Class of boys & girls

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GmatGreen: Senior | Next Rank: 100 Posts; Posts: 85; Joined: Tue Mar 31, 2009 10:58 am

by GmatGreen » Wed Dec 11, 2013 8:40 am

In a certain class consisting of 36 students, some boys and some girls, exactly 1/3 of the boys and exactly 1/4 of the girls walk to school. What is the greatest possible number of students in this class who walk to school?

A) 9

B) 10

C) 11

D) 12

E) 13

Can this be solved in under 2 minutes?

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Source: Beat The GMAT — Problem Solving | Wed Dec 11, 2013 8:40 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Wed Dec 11, 2013 8:57 am

GmatGreen wrote:In a certain class consisting of 36 students, some boys and some girls, exactly 1/3 of the boys and exactly 1/4 of the girls walk to school. What is the greatest possible number of students in this class who walk to school?

A) 9
B) 10
C) 11
D) 12
E) 13

The important thing here to recognize here is that the number of girls and the number of boys must be positive INTEGERS. For example, we can't have 5 1/3 boys.

Also recognize that we're told that we have some boys and some girls
Since "some" means 1 OR MORE, we cannot have zero boys or zero girls.

Okay, now onto the question...

We want to MAXIMIZE the number of students who walk to school. Since a greater proportion of boys walk to school, we want to MAXIMIZE the number of boys in the class.
The greatest number of boys is 35 (since 36 boys would mean 0 girls, and we must have at least 1 girl)

35 boys
This is no good, because 1/3 of the boys walk to school, and 35 is not divisible by 3.

So, let's try ...
34 boys
This is no good, because 1/3 of the boys walk to school, and 34 is not divisible by 3.

As you can see, we need only consider values where the number of boys is divisible by 3. So, that's what we'll do from here on...

33 boys
If 1/3 of the boys walk to school, then 11 boys walk. Fine.
HOWEVER, if there are 33 boys, then there must be 3 girls.
If 1/4 of the girls walk to school, then there can't be 3 girls, since 3 is not divisible by 4.

30 boys
This means there are 10 girls
If 1/4 of the girls walk to school, then there can't be 10 girls, since 10 is not divisible by 4.

27 boys
This means there are 13 girls
If 1/4 of the girls walk to school, then there can't be 10 girls, since 13 is not divisible by 4.

24 boys and 12 girls
1/3 of the boys walk to school, so 8 boys walk
1/4 of the girls walk to school, so 3 girls walk

So, a total of 11 children walk

Answer: C

Cheers,
Brent

Last edited by Brent@GMATPrepNow on Wed Dec 11, 2013 10:20 am, edited 1 time in total.

Brent Hanneson - Creator of GMATPrepNow.com

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Patrick_GMATFix: GMAT Instructor; Posts: 1052; Joined: Fri May 21, 2010 1:30 amFollowed by: 98 members; Thanked: 335 times

by Patrick_GMATFix » Wed Dec 11, 2013 9:06 am

This solution is taken from the GMATFix App. My signature below has more info.

-Patrick

Check out my site: GMATFix.com

To prep my students I use this tool >> (screenshots, video)

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Abhishek009: Master | Next Rank: 500 Posts; Posts: 359; Joined: Wed Mar 11, 2009 4:37 amFollowed by: 2 members; Location: Kolkata, India; Thanked: 50 times

by Abhishek009 » Wed Dec 11, 2013 10:32 am

GmatGreen wrote:In a certain class consisting of 36 students, some boys and some girls, exactly 1/3 of the boys and exactly 1/4 of the girls walk to school. What is the greatest possible number of students in this class who walk to school?

A) 9

B) 10

C) 11

D) 12

E) 13

Can this be solved in under 2 minutes?

Class consisting of 36 students, some boys and some girls..

So, B + G = 36

1/3 of the boys and exactly 1/4 of the girls walk to school...

So , total no of Boys in the class must be multiple of 3 and total number of girls in the class must be multiple of 4

Start forming sets ( B , G ) satisfying the above condition and total Number of students (36 in this case)...

24 , 12 can be the only possible combination...

So Students walking to class is 1/3 * 24 + 1/4*12 => 8 + 3 = 11...

You simple have to form set of possible values of ( B, G) satisfying the condition that it must be Multiple of 3 and 4 and Sum of ( B + G) will be 36 to reach the answer...

Plz revert in case of any doubt..

Abhishek

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