Understanding the Marks Distribution

The marks obtained by seven students are given as: 4, 15, 6, 7, 5, $a$, and $b$. We are also given specific conditions for $a$ and $b$:

• $a$ is a positive multiple of 4 ($a > 0$).
• $b$ is a prime number.
• The maximum possible mark is 30.

Our goal is to find the maximum possible value of the difference between the highest and lowest marks among these seven students.

Identifying Minimum and Maximum Marks

The known marks are {4, 5, 6, 7, 15}. The minimum among these is 4, and the maximum is 15.

To maximize the difference (Maximum Mark - Minimum Mark), we need to make the minimum mark as small as possible and the maximum mark as large as possible by choosing appropriate values for $a$ and $b$.

Finding the Smallest Possible Minimum Mark

The minimum mark in the set {4, 5, 6, 7, 15, $a$, $b$} depends on the values chosen for $a$ and $b$.

• Possible values for $a$ (positive multiple of 4, $\le 30$): {4, 8, 12, 16, 20, 24, 28}. The smallest possible value for $a$ is 4.
• Possible values for $b$ (prime, $\le 30$): {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}. The smallest possible value for $b$ is 2.

To achieve the smallest possible minimum mark for the entire group, we should select the smallest possible value for $b$, which is $b=2$. This makes the minimum mark in the set 2.

Finding the Largest Possible Maximum Mark

The maximum mark in the set {4, 5, 6, 7, 15, $a$, $b$} depends on the values chosen for $a$ and $b$.

• The largest known mark is 15.
• The largest possible value for $a$ is 28.
• The largest possible value for $b$ is 29.

The overall maximum mark will be the largest among {15, $a$, $b$}. To maximize this, we can potentially choose $a=28$ or $b=29$. The largest possible value is 29.

Calculating the Maximum Difference

We need to combine the choices for $a$ and $b$ to maximize the difference (Max - Min).

Scenario 1: Minimize the Minimum Mark first

• Choose $b=2$ to make the minimum mark 2. The marks set is {4, 5, 6, 7, 15, $a$, 2}. Minimum is 2.
• Now, maximize the remaining marks. The maximum will be $\max(15, a)$. To maximize this, we choose the largest possible value for $a$, which is $a=28$.
• The maximum mark is 28.
• The difference is Max - Min = $28 - 2 = 26$.
• The set of marks is {4, 5, 6, 7, 15, 28, 2}.

Scenario 2: Maximize the Maximum Mark first

• Choose $b=29$ to make the maximum mark 29. The marks set is {4, 5, 6, 7, 15, $a$, 29}. Maximum is 29.
• Now, minimize the remaining marks. The minimum will be $\min(4, a)$. To minimize this, we choose the smallest possible value for $a$, which is $a=4$.
• The minimum mark is $\min(4, 4) = 4$.
• The difference is Max - Min = $29 - 4 = 25$.
• The set of marks is {4, 5, 6, 7, 15, 4, 29}.

Comparing the differences calculated in both scenarios (26 and 25), the maximum possible difference is 26.