There are N students in a class. When the students are distributed int

Hi,

Is this qs representative of a hard GMAT quant qs?
Also, can you please help me understand the logic behind taking A=3?

TIA!­

\(\frac{N}{4A}\) = Remainder 3.

Therefore, N is not divisible by A and N is not divisible by 4.


N divided by \(\frac{A}{3}\) = Remainder 0. 3N is divisible by A.

Since N is not divisible by A, 3 has to be divisible by A.

So, A=3. or A=1.

But A cannot be 1 as a group cannot have 1/3 students. Therefore, A=3.


Substituting A=3 :

\(\frac{N}{4A}\) = \(\frac{N}{4(3)}\) = \(\frac{N}{12}\) = Remainder 3.


So N=15, 27, 39,....






l. 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

A=3. So, A+1 = 4.

Take any of the values of N.

N=15 : \(\frac{N}{A+1} = \frac{15}{4}\) = Remainder 3

N=27 : \(\frac{ N}{A+1} = \frac{27}{4}\) = Remainder 3

Hence, not true.




ll. If the students are distributed into groups that contain 3 students each, no students are left without a group.

Take any of the values of N.

N=15 : \(\frac{15}{3}\) = Remainder 0

N=27 : \(\frac{27}{3}\) = Remainder 0

Hence, true.




lll. If the students are distributed into groups that contain 12 students each, 9 students are left without a group.

Take any of the values of N.

N=15 : \(\frac{15}{12}\) = Remainder 3

N=27 : \(\frac{27}{3} \)= Remainder 3

Hence, not true.

Therefore, Option B is the correct answer choice.

 

I'm a bit confused by your approach in the first step. ­Shouldn't the conclusion for the first step be N is not divisible by 4A? I'm not sure how you concluded that N is not divisible by A and N is not divisible by 4. For example, if N =15 and A =3 as you stated, N is clearly divisible by A in this case.