Is there an easier way to solve this problem?

[superseema]

The problem below was the last question I got on my first CAT test. I ended up running out of time solving this problem because it involves writing a lot of long equations and solving them. This problem would take at least 3-4 minutes using the method demonstrated in the answer explanations. Is there a quicker way to do it? If you can narrow the answer down to choice a. or b. in the first 30 seconds using, does it just make more sense to guess rather than waste precious minutes on solving 3 or 4 equations with 6 variables apiece? Or is there another simpler way to solve the problem that I could have used?

In a group of 68 students, each student is registered for at least one of three classes - History, Math, and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?

A. 13
B. 10
C. 9
D. 8
E. 7

[superseema]

For your reference, this is the answer from the answer key...


For an overlapping set problem with three subsets, we can use a Venn diagram to solve.

Each circle represents the number of students enrolled in the History, English and Math classes, respectively. Notice that each circle is subdivided into different groups of students. Groups a, e, and f are comprised of students taking only 1 class. Groups b, c, and d are comprised of students taking 2 classes. In addition, the diagram shows us that 3 students are taking all 3 classes. We can use the diagram and the information in the question to write several equations:

History students: a + b + c + 3 = 25
Math students: e + b + d + 3 = 25
English students: f + c + d + 3 = 34
TOTAL students: a + e + f + b + c + d + 3 = 68

The question asks for the total number of students taking exactly 2 classes. This can be represented as b + c + d.

If we sum the first 3 equations (History, Math and English) we get:

a + e + f + 2b +2c +2d + 9 = 84.

Taking this equation and subtracting the 4th equation (Total students) yields the following:

a + e + f + 2b + 2c +2d + 9 = 84
-[a + e + f + b + c + d + 3 = 68]
b + c + d = 10

The correct answer is B.

[karan13]

hi.... let me try this...

there is a formula for 3 group problem
t=g1+g2+g3- (in 2)-2(all 3)+n

t is your total, n is neither in any group, g1 g2 g3 are the 3 groups, in 2 means that people present in any 2 groups and all 3 means people in all 3 groups.

just apply this and you get the ans in less than 30 sec

68=25+25+34-(in 2)-2(3)
in 2=10

ans is B...

FYI for 2 groups its T=g1+g2-both+n

hope it helps....

[jnelson0612]

Wow, nice formula! I have never seen that before.

[karan13]

thanks:)

[jnelson0612]

:-)

[rkdn]

Hi, I used a different method which failed, and I'm wondering why it failed. Instead of using a, e, f, I basically only had b,c,d (or x,y,z in my case). So my formulas were:

History=> 25-(x+y+3)
Math=> 25-(x+z+3)
English=> 34-(y+z+3)

I added them all up and set them to 68.

25-(x+y+3) + 25-(x+z+3) + 34-(y+z+3) = 68
84 -2x-2y-2z-3(3) = 68
75 = 68 + 2(x+y+z)
3.5=x+y+z

Obviously this was wrong :(

Any help would be greatly appreciated, thanks!

[messi10]

Hi rkdn,

The problem with your method is that you are adding up only those people who are enrolled in 1 class.

When you say: History=> 25-(x+y+3) , you are counting the students who are enrolled only in history. You are excluding the students who are also in two or three classes. This is probably what you meant to do, but the problem comes in when you add the three together to equate to 68. This is where you are going wrong. The single subject students do not add up to 68. Remember that in the Venn diagram with three circles, you have to consider all three cases:

Outer part: singles/individuals
Middle part: overlap of two
Inner part: overlap of all three

So if we work backwards from the answer which is 10. Lets break this 10 down into three parts - 5, 4, 1 Lets say 5 students take history and math, 4 students take maths and english and 1 student takes english and history. In your question, this equates to:
x = 5
y = 4
z = 1

So according to your equation:

History=> 25-(5 + 4 + 3) = 13
Math=> 25-(5 + 1 + 3) = 16
English=> 34-(4 + 1 + 3) = 26

If add the above, you get 55 not 68. What is this 55? It is the number of students enrolled only in 1 course.

If you need to use your method to solve the problem, you need to add another x + y + z + 3 to your equation:

25-(x+y+3) + 25-(x+z+3) + 34-(y+z+3)+ x + y + z + 3 = 68

=> 84 - 9 + 3 - 2x -2y -2z + x + y + z = 68
=> 78 - x - y - z = 68
=> 78 - 68 = x + y + z
=> x + y + z = 10

Hope this makes sense

Regards

Sunil

[man.undefeated]

Hi,
I followed the following approach and got the right ans. not sure whether i will end up with the correct ans every time using this approach.

1st of all, I assumed all the students took one course each.
so the number of students becomes= 84
but the given number of students=68
(84-68)=16, these are the students who took more than one courses.
3 students took 3 courses each (2 additional courses each)
So they took 6 additional courses in total
16-6=10
this is the number of students who took one additional course (that means who took exactly two courses)

[jnelson0612]

Hi,
I followed the following approach and got the right ans. not sure whether i will end up with the correct ans every time using this approach.

1st of all, I assumed all the students took one course each.
so the number of students becomes= 84
but the given number of students=68
(84-68)=16, these are the students who took more than one courses.
3 students took 3 courses each (2 additional courses each)
So they took 6 additional courses in total
16-6=10
this is the number of students who took one additional course (that means who took exactly two courses)

I like it! Great reasoning!

[maria.segal.1]

Hi, I was using the following approach and made a mistake somewhere along the way. Would be very grateful if any one were to help me figure this out.
Thanks,
Maria

68= A+B+C - A&B-B&C-A&C + A&B&C= 25+25+34 - (all students in 2 classes)+3

thus: 68-84-3=all students in 2 classes -> all students in 2 classes=19

?

[maria.segal.1]

Hi,
I followed the following approach and got the right ans. not sure whether i will end up with the correct ans every time using this approach.

1st of all, I assumed all the students took one course each.
so the number of students becomes= 84
but the given number of students=68
(84-68)=16, these are the students who took more than one courses.
3 students took 3 courses each (2 additional courses each)
So they took 6 additional courses in total
16-6=10
this is the number of students who took one additional course (that means who took exactly two courses)

wonderful solution! Thanks :)

[jnelson0612]

:-)

[amish.v]

[quote="jnelson0612"]:-)[/quote]

N(AUBUC) = N(A)+N(B)+N(C)+N(A&B&C "Y") - (ALL STUDENTS IN 2 CLASS "X")
N(A&B&C "Y") ; Y = 3
N(A) = 25 - Y = 22
N(B) = 25-Y= 22
N(C) = 34-Y=31

68 = 22+22+31+3 - X
X = 78-68 = 10
ANS IS 10.

Thanks
Amish123.

[tim]

as you can see, there are many ways to solve this problem. when you're studying, you should try as many different solution methods as you can..