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Music high school

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by j_shreyans » Tue Jun 16, 2015 8:27 am
All of the students of Music High School are in the band, the orchestra, or both. 80 percent of the students are in only one group. There are 119 students in the band. If 50 percent of the students are in the band only, how many students are in the orchestra only?

A)30
B)51
C)60
D)85
E)119

OAB

I was trying to solve this by Double matrix method, but couldn't succeed .

Please advise.
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by talaangoshtari » Tue Jun 16, 2015 8:57 am
My answer is 180!
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by DavidG@VeritasPrep » Tue Jun 16, 2015 9:02 am
All of the students of Music High School are in the band, the orchestra, or both. 80 percent of the students are in only one group. There are 119 students in the band. If 50 percent of the students are in the band only, how many students are in the orchestra only?

A)30
B)51
C)60
D)85
E)119
Who doesn't love a good matrix problem?

First, note two things: everyone is in band or orchestra or both, so the NOT BAND/NOT ORCHESTRA cell will be 0. Second, we know that 80% are in exactly one group. That means BAND ONLY + ORCHESTRA ONLY = 80%. But we know that 50% are in BAND ONLY, so 30% will be ORCHESTRA ONLY. If we call the total T, we'll have the following to start:


Image

Fill in the remaining cells to get:


Image

The last row tell us that 119 + .3T = T
119 = .7T
T = 170

Orchestra only is .3T, which is .3 * 170 = 51.
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by Brent@GMATPrepNow » Tue Jun 16, 2015 9:16 am
Note: David's approach (the Double Matrix Method) can be used to solve most questions featuring a population in which each member has two characteristics associated with it.
Here, we have a population of students, and the two characteristics are:
- in the band or not in the band
- in the orchestra or not in the orchestra

This question type is VERY COMMON on the GMAT, so be sure to master the technique.

To learn more about the Double Matrix Method, watch our free video: https://www.gmatprepnow.com/module/gmat- ... ems?id=919

Once you're familiar with this technique, you can attempt these additional practice questions:

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Easy Data Sufficiency questions
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Medium Data Sufficiency questions
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Difficult Data Sufficiency questions
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- https://www.beatthegmat.com/mba/2011/05/ ... question-3

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by [email protected] » Tue Jun 16, 2015 9:17 am
Hi j_shreyans,

This prompt is one big 'logic' problem with a little bit of arithmetic thrown in.

We're given a number of facts to work with:
1) ALL students are in the band, the orchestra or BOTH.
2) 80% of students are ONLY in 1 group.
3) There are 119 students in the band.
4) 50% of the students are in the band ONLY.

We're asked how many students are in the orchestra ONLY.

From facts 2 and 4, we can break the students down into groups (by percent):

80% are in ONLY 1 group and 50% are in the band ONLY.

This means that 100% - 80% = 20% are in BOTH groups.
This also means that 80% - 50% = 30% are in the orchestra ONLY.

From fact 3, we know that 119 students are in the band (which includes the students in band ONLY and the students in BOTH)

Band only = 50%
Both = 20%

50% + 20% = 70% = 119 students

.7(Total) = 119
Total = 119/.7
Total = 170

Finally, we're asked for the number of students in the orchestra:

30% of 170 = 51

Final Answer: B

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by GMATGuruNY » Tue Jun 16, 2015 10:16 am
All of the students of Music High School are in the band, the orchestra, or both. 80 percent of the students
are in only one group. There are 119 students in the band. If 50 percent of the students are in the band only,
how many students are in the orchestra only?

a. 30 b. 51 c. 60 d. 85 e.119
(Band only) + (Band and Orchestra) + (Orchestra only) = Total.

The TOTAL number of students in the band = 119.
Thus:
119 + (Orchestra only) = Total.

We can PLUG IN THE ANSWERS, which represent the number of students in the orchestra only.
Since 80% of the students are in only one group -- and 80% = 4/5 -- the total number of students must be a MULTIPLE OF 5.
Thus, when the correct answer choice is added to 119, the result must be a multiple of 5.
Only B works:
119 + 51 = 170.

The correct answer is B.
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by j_shreyans » Wed Jun 17, 2015 8:24 am
Hi Mitch ,


I am confused.

The formula should be (Brand only)-(Brand and Orchestra)+(Orchestra only) + Neither = Total right?

Then how (Band only) + (Band and Orchestra) + (Orchestra only) = Total.

Please advice and correct me if I am wrong.

Thanks
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by [email protected] » Wed Jun 17, 2015 8:52 am
Hi j_shreyans,

The standard Overlapping Sets Formula IS not quite what you've stated. It would be...

(Band) + (Orchestra) - (BOTH Band and Orchestra) + (Neither Band nor Orchestra) = Total

The formula "works" because some of the students are in the Band (so they're counted as part of that group) AND they're in the Orchestra (so they're counted as part of that group ALSO). Since you're not allowed to "count" a student twice when calculating a total, you have to SUBTRACT the "BOTH" group (doing so removes one of the 'counts').

In this question, there is NO "neither" group. The wording in the first sentence of the prompt gives us a way to "set up" the question...

(Band ONLY) + (Orchestra ONLY) + (BOTH Band and Orchestra) = Total

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by GMATGuruNY » Thu Jun 18, 2015 2:10 am
j_shreyans wrote:Hi Mitch ,


I am confused.

The formula should be (Brand only)-(Brand and Orchestra)+(Orchestra only) + Neither = Total right?

Then how (Band only) + (Band and Orchestra) + (Orchestra only) = Total.

Please advice and correct me if I am wrong.

Thanks
For overlapping sets G� and G₂, there are two useful formulae.

Formula 1:
Total = G� + G₂ - Both + Neither.
In this formula, G� and G₂ each include the members who are in BOTH groups:
G� = (only G�) + (both G� and G₂).
G₂ = (only G₂) + (both G� and G₂).

Formula 2:
Total = Only G� + Only G₂ + Both + Neither.
In this formula, Only G� and Only G₂ do NOT include members who are in both groups.

In my solution above, I used Formula 2:
Total = (Band Only) + (Orchestra Only) + (Both Band and Orchestra) + Neither.

Since all of the students must be in the band, the orchestra, or both, the value of Neither is 0:
Total = (Band Only) + (Orchestra Only) + (Both Band and Orchestra) + 0.
(Band Only) + (Both Band and Orchestra) + (Orchestra Only) = Total.

Since a total of 119 students are in the band, we get:
119 + (Orchestra Only) = Total.
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by Matt@VeritasPrep » Tue Jun 23, 2015 6:32 pm
*DON'T USE THE MATRIX FOR THIS PROBLEM* It's much easier without it!

Let t = the number of students at Music High School.

.5t = band only
.3t = orchestra only
.2t = both

So .7t = 119. Thus t = 170 and orchestra only = .3 * 170 = 51. Piece of cake!

This is much faster, and I have to think the problem was designed to be solved this way.
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