By forming a group of 4 members, 3 students are left.
Therefore, total students = 4 (n) + 3 where n > 0
By forming group of 9 member, 4 student left
Total students = 9 (n) + 4 where n > 0
Total student = 2 and 2 must satisfy 4 (a) + 3 and 9 (n) + 4
Question => what is the total number of student in the class?
Statement 1 => the number of students is a two digit number le than 70
Total students = z
9 < z < 70
4 (n) + 3 where 0 < n < 17 =. 1,2,3.......16
4 (n) + 3 will yield 7,11,15,19,23,27,31,35,39,43,47,51,55,59,63,67.
For 9 (n) + 4 where 0 < n < 8 => 1,2,3.......7
9 (n) + 4 will yield 13,22,31,40,49,58,67.
Total student = z and z must satisfy 4(n)+3 and 9(n)+4.
From the yield result of both 4(n)+3 and 9(n)+4, only 31 and 67 satisfy the condition attached to z.
(i.e both can form a group of 4 members, and 3 students will be left and can also form a group of 9 members in which 4 students will be left but unfortunately, both of them are less than 70.)
So, the information provided is not enough to get the exact number of students. Hence, statement 1 is NOT SUFFICIENT.
Statement 2 => The number of students is a two-digit number greater than 50.
Total student (s) = z; 50<z<100
For 4(n)+3 where 11<n<25 => 12, 13, 14, 15,...24.
So, 4(n)+3 will yield 51, 55, 59, 63, 67, 71, 75, 79, 83, 87, 91, 95, 99
For 9(n)+4 where 5<n<11 => 6,7,8,...
So, 9(n)+4 will yield 58, 67, 76, 85, 94.
Total student = z; and z must satisfy 4(n)+3 and 9(n)+4. From the yield result of both 4(n)+3 and 9(n)+4, 67 satisfies the condition attached to Z.
i.e 67 can form a group of 4 members with 3 students left it can also form a group of 9 members with 4 students left.
Hence, the total number of students in the class = z; and z = 67. Therefore, statement 2 is SUFFICIENT.
Conclusively, only statement 2 alone is SUFFICIENT. Option B is the correct option.
