Let E represent Economics, M represent Mathematics, and G represent Geography.

Here's the solution to the problem. Let $E$ represent Economics, $M$ represent Mathematics, and $G$ represent Geography. The total number of students is $n(U) = 70$. All students offer at least one subject, so $n(E \cup M \cup G) = 70$. We are given the following information: 1. $n(E \text{ only}) = 6$ 2. $n(E \cap M^c) = 18$ (Economics but not Mathematics) 3. $n(E \cap G) = 38$ (Economics and Geography) 4. $n(E) = 53$ (Economics) 5. $n(G) = 50$ (Geography) 6. $n(M \cap G) = 34$ (Mathematics and Geography) Let's denote the number of students in each region of the Venn diagram as follows: • $a = n(E \text{ only})$ • $b = n(M \text{ only})$ • $c = n(G \text{ only})$ • $d = n(E \cap M \cap G^c)$ (Economics and Mathematics only) • $e = n(E \cap G \cap M^c)$ (Economics and Geography only) • $f = n(M \cap G \cap E^c)$ (Mathematics and Geography only) • $g = n(E \cap M \cap G)$ (Economics, Mathematics, and Geography) Using the given information to find the values for each region: Step 1: From (1), $a = 6$. Step 2: From (2), $n(E \cap M^c) = a + e = 18$. Since $a=6$, we have $6 + e = 18 \implies e = 12$. Step 3: From (3), $n(E \cap G) = e + g = 38$. Since $e=12$, we have $12 + g = 38 \implies g = 26$. Step 4: From (4), $n(E) = a + d + e + g = 53$. Since $a=6, e=12, g=26$, we have $6 + d + 12 + 26 = 53 \implies d + 44 = 53 \implies d = 9$. Step 5: From (6), $n(M \cap G) = f + g = 34$. Since $g=26$, we have $f + 26 = 34 \implies f = 8$. Step 6: From (5), $n(G) = c + e + f + g = 50$. Since $e=12, f=8, g=26$, we have $c + 12 + 8 + 26 = 50 \implies c + 46 = 50 \implies c = 4$. Step 7: Since all students offer at least one subject, $n(E \cup M \cup G) = a + b + c + d +