Data Sufficiency

siddhans wrote:

Geva@MasterGMAT wrote:

siddhans wrote:How do you know if m should divide n OR n should divide m?

Putting aside the fact that you need to assign n students to m classrooms (so divide n by m), there is also the fact that n is greater than m (3 < m < 13 < n) and in fact is the greatest number. If the question was phrased so that n must divide m, the result would always be a fraction (dividing a small positive number by a greater positive number will always yield a fraction), and the answer to the question stem would be "no" regardless of what the statements say. In a DS question, the question stem is never sufficient to answer the question alone - you will always need the statement(s) to have any hope of answering the question.

Okay, also how in some of the exmples they have chosen n to be < 13. Dont we need to always choose n>13 to test statement 1 and 2? since 3<m<13<n which implies n is always greater than 13?

You're right - any example chosen must first and foremost satisfy the requirements of the question stem.

The key is to show that there are values of n and m that satisfy both the question stem and the statement, but still allow both a yes and a no answer - an n that is and is not divisible by m. The best way to go about this is to fix one and change the other:

stat. (1) we know that 3n/m is an integer. Let n=20 so that 3n=60. There are several possible values of m that will satisfy 60/m=integer: m could equal 4, 6, 10, 12. But is n=20 itself divisible by m? the answer is yes where m=4 or 10, no where m is 6 or 12. Insufficient.

Stat. (2) the 13n should deter you from trying the same trick here - with an n>13, you are bound to run into some uncomfortably high numbers. You can try the other way : fix m and play with n:

if m=4, and 13n/4 is an integer, it means that n must be a multiple of 4 (since 13 is not a multiple of 4).

if m=5, and 13n/5 is an integer, it means that n must be a multiple of 5 (since 13 is not a multiple of 4).

etc. etc. Because 13 is prime, it is divisible only by 1 and itself. Because m that is greater than 1 but smaller than 13, we will not be able to reduce the m in the denominator with the 13, and will therefore have to reduce it with the n - implying that n is a multiple of m, and the answer is a definite yes.

Hi,

But the question here asks

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 assigned
to it?

The question doesn't say each of the m classroom. So according to condition 1, we have atleast one value of m for each value of n. Therefore, 1 should be sufficient! Please, let me know if I am mistaken in my interpretation.

Sanjay

sanjaymittal wrote:Hi,

But the question here asks

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 assigned
to it?

The question doesn't say each of the m classroom. So according to condition 1, we have atleast one value of m for each value of n. Therefore, 1 should be sufficient! Please, let me know if I am mistaken in my interpretation.

Sanjay

Not sure I understand your reasoning, but I think that you are attaching too much significance to the phrase "one of the m classrooms". the phrase merely says that each student goes in one classroom - i.e. that it is not possible to assign the same student to more than once classroom.
It doesn't mean that you can assign all of the students to the same classroom, since the question (and statements) also specify that each classroom should have the same number of students at the end - which means that the number of students n needs to be divided equally among m classrooms.

Hi Geva,

Thanks for your prompt response. Let me try to make my explanation better. So according to condition 1, It is possible to assign each of 3n students to
one of m classrooms so that each classroom
has the same number of students assigned to it.

I am not assigning all the students in one classroom and I am also distributing students equally into each of m classrooms.

Now let's take n= 14. Therefore 42(3n) students can be accommodated in either m=6 or m=7 classrooms. Now, according to question, we can accommodate 14(n) students in m(7) classrooms.

2) When n= 15 ---> 3n=45. These students can be accommodated in 5 or 9 classrooms. Which gives us m to be either 5 or 9. Going back to the question stem we can accommodate n=15 students in 5 classrooms.

I hope this explanation makes my point clearer.

sanjaymittal wrote:Hi Geva,

Thanks for your prompt response. Let me try to make my explanation better. So according to condition 1, It is possible to assign each of 3n students to
one of m classrooms so that each classroom
has the same number of students assigned to it.

I am not assigning all the students in one classroom and I am also distributing students equally into each of m classrooms.

Now let's take n= 14. Therefore 42(3n) students can be accommodated in either m=6 or m=7 classrooms. Now, according to question, we can accommodate 14(n) students in m(7) classrooms.

2) When n= 15 ---> 3n=45. These students can be accommodated in 5 or 9 classrooms. Which gives us m to be either 5 or 9. Going back to the question stem we can accommodate n=15 students in 5 classrooms.

I hope this explanation makes my point clearer.

Differentiate between what you know and what you need to prove. If we go with n=14, we know that 42 is divisible by m: this still allows m=6 or m=7. Both of these values satisfy stat. (1). Now, what does that do to the question stem?
If n=14 and m=7, then the answer is yes.
If n=14 and m=6, the answer is no.

Therefore, stat.(1) does not limit the answer to the question stem to a single, definite answer, and is thus insufficient.

Thanks Geva! I did consider that but since it was satisfying the condition 'one of the classroom' so I eliminated two answers. But, I guess I am fretting too much about the phrase 'one of the classroom'..

Thanks

S

I googled this question because I had the same exact issue sanjay was having.

I think this problem is worded incorrectly.

It is asking: "Is it possible to assign..."

As shown in the previous posts, both (1) and (2) present enough information to show that it is at least possible that m/n = an integer with certain numbers.

While (2) shows that it always the case, (1) shows it is sometimes the case. Either way, the equation: m/n=integer is possible with both (1) and (2).

Is my reasoning wrong somewhere?

stompy wrote:I googled this question because I had the same exact issue sanjay was having.

I think this problem is worded incorrectly.

It is asking: "Is it possible to assign..."

As shown in the previous posts, both (1) and (2) present enough information to show that it is at least possible that m/n = an integer with certain numbers.

While (2) shows that it always the case, (1) shows it is sometimes the case. Either way, the equation: m/n=integer is possible with both (1) and (2).

Is my reasoning wrong somewhere?

This line of thinking actually has merit and made me pause for a second. The wording is the same as the official guide question. I think your line of reasoning would apply if the question asked "is it possible to find a set of values n and m that will match the conditions?", which is not the same question as "is it possible to assign n students to m classrooms under these conditions"?

At the end of the day, is someone walks up to you and presents the problem and asks you "is it possible to assign n students to m classrooms?", your reply based on stat. (1) alone would still be "it may or may not be possible to divide the students in such a way - depends on the value of m and n. Which is why stat. (1) is insufficient.

hard question
this is a lession that we should closely follow the og questions which is not easy at all.

for dividibility, if 2 numbers have common factors, we have to be careful.