In a certain class consisting of 36 students, some boys and

Question

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

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

let number of boys in a class are x
then number of gals become 36-x

1/3x+1/4(36-x)=9+x/12

x cannot be 36 as there are some gals in class
so maximum value of x/12 can be 2

Answer C

any other method to solve this question.

Bunuel · Accepted Answer

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

Let the number of boys be \(b\), then the number of girls will be \(36-b\). We want to maximize \(\frac{b}{3}+\frac{36-b}{4}\).

\(\frac{b}{3}+\frac{36-b}{4}=\frac{4b+3(36-b)}{12}=\frac{b+108}{12}=\frac{b}{12}+9\)

So we should maximize \(b\), but we must also ensure that \(\frac{b}{12}+9\) remains an integer, since it represents the number of students. The greatest possible value of \(b\) for which \(\frac{b}{12}\) is an integer is \(24\) (\(b\) cannot equal 36 because the problem states that there are some girls in the class).

Thus,

\(\frac{b}{12}+9=2+9=11\)

Answer: C.

Or: since a greater fraction of boys walk to school, we should maximize the number of boys, while ensuring that \(\frac{b}{3}\) and \(\frac{36-b}{4}\) are both integers. So \(b\) should be the greatest multiple of 3 for which \(36-b\) is a multiple of 4, which occurs when \(b=24\).

Thus,

\(\frac{b}{3}+\frac{36-b}{4}=11\)

Answer: C.

Similar question: https://gmatclub.com/forum/least-number ... 06175.html

Hope it's clear.

Mackieman · Answer

Since 1/3 boys > 1/4 girls you want to maximize no of boys

The no of girls has to be a multiple of 4 since we cannot have half girls and no of boys multiple of 3

Working backwards.

36-4 girls = 32 boys not divisible by three
36-8 girls = 28 boys not divisible by three
36 - 12 girls = 24 boys which is divisible by three

so 24 boys and 12 girls

and a total of 24*1/3 + 12*1/4 walking to school = 11

vikrantgulia · Answer

Answer is C.

It can be easily solve by using a number which is multiple of 3 & 4 together and less than 36.
So the number would be 12 & 24 only. Consider one of the number as count of boys or girls.

Say B=12 then G=24 which means 1/3 of 12 + 1/4 of 24=10
Now try for 24. Say B=24 then G=12 which means 1/3 of 24 + 1/4 of 12=11

So answer is C in less than a minute.
+1 for me. cheers.

infymys · Answer

Lets apply process of elimination

option A: 
See, 9 cannot be expressed as sum of multiple of 3 and 4.
neither can 10, 12 or 13.
only 11 can be expressed a sum of multiples of 3 and 4.

11= 3+ 2(4)

since we need 1/3 of boys, 1/4th of girls. and number of boys and girls have to be integers.

11 is the only option that satisfies the situation.

BrentGMATPrepNow · Answer

The important thing here to recognize here is that the number of girls and the number of boys who walk 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 6 girls
If 1/4 of the girls walk to school, then there can't be 6 girls, since 6 is not divisible by 4.

27 boys  
This means there are 9 girls
If 1/4 of the girls walk to school, then there can't be 9 girls, since 9 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
PERFECT - it works!!
So, a total of  11  children walk

Answer: C

Cheers,
Brent

ScottTargetTestPrep · Answer

We can let n = the number of boys; thus 36 - n = the number of girls. We are given that (1/3)n boys walk to school and (1/4)(36 - n) = 9 - (1/4)n girls walk to school.

Since (1/3)n and 9 - (1/4)n must be an integer, we see that n must be divisible by 3 and 4. In other words, n must be divisible by 12. Thus n can be either 12 or 24 (we exclude 0 and 36 since if n = 0, there will be no boys in the class, and if n = 36, there will be no girls in the class).

If n = 12, then (1/3)(12) = 4 boys and 9 - (1/4)(12) = 9 - 3 = 6 girls walk to school. That is, a total of 4 + 6 = 10 students walk to school.

If n = 24, then (1/3)(24) = 8 boys and 9 - (1/4)(24) = 9 - 6 = 3 girls walk to school. That is, a total of 8 + 3 = 11 students walk to school.

Thus the greatest possible number of students in this class who walk to school is 11.

Answer: C

pragya007 · Answer

Easy way to think about it 
We need check the multiple of 12 as person can't be in fraction so it should ne multiple of 3 and 4 so that we will get absolute value of no of boys and girls walk to school so the multiple of 12 under 36 is 12, 24 ... Cant be 36 as there must be atleast one boy or girl as its mentioned some boys and girls. Hence you left woth 24 and 12 . Now as u want to max we need to give boys 24 as it is 1/3 boys walk hence 8 and the remaining 12 girls give 3 . So total 11. 
This whole process should take about 30 sec max.

Posted from my mobile device

SS.56 · Answer

We can solve this using the Diophantine equation,

3b+4g=36
I. 3*12+4*0=36 (12+0=12) 
II. 3*8+4*3=36 (8+3=11)
III. 3*4+4*6=36 (4+6=10)
IV. 3*0+4*9=36 (0+9=9)

I and IV are not possible because there are some boys and girls. So, between II and III, 11 is greatest.

Answer C

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=82999 Harsha GMAT Tutor | gmatanchor.com GMAT-Club-Forum-bz6753tn.png

KevinMercurio · Answer

How is the answer not D? When rearranging the equation 1/3B + 1/4G = Max and B+G=36 I Get 12-G/12 = Max, thus with 0 girls in the class we get a value of 12. The question does not explicitly state there MUST be girls and boys or that we cannot use the value of 0.

Bunuel · Answer

The stem says “some boys and some girls,” so both groups must be present. Therefore, the number of girls cannot be 0. This is mentioned in the solutions above by the way.

In a certain class consisting of 36 students, some boys and

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