If the average of 7 numbers is 16, then the sum of numbers must be,  112. If average of first 3 numbers is 14, then the sum of numbers must be 42. And if the average of the next 3 numbers is 18, the sum of numbers must be 54. Now, 112 - 42 - 54 = 16. Hence, the last number is 16.

 Suppose the 3 numbers are A, B and C, so A + B = 2 x 4 = 8 ----------------- (1) B + C = 2 x 5 = 10 --------------- (2) A + C = 2 x 6 = 12 --------------- (3) On adding (1), (2) and (3), we get A + B + C = 15. Therefore, average = 15/3 = 5 

 Let the total number of employees in the factory be 100. Number of workers = 60 and number of executives = 40 We know that Average = Sum of all observations/Number of observations Therefore, sum of salaries of all the employees = 60 x Rs. 400 + 40 x Rs. 420 = Rs. 40800 The average salary of all the employees = Rs. 40800/100 = Rs. 408

Total sum of 43 numbers = (45 × 32) – (30 + 45) = 1440 – 75 = 1365 Average = 1365/43 = 31.7

Sum of remaining 5 numbers = 8 x 65 - (75 + 65 + 50) = 330 The average of remaining 5 numbers = 330/5 = 66

Total weight of (33 + 47) students = (33 x 42 + 47 x 32) kg = 2890 kg Therefore, thev average weight of the whole class = (2890/80) = 36.125 kg.

Fourth number = (5 x 6 + 5 x 6 – 5 x 9) = (30 + 30 - 45) = 15 Incorrect option

 Total score of students = 4 x 80 + 10 x 70 + 4 x 75 + 2 x 50 = 1420 Required average = 1420/20 = 71

Average speed for the first 240 km = 2*60*75/(60+75) = 66.67 km/hr

Sum of the ages of 6 members = 45 x 6 = 270 years Sum of the ages of the remaining 5 members = 270 – 30 = 240 years New sum of the ages of 6 members = 240 + 60 = 300 years Average age of new committee = 300/6 = 50 years

The weight of the new man = weight of the man who is replaced + 1.5 x 9 = 54 + 13. 5 = 67.5 kg.

 Let the number of passed candidates be x.Then, total marks = 125 x 40 = 40x + (125 – x) x 20 = 5000 = 40x + 2500 – 20x = 20x = 2500Therefore, x = 125

Total ages of 40 boys = 15 x 40 = 600 years. Total ages when class teacher is included = 16 x 41 = 656 years Age of class teacher = 656 – 600 = 56 years.

The total of 9 results = 9 x 50 = 450 The total of first 5 results = 5 x 49 = 245 The total of last 5 results = 5 x 52 = 260 The 5th result is common to both; 5th result = 245 + 260 - 450 = 55.

Let his original average of 17 innings = x So, the some of his runs of 17 innings = 17x Now his new average = x + 2 So, the new sum of his runs of 18 innings = 18(x + 2) Therefore difference of his 17 and 18 innings = 95 So, 18(x + 2) - 17x = 95 x = 59 So, his original average is 59 and his average after 18th innings = 59 + 2 => 61 Therefore, 61 is correct answer.