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. Find the number of traders who sell gari only

Let $G$ be the set of traders who sell gari, $R$ the set who sell rice, and $M$ the set who sell maize. We are given:

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

Let $x$ be the number of traders who sell gari and maize only, $y$ be the number of traders who sell rice and gari only, and $z$ be the number of traders who sell rice and maize only. Let $w$ be the number of traders who sell all three items, $a$ be the number of traders who sell only gari, $b$ be the number of traders who sell only rice, and $c$ be the number of traders who sell only maize.

We know that $b = 8$ and $c = 15$. We are also give that the number of traders who sell gari and rice = 10, so $y + w = 10$. The number of traders who sell rice and maize = 16, so $z + w = 16$. We also know that $a + b + c + y + z + x + w = 59$ From the given information: $|G|= 26 $, $|M| = 42$ $|G| = a + y + x + w = 26$ $|M| = c + x + z + w = 42$

2. Solve for $w$ (the number of traders who sell all three items)

The total number of traders is 59, and each sells at least one item. So, $59 = a + 8 + 15 + (10 - w) + (16 - w) + x + w$ $59 = a + 8 + 15 + 10 + 16 - w + x$ $59 = a + x - w + 49$ $10 = a + x - w$

We also have $a + (10-w) + x + w = 26$, so $a + x - w + 10 = 26$, which implies $a + x - w = 16$. From $|M|=42$, we have $15 + x + (16 - w) + w = 42$, so $x + 31 = 42$, which implies $x = 11$. Substituting $x=11$ into $a + x - w = 16$, we get $a + 11 - w = 16$, so $a - w = 5$, or $a = w + 5$. Substituting $x=11$ into $a+x-w = 10$ should give us a correct result. We get $a + 11 - w = 10$, so $a - w = -1$. However, $a = w + 5$ so $(w+5) - w = 5$ which does not equal to -1 obtained previously.

Let's represent it differently like this: $|G \cup R \cup M| = |G| + |R| + |M| - |G \cap R| - |G \cap M| - |R \cap M| + |G \cap R \cap M|$ $59 = 26 + (8 + 10 + 16) + 42 - 10 - |G \cap M| - 16 + w$ Notice that the $8 + 10 + 16$ gives the full size of R, and the 10 and 16 are the intersections. $59 = 26 + |R| + 42 - 10 - |G \cap M| - 16 + w$ $|R|= 8 + (10 - w) + (16 - w) + w = 34 - w$ $59 = 26 + 34 - w + 42 - 10 - |G \cap M| - 16 + w$ $59 = 26 + 34 + 42 - 10 - 16 - |G \cap M|$ $59 = 76 - |G \cap M|$ $|G \cap M|= 17$

$|G \cap M| = x + w = 17$ Since $M = 42 = 15 + x + 16 - w + w$, then $42 = 31 + x$, $x = 11$ So, $11 + w = 17$, Thus $w = 6$

3. Calculate $a$, $y$, $z$. $a = 26 - (10 - w) - w - 11 = 26 - 10 + w - w - 11 = 5$ $z = 16 - w = 16 - 6 = 10$ $y = 10 - w = 10 - 6 = 4$

4. Solve part a: Gari or Maize

We want to find $|G \cup M|$ $|G \cup M| = |G| + |M| - |G \cap M| = 26 + 42 - (x + w) = 26 + 42 - 17 = 51$

5. Solve part b: Gari and Maize

We want to find $|G \cap M| = x + w = 17$

6. Solve part c: Exactly two items

We want to find $x + y + z = 11 + 4 + 10 = 25$