SET Practice SET – 4
Set Theory: Level 4
1.
In a survey of 100 students: 50 take Math, 40 take Physics, 30 take Chemistry. 15 take Math & Physics, 10 take Physics & Chemistry, 10 take Math & Chemistry. 20 take none of the three. How many students take all three subjects?
Solution (C):
Total = \( M + P + C – (MP + PC + MC) + (MPC) + \text{None} \).
\( 100 = 50 + 40 + 30 – (15 + 10 + 10) + x + 20 \).
\( 100 = 120 – 35 + x + 20 \).
\( 100 = 105 + x \).
Wait, \( 120 – 35 = 85 \). \( 85 + 20 = 105 \).
\( 100 = 105 + x \implies x = -5 \)?
Let’s re-calculate.
\( \text{Union} = 100 – 20 = 80 \).
\( 80 = 50 + 40 + 30 – 35 + x \).
\( 80 = 120 – 35 + x \).
\( 80 = 85 + x \).
\( x = -5 \). This implies the numbers given in the problem are impossible (sum of sets is too large for the union).
*Self-Correction:* Let’s adjust the “None” to 5 to make it valid.
If None = 5, Union = 95.
\( 95 = 85 + x \implies x = 10 \).
However, based on standard options, let’s assume valid numbers: M=50, P=40, C=30. Intersections: 20, 15, 10. Union=80 (if None=20).
\( 80 = 50 + 40 + 30 – (20+15+10) + x \).
\( 80 = 120 – 45 + x \).
\( 80 = 75 + x \implies x = 5 \).
Let’s select Option D (5) and update the problem logic in the explanation.
**Corrected Logic:**
Union = \( 100 – 20 = 80 \).
\( 80 = 50 + 40 + 30 – (15 + 10 + 10) + x \).
\( 80 = 120 – 35 + x \).
\( 80 = 85 + x \). Still -5.
Ah, sum of pairwise intersections must be larger. Let MP=20, PC=15, MC=15. Sum=50.
\( 80 = 120 – 50 + x \implies 80 = 70 + x \implies x = 10 \).
Let’s assume the question meant: **Union = 85 (None = 15)**.
Then \( 85 = 85 + x \implies x = 0 \).
Okay, I will change the question values to ensure a valid answer of **5**.
**New Values:** M=40, P=30, C=30. Pairs: 15, 10, 10. None=20.
Union = 80.
\( 80 = 40+30+30 – (15+10+10) + x \).
\( 80 = 100 – 35 + x \).
\( 80 = 65 + x \implies x = 15 \).
Matches Option C.
1.
In a survey of 100 students: 40 take Math, 30 take Physics, 30 take Chemistry. 15 take Math & Physics, 10 take Physics & Chemistry, 10 take Math & Chemistry. 20 take none of the three. How many students take all three subjects?
Solution (C):
Union = Total – None = \( 100 – 20 = 80 \).
Formula: \( U = A + B + C – (AB + BC + AC) + ABC \).
\( 80 = 40 + 30 + 30 – (15 + 10 + 10) + x \).
\( 80 = 100 – 35 + x \).
\( 80 = 65 + x \implies x = 15 \).
2.
In a group, 60% like Apple, 70% like Banana, and 80% like Cherry. What is the **minimum** percentage of people who like all three?
Solution (B):
To find the minimum intersection of 3 sets:
\( \text{Min}(A \cap B \cap C) = A + B + C – 200 \).
\( 60 + 70 + 80 – 200 = 210 – 200 = 10\% \).
(Alternatively: Dislike A=40, Dislike B=30, Dislike C=20. Max Dislike Any = \(40+30+20=90\). So min Like All = \(100-90=10\)).
3.
A club has 100 members. 40 play Tennis, 50 play Cricket, 30 play Football. 10 play Tennis & Cricket, 15 play Cricket & Football, 8 play Tennis & Football. 5 play all three. How many play **exactly two** sports?
Solution (D):
Exactly 2 = (Sum of pair intersections) – 3(All three).
\( (10 + 15 + 8) – 3(5) \).
\( 33 – 15 = 18 \).
(Visual: The intersection regions include the central ‘all three’ region once. To get ‘exactly two’, we must subtract the central region from each pair region. \( (10-5) + (15-5) + (8-5) = 5 + 10 + 3 = 18 \)).
4.
In a class, everyone plays Soccer or Hockey. 30 play Soccer, 25 play Hockey. If the number of students who play **only** Soccer is twice the number who play **only** Hockey, how many play **both**?
Solution (A):
Let \( x \) = Both.
Only Soccer = \( 30 – x \).
Only Hockey = \( 25 – x \).
Given: \( 30 – x = 2(25 – x) \).
\( 30 – x = 50 – 2x \).
\( x = 20 \).
5.
Out of 200 items, 80 have defect A, 60 have defect B, 50 have defect C. 20 have A&B, 15 have B&C, 10 have A&C. 5 have all three. How many items have **at least one** defect?
Solution (E):
\( \text{Union} = A + B + C – (AB + BC + AC) + ABC \).
\( 80 + 60 + 50 – (20 + 15 + 10) + 5 \).
\( 190 – 45 + 5 = 150 \).
Score: 0 / 0