The average of nine numbers is 50. The average of the first five numbers is 54 and that of the last three numbers is 52. Then the sixth number is

Average of all prime numbers between 60 and 90
= (61 + 67 + 71 + 73 + 79 + 83 + 89)/7
= 523/7 = 74.7

Required average = ( Sum of scores) / (No of scores)
= (124 + 856 + 331 + 227 + 963 + 338 + 259 + 662)/8
= 3760/8 = 470

Required average = (Sum of all scores) / (Number of scores)
= (235 + 124 + 255 + 534 + 836 + 375 + 101 + 433 + 760)/9
= 3681/9= 409

Sum of 7 numbers = sum of first 3 numbers + N₄ + sum of last 3 numbers.
(23 x 3) + N₄ + (42 x 3) = 235
? N₄ = 40
? 4_th number = 40

Let the numbers be 2y, y and 4y
? Average = (2y + y + 4y)/3
? 7y/3 = 56
? y = (3 x 56) / 7 = 24
Hence , the number in order are 48, 24 and 96

Total of 6 observation = 6 x 12 = 72
Total of 7 observation (including the 7th) = 7 x (12 - 1) = 77
? 7th observation = 77 - 72 = 5

Here initial average = 7
As we know that, if all the numbers are multiple by a certain number, then their average must be a multiple of that number.
? New average = 7 x 12 = 84

Total correct weight = (40 x 60) - 36 + 33 = 2400 - 3 = 2397 kg
? Required average weight = 2397/60 = 39.95 kg

Let the average of 14th innings = x
? Average of 15th innings = x+2
Then, 14x+72 = 15(x+2)
? 14x+72 = 15x +30
? x = 42
? Average for 15th innings = x+2 = 42 + 2 = 44

Total of 5 observations = 5 x 15 = 75

New sum = 75 + 16.5 + 18 + 14.5 = 124

New Average = 124 ? 8 = 15.5