Given that P = {p: 1< p < 20}, where p is an integer and R = {r : 0 $\leq$ r $\leq$ 25, where r is a multiple of 4}. Find P ∩ R

Analyze the question: We need the elements common to both sets P and R.

Concept: The intersection (P ∩ R) contains only the elements that belong to both sets.

Step 1: List P P = {p: 1 < p < 20}, with p an integer. The strict inequalities exclude both 1 and 20, so P = {2, 3, 4, ..., 19}

Step 2: List R R = {r: 0 ≤ r ≤ 25, r a multiple of 4}. The inequalities here include 0 and 25, so R = {0, 4, 8, 12, 16, 20, 24}

Step 3: Find the common elements

• 0 is not in P (since 0 is not greater than 1)
• 4, 8, 12 and 16 are all between 1 and 20, so they are in P
• 20 is not in P (since p < 20 is strict)
• 24 is not in P

So P ∩ R = {4, 8, 12, 16}

Answer: B

Tip: Always check the inequality signs at the boundaries. "<" excludes the end value, while "≤" includes it. Here, the strict "p < 20" is what removes 20 from the answer, and this is exactly where option C tempts students.