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

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In a certain class consisting of 36 students, some boys and  [#permalink]

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08 Feb 2011, 03:33
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Question Stats:

62% (01:57) correct 38% (02:22) wrong based on 753 sessions

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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

any other method to solve this question.

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Re: In a certain class  [#permalink]

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08 Feb 2011, 06:50
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GMATD11 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

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) --> $$\frac{b}{12}+9=2+9=11$$.

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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Re: In a certain class  [#permalink]

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08 Feb 2011, 06:11
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3
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
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Re: In a certain class consisting of 36 students, some boys and  [#permalink]

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14 Oct 2013, 22:07
2
GMATD11 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

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

any other method to solve this question.

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.
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Re: In a certain class consisting of 36 students, some boys and  [#permalink]

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13 Dec 2013, 10:27
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3

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.
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Re: In a certain class consisting of 36 students, some boys and  [#permalink]

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26 Nov 2017, 14:17
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Top Contributor
GMATD11 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 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

Cheers,
Brent
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Re: In a certain class consisting of 36 students, some boys and  [#permalink]

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29 Nov 2017, 10:04
GMATD11 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

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.

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Re: In a certain class consisting of 36 students, some boys and  [#permalink]

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01 Dec 2018, 14:10
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Re: In a certain class consisting of 36 students, some boys and   [#permalink] 01 Dec 2018, 14:10
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