Question

If the mean of the following data is k(> 0), then the value of k is

Marks	K	4	K	8	3
Frequency	K	2K	3K	4K	5K

• A. 10
• B. 8
• C. 5
• D. 4

Answer 1

The Correct Option is C

The given table provides the following data: \[ \text{Marks} \quad k, 4, k, 8, 3 \quad \text{(with corresponding frequencies)} \quad k, 2k, 3k, 4k, 5k. \] To find the mean, we use the formula for the mean of grouped data: \[ \text{Mean} = \frac{\sum (f \cdot x)}{\sum f}, \] where \( f \) represents the frequency and \( x \) represents the corresponding marks. Let's first calculate the sum of \( f \cdot x \) for each group:
For \( k \), the contribution is \( k \cdot k = k^2 \). 
For \( 4 \), the contribution is \( 2k \cdot 4 = 8k \).
For \( k \), the contribution is \( 3k \cdot k = 3k^2 \). 
For \( 8 \), the contribution is \( 4k \cdot 8 = 32k \).
For \( 3 \), the contribution is \( 5k \cdot 3 = 15k \).
Thus, \[ \sum (f \cdot x) = k^2 + 8k + 3k^2 + 32k + 15k = 4k^2 + 55k. \] Now, let's calculate the sum of frequencies: \[ \sum f = k + 2k + 3k + 4k + 5k = 15k. \] The mean is given as \( k \), so: \[ k = \frac{4k^2 + 55k}{15k}. \] Simplifying: \[ k = \frac{4k + 55}{15}. \] Now, multiply both sides by 15: \[ 15k = 4k + 55. \] Solving for \( k \): \[ 15k - 4k = 55 \quad \Rightarrow \quad 11k = 55 \quad \Rightarrow \quad k = 5. \]

The correct option is (C): \(5\)