All of the students of Music High School are in the band

[mridul12]

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?


Can someone / Instructor make a Overlapping Set Table ( a strategy mentioned in MGMAT Guide) and than explain /solve the problem?

[Harish Dorai]

Not sure it is legible on the screen, but here is the Table and explanation.

Let us assume X be the total number of students.

The given facts in the question.

1) Total # of students in the Band = 119 (This is equal to # of students in Band only + # of students in Band & Orchestra).
2) 80% of the students are in only one group = 0.8X.
3) 50% of the students are in Band Only = 0.5X. So the number of students in Orchestra only is 0.3X (From statement (2) above 0.8X - 0.5X).
4) Also please note that all of the students are in either Band, Orchestra OR both. So there are 0 Students who neither in Band nor Orchestra.

Refer the table below and the rest of the explanation.



| Band Not Band |
---------------------------------------------
Orchestra | | 0.3x |
---------------------------------------------
Not orchestra | 0.5x | 0 |
---------------------------------------------
119 | 0.3x | x


119 + 0.3x = x
0.7x = 119

x = 119/0.7 = 170

So number of students in Orchestra only = 0.3x
= 170 x 0.3
= 51

[StaceyKoprince]

Thanks, Harish - it's tough to get the formatting to work on the forums.

mridul12 - please remember to post the entire text of the question, including answer choices.

[Guest]

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?


Can someone / Instructor make a Overlapping Set Table ( a strategy mentioned in MGMAT Guide) and than explain /solve the problem?

Can someone / Instructor make a Overlapping Set Table ( a strategy mentioned in MGMAT Guide) and than explain /solve the problem?[/quote]

100% = Only B + Only O + Both B&O
100% = 50% + Only O + Both B&O
100% = 50% + 30% +20% [Since 80% are only in one group, Only B + Only O = 80%.]

So, 70% in Band. Now 70% is equivalent to 119 students. So, 30% is equivalent to 51 students. [ANS]

[RonPurewal]

here you go

[relentlesspursuito700plus]

First of all, thank you to MGMAT for explaining this so clearly in your book. As someone who has trouble organizing numbers in complicated problems, such as this one, I find your method especially helpful. That said, I used a VD. It's kind of hard to draw it on the post but once you get the numbers straight it will only take a few seconds. I was able to solve this in about 45 seconds.

Draw two cricles that overlap in the middle

Band Orchestra

Over the circles, I labeled 0.8x and drew two arrows. One going to band and one going to orchestra. That means the middle piece, the area representing students who are in both) is 0.2x (0.2x + 0.8x=1 or 100%).

Now in the left circle, write 0.5x because the problem tells us 50% of the students are in band ONLY. This means .3x are in orchestra only. So in your cricles, going from left to right, you should have 0.5X, 0.2x, and 0.3x, which adds to 1X.

Now it's just simple multiplication. The problem tells us 119 students are in band. That's 0.7x=119, which means x equals 170. So .3x or 30% of 170 would be 17+17+17= 51.

Thank you MGMAT. You guys rock. I am a visual person so your method helped significantly.

[rfernandez]

Good method, too. We're glad you're finding our curriculum helpful.

[andrewthai2000]

Ron:

Can you post the overlapping sets in another format? I clicked on your link above and the data is not showing.

Thanks.

[RonPurewal]

Ron:

Can you post the overlapping sets in another format? I clicked on your link above and the data is not showing.

Thanks.

It looks like this thread is more than 6 years old, so it's not a surprise that the links are broken.

In any case, this thread still doesn't follow the forum rules: The original text of the problem, with all answer choices, should be posted.

If you have questions about this problem, please correct this issue by posting the complete problem as it originally appears, with all answer choices.

[RonPurewal]

Also, please include some comments about your current progress with the problem.
What do you already understand?
What don't you understand?
Etc.

Thanks.

[andrewthai2000]

Ron:

Can you post the overlapping sets in another format? I clicked on your link above and the data is not showing.

Thanks.

It looks like this thread is more than 6 years old, so it's not a surprise that the links are broken.

In any case, this thread still doesn't follow the forum rules: The original text of the problem, with all answer choices, should be posted.

If you have questions about this problem, please correct this issue by posting the complete problem as it originally appears, with all answer choices.

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

Is my overlapping sets diagram as efficient as it should be (sorry if the spacing is off)?

Band No Band Total

Orchestra .2X .3X<--(what're we looking for)
No Orchestra .5X 0%
Total .7X .3X X

X = total number of students in the school
The "Total" column should sum as .5X and .5X for the first and second rows, respectively.
Notes:
.5X+.3X=.8X (80% of the students are in one group only)
The remaining would be the 20% who are in both Band and Orchestra (since there are no students in neither).

From the table, 70% of the students are in the band. From the problem, there are 119 students in the band. We could use this information to figure out the total number of students in the school.
.2X + .5X = 119
.7X =119
X = 170

From the table, 30% of the students are in the orchestra only.
.3(170)=51
Answer is B

For problems where there is a "neither" scenario, such as this one, would a Venn Diagram be more efficient?

Thanks.

[tim]

Your approach looks solid. I would still recommend a double set matrix over a Venn diagram regardless of the specifics of the problem.

[RonPurewal]

For problems where there is a "neither" scenario, such as this one, would a Venn Diagram be more efficient?

Thanks.

Those are precisely the situations to which Venn diagrams are LEAST well adapted.

Think about it. (Let's say that the two circles are "this" and "that".)

If NO ONE is "neither this nor that", then ...
... "not this" is just one region. (it's the part of the "that" circle outside the "both" region.)
... "not that" is just one region (same idea).

If some people/items/animals ARE "neither this nor that", though, then ...
... "neither this nor that" is the awkward region outside both circles.
... each of "not this" and "not that" is the combination of TWO regions——the one mentioned above, and the awkward outside region.

Etc.

[RonPurewal]

As Tim said——Regardless of whether anyone/anything is "neither", a two-way table ("double set matrix") will just about always be more convenient.
If anyone/anything IS "neither", the difference is only amplified.

If you have an introductory statistics textbook, find the chapter that discusses "Bayesian" or "conditional" probability.
• You'll notice a TON of two-way tables ("double set matrices").
• You probably won't notice ANY Venn diagrams. If Venn diagrams are present, then Step One is almost certainly to convert them into two-way tables.

This is not a coincidence. For this specific kind of task, using a Venn diagram is like threshing corn by hard, while using the two-way table / double-set matrix is like using a threshing machine. In terms of productivity/efficiency, it's not even close.

[RonPurewal]

Finally—
If you are presented with three overlapping criteria——a situation that is rare, but does exist——then you should use a Venn diagram.

Not out of preference, but out of pure necessity.
The 3-set equivalent of a two-way table / "double set matrix" would be a three-dimensional, 3x3x3 cube. Not the kind of thing you could draw in a short time.

If a 3-set problem were invented at random, then Venn diagrams would be just as bad as always.
But the test writers are not jerks, and they don't write problems that can't be solved in a reasonably short time. So, any 3-set problem on the GMAT will be specifically engineered to suit a Venn diagram.

(Readers: If any of you like to make up your own problems, take note. If you invent a random set of specifics for 3 sets, a Venn diagram will almost certainly cause you a world of pain. The GMAC problems are very carefully engineered.)

The 2-set problems are not engineered in such a way, so the best advice for 2-set Venn diagrams is "Don't."