A school has a students and b teachers. If a

Question

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

(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

CounterSniper · Answer

Here is my take on this .

(1) It is possible to divide the students evenly into groups of 2, 3, 5, 6, 9, 10, or 15
we have to find LCM of the given numbers
We therefore have 90 students

case 1
say number of teachers = 15 (b<25) 
then yes students can be evenly distributed

case 2
when no of teachers = 24 
then no

(2) The greatest common factor of a and b is 10
a and b can be 100 and 10
or
  90 and 20  
or
90 and 10

insufficient

using both still insufficient

E

fskilnik · Answer

2 \leqslant A \leqslant 149\,\,\,\operatorname{int} \,\,\,\,\left( * ight)\,\,\,\,\left( {A\,\, = \,\,\# \,\,{	ext{students}}} ight)

2 \leqslant B \leqslant 24\,\,\,\operatorname{int} \,\,\,\,\,\,\,\,\,\left( {B\,\, = \,\,\# \,\,{	ext{classes}}\,\,{	ext{ = }}\,\,\,\# \,\,{	ext{teachers}}} ight)

\left( {\frac{{n\,\,{	ext{students}}}}{{1\,\,\,{	ext{class}}}}} ight)\,\,\,\left( {B\,\,{	ext{classes}}} ight)\,\,\,\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 ight) \cap \left( * ight)\,\,\,\left\{ \begin{gathered}
 A\,\,{	ext{is}}\,\,{	ext{a}}\,\,{	ext{multiple}}\,\,{	ext{of}}\,\,LCM\left( {2,3,5,6,9,10,15} ight) = 90 \hfill \
 \left( * ight)\,\,\,\,2 \leqslant A \leqslant 149 \hfill \ 
\end{gathered} ight.\,\,\,\,\,\, \Rightarrow \,\,\,\,A = 90

\left( {1 + 2} ight) \cap \left( * ight)\,\,\,\left\{ \begin{gathered}
 A = 90\,\,\,\,\,;\,\,\,\,\,2 \leqslant B \leqslant 24\,\,\,\operatorname{int} \,\,\, \hfill \
 GCD\left( {A,B} ight) = 10 \hfill \ 
\end{gathered} ight.

\left\{ \begin{gathered}
 \,{	ext{Take}}\,\,\left( {{	ext{A,B}}} ight) = \left( {90,10} ight)\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{	ext{YES}}} ightangle \,\,\,\,\,\,\,\,\,\left( {\frac{{90}}{{10}} = 9} ight) \hfill \
 \,{	ext{Take}}\,\,\left( {{	ext{A,B}}} ight) = \left( {90,20} ight)\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{	ext{NO}}} ightangle \,\,\,\,\,\,\,\,\,\left( {\frac{{90}}{{20}} 
e \operatorname{int} } ight) \hfill \ 
\end{gathered} ight.\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,{	ext{INSUFF}}.

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

Regards,
fskilnik.

Mo2men · Answer

The question asks if we can evenly divide the students into classes.

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

The least common multiple =90 students

The number teachers < 25

Let b= 20

\frac{a}{b}=\frac{90}{20}=\frac{9}{2} .............Answer is No

Let b= 10

\frac{a}{b}=\frac{90}{10}=\frac{9}{1} .............Answer is yes

Insufficient

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

Use same examples above

Let a=90 & b=20

Let a=90 & b=10

Insufficient

Combine 1 & 2

Use same examples above........No conclusive answer

Insufficient

Answer: E

Kezia9 · Answer

Hi,

I have a doubt. When we combine the statements. 90/20 is not a whole number. Hence we can eliminate it and 90/10 could be the solution. Please clarify.

BijayKru · Answer

a student and b teacher. we know that a < 150 and b <25
Max number of student can possible in a class is 15

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

Is it possible? think, May be yes but we don't have the clue about the teachers. We can't find our answer with only the value we have is for students.

So possible Answer : B, C, E

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

Can we get our answer with this piece of information : Straight NO. why

May be HCF ( 10,10 ) = 10
HCF ( 20,10) = 10 So how can we find the value of X and Y. So this statement alone is also not possible . So remaining ans. C and E

Let's combine both the statement. 1 and 2

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

So lets take the LCM ( Why LCM as we need to find the evenly in the student) means less then 150 how many student group we can find , So LCM of (2, 3, 5, 6, 9, 10, 15) = 90 and yes 90<150 ---------------1

From statement 2 HCF ( X and Y) = 10, Now we know number of student =90( from statement 1) lets put the value
HCF ( 90, Y ) = 10 , So Y can be 10 ( yes possible ) BUT
HCF ( 90, 20)= 10 , Y can be 20 as well. So we can't find the actual value of Y here.

So combine statement is also not sufficient.   SO THE ANSWER IS "E"

"E"

bumpbot · Answer

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