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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]  08 Feb 2011, 03:33
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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

[Reveal] Spoiler:
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.
[Reveal] Spoiler: OA

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Re: In a certain class [#permalink]  08 Feb 2011, 06:11
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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
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Re: In a certain class [#permalink]  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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Kudos [?]: 2 [0], given: 9

Re: In a certain class consisting of 36 students, some boys and [#permalink]  14 Oct 2013, 22:07
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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Kudos [?]: 12 [1] , given: 36

Re: In a certain class consisting of 36 students, some boys and [#permalink]  13 Dec 2013, 10:27
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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]  28 Jan 2015, 05:18
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Re: In a certain class consisting of 36 students, some boys and   [#permalink] 28 Jan 2015, 05:18
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