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In 2003, the number of girls attending Jefferson High School [#permalink]
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29 Dec 2010, 11:59
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In 2003, the number of girls attending Jefferson High School was equal to the number of boys. In 2004, the population of girls and the population of boys both increased by 20 percent. Which of the following could be the total student population at Jefferson High School in 2004? A. 4832 B. 5034 C. 5058 D. 5076 E. 5128
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Last edited by Bunuel on 20 Sep 2013, 03:35, edited 1 time in total.
Renamed the topic and edited the question.



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Re: % change [#permalink]
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29 Dec 2010, 15:59
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Total number of students Y 2004 (B = boys per 2003 and G = girls per 2003) \(Y = B*1.2 +G*1.2\) \(Y = 2.4B\) (since B=G) \(B = Y/2.4 => B= 10Y/24\) \(Factors = 24, 12, 8, 6, 4, 3, 1\) Our answer will be divisible by all these factors without leaving a remainder First, \(A)4+8+3+2 = 17\) (not divisble by three = out) \(B)5+0+3+4 = 12\) \(C)5+0+5+8 = 18\) \(D)5+0+7+6 = 18\) \(E)5+1+2+8 = 16\) (not divisible by three = out) B,C,E left If the two last digits are divisible by four so will the whole number \(34/4 = not integer\) \(58/4 = not integer\) \(76/4 = 60 + 16 = 15*4 + 4*4 = 19\) D = correct
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Re: % change [#permalink]
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29 Dec 2010, 20:57
the fig of 2004 is 120 (60+60) ( if we consider the fig of 2003 as 100 ( 50+50) ) now ideally 120 is ( 6 parts of 20 each , we require 5 parts of 20 to make it 100) , so the answer must be divisible by 6 B,C and D are left , dividing each with 6 brings (839,843 and 846) which is one part (20%) and we require five parts to make it to 100 ( 2003 population) the no must be even since it is boys + girls  839*5 and 843*5 are singular, so the best solution is D
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Re: % change [#permalink]
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29 Dec 2010, 21:29
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shrive555 wrote: In 2003, the number of girls attending Jefferson High School was equal to the number of boys. In 2004, the population of girls and the population of boys both increased by 20 percent. Which of the following could be the total student population at Jefferson High School in 2004?
4832 5034 5058 5076 5128 Let number of boys = number of girls = a (An integer) Total number of students = 2a New total number of students = (6/5)*2a (Since each a increased by 20%, the total will increase by 20% i.e. an increase of 1/5) Now (12/5)*a is the answer i.e. (12/5)*a = 4832 or 5034 or 5058 etc.... or a = Answer Option * 5/12 The only restriction is that 'a' should be an integer. So the answer option must be divisible by 12. Check the options: Which one is divisible by both 3 and 4? Only 5076.
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Re: In 2003, the number of girls attending Jefferson High School [#permalink]
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18 Jul 2014, 00:02
Here I did it in a following way.
Lets assume in 2003 we have 100 students and both boys and girls are equal.
total = 100 Boys = 50 girls = 50
now 20% increment in both boys and girls
boys = 60 girls = 60
here total student = 120. we can say total 20% increment on 100
now we can say 6/5 x = value from options or x= 5/6 (value from option. Should be multiple of 6. only D fits in it)



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Re: In 2003, the number of girls attending Jefferson High School [#permalink]
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22 Jul 2014, 09:38
VeritasPrepKarishma wrote: shrive555 wrote: In 2003, the number of girls attending Jefferson High School was equal to the number of boys. In 2004, the population of girls and the population of boys both increased by 20 percent. Which of the following could be the total student population at Jefferson High School in 2004?
4832 5034 5058 5076 5128 Let number of boys = number of girls = a (An integer) Total number of students = 2a New total number of students = (6/5)*2a (Since each a increased by 20%, the total will increase by 20% i.e. an increase of 1/5) Now (12/5)*a is the answer i.e. (12/5)*a = 4832 or 5034 or 5058 etc.... or a = Answer Option * 5/12 The only restriction is that 'a' should be an integer. So the answer option must be divisible by 12. Check the options: Which one is divisible by both 3 and 4? Only 5076. HI Karishma, I have one query here. say we have x students (x= integer and x/2 is also an integer) now girls and boys are x/2 each. if we say 20 % increment means 6/5 x or we can say x = 5/6*options. Here c and d both are divided by 6. then how we will choose correct answer.? Thanks



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Re: In 2003, the number of girls attending Jefferson High School [#permalink]
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22 Jul 2014, 22:37
PathFinder007 wrote: say we have x students (x= integer and x/2 is also an integer)
now girls and boys are x/2 each.
if we say 20 % increment means 6/5 x or we can say x = 5/6*options.
Here c and d both are divided by 6. then how we will choose correct answer.?
Thanks
What you forgot to consider here is that x and x/2, both, should be integers. This means, after you divide the option by 6 and get x (after multiplying by 5 which doesn't matter), it should again be divisible by 2. Hence, the option must be successively divisible by 6 and 2 i.e. it must be divisible by 12 i.e by 3 as well as 4. We see that (C) and (D) are both divisible by 3. But only (D) is divisible by 4. Answer (D)
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In 2003, the number of girls attending Jefferson High School [#permalink]
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22 Jul 2014, 23:47
shrive555 wrote: In 2003, the number of girls attending Jefferson High School was equal to the number of boys. In 2004, the population of girls and the population of boys both increased by 20 percent. Which of the following could be the total student population at Jefferson High School in 2004?
A. 4832 B. 5034 C. 5058 D. 5076 E. 5128 There is 1 key hint in this questions that will help you narrow down the answer. "In 2004, the population of girls and the population of boys both increased by 20 percent."This means that \(New Population=\frac{6}{5}*Original population=New Population*\frac{5}{6}=*Original population\) Since the number we are looking for is an integer, the number MUST BE divisible by 6.Only Answer that will fulfill this requirement is D.



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In 2003, the number of girls attending Jefferson High School [#permalink]
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23 Jul 2014, 00:11
PathFinder007 wrote: VeritasPrepKarishma wrote: shrive555 wrote: In 2003, the number of girls attending Jefferson High School was equal to the number of boys. In 2004, the population of girls and the population of boys both increased by 20 percent. Which of the following could be the total student population at Jefferson High School in 2004?
4832 5034 5058 5076 5128 Let number of boys = number of girls = a (An integer) Total number of students = 2a New total number of students = (6/5)*2a (Since each a increased by 20%, the total will increase by 20% i.e. an increase of 1/5) Now (12/5)*a is the answer i.e. (12/5)*a = 4832 or 5034 or 5058 etc.... or a = Answer Option * 5/12 The only restriction is that 'a' should be an integer. So the answer option must be divisible by 12. Check the options: Which one is divisible by both 3 and 4? Only 5076. HI Karishma, I have one query here. say we have x students (x= integer and x/2 is also an integer) now girls and boys are x/2 each. if we say 20 % increment means 6/5 x or we can say x = 5/6*options.Here c and d both are divided by 6. then how we will choose correct answer.? Thanks Not Karishma, but let's try \(x = \frac{5}{6} * (options)\) also \(\frac{x}{2} = \frac{5}{12} * (options)\) (Both conditions should satisfy)Option C \(= \frac{5058}{6} = 843\) >> 843 is a odd number; so it cannot be divided equally (No. of girls = No. of boys)Also \(\frac{x}{2}\) condition is not fulfilled So option C = 5058 has to be ignored. Answer = D = 5076
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Re: In 2003, the number of girls attending Jefferson High School [#permalink]
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23 Jul 2014, 00:15
justbequiet wrote: shrive555 wrote: In 2003, the number of girls attending Jefferson High School was equal to the number of boys. In 2004, the population of girls and the population of boys both increased by 20 percent. Which of the following could be the total student population at Jefferson High School in 2004?
A. 4832 B. 5034 C. 5058 D. 5076 E. 5128 There is 1 key hint in this questions that will help you narrow down the answer. "In 2004, the population of girls and the population of boys both increased by 20 percent."This means that \(New Population=\frac{6}{5}*Original population=New Population*\frac{5}{6}=*Original population\) Since the number we are looking for is an integer, the number MUST BE divisible by 6.Only Answer that will fulfill this requirement is D. It must be divisible by 12; 5058 is divisible by 6; but it yields a wrong answer. Refer explanation above
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Re: In 2003, the number of girls attending Jefferson High School [#permalink]
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Be quick D
Girls = Boys = a
Number of students in 2014: 1,2*a + 1,2*a = 2,4*a
Now, we need an answer that could be divisible by 24= 3*2*2*2
Only B , C, D is divisible by 3, and only D is divisible by 4
Therefore, D is a correct answer for this question



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Re: In 2003, the number of girls attending Jefferson High School [#permalink]
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03 Aug 2014, 00:21
shrive555 wrote: In 2003, the number of girls attending Jefferson High School was equal to the number of boys. In 2004, the population of girls and the population of boys both increased by 20 percent. Which of the following could be the total student population at Jefferson High School in 2004?
A. 4832 B. 5034 C. 5058 D. 5076 E. 5128 in 2003: G = B = x in 2004: 1.2G = 1.2B = 1.2x total student = 2.4x = 24*0.1*x = 3*4*2*0.1*x so 3,4,2 must be the divisor of total number of students A and E are eliminated since they cannot be divided by 3 B and C are eliminated since they cannot be divided by 4 D is the answer. +1 Kudos if you guys like my answer. THanks
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Re: In 2003, the number of girls attending Jefferson High School [#permalink]
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27 Sep 2017, 18:37
shrive555 wrote: In 2003, the number of girls attending Jefferson High School was equal to the number of boys. In 2004, the population of girls and the population of boys both increased by 20 percent. Which of the following could be the total student population at Jefferson High School in 2004?
A. 4832 B. 5034 C. 5058 D. 5076 E. 5128 Given that 2x is the student population in 2003 (as there are equal amounts of both boys and girls) then 2004 is: 2x*(6/5) = total 12x/5 = total x = [(total)*5]/12 So the total has to be divisible by 12, which is the same as being divisible by 4 and 3. B, C, D are all divisible by 3 and D is the only one that is divisible by 4 (the last two digits are a multiple of 4). (D)




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