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Of all the students in a certain dormitory, 1/2 are first-year student

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Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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Of all the students in a certain dormitory, 1/2 are first-year students and the rest are second-year students. If 4/5 of the first-year students have not declared a major and if the fraction of second-year students who have declared a major is 3 times the fraction of first-year students who have declared a major, what fraction of all the students in the dormitory are second-year students who have not declared a major?

A. 1/15
B. 1/5
C. 4/15
D. 1/3
E. 2/5

Originally posted by msand on 27 Dec 2009, 09:47.
Last edited by Bunuel on 22 Aug 2017, 10:43, edited 3 times in total.
Renamed the topic and edited the question.
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Re: Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post 27 Dec 2009, 17:09
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msand wrote:
Of all the students in a certain dormitory, 1/2 are first-year students and the rest are second-year students. If 4/5 of the first-year students have not declared a major and if the fraction of second-year students who have declared a major is 3 times the fraction of first-year students who have declared a major, what fraction of all the students in the dormitory are second-year students who have not declared a major?

A. 1/15
B. 1/5
C. 4/15
D. 1/3
E. 2/5

OA :


First-year=Second-year=x --> all students \(2x\);

\(\frac{4}{5}x\) of the first-year students have not declared a major --> \(\frac{1}{5}x\) declared;

Second-year students who have declared a major is 3 times the fraction of first-year students who have declared a major =\(\frac{3}{5}x\) --> who have not \(\frac{2}{5}x\);

Hence: \(\frac{\frac{2}{5}x}{2x}=\frac{1}{5}\)

Answer: B.
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Re: Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post 27 Dec 2009, 21:10
tot students=x

1st year student=x/2----> non majaor=4/5(x/2)-----> maj=1/5(x/2)

2nd year student=x/2---->maj=3(1/5(x/2))=3/10(x)--->non major=x/2-3/10(x)=1/5(x)

hence 1/5
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Re: Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post 12 Aug 2011, 14:23
I think the answer choices you have provided because the answer cannot be greater than one.

Here is an easy way to think about this is by picking "smart" numbers.
Let's pick 50 for the first half of students who are 1st years and 50 for the other half who are 2nd year.
We know that 4/5 of the 1st years have not declared a major so that gives us 50*4/5=40, so 10 have declared a major.
We also know that the number of 2nd year that have not declared a major = 3 times the # of 1st year who have declared a major. This gives us 3*10=30

The question asks for fraction of 2nd years who have not declared a major over the total students.
30 have declared a major out of the 2nd year so 20 have not. Since we have 100 students (50+50), the answer is 20/100 or 1/5.

You can run a search on the web and you will find this question without the correct answer choices.

Thanks
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Re: Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post 30 Jan 2013, 06:25
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1
You can also do this problem sans algebra (in your head):

100 Students Total:

1st yr = 50 students
4/5x50 = 40 = No Major
Therefore: 10 = Major

2nd yr = 50 students
10x3 = 30 = Major
Therefore: 20 = No Major
20/100 = 1/5 = B
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Re: Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post 08 Jun 2013, 06:34
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In this kind of problem table structure is the weapon for me...

Let 2X would be total student which revealed that 1 yrs student is X and 2nd Yr is X as well.

No form the table




MAJOR NO MAJOR Total Student
1st yr Student x/5 4x/5 x
2nd yr Student 3x/5 2x/5 x
Total Student 4x/5 6x/5 2x

From the table 2nd yr student with major is 2x/5 so the answer would be =2x/5/2x =>1/5 Ans

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Re: Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post 03 Sep 2014, 01:26
First ............................................. Second ........................................ Total

x ........................................................ x ................................................ 2x

Major........... NotMajor ................... Major ............ NotMajor ..........

\(\frac{4x}{5}\) ................ \(\frac{x}{5}\) ........................ \(\frac{3x}{5}\) ................... \(\frac{2x}{5}\) ...................... 2x


Fraction \(= \frac{\frac{2x}{5}}{2x} = \frac{1}{5}\)

Answer = B
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Re: Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post Updated on: 07 May 2016, 20:08
1
Attached is a visual that should help.

FY = First Year
SY = Second Year
NDM = Not Decided on Major
YDM = Yes Decided on Major
SYY = Senior Year Yes
SYN = Senior Year No


Attachments

Screen Shot 2016-05-05 at 6.59.42 PM.png
Screen Shot 2016-05-05 at 6.59.42 PM.png [ 68.47 KiB | Viewed 5707 times ]


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Originally posted by mcelroytutoring on 05 May 2016, 18:02.
Last edited by mcelroytutoring on 07 May 2016, 20:08, edited 6 times in total.
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Re: Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post 05 May 2016, 20:38
mojorising800 wrote:
Of all the students in a certain dormitory, 1/2 are first-year students and the rest are second-year students. If 4/5 of the first-year students have not declared a major and if the fraction of second-year students who have declared a major is 3 times the fraction of first-year students who have declared a major, what fraction of all the students in the dormitory are second-year students who have not declared a major?

A. 151
B. 51
C. 154
D. 31
E. 52


Say there are in all 100 students. Use a matrix:
50 are first year and 50 are second year students.

........................Major Dec................Major not dec..............Total
First Year............................................................................50
Second Yr...........................................................................50
Total

4/5 * 50 = 40 not declared major first year students.

........................Major Dec................Major not dec..............Total
First Year................10..................................40....................50
Second Yr...........................................................................50
Total

second-year students who have declared a major is 3 times the fraction of first-year students who have declared a major

........................Major Dec................Major not dec..............Total
First Year................10..................................40.....................50
Second Yr...............30..................................20......................50
Total

So second year students who've not declared a major is 20/100 = 1/5

Answer (B)
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Re: Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post 27 Jun 2016, 07:04
\((1-(1-4/5)*3)*(1-1/2)=1/5\)

Answer choice B.
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Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post Updated on: 07 Oct 2017, 06:33
This problem is 99 in og 2017.

T = total students in dorm
4/5 first year no declared major = 4/5(.5T) = .4T = # first year no major
.5T - .4T = .1T = 1st year w/ major
.1T x 3 = .3T
2nd year no major = .5T - .3T = .2T = 2nd no declared
.2T does not equal 2/5!

.2T / T = .2 = 1/5

Originally posted by CyberStein on 22 Aug 2017, 10:32.
Last edited by CyberStein on 07 Oct 2017, 06:33, edited 1 time in total.
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Re: Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post 22 Aug 2017, 15:05
msand wrote:
Of all the students in a certain dormitory, 1/2 are first-year students and the rest are second-year students. If 4/5 of the first-year students have not declared a major and if the fraction of second-year students who have declared a major is 3 times the fraction of first-year students who have declared a major, what fraction of all the students in the dormitory are second-year students who have not declared a major?

A. 1/15
B. 1/5
C. 4/15
D. 1/3
E. 2/5

Attachment:
matrix111.png
matrix111.png [ 17.63 KiB | Viewed 3914 times ]

Double matrix works well here, at least for me. I assumed 60 total students.

1. Number of dormitory students = 60. Half are first-year students = 30. So the other half, second-year students = 30

2. \(\frac{4}{5}\) of first-year students have NOT declared a major: \(\frac{4}{5}\) * 30 = 24

3. First-year students who HAVE declared a major = 30 total - 24 not = 6 who have

4. The fraction of second-year students who have declared a major is three times the fraction of first-year students who have declared a major. First-year declared = 6. Second-year declared = 3 * 6 = 18

5. Second-year students who have NOT declared a major: 30 total second-year students - 18 who have declared = 12 second-year students who have not

6. What fraction of ALL students are second-year students who have not declared a major? \(\frac{12}{60} = \frac{1}{5}\)

Answer B
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Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post 22 Aug 2017, 23:45
msand wrote:
Of all the students in a certain dormitory, 1/2 are first-year students and the rest are second-year students. If 4/5 of the first-year students have not declared a major and if the fraction of second-year students who have declared a major is 3 times the fraction of first-year students who have declared a major, what fraction of all the students in the dormitory are second-year students who have not declared a major?

A. 1/15
B. 1/5
C. 4/15
D. 1/3
E. 2/5


This question is pretty easy if we assume numbers. Used numbers because the question does not have fixed values and asks for percentages.

Assuming the total number of students in the dormitory to be 100

Given 1/2 are 1st year and other 1/2 are 2nd year. Hence 50 students each are in either year.

Given 1st year (Not decided Major)= 1stNDM=4/5th of 1st year=40

Hence 1st year (Decided Major)= 1stDM=10

Given 2nd year (Decided Major)= 2ndDM=3x1stDM=30

Hence 2nd year (Not decided Major)= 2ndNDM=100-(40+10+30)=20

Fraction of 2nd year students who havent declared a major = 20/100=1/5
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Re: Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post 07 Jun 2018, 16:10
msand wrote:
Of all the students in a certain dormitory, 1/2 are first-year students and the rest are second-year students. If 4/5 of the first-year students have not declared a major and if the fraction of second-year students who have declared a major is 3 times the fraction of first-year students who have declared a major, what fraction of all the students in the dormitory are second-year students who have not declared a major?

A. 1/15
B. 1/5
C. 4/15
D. 1/3
E. 2/5


Since 1/2 are first-year students, the other 1/2 are second-year students. Since 4/5 of the first-year students have not declared a major, so 1/5 of them have declared a major. Since the fraction of second-year students who have declared a major is 3 times the fraction of first-year students who have declared a major, 3/5 of the second-year students have declared a major. Therefore, 2/5 of the second-year students have not declared a major. Since the second-year students are 1/2 of all the students, 1/2 x 2/5 = 1/5 of all students are second-year students who have not declared a major.

Answer: B
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Re: Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post 05 Jul 2018, 18:47
pushpitkc niks18 Abhishek009 pikolo2510

Is there any better approach to find LCM of ordered set {2,3,5}?

I normally find the smallest multiple of largest no (here: 5) and see if it is divisible by
other two elements in set.

Also VeritasPrepKarishma made calculations bit easier by using starting number as 100
instead of taking LCM in spite of no percentage mentioned in problem. How do we get a hint about it?
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Re: Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post 05 Jul 2018, 23:14
adkikani wrote:
pushpitkc niks18 Abhishek009 pikolo2510

Is there any better approach to find LCM of ordered set {2,3,5}?

I normally find the smallest multiple of largest no (here: 5) and see if it is divisible by
other two elements in set.

Also VeritasPrepKarishma made calculations bit easier by using starting number as 100
instead of taking LCM in spite of no percentage mentioned in problem. How do we get a hint about it?


Hi adkikani

What do you mean by a better approach to finding the LCM of the ordered set?

IMO, using the LCM of the numbers is always a better choice instead of going with a random
number. When you use a random number, there is a chance of reaching a stage where you
might reach a fraction and have to restart the process.

For this problem, you can see the method used by generis to assume 30(LCM of 2,3,5).
The more questions you practice using this method, you will get better at choosing numbers.

Hope that helps you
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Re: Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post 05 Jul 2018, 23:36
adkikani wrote:
Is there any better approach to find LCM of ordered set {2,3,5}?

I normally find the smallest multiple of largest no (here: 5) and see if it is divisible by
other two elements in set.


Hi adkikani,

Rather taking multiple of 5 and checking every time, you can simply multiply 2*3*5 = 30.

Now, suppose we have another 3 numbers say 12, 30, and 35. To find LCM you can also use the prime factorization method.

\(12 = 2^2 *3\)
30 = 5*3*2
35 = 5*7

Now, take the maximum power of each prime number and multiply.

LCM = 2^2*3*5*7 = 420.

Hope this helps.

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Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post 11 Aug 2018, 17:48
1
If you are looking for more questions to practice the matrix box (also called double matrix) approach, I have compiled a list of questions from the Official Guide and Veritas Prep books at this link:

Matrix Boxes

For time efficiency, I generally recommend using a matrix box whenever you see a question that divides up a group into two categories in two different ways. I recommend drawing a matrix box that looks essentially like the one that generis posted in this thread. To me, matrix boxes improve efficiency because they help you to see the information in an organized way. You can also mark the boxes that you need to fill in to answer the question, so that you know exactly when you can stop because you have the information you need.

Please let me know if you have any questions, and if you want me to post a video solution!
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Re: Of all the students in a certain dormitory, 1/2 are first-year student  [#permalink]

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New post 11 Aug 2018, 18:09
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msand wrote:
Of all the students in a certain dormitory, 1/2 are first-year students and the rest are second-year students. If and if the fraction of second-year students who have declared a major is 3 times the fraction of first-year students who have declared a major, what fraction of all the students in the dormitory are second-year students who have not declared a major?

A. 1/15
B. 1/5
C. 4/15
D. 1/3
E. 2/5


I would use a double-matrix to organize the data.
That said, here is an alternate approach:

Since \(\frac{4}{5}\) of the first-year students have not declared a major, \(\frac{1}{5}\) of the first-year students HAVE declared a major.

Since the fraction of second-year students who have declared a major is 3 times the fraction of first-year students who have declared a major, \(\frac{3}{5}\) of the second-year students have declared a major, implying that \(\frac{2}{5}\) of the second-year students have NOT declared a major.

Since \(\frac{1}{2}\) of the students are second-year students, and \(\frac{2}{5}\) of these second-year students have not declared a major, \(\frac{1}{2} * \frac{2}{5} = \frac{1}{5}\) of the students are second-year students who have not declared a major.


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