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

Let # of boys be \(b\), then # of girls will be \(36-b\). We want to maximize \(\frac{b}{3}+\frac{36-b}{4}\) --> \(\frac{b}{3}+\frac{36-b}{4}=\frac{b+3*36}{12}=\frac{b}{12}+9\), so we should maximize \(b\), but also we should make sure that \(\frac{b}{12}+9\) remains an integer (as it represent # of people). Max value of \(b\) for which b/12 is an integer is for \(b=24\) (b can not be 36 as we are told that there are some # of girls among 36) --> \(\frac{b}{12}+9=2+9=11\).

Answer: C.

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 \(b=24\) --> \(\frac{b}{3}+\frac{36-b}{4}=11\).

Similar question: least-number-of-homeowners-106175.html

Hope it's clear.
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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

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.

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.

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 [spoiler]11[/spoiler] children walk

Answer: C

Cheers,
Brent

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
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can't we consider the group full of boys? Because the core doesn't give us information on having/not having girls in the class, so as to make the ques tricky.

b + g = #walking
3b + 4g = 36

1) Take extremes: g=0 means 12 walked, b=0 means 9 walked. This eliminates A, D, E -- answer is either B) 10 or C) 11 since we must have a positive integer constraint on the number of boys and girls.

2) Note that taking g=0 gives a larger amount (12), so we want to maximize the number of boys and minimize the number of girls, we can test C) first

3) Also note that 11 is Odd, which means b + g is E+O or O+E. HOWEVER, we cannot have b as Odd because 3b + 4g = 36 (Even). This means b is even and g is odd.

Take g=1, 36-4 = 32, not div by 3
Take g=3, 36-12 = 24, div by 3
So there are 24/3 = 8 boys and 12/4 = 3 girls for a total of 11.

Given: 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.

Asked: What is the greatest possible number of students in this class who walk to school?

Let the number of boys be x
number of girls = 36-x

Number of students walking to school = y = x/3 + (36-x)/4 = 9 + x(1/3 - 1/4) = 9 + x/12

Greatest x = 24 < 36
Greatest number of students walking to school = 9 + 2 = 11

IMO C

The sum although a little difficult is purely number manipulation.

The question asks how do i manipulate the numbers to get maximum number of people walking..

The class is divided into boys and girls and from these boys and girls we have 1/3 boys (33% boys) that walk and 1/4 girls (25%) girls that walk.

So its clear to me that, i want a larger number of boys than girls because they will enroll a higher percentage of walking people ( 33% of Boys vs 25% of girls)

The question now becomes how many boys should be considered. It cannot be a random number between 1 and 36 since 1/3 of that number may be in decimals and we cannot have decimals representing a boy (it would mean that we are only covering some random body part of the boy..lol...which wouldn't be correct. It has to be a whole number)

If i consider 36 boys, which is divisible by 3, that would mean that 12 boys walk however if all 36 students are boys then where will we accommodate girls (the question mentions both boys and girls are present in the class of 36 students).

In order to derive a number where the number of boys are a multiple of 3 and the number of girls are a multiple of 4, we need to take the lowest common factor, which in this case is 12. the different possible numbers could be 12 , 24 and 36 (36 is ruled out however as mentioned above)

The next highest is 24 boys and hence 12 girls, since 24 boys and 12 girls = 36 students

1/3 * 24 = 8
1/4 * 12 = 3

Total = 8+3 = 11 (highest number of walking students)

Answer is C

Hope this helps a little

Since Boys who walk are 1/3 rd and Girls who walk are 1/4th , there are more boys who walk, and we should maximize boys.

Now 1/3 of Girls who walk should come as an integer and 1/4th of Boys should alos be an interger. Minimum value which will give both as interger is LCM (3,4) = 12

Ths girls = 12
boys = 36-12 = 24

thus total students who walk is 12/4 + 24/3 = 11

Kudos if you liked the approach !

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.

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since \(\frac{1}{3} > \frac{1}{4}\) ,
therefore we want to maximize the number of boys --> which also means we want to minimize number of girls

We also know that number of boys and girls would be integers

Step 1: 32 boys + 4 girls
Not possible as 32/3 not an integer

Step 2: 28 boys + 8 girls
Still not possible as \(\frac{28}{3}\) is not an integer

Step 3: 24 boys +12 girls
Both are integers now, all we have to do is calculate it
\(\frac{24}{3} + \frac{12}{4 }= 8+3=11\)

Answer: C

Number of boys = B
Number of girls = 36 - B

Number of boys who walk to school = B/3
Number of girls who walk to school = (36 - B)/4

M = B/3 + (36 - B)/4

To maximize: M
M = B/3 + 9 - B/4 = B/12 + 9 (B must be divisible by 12)
=> Maximum value of B = 24

=> M = 2 + 9 = 11

Answer: C
_________________

Since 1/3 > 1/4

so, number of students in this class who walk to school will be maximum if boy students > girl students
LCM of 3 & 4 is 12

The number of Boy students Or Girl students must be a multiple of 12

So, the number of boys students is 24 for it needs to be greater and a multiple of 12
The number of girls is 12

number of students in this class who walk to school 24/3 + 12/4 = 8+3=11

Best thing to do here is to set up the equation and then randomly select the answer choices.

Let x be boys and y be girls

x + y = 36
y = 36 - x

x/3 + y/4 = 11
x/3 + 36 - x /4 = 11
x = 24

y = 12

Answer is C.

Just to be sure we can try D too.

x/3 + y/4 = 12
x = 36 <----Can't be right since there are 36 males AND females so you can't have 36 males alone.
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but the Q stem gave info regarding "some boys and some girls" so we cant consider full or zero

Boys + Girls = 36
Question asked: What's max of (1/4)G + (1/3) B?

Ans: Max of (1/4)G + (1/3) B = A
(1/4)*(B+G) < A < (1/3)*(B+G)
9< A <12
Hence A must be 11.

Is this approach correct Bunuel

Total No. of students = 36
Now that we know 1/3 of boys will walk to school. Therefore, no of boys should be a multiple of 3
Now that we know 1/4 of girls will walk to school. Therefore, no of girls should be a multiple of 4

3B+4G=36
always remember when both the variables in the single equation are of positive sign, value of one variable decreases as the value of another increases. Also, B will increase/decrease by a factor of 4 and G will decrease/increase by a factor of 3 i.e. the factor will the constant with other variable.

Therefore, possible values of B & G are as follows:
B=12, G=0 [Not Possible as there are some boys and some girls]
B=8, G=3 (Since, B is decreasing by 4 (constant of G), therefore, G will increase by 3 (constant of B)) B+G=11 is the correct answer.
B=4, G=6 B+G=10
B=0, G=9 [Not Possible as there are some boys and some girls]