Symbiosis National Aptitude Test SNAP 2019 Paper

Let number of students in n, where n is a two digits prime number, whose both digits are also prime.
Again minimum earning of student is x, then maximum earning of the student is (x + 45).
Now, total stipend of n students = 50n
and total stipend of minimum and maximum earning students
= x + x + 45
= (2x + 45)
Total earning of remaining students = 50n – (2x + 45)
ATQ,
Average stipend of reamining students = 50 – 1 = 49
Total earning of remaining students = 49(n – 2)
∴ 50n – (2x + 45) = 49(n – 2)
2x – n = 53
n = 2x – 53
As, 42 ≤ x ≤ 47
for x = 45,
x = 2 × 45 – 53 = 90 – 53 = 37
Here '37' is a prime number, also its two digits 3 and 7 are prime.
So, number of students = 37.