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In 2003, the number of girls attending Jefferson High

Expert replies
by M7MBA » Wed May 23, 2018 4:41 am

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

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

The OA is the option D.

How can I find the population in 2004 if I don't know the population in 2003? Could anyone clarify this for me? Thanks.
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Source: — Problem Solving |

by GMATGuruNY » Wed May 23, 2018 4:57 am
M7MBA 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
Test the SMALLEST POSSIBLE CASE.

Since both populations increase by 20% = 1/5, the original number of boys and the original number of girls must each be a multiple of 5.
Since in 2003 there must be an equal number of boys and girls, the smallest possible case is as follows:
2003: boys = 5 and girls = 5
2004: boys = 5 + (1/5)5 = 6 and girls = 5 + (1/5)5 = 6, for a total of 12 students.

The result in blue implies that the total number of students in 2004 must be a MULTIPLE OF 12.
For a number to be a multiple of 12, it must be divisible by both 4 and 3.

A number is divisible by 4 only if its last two digits form an integer that can divided twice by 2.
Of the five answer choices, only A, D and E satisfy this condition:
A: 32 --> 16 --> 8
D: 76 --> 38 --> 19
E: 28 --> 14 --> 7
Eliminate B and C.

A number is divisible by 3 only its digits sum to a multiple of 3.
Of the three remaining answer choices, only D works:
5+0+7+6 = 18, which is a multiple of 3.
Eliminate A and E.

The correct answer is D.
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by Scott@TargetTestPrep » Tue May 29, 2018 6:41 am
M7MBA 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
We can let the number of girls in the school in 2003 = x = the number of boys in the school in 2003.

Therefore, the number of students in the school in 2004 is 1.2x + 1.2x = 2.4x. In order for this to be a whole number, x must be a multiple of 5, so we can let x = 5k where k is a positive integer, and 2.4x = 2.4(5k) = 12k. Therefore, we see that the total number of students in 2004 must be a multiple of 12.

Notice that 12 = 3 x 4; therefore, that number has to be divisible by 3 and 4. This eliminates choices B, C since they are not divisible by 4 (recall that a number is divisible by 4 if the last two digits of the number are divisible by 4). It also eliminates choices A and E since they are not divisible by 3 (recall that a number is divisible by 3 if the sum of the digits of the number is divisible by 3).

Thus, the only number that could be the total number of students in 2004 is choice D since it's divisible by both 3 and 4 (notice that the sum of the digits = 5 + 0 + 7 + 6 = 18 and the last two digits are 76).

Answer: D

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by swerve » Tue May 29, 2018 11:00 am
Let's 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 students = 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).

Regards!
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