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by gmat_for_life » Sun Jun 12, 2016 6:12 am

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A school has a students and b teachers. If a < 150, b < 25, and classes have a maximum of 15 students, can the a students be distributed among the b teachers so that each class has the same number of students (Assume that any student can be taught by any teacher.)
(1) It is possible to divide the students evenly into groups of 2, 3, 5, 6, 9, 10, or 15.
(2) The greatest common factor of a and b is 10.
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Source: — Data Sufficiency |

by [email protected] » Sun Jun 12, 2016 10:48 am
Hi gmat_for_life,

When posting GMAT questions, you should make sure to include the answer.

This prompt is essentially just a big 'prime factorization' question, although you could 'brute force' the solution if you didn't know how to do the technical math involved.

We're told a number of facts to start:
1) There are fewer than 150 students
2) There are fewer than 25 teachers
3) The MAXIMUM class size is 15 students.

We're asked if the number of students can be distributed so that each teacher has the SAME number of students. This is a YES/NO question.

1) It is possible to divide the students evenly into groups of 2, 3, 5, 6, 9, 10, or 15.

With this list of 7 numbers, we can use the prime-factorization of each number to 'narrow down' what the total number of students could be...

With a 2, two 3s and a 5, we can create all 7 of the numbers involved. This means that the total number of students MUST be a multiple of (2)(3)(3)(5) = 90.

When we combine this information with the fact that there are FEWER than 150 students, we know that there can only be 90 students (not 180, 270, etc. since those numbers are all greater than 150).

Unfortunately, we don't know the number of teachers...

IF there are 9 teachers, then the answer to the question is YES.
IF there are 8 teachers, then the answer to the question is NO.
Fact 1 is INSUFFICIENT.

2) The greatest common factor of a and b is 10.

This Fact tells us that both the number of students and the number of teachers are multiples of 10, but those two numbers only have a GCF of 10.

IF the number of students is 90 and the number of teachers is 10, then the answer to the question is YES.
IF the number of students is 90 and the number of teachers is 20, then the answer to the question is NO.
Fact 2 is INSUFFICIENT.

Combined, there's no additional work needed. We already have two TESTs that fit both Facts, but provide different answers:

IF the number of students is 90 and the number of teachers is 10, then the answer to the question is YES.
IF the number of students is 90 and the number of teachers is 20, then the answer to the question is NO.
Combined, INSUFFICIENT.

Final Answer: E

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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by gmat_for_life » Sun Jun 12, 2016 11:20 am
Thank you Rich! Your explanation was really helpful. I'll ensure to provide the answers going forward.

Regards,
Amit
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DS 700

by fskilnik@GMATH » Sun Sep 09, 2018 11:17 am
A school has A students and B teachers. If A < 150, B < 25, and classes have a maximum of 15 students, can the A students be distributed among the B teachers so that each class has the same number of students?
(Assume that any student can be taught by any teacher, and any class has exactly one teacher associated with it.)

(1) It is possible to divide the students evenly into groups of 2, 3, 5, 6, 9, 10, or 15.
(2) The greatest common factor of a and b is 10.
\[2 \leqslant A \leqslant 149\,\,\,\operatorname{int} \,\,\,\,\left( * \right)\,\,\,\,\left( {A\,\, = \,\,\# \,\,{\text{students}}} \right)\]
\[2 \leqslant B \leqslant 24\,\,\,\operatorname{int} \,\,\,\,\,\,\,\,\,\left( {B\,\, = \,\,\# \,\,{\text{classes}}\,\,{\text{ = }}\,\,\,\# \,\,{\text{teachers}}} \right)\]
\[\left( {\frac{{n\,\,{\text{students}}}}{{1\,\,\,{\text{class}}}}} \right)\,\,\,\left( {B\,\,{\text{classes}}} \right)\,\,\,\mathop = \limits^? \,\,A\,\,\,\mathop \Leftrightarrow \limits^{1\,\, \leqslant \,\,n\,\,\operatorname{int} \,\, \leqslant \,\,15} \,\,\,\,\,\boxed{\,\,?\,\,\,\,:\,\,\,\,1 \leqslant \,\,\frac{A}{B}\,\,\, = \,\,\operatorname{int} \,\, \leqslant 15\,\,\,}\]
\[\left( 1 \right) \cap \left( * \right)\,\,\,\left\{ \begin{gathered}
A\,\,{\text{is}}\,\,{\text{a}}\,\,{\text{multiple}}\,\,{\text{of}}\,\,LCM\left( {2,3,5,6,9,10,15} \right) = 90 \hfill \\
\left( * \right)\,\,\,\,2 \leqslant A \leqslant 149 \hfill \\
\end{gathered} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,A = 90\]
\[\left( {1 + 2} \right) \cap \left( * \right)\,\,\,\left\{ \begin{gathered}
A = 90\,\,\,\,\,;\,\,\,\,\,2 \leqslant B \leqslant 24\,\,\,\operatorname{int} \,\,\, \hfill \\
GCD\left( {A,B} \right) = 10 \hfill \\
\end{gathered} \right.\]
\[\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {{\text{A,B}}} \right) = \left( {90,10} \right)\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\left( {\frac{{90}}{{10}} = 9} \right) \hfill \\
\,{\text{Take}}\,\,\left( {{\text{A,B}}} \right) = \left( {90,20} \right)\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\,\,\,\,\left( {\frac{{90}}{{20}} \ne \operatorname{int} } \right) \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,{\text{INSUFF}}.\]

The right answer is therefore [spoiler]__(E)__[/spoiler].

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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