is 10 * 50 = 500. Replacing one number with 70 gives a new sum of 500 - x + 70. The new average is (500 - x + 70) / 10. Since x is 50, the new average is (500 - 50 + 70) / 10 = 52.

Let the integers be x, x+1, x+2, x+3. Average = (4x + 6) / 4 = 25. Solving gives x = 24. Largest integer = x + 3 = 27.

Let the integers be x, x+1, x+2, x+3. Average = (4x + 6) / 4 = 25. So, 4x + 6 = 100. Therefore, x = 23.

Total of 4 numbers = 4 * 30 = 120. Total of other three numbers = 120 - 10 = 110. Average of other three = 110 / 3 = 36.67, approximately 36.

Total = 30 * 4 = 120. New total = 120 + 10 = 130. New average = 130 / 4 = 32.5.

Total of 5 numbers = 5 * 12 = 60. Total of 4 numbers = 4 * 10 = 40. Removed number = 60 - 40 = 20.

The total of the 5 numbers is 5 * 40 = 200. The sum of the other four numbers is 200 - 60 = 140. The average of the other four numbers is 140 / 4 = 35.

Average = (5 + 10 + 15 + 20 + x) / 5 = 12. Therefore, 50 + x = 60, so x = 10.

The total of the 8 numbers is 8 * 12 = 96. The total of the remaining 7 numbers is 7 * 10 = 70. Therefore, the removed number is 96 - 70 = 26.

Total of 8 numbers = 8 * 40 = 320. Total of 7 numbers = 7 * 38 = 266. Removed number = 320 - 266 = 54.

Total of 8 numbers = 8 * 50 = 400. New total = 400 + 10 = 410. New average = 410 / 8 = 51.25, approximately 51.

Total of five numbers = 5 * 20 = 100. Total of four numbers = 4 * 18 = 72. Excluded number = 100 - 72 = 28.

Let the integers be x, x+1, x+2. Average = (x + (x+1) + (x+2)) / 3 = 20. Solving gives x = 19, largest integer = 21.

Total of three numbers = 3 * 15 = 45. Sum of other two numbers = 45 - 12 = 33. Average = 33 / 2 = 16.5.

Total of three numbers = 20 * 3 = 60. The sum of the other two = 60 - 10 = 50. Average = 50 / 2 = 25.

Total of three numbers = 3 * 30 = 90. Sum of other two numbers = 90 - 20 = 70. Average = 70 / 2 = 35.

Total of three numbers = 3 * 30 = 90. New total = 90 + 10 = 100. New average = 100 / 3 = 33.33.

Let the three numbers be a, b, and c. We have (a + b + c) / 3 = 30. Thus, a + b + c = 90. If c = 10, then a + b = 90 - 10 = 80. The average of a and b is 80 / 2 = 40.

Let the other number be x. (30 + x) / 2 = 40. So, 30 + x = 80. Therefore, x = 50.

Total score = 30 * 75 = 2250. New total = 2250 - 90 = 2160. New average = 2160 / 29 = 74.55.

The total score in 6 subjects is 6 * 75 = 450. Adding the score of 90 gives a new total of 450 + 90 = 540. The new average is 540 / 7 = 77.14, which rounds to 77.

Total score in 5 matches = 5 * 75 = 375. New total = 375 + 90 = 465. New average = 465 / 6 = 77.5.

Let P's age be x. Then O's age is x + 10. So, x + (x + 10) = 80. Solving gives x = 35, hence O = 45.

If the grandson is 5 years old, then the grandfather is 4 * 5 = 20 years old.

If the ratio is 3:2, then for every 3 boys, there are 2 girls. Therefore, if there are 30 boys, the number of girls = (2/3) * 30 = 20

If G is 30 years old, then H's age is (2/3) * 30 = 20 years.

If H is 32 years, then G's age = (3/4) * 32 = 24 years.

Let their ages be 2x and 3x. Then, 2x + 3x = 50, 5x = 50, x = 10. Older friend's age = 3x = 3*10 = 30.

Let the sides be 3x and 4x. Then, 3x + 4x = 70, 7x = 70, x = 10. Shorter side = 3x = 3*10 = 30.