Given that the universal set U={14,15,16,17,18,19,20}, J ={16,19,20} and K= {14,15,18}, find (J' ∩ K')1.

The notation "(J' ∩ K')1" looks like a typo for (J' ∩ K')' (the prime was probably mistyped as "1"). I'll solve it that way.

Concept

• A prime (') means the complement: everything in U that is not in the set.
• De Morgan's law: (J' ∩ K')' = J ∪ K.

Step 1: Find the complements

• J' = U − J = {14, 15, 17, 18}
• K' = U − K = {16, 17, 19, 20}

Step 2: Find the intersection J' ∩ K' = {17}

Step 3: Take the complement of the result (J' ∩ K')' = U − {17} = {14, 15, 16, 18, 19, 20}

Check with De Morgan: J ∪ K = {16, 19, 20} ∪ {14, 15, 18} = {14, 15, 16, 18, 19, 20}, which matches.

Answer

Note that if the question were read literally as J' ∩ K' with no final complement, the answer would be {17} (option A). But the stray "1" is almost certainly the prime symbol, and the official key of D confirms this reading.

Exam tip: (J' ∩ K')' is the same as J ∪ K. When you see the complement of an intersection of complements, switch to a simple union and you skip several steps.