IMO : 20

Used the Venn Diagram Approach,

All 3 = 5

Let,
Number taken only A & B = X
Number taken only B & C = Y
Number taken only C & A = Z

Hence,
Number taken only A : 60-(X+Y-5)=55-(X+Y)
Number taken only B : 40-(X+Z-5)=35-(X+Z)
Number taken only A : 20-(Y+Z-5)=15-(Y+Z)

We need X+Y+Z

Total is 90 students,
Therefore :
X+Y+Z+5+55-(X+Y)+35-(X+Z)+15-(Y+Z)=90
Solving , X+Y+Z = 20

Whats the OA?

Sanjana,

I'm getting 40 here. See if my approach is wrong

This is the equation i have formulated,
90=60+40+20-X-Y-Z+2*5, where x=A&B, y=B&C, z=A&C

Therefore, X+Y+Z=40

papgust wrote: 90=60+40+20-X-Y-Z+2*5

My sincere apologies.. Made a silly mistake in formula. It shud be (-2*5).

IMO 20.

Yes!
True # of items = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in only 1/2) - (# in only 1/3) - (# in only 2/3) - 2(# in 1/2/3)

engg.manik wrote:9. 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?

Remember these following general formulas for three-component set problems to knock down most of the set problems:

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)

using the above 2 formulas u get the answer to be 20 for the given problem.
90=60+40+20-X +5 where X=P(AnB) + P(AnC) + P(BnC)
X=35
then use the following formula for exactly two of the sets.
Number of people in exactly two of the sets =
P(AnB) + P(AnC) + P(BnC) - 3P(AnBnC)
= 35 -3(5)= 20

Let me know if you have any query in the above formulas.