simple set question

[kaushikt]

This semester, each of the 90 students in a certain class took at least one course from A, B, and C. If 60 students took A, 40 students took B, 20 students took C, and 5 students took all the three, how many students took exactly two courses?


Is this formula correct?

AUBUC = A + B + C - {A n B + B n C + C n A} + A n B n C

[kannn]

That is correct. Ans is 35

http://www.urch.com/forums/gmat-prob...n-diagram.html (Venn Diagram)
If there are three sets A, B, and C, then
P(AuBuC) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)

Number of people in exactly one set =
P(A) + P(B) + P(C) – 2P(AnB) – 2P(AnC) – 2P(BnC) + 3P(AnBnC)

Number of people in exactly two of the sets =
P(AnB) + P(AnC) + P(BnC) – 3P(AnBnC)

Number of people in exactly three of the sets =
P(AnBnC)

Number of people in two or more sets =
P(AnB) + P(AnC) + P(BnC) – 2P(AnBnC)

No of people in atleast 1 set =
P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + 2 P(AnBnC)

n(AUB)=n(A)+ n(B)-n(AnB) +neither

[vbhup2]

Isn't the answer to this one 20? I'm getting a 20.

I generally don't use a formula for these. I visualise a Venn diagram and do it. So, I guess I can't explain it easily. Basically, I start off with an assumption that none of them took exactly two courses.

The statements that follow are based on the assumption

Since 5 students took all the three,
Students who took only A = 55
Students who took only B = 35
Students who took only C = 15

Add all the students to get the total no. of students = 5 + 55 + 35 + 15 = 110

But it is given that the no. of students is 90. Hence, the difference is the number of people who took exactly two courses.
=> 110 - 90 = 20

How is the answer 35?

EDIT: I guess using the formula Kann mentioned above, we get 20 as well

Number of people in exactly two of the sets =
P(AnB) + P(AnC) + P(BnC) – 3P(AnBnC)
35 - 3*5 = 20

[v-rap]

I think

AUBUC = NA+NB+NC-ANB-ANC-CNB+ANBNC
i.e
90=60+40+20-(ANB+ANC+CNB)+5
so (ANB+ANC+CNB) = 35

[narinder82]

The Correct answer is 35.
n(AUBUC) = n(A) + n(B) +n(C) - [ n(A intersection B) + n ( B intersection C) + n ( A interction c)] + n ( A intersection B Intersection C)

90 = 60 + 40 + 20 -[x] +5

[x] = 35

[soulasylum]

Since 5 students took all the three,
Students who took only A = 55
Students who took only B = 35
Students who took only C = 15

Add all the students to get the total no. of students = 5 + 55 + 35 + 15 = 110

But it is given that the no. of students is 90. Hence, the difference is the number of people who took exactly two courses.
=> 110 - 90 = 20

How is the answer 35?

Vbhup2, I guess, the mistake in your above approach was that you took
Students who took ONLY x as (Students who took x - Number who took all three).

55 (60 - 5) would include students who took AnB and AnC also..Using the formula does help !..

[vbhup2]

But I included the statements that follow are based on an assumption. Then, I proved that the assumption was wrong. However, I may be wrong.

But, consider this scenario:

The students who took only A: 45
The students who took only B: 20
The students who took only C: 0
The students who took both AnB (but not all 3 and not {only A or only B}): 5
The students who took both BnC (but not all 3 and not {only B or only C}): 10
The students who took both CnA (but not all 3 and not {only C or only A}): 5
The students who took AnBnC (given): 5

Total (Since all are mutually exclusive) = 45 + 20 + 0 + 5 + 10 + 5 + 5 = 90

None of the conditions in the original question has been contradicted. Yet, the value of (AnB, BnC, CnA) = 20. Any contradictions, please let me know.Yes, I can give a similar example for 35. I guess these can have multiple scenarios, but shouldn't the number in each specific group (not given in the stem) be the same at the least?

[vbhup2]

Kaushikt, can you post the OA for this one?

Just saw this explanation for one similar problem on Manhattan GMAT but with different numbers. It uses a Venn Diagram. But I will represent each mutually exclusive part of the Venn diagram below and then use that explanation.

a = Only A
e = Only B
f = Only C
b = AnB
c = AnC
d = BnC

A students: a + b + c + 5 = 60
B students: e + b + d + 5 = 40
C students: f + c + d + 5 = 20
TOTAL students: a + e + f + b + c + d + 5 = 90

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 (A,B and C) we get:

a + e + f + 2b +2c +2d + 15 = 120.

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


a + e + f + 2b + 2c +2d + 15 = 120
–[a + e + f + b + c + d + 5 = 90]
b + c + d = 20

[Name User]

I get 20.

Let x= aUb, y=aUc, z=bUc

With the givens, we get following equation

x+y+z+5+60-(x+y+5)+40-(x+z+5)+20-(y+z+5)=90

We are looking for value of x+y+z

solving the equation above, you get 110-x-y-z=90
x+y+z=20

[kaushikt]

i thought the answer is 35. I dont have a OA for this.. is this wrong...
AUBUC = A + B + C - {A n B + B n C + C n A} + A n B n C