All questions of Averages for RRB NTPC/ASM/CA/TA Exam

For the 1st 10 overs the score will be 10*3.2=32so the remaining score is 282-32=250remaining overs = 40so the run rate 250/40=6.25ans is 6.25

1. Let the area of the first farm be AAA acres.
2. Therefore, the area of the second farm is A−12A - 12A−12 acres.

• The first farm gets an average harvest of 21 tons of wheat per acre.
• The second farm gets an average harvest of 25 tons of wheat per acre.
• The second farm harvested 300 tons more wheat than the first.

1. The total harvest of the first farm: 21A21A21A tons
2. The total harvest of the second farm: 25(A−12)25(A - 12)25(A−12) tons
3. The difference in harvest: 25(A−12)−21A=30025(A - 12) - 21A = 30025(A−12)−21A=300

1. The first farm: 21×150=315021 \times 150 = 315021×150=3150 tons
2. The second farm: 25×138=345025 \times 138 = 345025×138=3450 tons

1. 3150, 3450

Understanding the Transfer Impact on Averages
When analyzing the transfer of people between two groups (E and R), it’s essential to consider how the averages of these groups are affected by the movement of individuals.
Initial Situation
- Let’s denote the average score of group E as AE and that of group R as AR.
- Each group has a certain number of people, which we’ll assume remains the same except for the transfers.
Transfer Details
- 5 people are transferred from E to R.
- An independent set of 5 different people is transferred from R back to E.
Impact on Group R’s Average
1. Decrease in Group E's Average:
- When 5 individuals are transferred from E to R, these individuals have an average score that is likely lower than the average of R (assuming R's average is higher).
- Consequently, when these individuals leave E, E's average may decrease.
2. Transfer Back to Group E:
- The second transfer involves 5 people moving from R back to E.
- The individuals moving back to E are not the same as those who came from E, meaning R retains its higher-performing members.
Conclusion on R's Average
- Since the individuals transferring from E to R likely have lower scores, and R is losing 5 members who probably have higher scores, R's average will decrease.
- Therefore, regardless of E's average changing, R’s average will consistently decrease after the transfers.
Final Answer
- The correct answer is B: Decrease always. The logic behind this is that R loses some of its higher-scoring members while gaining lower-scoring individuals from E, leading to a net decrease in its average score.

Question Analysis:
The question provides information about the transfer of 5 people from E to R and another independent set of 5 people from R to E. We have to find the maximum possible average achieved by class E at the end of the operation.

Given Information:
Transfer of 5 people from E to R.
Transfer of another independent set of 5 people from R to E.
The set transferred from R to E contains none from the set of students that came to R from E.

Solution:
Let us assume that there are n students in class E before the transfer. Then, the average of class E before the transfer can be calculated as:

Average of class E = Sum of Marks of all students in E / Total number of students in E

After transferring 5 students from E to R, the new number of students in class E will be (n-5). Similarly, after transferring 5 students from R to E, the new number of students in class E will be (n-5+5) = n.

Let us assume that the sum of marks of all students in class E before the transfer is S. Then, the sum of marks of remaining students in class E after the transfer will be (S - 5x), where x is the average marks of the 5 students transferred from E to R.

Similarly, let us assume that the sum of marks of all students in class R before the transfer is T. Then, the sum of marks of remaining students in class R after the transfer will be (T + 5y), where y is the average marks of the 5 students transferred from R to E.

Now, the new average of class E after the transfer can be calculated as:

New average of class E = Sum of Marks of remaining students in E / Total number of students in E

= (S - 5x) / n

We have to find the maximum possible value of this expression. To maximize this expression, we have to minimize the value of x (the average marks of the 5 students transferred from E to R) and maximize the value of y (the average marks of the 5 students transferred from R to E).

Since the set transferred from R to E contains none from the set of students that came to R from E, we can assume that the average marks of the 5 students transferred from R to E is greater than or equal to the average marks of the 5 students transferred from E to R.

Therefore, to maximize the new average of class E, we have to assume that the average marks of the 5 students transferred from E to R is minimum, i.e., 1. In this case, the sum of marks of remaining students in E after the transfer will be (S - 5), and the sum of marks of remaining students in R after the transfer will be (T + 5y).

To maximize the new average of class E, we have to assume that all the 5 students transferred from R to E have scored 100 marks (maximum possible marks). In this case, the sum of marks of remaining students in R after the transfer will be (T + 500).

Therefore, the new average of class E after the transfer will be:

New average of class E = (S - 5) / n

= (S / n) - 5/n

Since we have to maximize this expression, we have to assume that the original average of

Problem:
The average of 20 numbers is zero. How many of them may be greater than zero, at the most?

Solution:
To find the maximum number of numbers that can be greater than zero, we need to understand the concept of average and the properties of numbers.

Understanding the Average:
The average of a set of numbers is found by summing all the numbers in the set and then dividing the sum by the total number of numbers.

Properties of Numbers:
1. The sum of positive numbers is always greater than zero.
2. The sum of negative numbers is always less than zero.
3. The sum of positive and negative numbers can be zero if the sum of positive numbers equals the sum of negative numbers.

Explanation:
Given that the average of 20 numbers is zero, we can conclude that the sum of these 20 numbers is also zero.

Let's assume that there are 'x' numbers greater than zero and 'y' numbers less than or equal to zero.

Since the sum of these 20 numbers is zero, we can write the equation:
(x * positive number) + (y * non-positive number) = 0

To maximize the number of numbers greater than zero, we need to minimize the number of non-positive numbers. The smallest non-positive number is zero. Therefore, we can rewrite the equation as:
(x * positive number) + (y * 0) = 0

Simplifying the equation, we get:
x * positive number = 0

In order for this equation to be true, the value of 'x' must be zero. This means that there can be zero numbers greater than zero in the set of 20 numbers.

Therefore, the maximum number of numbers that can be greater than zero is 0.

Hence, the correct answer is option 'C' - 0.

Understanding the Problem
To find the largest possible average age of people below 51 years, we first gather the provided data:
- Number of people aged 51 years and above: 30
- Maximum number of people below 51 years: 39
- Overall average age: 38 years
Calculating Total Population and Age Sum
1. Total Population:
- Let N be the total number of people.
- N = 30 (aged 51+) + 39 (aged < 51)="" />
2. Total Age Sum:
- Average age = Total Age Sum / N
- 38 = Total Age Sum / 69
- Total Age Sum = 38 * 69 = 2622 years
Age Sum of People Aged 51 and Above
To maximize the average age of those below 51, we need to minimize the total age of the people aged 51 and above.
1. Minimum Age Calculation:
- If we assume all 30 people aged 51 and above are exactly 51 years old:
- Total Age of people aged 51 and above = 30 * 51 = 1530 years
Age Sum of People Below 51 Years
1. Age Sum Calculation:
- Total Age of people below 51 = Total Age Sum - Total Age of people aged 51+
- Total Age of people below 51 = 2622 - 1530 = 1092 years
Average Age of People Below 51 Years
1. Finding Average:
- Average age of people below 51 = Total Age of people below 51 / Number of people below 51
- Average age = 1092 / 39 = 28 years
Thus, the largest possible average age of the people whose ages are below 51 years is 28 years (option D).

Given information:
- Three years ago, the average age of a family of 5 members was 17 years.
- Presently, the family still consists of 5 members.
- The present average age of the family remains the same even after the birth of a child.

To find:
- The present age of the child.

Solution:
Let's assume that the present age of the 5 family members is x.
- Three years ago, their age was (x-3) each.
- So, the total of their ages three years ago was 5(x-3).
- According to the question, the average age of the family remains the same even after the birth of a child.
- So, the present total age of the 6 members (5 original members + 1 child) is 6x.
- The difference between the total ages of the 6 members and the 5 original members will give us the age of the child.
- So, the present age of the child will be:

Present age of child = Present total age of 6 members - Total age of 5 original members
= 6x - 5(x-3)
= x+15

Given that the average age of the family remains the same. So, we can write:
Total age / Number of members = Average age
=> 5x / 5 = x
=> x = 5

Substituting x=5 in the above equation, we get:
Present age of child = x+15 = 5+15 = 20

Therefore, the present age of the child is 2 years (option C).