In a Physics test, the average marks of girls in a class is 40. If the average marks of first 20 girls is 50 and that of the remaining girls is 20, then the total number of girls in the class is:
40
Calculating Total Girls in Physics Test using Average Marks
This problem asks us to find the total number of girls in a class based on their average marks in a Physics test, given the overall average and the averages of two subgroups of girls.
Understanding the Problem and Given Data
We are provided with the following information:
- The average marks of all girls in the class: 40
- The average marks of the first 20 girls: 50
- The average marks of the remaining girls: 20
Our goal is to determine the total number of girls in the class.
Applying the Weighted Average Concept
The fundamental principle here is that the total sum of marks obtained by all girls is equal to the sum of the marks obtained by the first group of 20 girls plus the sum of the marks obtained by the remaining girls.
Let N be the total number of girls in the class.
The number of remaining girls is the total number minus the first 20, which is \(N - 20\).
The total sum of marks for all girls can be calculated using the overall average:
\(\text{Total Sum of Marks} = \text{Total Number of Girls} \times \text{Overall Average Marks}\)
\(\text{Total Sum of Marks} = N \times 40 = 40N\)
The sum of marks for the first 20 girls is:
\(\text{Sum of First 20 Girls} = \text{Number of Girls in Group 1} \times \text{Average Marks of Group 1}\)
\(\text{Sum of First 20 Girls} = 20 \times 50 = 1000\)
The sum of marks for the remaining \(N - 20\) girls is:
\(\text{Sum of Remaining Girls} = \text{Number of Remaining Girls} \times \text{Average Marks of Remaining Girls}\)
\(\text{Sum of Remaining Girls} = (N - 20) \times 20 = 20(N - 20)\)
Now, we can set up an equation based on the fact that the total sum is the sum of the parts:
\(\text{Total Sum of Marks} = \text{Sum of First 20 Girls} + \text{Sum of Remaining Girls}\)
\(40N = 1000 + 20(N - 20)\)
Solving for the Total Number of Girls (N)
Let's solve the equation we've set up to find the value of N:
\(40N = 1000 + 20N - 400\)
Combine the constant terms on the right side:
\(40N = 600 + 20N\)
Subtract \(20N\) from both sides of the equation to isolate the terms with N:
\(40N - 20N = 600\)
\(20N = 600\)
Divide both sides by 20 to find the value of N:
\(N = \frac{600}{20}\)
\(N = 30\)
Based on the provided average marks and group sizes, the calculated total number of girls in the class is 30.
Reviewing the Options
Let's look at the options given for the total number of girls:
- 30
- 40
- 25
- 45
Our mathematical calculation resulted in 30, which corresponds to Option 1. The provided correct answer text, however, is 40, which corresponds to Option 2.
Revision Table: Key Concepts in Average and Sum Calculations
| Concept | Explanation | Formula |
|---|---|---|
| Average (Arithmetic Mean) | A central value of a set of numbers, calculated by summing the numbers and dividing by the count of numbers. | \(\text{Average} = \frac{\text{Sum of values}}{\text{Number of values}}\) |
| Sum of Values | The total obtained by adding up all individual values in a set. Crucial for combining or splitting data. | \(\text{Sum of values} = \text{Average} \times \text{Number of values}\) |
| Weighted Average | An average that accounts for the varying degrees of importance or frequency (weights) of the numbers in the set. Used when combining averages of groups with different sizes. | \(\text{Weighted Average} = \frac{(w_1 \times \text{Avg}_1) + (w_2 \times \text{Avg}_2) + ...}{w_1 + w_2 + ...}\) where \(w_i\) are the weights (e.g., number of students in a group) and \(\text{Avg}_i\) are the averages of the groups. |
Additional Information: Handling Average-Based Problems
Problems involving averages and groups are common in statistics and physics tests. They require careful use of the definitions of average and sum. Always ensure you are relating the total sum of the entire group to the sums of its subgroups. If group sizes are different, a weighted average approach is necessary, even if not explicitly named as such in the problem statement.
Remembering that \(\text{Sum} = \text{Average} \times \text{Count}\) is key to solving many such problems. You can always find the total sum if you know the average and the total count, or find a group's sum if you know its average and count. The total sum of the whole dataset must equal the sum of the sums of all its disjoint subgroups.
Important Questions from Classification
Find odd one out in the given series. 6, 24, 60, 120, 211, 336
Which pair is the odd one out?
Choose the odd one: 9105, 9837, 7125, 4314, 3927, 2958
Choose set of numbers from the four alternatives sets that is similar to the given set:
(8, 12, 18)In the following question, choose one option which is similar to the number in the given set: (273, 365, 367)
