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BTG-PQ "Jefferson High School"

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BTG-PQ "Jefferson High School"

by gnod » Tue Jun 21, 2011 7:40 pm
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

this is from the Beat the Gmat practice questions but the video explanation didn't cut it for me.

Let girls = x, and boys = x.
2003 population = 2x.

for 2004, let girls = 1.2x and boys = 1.2x
2004 population = 2.4x

2.4x = 24/10(x) = 12/5 (x)

I came to this point during the practice quiz but got stuck.
After trying to figure it out, and reviewing the solution, still don't understand how and why 2004 needs to be divisible by 12 like the video states if 2004 population = 12/5 (x).

can someone explain this part?

thank you
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by Frankenstein » Tue Jun 21, 2011 8:18 pm
Hi,
Hi, as x indicates number of girls/boys, it should be integer right?
Similarly 12x/5 should be integer right?
12x/5 will be integer when x is of the form 5k right?
As x is integer, k will be integer.
So, 12x/5 will be 12k where k is an integer
Hence, total population should be a multiple of 12
If you still don't understand my reasoning, consider and example
Let us take value not divisible by12
Let12x/5 be 6
Then x = 5/2 right?
But x is the number of girls/boys and it cannot be a fraction. It has to be an integer.
So, the number should be divisible by 12.
Cheers!

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by Tani » Tue Jun 21, 2011 8:30 pm
One approach is simply to figure out what the prior year's population would have been. Let the current number of students equal C, and the prior year's equal P. Since the population grew by 20%, C = 1.2 *P, or, restating C/(1.2) = P. Try dividing the answers (C) by 1.2 and you will find only one that gives you a whole number.
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by yjeezle » Tue Jun 21, 2011 8:33 pm
starting where you left off:

24/10 = 12/5;
12/5(2003 pop) = 2004 population
2003 population = 5x2004population/12

also we know that people need to be full integers (can't have 1/2 a person) so it needs to be divisible by 12.
test numbers by using this method: we know that 12 = 2^2 x 3; if it's divisible by 12, then 2 2's & 1 3 will be divisible by that number... save time by looking for a factor divisible by 3 first.
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by Ashley@VeritasPrep » Tue Jun 21, 2011 8:33 pm
Hey there,

A clearer (if slightly less direct) way to think of this might be to start with that logic closer to the beginning of the problem. The key is really that since 20% = 1/5, so a 20% increase means that for every five students there were originally, you gain one new student. Now if we were talking approximately (which it's key to this problem that we're *not*) or if we were talking about say, a length increasing by 20%, this wouldn't be very restrictive, because I can chop lengths into tinier and tinier decimal parts if need be, so if I needed to increase a length of 16 inches (for example) by 20%, I'd need to increase it by 1/5 of 16 inches -- in other words, by 3.2 inches. That'd be totally fine. But when we're talking about a population, we run up against the fact that it's physically impossible to have 3.2 children... I can ONLY have whole numbers of children. So if 20% (i.e. 1/5) of something needs to wind up giving me a whole number, the thing that I'm taking 1/5 of MUST BE a multiple of 5. Try to take 1/5 of anything that's not divisible by 5, and your answer will never wind up whole.

Now that we know that, we can turn to the fact that the number of girls after the increase will be 6/5 of the original number of girls (1 + 1/5 = 5/5 + 1/5 = 6/5). That means this new number MUST BE a multiple of 6, since it's 6*1/5 of the original number, and we know that 1/5 of the original number was a whole number. But now here comes the additionally tricky part -- we're told that # of boys = # of girls. So if I've got that the number of girls is a multiple of 6 and the number of boys is that same multiple of 6, then the total number of boys + girls = 2*a multiple of 6. Taking any multiple of 6 and multiplying it by 2 will yield a multiple of (6)(2), i.e. a multiple of 12. So there you go!
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by sanabk » Tue Jun 21, 2011 8:43 pm
Students 2003 2004
Girls x 1.2x
Boys x 1.2x
_____________________
Total 2x 2.4x

A) 2.4x=4832 x=4832/2.4 => x=48320/24
B) 2.4x=5034 x=5034/2.4 => x=50340/24
C) 2.4x=5058 x=5058/2.4 => x=50580/24
D) 2.4x=5076 x=5076/2.4 => x=50760/24
E) 2.4x=5128 x=5128/2.4 => x=51280/24

E
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by yjeezle » Tue Jun 21, 2011 8:48 pm
sanabk wrote:Students 2003 2004
Girls x 1.2x
Boys x 1.2x
_____________________
Total 2x 2.4x

A) 2.4x=4832 x=4832/2.4 => x=48320/24
B) 2.4x=5034 x=5034/2.4 => x=50340/24
C) 2.4x=5058 x=5058/2.4 => x=50580/24
D) 2.4x=5076 x=5076/2.4 => x=50760/24
E) 2.4x=5128 x=5128/2.4 => x=51280/24

E
I think you mean D...
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by GMATGuruNY » Tue Jun 21, 2011 8:55 pm
gnod 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
Easier example:
Let students in 2003 = 100.
Then students in 2004 = 120.
100/120 = 5/6.
This shows us that the number of students in 2003 is 5/6 of the number of students in 2004.

We can plug in the answers, which represent the number of students in 2004.
5/6 of the correct answer will yield the number of students in 2003.
The resulting number of students in 2003 must be even so that it can be split equally between the boys and the girls.
Thus, we need to be able to divide the correct answer by 6 and then by 2, implying that the correct answer is a multiple of 12.

A multiple of 12 must be a multiple both of 3 and of 4.

The sum of the digits of a multiple of 3 must be a multiple of 3.
Answer choice A: 4+8+3+2= 17, which is not a multiple of 3.
Answer choice E: 5+1+2+8 = 16, which is not a multiple of 3.
Eliminate A and E.

The last two digits of a multiple of 4 must form a multiple of 4.
Answer choice B: the last two digits form 34, which is not a multiple of 4.
Answer choice C: the last two digits form 58, which is not a multiple of 4.
Eliminate B and C.

The correct answer is D.
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by gnod » Wed Jun 22, 2011 2:46 am
thank you EVERYONE!
i loved each and every different method posted.
good lesson.

thanks again!!! :D
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