•   As per the question B = F.

•   Average weight of B, D, E is 42. Replace B by F here.

•   We have, average weight of F, D and E is 42. (F+D+E/3)

•   Again, It is given that average weight of A, B, C is 40.

•   A+B+C = 120 and  D +E +F = 126.

•   Total of A to F (A+B+C+D+E+F) = 120 + 126 = 246.

•   So, average = 246/6 = 41.

Alternately you can do it by allegation method.

•   Average = Sum of participants/ number of participants.

•   Since the average of students of Class-A is lower than Class-B, when the 4 students will move they will reduce the average of the Class-B as the number of participant will increase but sum will not increase by much. Thus statement 1 is correct.

•   If the 4 student moved to other class were of lower range of score, then the average of the Class-A will increase. If the one’s that move will be of higher range, then the average will reduce. Thus, we cannot be sure of the average of Class-A.

Let the number of girls be a and number of boys be b.

Overall average marks in English = Total marks in English / Total Students = (9a + 8b) / (a + b) = 8.8

or 9a + 8b = 8.8a + 8.8b

or 0.2a = 0.8b

or a = 4b …...(1)

Similarly, Overall average marks in Hindi = Total marks in Hindi / Total Students = (8a + 7b) / (a + b) = x

Putting value of g from (1), we get:

(8 × 4b + 7b) / (4b + b) = x

or 39b/5b = x

or x = 7.8

So, overall average marks in Hindi is 7.8.

Average weight of the students of the class = Total weight of students of the class / Total number of students in the class

As no student left the class, nor any new student joined, so both the total weight and number of students remain the same.

•  Average score after 50 innings = 46.4

So, total score after 50 innings = 50 × 46.4 = 2320

Average score after 60 innings = 46.4 + 2.6 = 49

So, total score after 60 innings = 60 × 49 = 2940

Hence, Average score of 10 innings = Runs scored in 10 innings/10 = (2940 – 2320)/10 = 620/10 = 62

•  Given, average marks of 100 students = 40.

•  So, total marks of 100 students = 40 × 100 = 4000.

•  But in this total the error is 83 – 53 = 30, more than the actual total.

•  So, Actual total = 4000 – 30 = 3970.

•  So the correct Mean = 3970/100 = 39.70.

•  Average of weight of two children and their mother (i.e. total 3 members) = 50.

•  So, the sum of the weight of two children and mother = C1 + C2 + M = 50 × 3 = 150 ... (1) (where C1 and C2 are the weights of two children and M is the weight of the Mother)

•  Again if the weight of the Father is F, we have C1 + C2 + F = 52 × 3 = 156 ... (2)

•  Now it is given that the weight of the father = F = 60.

•  By putting this value in equation (2), we have C1 + C2 = 156 - 60 = 96.

•  Again by using equation (1) we have M = 150 - 96 = 54 kg.

•  The only three single digit odd prime numbers are 3, 5 and 7. So the length of the first piece = (3+ 5 + 7)/3 = 5. Now let the length of the second piece = X units.

•  Then by the condition given in question we have – Length of the third piece = X + 5. Length of the second piece = X = 5 + (X + 5)/3.

•  Solve this to get X = 10. So, length of the first piece = 5, second piece = 10 and third piece = 15. So length of the original sheet = 5 + 10 + 15 = 30 units.

•   The average weight of 9 persons is 50 kg. So, total weight of all the persons = 50 × 9 = 450 kg

•   The average weight of the first 5 persons is 45 kg. So, their total weight = 45 × 5 = 225 kg

•   The average weight of the last 5 persons is 55 kg. So, their total weight = 55 × 5 = 275 kg

•   The fifth person has been included in both the last cases.

•   Hence, the weight of the fifth person = (275 + 225) – 450 = 50 kg

•   The average rainfall in six days = 0.5 inch.

•   Hence, total rainfall in six days = 0.5 × 6 = 3 inch

•   The average rainfall in a city for the first four days = 0.4 inch

•   So, total rainfall in these four days = 0.4 × 4 = 1.6 inch

•   Hence, rainfall in the last two days = 3 – 1.6 = 1.4 inch

•   Now, it’s given that the rainfall on the last two days was in the ratio of 4:3.

•   Hence, rainfall on the fifth day = 1.4 × [4/(4 + 3)] = 1.4 × (4/7) = 0.8 inch

•   Let the total number of male employees be M & that of female be F.

•   Total salary (males) = Rs 5200 x M

•   Total salary (females) = Rs 4200 x F

•   So, we have 5200M + 4200F = 5000 (M+F)

•   Solving, we get 200M = 800F

•   Hence, M = 4F.

•   Percentage of males employed = 4F/5F *100 = 80 %.

•   So answer is (b).

•   Average monthly income = Rs.10,000

•   So, total monthly income of the family of 5 = Rs.50,000

•   If one person's income increases by Rs.1,20,000 per year, his/her monthly income increases by Rs.10,000.

•   So, total monthly income now becomes = Rs.60,000

•   So, average monthly income now becomes 60/5 = Rs.12,000.

•   Hence, answer is (a).

1. Correct Answer: Option B: Test II
 2. Explanation:
     To determine in which test the student's performance is the best comparatively, we need to consider how the score of 74 compares to the average and the range of marks for each test.
     ● Test I:  
           ○ *Average Marks*: 60
           ○ *Range of Marks*: 30 to 90
           ○ *Comparison*: The score of 74 is 14 marks above the average (74 - 60 = 14).
     ● Test II:  
           ○ *Average Marks*: 60
           ○ *Range of Marks*: 45 to 75
           ○ *Comparison*: The score of 74 is 14 marks above the average (74 - 60 = 14). However, since the maximum score is 75, scoring 74 is very close to the top of the range.
     ● Test III:  
           ○ *Average Marks*: 70
           ○ *Range of Marks*: 20 to 100
           ○ *Comparison*: The score of 74 is 4 marks above the average (74 - 70 = 4).
     ● Test IV:  
           ○ *Average Marks*: 80
           ○ *Range of Marks*: 0 to 100
           ○ *Comparison*: The score of 74 is 6 marks below the average (74 - 80 = -6).
     Conclusion: The student's score of 74 is closest to the maximum score in Test II and is significantly above the average, making it the best comparative performance.

The average score of student in 3 tests is N points. Hence the total score = 3N points.

Given the score in fourth test = N + 20, the average score of student in four tests will be = (3N +N+20)/4 = N+5