Question:medium

If the mean of 3, 4, 9, 2k, 10, 8, 6 and $(k + 6)$ is 8, and the mode of 2, 2, 3, $2p$, $(2p + 1)$, 4, 4, 5 and 6 (where $p$ is a natural number) is 4, then the value of $k - 2p$ is:

Updated On: Jan 16, 2026
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The Correct Option is A

Solution and Explanation

To determine the value of \(k - 2p\), it is necessary to first ascertain the values of \(k\) and \(p\).

Step 1: Determine \(k\)

The provided set of numbers is 3, 4, 9, 2k, 10, 8, 6, and \(k + 6\). Their mean is 8. The calculation of the mean is defined as:

\(\text{Mean} = \frac{\text{Sum of all elements}}{\text{Number of elements}}\)

Applying this formula to the given numbers yields:

\(\frac{3+4+9+2k+10+8+6+(k+6)}{8} = 8\)

Simplifying this equation to solve for \(k\):

\(46 + 3k = 64\)

\(3k = 64 - 46\)

\(3k = 18\)

\(k = \frac{18}{3}\)

\(k = 6\)

Step 2: Determine \(p\)

The set of numbers is 2, 2, 3, \(2p\), \(2p + 1\), 4, 4, 5, and 6. The mode is stated to be 4. The mode represents the value that occurs with the highest frequency.

For 4 to be the mode, its frequency must exceed that of any other number. Analyzing the possible values of \(2p\):

If \(2p = 4\)This implies \(p = 2\), and the sequence becomes 2, 2, 3, 4, 5, 4, 4, 5, 6. In this case, 4 appears 3 times, which is the highest frequency.
If \(2p eq 4\)For instance, if \(2p = 3\), the sequence would be 2, 2, 3, 3, 4, 4, 5, 6. Here, 2 and 3 would both have a frequency of 2, and 4 would also have a frequency of 2, leading to multiple modes or a different mode if \(2p\) made another number more frequent. Therefore, the mode would not be uniquely 4.

Given that 4 is the mode, it must be that \(2p = 4\), which leads to \(p = 2\).

Step 3: Compute \(k - 2p\)

Using the determined values of \(k = 6\) and \(p = 2\), the calculation is as follows:

\(k - 2p = 6 - 2 \times 2\)

\(= 6 - 4\)

\(= 2\)

Consequently, the value of \(k - 2p\) is 2.

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