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Overlapping sets

by karthikpandian19 » Tue Jan 10, 2012 1:16 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?

30

51

60

85

119
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by ronnie1985 » Tue Jan 10, 2012 2:08 am
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by LalaB » Wed Jan 11, 2012 2:14 am
band ; Orchestra ;Both
119 ; ?


it is stated,that only one of the groups =0.8*all and only band=0.5*all
so, only orchestra= only one of the groups-only band=0.3 all

if only one of the groups =0.8*all, then 0.2*all= not only one group

so, 0.2*all=(119-0.5all)

all=170
only orchestra= only one of the groups-only band=0.3 all =0.3*170=51

the answ is 51
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by Neo Anderson » Wed Jan 11, 2012 5:53 am
it is stated,that only one of the groups =0.8*all and only band=0.5*all
so, only orchestra= only one of the groups-only band=0.3 all

if only one of the groups =0.8*all, then 0.2*all= not only one group

so, 0.2*all=(119-0.5all)

all=170
only orchestra= only one of the groups-only band=0.3 all =0.3*170=51

the answ is 51
Will go with you! answer should be 51
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by ArunangsuSahu » Thu Jan 12, 2012 4:30 pm
(100-80)%= 119 - 50%

70% =119
Only Orchestra = (80-50)%=30%= 119*3/7=51

(B) is the answer
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by GMATGuruNY » Thu Jan 12, 2012 8:15 pm
karthikpandian19 wrote: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?

30

51

60

85

119
We can plug in the answers, which represent the number of students in the orchestra only.
Given that the percentages are all multiples of 10, when the correct answer is added to 119 -- the total number of students in the band -- the resulting sum almost certainly will be a multiple of 10.
Thus, the most likely answer choice is B.

Answer choice B: 51 in the orchestra only.
Thus, the total number of students = 119+51 = 170.
Total in the band only = .5(170) = 85.
Students in only one group/Total = (51+85)/170 = 136/170 = 8/10 = 80%. Success!

The correct answer is B.

An algebraic solution:
Since 80% are in only one group, and 50% are in the band only, 30% are in the orchestra only.
Since 30% are in the orchestra only, 70% are in the band.
Thus, 119 is 70% of the total:
119 = .7x
x = 119/.7 = 170.
Thus, the number in the orchestra only = .3(170) = 51.

The correct answer is B.
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by buzzdeepak » Sat Feb 09, 2013 8:19 pm
Given that the percentages are all multiples of 10, when the correct answer is added to 119 -- the total number of students in the band -- the resulting sum almost certainly will be a multiple of 10.
Thus, the most likely answer choice is B.

Nice tip and could be extremely handy when you don't have much time! Thanks Mitch...
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by DCS80 » Mon Feb 11, 2013 10:43 am
GMATGuruNY wrote:
karthikpandian19 wrote: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?

30

51

60

85

119
We can plug in the answers, which represent the number of students in the orchestra only.
Given that the percentages are all multiples of 10, when the correct answer is added to 119 -- the total number of students in the band -- the resulting sum almost certainly will be a multiple of 10.
Thus, the most likely answer choice is B.

Answer choice B: 51 in the orchestra only.
Thus, the total number of students = 119+51 = 170.
Total in the band only = .5(170) = 85.
Students in only one group/Total = (51+85)/170 = 136/170 = 8/10 = 80%. Success!

The correct answer is B.

An algebraic solution:
Since 80% are in only one group, and 50% are in the band only, 30% are in the orchestra only.
Since 30% are in the orchestra only, 70% are in the band.
Thus, 119 is 70% of the total:
119 = .7x
x = 119/.7 = 170.
Thus, the number in the orchestra only = .3(170) = 51.

The correct answer is B.
might be the dumbest question ever, but eating lunch @ work and trying to figure your first approach...

how did you, based on using ONLY 'multiples of 10' as a guideline, eliminate 60 and 30 as options?
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by GMATGuruNY » Mon Feb 11, 2013 1:39 pm
DCS80 wrote:
GMATGuruNY wrote: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?

30

51

60

85

119

might be the dumbest question ever, but eating lunch @ work and trying to figure your first approach...

how did you, based on using ONLY 'multiples of 10' as a guideline, eliminate 60 and 30 as options?
The answer choices represent the number of students in the orchestra only.
This value does NOT need to be a multiple of 10.
Rather, the TOTAL number of students must be a multiple of 10, so that 80% of the total (the number of students in only one group) and 50% of the total (the number of students in the band only) are integer values.

Thus, when the number of students in the orchestra only (the values in the answer choices) is added to 119 (the number of students either in the band only or in both the band and the orchestra), the sum must be a multiple of 10.
A: 30+119 = 149.
C: 60+119 = 179.
Since neither 149 nor 179 is a multiple of 10, eliminate A and C.

Only answer choice B yields a multiple of 10 for the total number of students:
51+119 = 180.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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