Q.32
Let U = {1, 2, …, 15}. Let P ⊆ U consist of all prime numbers, Q ⊆ U consist of all even numbers and R ⊆ U consist of all multiples of 3. Let T = P – Q. Then, which of the following is/are CORRECT?
- (A) |T| = 5 and |T ∪ R| = 9
- (B) |T| = 6 and |T ∪ R| = 9
- (C) |T| = 5 and |T ∩ R| = 1
- (D) |T| = 6 and |T ∩ R| = 1
Set Theory Problem: Primes, Evens, and Multiples of 3 in U = {1,2,…,15}
This set theory question tests understanding of subsets, set operations, and cardinalities for competitive exams like CSIR NET. T represents odd primes (primes excluding the even prime 2), with specific sizes for T, T ∪ R, and T ∩ R verified through explicit calculations.
Define the Sets
U = {1, 2, …, 15} contains 15 elements. P includes all primes: {2, 3, 5, 7, 11, 13} (6 elements). Q covers even numbers: {2, 4, 6, 8, 10, 12, 14} (7 elements). R lists multiples of 3: {3, 6, 9, 12, 15} (5 elements).
T = P – Q excludes even primes from P, yielding T = {3, 5, 7, 11, 13}. Thus |T| = 5.
Option Analysis
Option (A): |T| = 5 and |T ∪ R| = 9.
T ∪ R = {3, 5, 6, 7, 9, 11, 12, 13, 15} has 9 elements, matching the calculation. Correct.
Option (B): |T| = 6 and |T ∪ R| = 9.
|T| = 5, not 6 (includes only odd primes). Incorrect despite correct union size.
Option (C): |T| = 5 and |T ∩ R| = 1.
T ∩ R = {3} (only odd prime multiple of 3), so |T ∩ R| = 1. Correct.
Option (D): |T| = 6 and |T ∩ R| = 1.
|T| = 5, not 6. Incorrect.
Both (A) and (C) hold true.
Article Excerpt (SEO-Optimized):
In set theory problems for CSIR NET, U={1,2,…,15} defines primes P={2,3,5,7,11,13}, evens Q={2,4,6,8,10,12,14}, multiples of 3 R={3,6,9,12,15}. T=P-Q={3,5,7,11,13} gives |T|=5. Verify |T∪R|=9 and |T∩R|=1 to select correct options (A) and (C).
