The Venn diagram shows two sets R and S such that RUS = E. On separate copies of the diagram, shade R'US, R'∩S.

Here are the solutions to the set theory and Venn diagram problems:

3. (b) The universal set is $\mathcal{E} = \{x : x \text{ is an integer } 10 \le x \le 100\}$. Set $A = \{x : x \text{ is a multiple of } 17\}$. Set $B = \{x : x \text{ divided by } 15 \text{ leaves a remainder of } 7\}$.

(i) Find $n(A)$

Step 1: List the elements of set A within the universal set $\mathcal{E}$. Multiples of 17 between 10 and 100 are: $17 \times 1 = 17$ $17 \times 2 = 34$ $17 \times 3 = 51$ $17 \times 4 = 68$ $17 \times 5 = 85$ $17 \times 6 = 102$ (This is greater than 100, so it's not in $\mathcal{E}$) So, $A = \{17, 34, 51, 68, 85\}$.

Step 2: Count the number of elements in set A. $n(A) = 5$. The number of elements in A is $\boxed{5}$.

(ii) Find the largest member of B

Step 1: List the elements of set B within the universal set $\mathcal{E}$. Numbers $x$ that leave a remainder of 7 when divided by 15 can be written as $x = 15k + 7$, where $k$ is an integer. For $x \ge 10$: If $k=0$, $x = 15(0) + 7 = 7$ (not in $\mathcal{E}$) If $k=1$, $x = 15(1) + 7 = 22$ (in $\mathcal{E}$) If $k=2$, $x = 15(2) + 7 = 37$ (in $\mathcal{E}$) If $k=3$, $x = 15(3) + 7 = 52$ (in $\mathcal{E}$) If $k=4$, $x = 15(4) + 7 = 67$ (in $\mathcal{E}$) If $k=5$, $x = 15(5) + 7 = 82$ (in $\mathcal{E}$) If $k=6$, $x = 15(6) + 7 = 97$ (in $\mathcal{E}$) For $x \le 100$: If $k=7$, $x = 15(7) + 7 = 105 + 7 = 112$ (not in $\mathcal{E}$) So, $B = \{22, 37, 52, 67, 82, 97\}$.

Step 2: Identify the largest element in set B. The largest member of B is $\boxed{97}$.

(iii) Find $n(A \cup B)$

Step 1: Use the formula for the number of elements in the union of two sets: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$. From part (i), $n(A) = 5$. From part (ii), $n(B) = 6$.

Step 2: Find the intersection of sets A and B ($A \cap B$). $A = \{17, 34, 51, 68, 85\}$ $B = \{22, 37, 52, 67, 82, 97\}$ There are no common elements between A and B. So, $A \cap B = \emptyset$.

Step 3: Calculate $n(A \cap B)$. $n(A \cap B) = 0$.

Step 4: Calculate $n(A \cup B)$. $n(A \cup B) = 5 + 6 - 0 = 11$. The number of elements in $A \cup B$ is $\boxed{11}$.

3. (c) The Venn diagram shows two sets R and S such that $R \cup S = \mathcal{E}$. This means the universal set $\mathcal{E}$ consists only of the elements within R or S (or both), and there are no elements outside both circles.

(i) Shade $R' \cup S$

Step 1: Understand the notation. $R'$ represents the complement of R (elements not in R). $S$ represents the set S. The union $\cup$ means "or". So, $R' \cup S$ means all elements that are either not in R, or are in S (or both).

Step 2: Apply the condition $R \cup S = \mathcal{E}$. Since $R \cup S = \mathcal{E}$, the region $R'$ (elements not in R) is equivalent to the region $S \setminus R$ (elements in S but not in R). Therefore, $R' \cup S = (S \setminus R) \cup S$. Since $S \setminus R$ is a part of $S$, the union of $S \setminus R$ and $S$ is simply $S$.

Step 3: Describe the shading. To shade $R' \cup S$, you should shade the entire circle S.

(ii) Shade $R' \cap S$

Step 1: Understand the notation. $R'$ represents the complement of R (elements not in R). $S$ represents the set S. The intersection $\cap$ means "and". So, $R' \cap S$ means all elements that are not in R AND are in S.

Step 2: Describe the shading. This corresponds to the region of S that does not overlap with R. To shade $R' \cap S$, you should shade the part of circle S that is outside circle R (the crescent-shaped region of S).