N=4Ax+3.......(1)
where x is a positive integer
Also, N= (A/3)*y.......(2)
y is positive integer
Also if it is possible to divide students in the group of A/3 students, A/3 must be an integer
From (1) and (2)
4Ax+3=(A/3)*y
(A/3)*(y-12x)=3
There are 2 cases possible
1) When A/3=1, and (y-12x)=3
We get A=3
and y=12x+3
Value of N = 1*(12x+3)=12x+3Case 2) When A/3=3 and (y-12x)=1
We get A=9 and y=12x+1
N= 3*(12x+1)=36x+31) A+1 is Even in both cases and N is odd.
If the students are distributed into groups that contain A+ 1 students each, the number of students that are left without a group
can never be 2II) N in both cases is multiple of 3
Hence, If the students are distributed into groups that contain 3 students each, no students are left without a group
III) N can be 12x+3 or 36x+3
12x+3= 3(mod 12)
36x+3= 3(mod 12)
In both cases only 3 students are left without group
Only statement II) is correct
kiran120680 wrote:
There are N students in a class. When the students are distributed into groups that contain 4A number of students each, 3 students are left without a group. When the students are distributed into groups that contain A/3 number of students each, no students are left without a group. Which of the following statements is correct?
I. If the students are distributed into groups that contain A+ 1 students each, the number of students that are left without a group can be 2
II. If the students are distributed into groups that contain 3 students each, no students are left without a group
III. If the students are distributed into groups that contain 12 students each, 9 students are left without a group
A. I only
B. II only
C. III only
D. I and II only
E. II and III only