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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] New post 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

Answer C

any other method to solve this question.
[Reveal] Spoiler: OA

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Re: In a certain class [#permalink] New post 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] New post 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.

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

Answer C

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] New post 13 Dec 2013, 10:27
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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.:P
Re: In a certain class consisting of 36 students, some boys and   [#permalink] 13 Dec 2013, 10:27
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