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

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

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

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

Rgds
Prasannajeet

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

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

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)

\((1-(1-4/5)*3)*(1-1/2)=1/5\)

Answer choice B.

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

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

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

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

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

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.

Thanks.

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!

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.

MATRIX BOX is the way to solve a question like this!