The problem states that a student's marks were recorded incorrectly. We need to find the actual difference between the correct marks and the incorrectly entered marks.
The total difference in marks is calculated as:
Difference = Correct Marks - Incorrect Marks
Difference = \(69 - 21 = 48\)
This difference of 48 marks affected the class average. The problem states the average marks decreased by \(\frac{6}{5}\).
Let '\(N\)' be the total number of students in the class.
The decrease in the average is the total difference in marks divided by the number of students.
Decrease in Average = \(\frac{\text{Total Difference in Marks}}{N}\)
We can now set up an equation using the information provided:
\(\frac{48}{N} = \frac{6}{5}\)
To solve for '\(N\)', we can cross-multiply:
\(48 \times 5 = 6 \times N\)
\(240 = 6N\)
Now, divide both sides by 6 to find the value of '\(N\)':
\(N = \frac{240}{6}\)
\(N = 40\)
Thus, there are 40 students in the class.
| Marks obtained | No. of students |
| 25 | 5 |
| 30 | 6 |
| 35 | 3 |
| 40 | 1 |
| Age Band | Students |
|---|---|
| 0 - 10 | 4 |
| 10 - 20 | 5 |
| 20 - 30 | 15 |
| 30 - 40 | 4 |
The median of the observations 4, 14, 11, 8, 7, 8, 9, 12, 4, and 6 is:
The weights (in kg) of 9 members of a school cricket team are 50, 63, x, 62, 64, 48, 51, 55, 48. If the mean weight is 56 kg, then the value of x is:
A group of people consists of men, women, and children. 40% of them are men, 35% are women, and the rest are children, and their average weights are 70 kg, 60 kg, and 30 kg, respectively. The average weight of the group is:
If the mean of 22, 25, 27, 24, and x is 26, then the value of x is:
The average of 70 values is 40. If each value is multiplied by 20, what will be the changed average?
| Marks obtained | No. of students |
| 25 | 5 |
| 30 | 6 |
| 35 | 3 |
| 40 | 1 |
The average runs scored by a batsman in 23 matches is 40. In the next 10 matches, the batsman scored an average of 16 runs. Find his average runs scored (rounded off to two decimals) in all the 33 matches.