SET Practice SET – 3
Set Theory: Level 3 (Hard)
1.
In a class of 60 students, 30 study Math, 25 study Physics, and 20 study Chemistry. 10 study M&P, 8 study P&C, 7 study M&C. 3 study all three. How many study **only** Math?
Solution (D):
Only M = \( M – (M \cap P) – (M \cap C) + (M \cap P \cap C) \).
Only M = \( 30 – 10 – 7 + 3 = 16 \).
2.
In a group, 70% like A, 80% like B. What is the **minimum** percentage that must like both?
Solution (B):
Min Both = \( A + B – 100 \).
\( 70 + 80 – 100 = 50\% \).
3.
Set S is the set of all prime numbers less than 20. Set T is the set of all odd integers less than 20. How many elements are in \( T – S \) (in T but not in S)?
Solution (C):
T (Odd < 20): {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}. (10 items).
S (Primes < 20): {2, 3, 5, 7, 11, 13, 17, 19}.
Primes in T: {3, 5, 7, 11, 13, 17, 19}. (7 items).
Elements in T not in S: {1, 9, 15}. (Note: 2 is in S but not T, but we want T-S).
Wait, I listed 3 items. Let's recount.
Odd numbers not prime: 1, 9, 15. Correct.
Option C is 4. My count is 3.
Is 21 < 20? No.
Let me check my arithmetic.
T: {1,3,5,7,9,11,13,15,17,19}.
S: {2,3,5,7,11,13,17,19}.
Intersection: {3,5,7,11,13,17,19}.
T minus S: {1, 9, 15}. Count is 3.
Correct answer is 3. I will map to Option B.
(Prompt labeled C as correct. I will fix data-answer to B).
3.
Set S is the set of all prime numbers less than 20. Set T is the set of all odd integers less than 20. How many elements are in \( T – S \) (in T but not in S)?
Solution (B):
Odd numbers < 20: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.
Primes < 20: 2, 3, 5, 7, 11, 13, 17, 19.
Odd numbers that are NOT prime: 1, 9, 15.
There are 3 elements.
4.
Out of 100 families, 50 have a car, 40 have a bike, and 40 have neither. How many have **both**?
Solution (A):
At least one = Total – Neither = \( 100 – 40 = 60 \).
\( 60 = 50 + 40 – \text{Both} \).
\( 60 = 90 – \text{Both} \implies \text{Both} = 30 \).
5.
If A and B are sets, \( n(A)=12, n(B)=15, n(A \cup B)=20 \). What is \( n(A \Delta B) \) (Symmetric Difference: elements in exactly one set)?
Solution (D):
First find intersection: \( 20 = 12 + 15 – Int \implies Int = 7 \).
Symmetric Difference = Only A + Only B.
Only A = \( 12 – 7 = 5 \).
Only B = \( 15 – 7 = 8 \).
Sum = \( 5 + 8 = 13 \).
(Alternative formula: \( n(A \cup B) – n(A \cap B) = 20 – 7 = 13 \)).
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