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 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.

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

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manutdtillidie

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Another Approach

# of boys = x
# of girls = y
since no. of boys should be a multiple of 3 and no. of girls should be multiple of 4 .
3x+4y=36

we can see the unit digit is 6. ... the only option to get unit digit as 6 is to get unit digit of 3x as 4 and unit digit of 4y as 2 . 3*8=24
4*3=12

= no of boys and girls = 8+3 = 11

is this correct Bunuel

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As per given scenario, the no. of girls must be divisible by 4 and the no. of boys must be divisible by 3
Also total no. of boys + girls =36
As per given scenario only two cases possible
1. boys = 12 and girls=24
therefore no. of students who walk to school in this case = (12/3) + (24/4)=10
2. boys = 24 and girls=12
therefore no. of students who walk to school in this case = (24/3) + (12/4)=11

Therefore maximum o. students who walk to school =11
hence C

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01 Aug 2023, 11:46

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GMATD11

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

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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.

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

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vikrantgulia

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.

I would like to add one thing that we don't need to try putting value here. Go by logic. Maximum num. who walk ? how can be maximize? b>g . 24 is only possible solution.

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Harsha
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23 Aug 2025, 07:18

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Bunuel

Or: as there are bigger percentage of boys who walk then we should maximize # of boys, but we should ensure that $\frac{b}{3}$ and $\frac{36-b}{4}$ are integers. So $b$ should be max multiple of 3 for which $36-b$ is a multiple of 4, which turns out to be for $b=24$ --> $\frac{b}{3}+\frac{36-b}{4}=11$.

Hi,

On the similar lines of your 2nd approach, if we try to answer the least possible number of students walking to school, the answer to it would be 10, right?

isaperelli

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So we have 36 pp. We don't know how many are Boys (B) and how many are girls (G). BUT we know the fraction of which of each walk.
So the easiest approach is:
36 pp--> usually to maximise we would divide in half the amount we have. so 36:2=18. But our fractions are 1/3 and 1/4. 18 isn't divisible by both 3 and 4 (18 is divisible only by 3). What we should do is find the smallest number that is divisible by both 3 and 4 --> that is 3x4=12. This will be our smallest number that will be assigned to the smallest fraction. Now we know that 36= 12 (we just found) + 24 (36-12).
Since we want to Maximise we need to make the greatest fraction have the greatest amount of pp. Given 1/3 > 1/4, we assign 24 to 1/3 and find that 1/3 of the boys means 8 boys.
1/4 of the girls which are 12=3.
Solution 8+3=11 (C)

GMATD11

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B+G=36
B/3+G/4= Max
For this to be max, B should be max as a greater proportion of boys walk to school.

As B & G can only be positive integers, and no. of students walking to school also has to be an integer, this means B will be multiple of 3 and G will be multiple of 4.
We can write it as:
3n+4m= 36
3n= 4*9-4m
3n= 4(9-m)

Now for n&m to be integers, n has to be a multiple of 4.
This means that B has to be a multiple of 12 (3&4 both).
To maximise B, B=24 as the next multiple is 36, which will mean no girls, hence not possible.

Putting B=24 & G=12, we get 11 as our answer.

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10 is possible as x=12, y=24 -> 4+6 = 10

infymys

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.