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In a certain class consisting of 36 students, some boys and [#permalink]
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08 Feb 2011, 04:33
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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 let number of boys in a class are x then number of gals become 36x
1/3x+1/4(36x)=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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Re: In a certain class [#permalink]
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08 Feb 2011, 07:11
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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. 364 girls = 32 boys not divisible by three 368 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]
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08 Feb 2011, 07:50
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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 \(36b\). We want to maximize \(\frac{b}{3}+\frac{36b}{4}\) > \(\frac{b}{3}+\frac{36b}{4}=\frac{b+3*36}{12}=\frac{b}{12}+9\), 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 \(\frac{36b}{4}\) are integers. So \(b\) should be max multiple of 3 for which \(36b\) is a multiple of 4, which turns out to be for \(b=24\) > \(\frac{b}{3}+\frac{36b}{4}=11\). Similar question: leastnumberofhomeowners106175.htmlHope it's clear.
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Re: In a certain class consisting of 36 students, some boys and [#permalink]
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14 Oct 2013, 23:07
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 36x
1/3x+1/4(36x)=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]
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13 Dec 2013, 11:27
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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.



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