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A school administrator will assign each student in a group

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Re: A school administrator will assign each student in a group  [#permalink]

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13 Dec 2017, 07:12
gwiz87 wrote:
A school administrator will assign each student in a group of N students to one of M classrooms. If 3<M<13<N, 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?

(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.

(2) It is possible to assign each of 13N students to one of M classrooms so that each classroom has the same number of students assigned to it.

We are given that each student in a group of n students is going to be assigned to one of m classrooms. We are being asked whether it is possible to assign each of the n students to one of the m classrooms so that each classroom has the same number of students.

Thus, we need to determine whether n/m = integer.

Statement One Alone:

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.

Statement one is telling us that 3n is evenly divisible by m. Thus, 3n/m = integer.

However, we still do not have enough information to answer the question. When n = 16 and m = 4, n/m DOES equal an integer; however, when n = 20 and m = 6, n/m DOES NOT equal an integer.

Statement Two Alone:

It is possible to assign each of 13n students to one of m classrooms so that each classroom has the same number of students assigned to it.

This statement is telling us that 13n is divisible by m. Thus, 13n/m = integer.

What is interesting about this statement is that we know that n is greater than 13 and that m is less than 13 and greater than 3. Thus, we know that m could equal any of the following: 4, 5, 6, 7, 8, 9, 10, 11, or 12. We see that none of those values (4 through 12) will divide evenly into 13.

Knowing this, we can say conclusively that m will never divide evenly into 13. Thus, in order for m to divide into 13n, m must divide evenly into n.

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Re: A school administrator will assign each student in a group  [#permalink]

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22 Aug 2018, 01:46
gwiz87 wrote:
A school administrator will assign each student in a group of N students to one of M classrooms. If 3<M<13<N, 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?

(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.

(2) It is possible to assign each of 13N students to one of M classrooms so that each classroom has the same number of students assigned to it.

That was a tricky question. Thanks for sharing.
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Re: A school administrator will assign each student in a group  [#permalink]

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21 Sep 2018, 11:51
gwiz87 wrote:
A school administrator will assign each student in a group of N students to one of M classrooms. If 3<M<13<N, 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?

(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.

(2) It is possible to assign each of 13N students to one of M classrooms so that each classroom has the same number of students assigned to it.

hahaa...this question gets me every time I try it, it is such an easy question with a complex language, which gets to everyone I suppose !
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21 Sep 2018, 13:53
gwiz87 wrote:
A school administrator will assign each student in a group of N students to one of M classrooms. If 3<M<13<N, 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?

(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.

(2) It is possible to assign each of 13N students to one of M classrooms so that each classroom has the same number of students assigned to it.

To assign the same number of students to each classroom, the number of students (n) must be divisible by the number of classrooms (m).

Question rephrased: Is $$\frac{n}{m}$$ an integer?

Statement 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.
In other words, the number of students here (3n) is divisible by the number of classrooms (m), implying that $$\frac{3n}{m}$$ is an integer.
Since we need to determine whether m will always divide into n, plug in EXTREME values for m.

m=4:
It's possible that m=4 and n=16, with the result that $$\frac{3n}{m} = \frac{3*16}{4} = 12$$.
In this case, then $$\frac{n}{m} = \frac{16}{4} = 4$$, which is an integer.

m=12:
It's possible that m=12 and n=16, with the result that $$\frac{3n}{m} = \frac{3*16}{12} = 4$$.
In this case, $$\frac{n}{m}= \frac{16}{12} = \frac{4}{3}$$, which is NOT an integer.
INSUFFICIENT.

Statement 2: It is possible to assign each of 13n students to one of m classrooms so that each classroom has the same number of students assigned to it.
In other words, the number of students here (13n) is divisible by the number of classrooms (m), implying that $$\frac{13n}{m}$$ is an integer.
It is not possible that m divides into 13, since the only factors of 13 are 1 and 13, and m must be BETWEEN 3 and 13.
Thus, for $$\frac{13n}{m}$$ to be an integer, m must divide into n, implying that $$\frac{n}{m}$$ is an integer.
SUFFICIENT.

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Re: A school administrator will assign each student in a group  [#permalink]

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07 Nov 2018, 11:09
GMATGuruNY wrote:
gwiz87 wrote:
A school administrator will assign each student in a group of N students to one of M classrooms. If 3<M<13<N, 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?

(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.

(2) It is possible to assign each of 13N students to one of M classrooms so that each classroom has the same number of students assigned to it.

To assign the same number of students to each classroom, the number of students (n) must be divisible by the number of classrooms (m).

Question rephrased: Is $$\frac{n}{m}$$ an integer?

Statement 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.
In other words, the number of students here (3n) is divisible by the number of classrooms (m), implying that $$\frac{3n}{m}$$ is an integer.
Since we need to determine whether m will always divide into n, plug in EXTREME values for m.

m=4:
It's possible that m=4 and n=16, with the result that $$\frac{3n}{m} = \frac{3*16}{4} = 12$$.
In this case, then $$\frac{n}{m} = \frac{16}{4} = 4$$, which is an integer.

m=12:
It's possible that m=12 and n=16, with the result that $$\frac{3n}{m} = \frac{3*16}{12} = 4$$.
In this case, $$\frac{n}{m}= \frac{16}{12} = \frac{4}{3}$$, which is NOT an integer.
INSUFFICIENT.

Statement 2: It is possible to assign each of 13n students to one of m classrooms so that each classroom has the same number of students assigned to it.
In other words, the number of students here (13n) is divisible by the number of classrooms (m), implying that $$\frac{13n}{m}$$ is an integer.
It is not possible that m divides into 13, since the only factors of 13 are 1 and 13, and m must be BETWEEN 3 and 13.
Thus, for $$\frac{13n}{m}$$ to be an integer, m must divide into n, implying that $$\frac{n}{m}$$ is an integer.
SUFFICIENT.

understand your explanation and appreciate for that...
just wonder when question ask about "possible", i assume it will be enough to prove if i can get the answer n/m = integer
in (1) = 16/4, even though 16/12 is not integer, but 48/12 makes me distribute 4 students to one of 12 classes.
in (2) = 14/7
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Re: A school administrator will assign each student in a group  [#permalink]

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16 Feb 2019, 04:40
I guess there is a language problem in this question. The solution will be valid only if the question stated that the M classrooms had 0 students prior to the administrator's assignment. If the M Classrooms already had a certain no.of students then the Answer to this question would be E.

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Re: A school administrator will assign each student in a group  [#permalink]

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25 Feb 2019, 15:18
manimgoindowndown wrote:
Bunuel wrote:
manimgoindowndown wrote:
The hardest part of this problem wasn't the math. It was the prompt.

Could someone break it down for me?

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?

"is it possible" to me implied whether we can choose an two values within the problem's given inequality to make n/m an integer, at all which we could for condition 1.

There are:
m classrooms where 3 < m < 13 < n.
n students where 3 < m < 13 < n.

We are asked to find whether we can divide equally n students in m classrooms, so whether n/m is an integer.

Does this make sense?

It is possible as you demonstrated in your solutions post to have a situation where premise I will make n/m an integer. However it is also possible that there is a situation where it wont be an integer. I guess I focused on the possibility part. Since its possible I would see I. As correct.

Hey guys! I've been an engineer dealing with physics for several years. Basicaly, the GMAT wording and logic are different from scientific ones. For exemple, in engineering if we ask for the possible value, an expression like value=x*6+y would be ok if given that x and y are positive integers. However, in GMAT, only the real number value can answer the question about 'what is the value of ***'. Otherwise, they will ask 'the value of ** could be expressed as'.

Another important thing is that there is no possible/pseudo value of problems. If you can find both 6 and 8 as the answer of a certain equation/inequation question, then the option is definitely not sufficient. Only the unique value is sufficient.

Back to this question, I did this process to justify the answer E although I had the wrong answer.
Option 1:
3n/m=int => situation 1: n/m=int sufficient; situation 2: 3n/m=int only with m=3k (k int, (3*k*n/k)/m=integer) then 3n/m=3n/3k=n/k has to be integer. Given that 3<m<13, we have 1<k<=4. there is k=2 or 3 or 4. But in the end we still can not make sure whether n/k will be integer or not.

so choose E
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Re: A school administrator will assign each student in a group  [#permalink]

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08 Apr 2019, 23:16
Given 3<m<13<n
we had to find that whether n is a multiple of m.

St-1 says 3n is a multiple of m.
checking multiple values of m
3n=18, n=6 and m=9 ; implies n is not a multiple of m
3n=27, n=9 and m=9 ; implies n is a multiple of m

St-2 says 13n is a multiple of m (given 3<m<13).
As 13 is a prime number, 13 has only two factors (1 and itself) i.e. it has no no factor b/w 1 and 13.
since m lies b/w 3 and 13. m doesn't divide 13.
henceforth, for 13n to be a multiple of m, n shd be divisible by m.
Suff
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Re: A school administrator will assign each student in a group   [#permalink] 08 Apr 2019, 23:16

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