A survey of 100 traders revealed that 70 sell Gari, 10 sell only Rice, 15 sell only Maize, 50 sell Gari and Rice, 45 sell Rice and Maize, and 65 sell Maize. If each trader sells at least one item, fin

Let $G$ be the set of traders who sell Gari, $R$ be the set of traders who sell Rice, and $M$ be the set of traders who sell Maize.
The total number of traders is $n(U) = 100$.
Each trader sells at least one item, so $n(G \cup R \cup M) = 100$.

Given information:

• $n(G) = 70$
• $n(R \text{ only}) = 10$ (traders who sell Rice but not Gari or Maize)
• $n(M \text{ only}) = 15$ (traders who sell Maize but not Gari or Rice)
• $n(G \cap R) = 50$ (traders who sell both Gari and Rice)
• $n(R \cap M) = 45$ (traders who sell both Rice and Maize)
• $n(M) = 65$

Let $x = n(G \cap R \cap M)$ be the number of traders who sell all three items.

We can represent the number of traders in each disjoint region of the Venn diagram:

• Traders selling only Gari: $n(G \text{ only})$
• Traders selling only Rice: $n(R \text{ only}) = 10$
• Traders selling only Maize: $n(M \text{ only}) = 15$
• Traders selling Gari and Rice only (not Maize): $n(G \cap R \text{ only}) = n(G \cap R) - n(G \cap R \cap M) = 50 - x$
• Traders selling Gari and Maize only (not Rice): $n(G \cap M \text{ only})$
• Traders selling Rice and Maize only (not Gari): $n(R \cap M \text{ only}) = n(R \cap M) - n(G \cap R \cap M) = 45 - x$
• Traders selling Gari, Rice, and Maize: $n(G \cap R \cap M) = x$