In a group of 59 traders, 26 sell gari, 8 sell only rice and 15 sell only maize, 10 sell gari and rice, 16 sell rice and maize and 42 sell maize. Each trader sells at least one of... In a group of 59 traders, 26 sell gari, 8 sell only rice and 15 sell only maize, 10 sell gari and rice, 16 sell rice and maize and 42 sell maize. Each trader sells at least one of the three items. Find the number of traders who sell: a. Gari or maize b. Gari and maize c. Exactly two items Read more

Steps to Solve

1. Define sets and variables

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. Let $|G|$ denote the number of traders who sell gari, $|R|$ the number who sell rice, and $|M|$ the number who sell maize. We are given the following information:

• Total number of traders = 59
• $|G| = 26$
• Number of traders who sell only rice = 8
• Number of traders who sell only maize = 15
• Number of traders who sell gari and rice = $|G \cap R| = 10$
• Number of traders who sell rice and maize = $|R \cap M| = 16$
• $|M| = 42$

2. Use the Principle of Inclusion-Exclusion

Let $T$ be the total number of traders. We have: $T = |G \cup R \cup M| = |G| + |R| + |M| - |G \cap R| - |G \cap M| - |R \cap M| + |G \cap R \cap M|$ We know that $T = 59$, $|G| = 26$, $|M| = 42$, $|G \cap R| = 10$, $|R \cap M| = 16$. We need to find $|R|$.

3. **Calculate $|R|$ **

The total number of rice sellers, $|R|$, is given by the sum of those who sell only rice, those who sell rice and gari, rice and maize, and all three items. Let $x$ = number of traders selling gari, rice, and maize. Then we know the number selling gari and rice is 10, and the number selling rice and maize is 16. Also, we know 8 sell ONLY rice, 15 sell ONLY maize We know $59 = |G \cup R \cup M|$. Breaking down R into those selling only rice + those selling rice and gari AND not maize + those selling rice and maize AND not gari + those selling rice, gari and maize, we have: $|R| = \text{only rice} + (|G \cap R| - x) + (|R \cap M| - x) + x = 8 + (10-x) + (16 - x) + x$ $|R| = 8 + 10 - x + 16 - x + x = 34 - x$

4. Find $|G \cap M|$

Substituting the values we have into the Inclusion-Exclusion Principle: $59 = 26 + (34-x) + 42 - 10 - |G \cap M| - 16 + x$ $59 = 26 + 34 - x + 42 - 10 - |G \cap M| - 16 + x$ $59 = 76 - |G \cap M|$ $|G \cap M| = 76 - 59 = 17$

5. Find the number of traders who sell Gari or maize

$|G \cup M| = |G| + |M| - |G \cap M| = 26 + 42 - 17 = 51$

6. (b) Find the number of traders who sell Gari and maize

$|G \cap M| = 17$

7. Find x (number of traders who sell Gari, Rice, and Maize)

Using $|M| = $ only maize $ + (|M \cap R| - x) + (|M \cap G| - x) + x$, we have: $42 = 15 + (16-x) + (17 - x) + x$ $42 = 15 + 16 - x + 17 - x + x$ $42 = 48 - x$ $x = 48 - 42 = 6$

8. Calculate $|R|$ again

$|R| = 34 - x = 34 - 6 = 28$

9. (c) Find the number of traders who sell exactly two items

The number of people who sell exactly two items is: $(|G \cap R| - x) + (|G \cap M| - x) + (|R \cap M| - x) = (10 - 6) + (17 - 6) + (16 - 6) = 4 + 11 + 10 = 25$