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20 Oct 2023, 06:25
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RiyazAhamed
“is it possible to assign each of the N students to one of the M classrooms so that each classroom has the same number of students” simply means is N divisible by M? That is, N/M? 1) given 3N is divisible by M Now, since M is between 3 and 13 and N is more than 13, this is not sufficient. Let’s say, M = 6, N = 60. 3N/M 3(60)/6 = Integer value. So, 60/6 = N/M = integer = yes possible to arrange each student to each class. However, if, M = 12 and N = 16. 3(16)/12 = 4 = integer value. But, N/M = 16/12 is not an integer. So, we may arrange or we may not arrange equally. Insufficient. B) given, 13N/M. We know 3<M<13<N. M has to be less than 13. That is M cannot be a multiple of 13 to cancel out that 13 in the numerator. Thus, N has to cancel out M in denominator to give us 13N/M as an integer. That is N/M is an integer. Thus, sufficient. Answer is B.
Where is this question from?Source?
20 Oct 2023, 06:28
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It is an OG question. Question number 487 in Official Guide 2022.
20 Oct 2023, 09:14
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RiyazAhamed
“is it possible to assign each of the N students to one of the M classrooms so that each classroom has the same number of students” simply means is N divisible by M? That is, N/M? 1) given 3N is divisible by M Now, since M is between 3 and 13 and N is more than 13, this is not sufficient. Let’s say, M = 6, N = 60. 3N/M 3(60)/6 = Integer value. So, 60/6 = N/M = integer = yes possible to arrange each student to each class. However, if, M = 12 and N = 16. 3(16)/12 = 4 = integer value. But, N/M = 16/12 is not an integer. So, we may arrange or we may not arrange equally. Insufficient. B) given, 13N/M. We know 3<M<13<N. M has to be less than 13. That is M cannot be a multiple of 13 to cancel out that 13 in the numerator. Thus, N has to cancel out M in denominator to give us 13N/M as an integer. That is N/M is an integer. Thus, sufficient. Answer is B.
Thankyou so much. It cleared up all the confusion. 
