Question

Average stipend of a group is \rupee50 per day. The difference between maximum and minimum stipend is \rupee45. If both these students are excluded, the average decreases by \rupee1. The minimum earning of any student lies between \rupee42 and \rupee47, and the number of students is a prime number whose both digits are also prime. Find the initial number of students.

• A. 33
• B. 35
• C. 37
• D. 39

Hint:

With two outliers removed and a new average given, use \(\;M+m = \text{old sum} - \text{new sum}\;\) over the same base \(n\), then apply the difference \(M-m\) to solve for \(M\) and \(m\). Finally enforce the side constraints.

Answer 1

The Correct Option is C

Let the initial number of students be \(n\), total sum \(S=50n\). Let minimum stipend be \(m\) and maximum be \(M\) with \(M-m=45\).
Excluding these two, the new average is \(49\) on \(n-2\) students: \[ 50n - (M+m) = 49(n-2) \quad\Rightarrow\quad M+m = n+98. \] Solve \[ M=\frac{(M+m)+(M-m)}{2}=\frac{(n+98)+45}{2}=\frac{n+143}{2},\qquad m=\frac{(n+98)-45}{2}=\frac{n+53}{2}. \] Given \(m\in[42,47]\): \[ 42 \le \frac{n+53}{2} \le 47 \ \Longrightarrow\ 31 \le n \le 41. \] Now \(n\) is a two-digit \emph{prime} whose both digits are prime (digits from \(\{2,3,5,7\}\)). The only such prime in \([31,41]\) is \(\boxed{37}\).
(Then \(m=(37+53)/2=45\) which lies in \([42,47]\), and \(M=(37+143)/2=90\); removing \(\{45,90\}\) changes the average from \(50\) to \(49\) as required.) \[ \boxed{37} \]