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A certain high school has 5,000 (OG2016)

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A certain high school has 5,000 (OG2016)

by boomgoesthegmat » Fri May 06, 2016 12:30 pm

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A certain high school has 5,000 students. Of these students, x are taking music, y are taking art, and z are taking both music and art. How many students are taking neither music nor art?

A) 5000 - z
B) 5000 - x - y
C) 5000 - x + z
D) 5000 - x - y - z
E) 5000 - x - y + z

OA: E
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by OptimusPrep » Sat May 07, 2016 7:50 pm
boomgoesthegmat wrote:A certain high school has 5,000 students. Of these students, x are taking music, y are taking art, and z are taking both music and art. How many students are taking neither music nor art?

A) 5000 - z
B) 5000 - x - y
C) 5000 - x + z
D) 5000 - x - y - z
E) 5000 - x - y + z

OA: E
Music = x,
Art = y
Both Music and Art = z

Therefore people who take atleast one = x + y - z
People who have taken none = 5000 - (x + y - z) = 5000 - x - y + z

Correct Option: E
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by Jeff@TargetTestPrep » Mon May 09, 2016 10:46 am
boomgoesthegmat wrote:A certain high school has 5,000 students. Of these students, x are taking music, y are taking art, and z are taking both music and art. How many students are taking neither music nor art?

A) 5000 - z
B) 5000 - x - y
C) 5000 - x + z
D) 5000 - x - y - z
E) 5000 - x - y + z

OA: E
Solution:

We are given an overlapping set with 2 categories: music and art. In any two-category set problem we can use the following formula:

Total number of members = # in category A + # in category B - # in both categories + # in neither category

Using the formula for the current problem we have:

5,000 = x + y - z + neither

5,000 - x - y + z = neither

Answer:E

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by 800_or_bust » Mon May 09, 2016 11:31 am
boomgoesthegmat wrote:A certain high school has 5,000 students. Of these students, x are taking music, y are taking art, and z are taking both music and art. How many students are taking neither music nor art?

A) 5000 - z
B) 5000 - x - y
C) 5000 - x + z
D) 5000 - x - y - z
E) 5000 - x - y + z

OA: E
In lieu of solving, I'll just post a reminder that you have to add the members of the combined set to avoid double counting. With that said, the answer should be fairly intuitive.
800 or bust!
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by Matt@VeritasPrep » Wed May 11, 2016 11:47 pm
You could also just pick numbers.

Suppose x = 100, y = 50, and z = 10.

Since z is counted in both x and y, we have 90 taking just music, 40 taking just art, and 10 taking both.

5000 - 90 - 40 - 10 = 4860 = E.
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by Scott@TargetTestPrep » Thu Dec 14, 2017 6:00 am
boomgoesthegmat wrote:A certain high school has 5,000 students. Of these students, x are taking music, y are taking art, and z are taking both music and art. How many students are taking neither music nor art?

A) 5000 - z
B) 5000 - x - y
C) 5000 - x + z
D) 5000 - x - y - z
E) 5000 - x - y + z
We can use the following formula:

Total students = # taking music + # taking art - # taking both + # taking neither

5,000 = x + y - z + neither

5,000 - x - y + z = neither

Answer: E

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by [email protected] » Thu Mar 15, 2018 7:06 pm
Hi All,

While this question can be solved Algebraically, it can also be solved by TESTing VALUES and taking some basic notes:

We're given a series of facts to work with:
1) A certain high school has 5,000 students.
2) Of these students:
X are taking music,
Y are taking art, and
Z are taking BOTH music and art.

We're asked how many students are taking NEITHER music nor art?

Let's TEST
X = 2
Y = 2
Z = 1

So, we have 2 students taking music, 2 taking art and 1 taking BOTH music and art. That 1 person has been counted TWICE though (once in the music 'group' and once in the art 'group'), so what we really have is...

1 student taking JUST music
1 student taking JUST art
1 student taking BOTH music and art
Total = 3 students

We're asked for the total number of students who are taking NEITHER Course. That is 5000 - 3 = 4997. So that's the answer that we're looking for when X=2, Y=2 and Z=1. There's only one answer that matches...

Final Answer: E

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Rich
Contact Rich at [email protected]
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