Set quest
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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?
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?
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.