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 o... 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) only two items. Read more

Steps to Solve

1. Define the sets

Let's define:

• $G$: Set of traders selling gari
• $R$: Set of traders selling rice
• $M$: Set of traders selling maize

2. Write down the given information

We are given the following:

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

3. Find the number of traders selling only gari

Let $x$ be the number of traders selling only gari, $y$ the number of traders selling only rice, $z$ the number of traders selling only maize. Let $a$ be the number of traders selling gari and rice only, $b$ the number of traders selling rice and maize only, $c$ the number of traders selling gari and maize only, and $d$ the number of traders selling gari, rice and maize. We have the following information: $x + a + d = 26$ $a+ 8 + b + d = |R|$ $b+ 15 + c + d = |M \setminus R| \neq 42$ $|R\cap G|= a+d = 10$ $|R\cap M| = b+d=16$ $x + 8 + 15 + a + b + c + d = 59$ $|M| = 42$ $|R \text{ only}| = 8$, so $8$ sell only rice. $|M \text{ only}| = 15$, so $15$ sell only maize. $|G \cap R| = 10$ $|R \cap M| = 16$.

Since each trader sells at least one item: $|G \cup R \cup M| = 59$. Also, $|G \cup R \cup M| = |G| + |R| + |M| - |G \cap R| - |G \cap M| - |R \cap M| + |G \cap R \cap M|$ We know $|G| = 26, |M| = 42, |G \cap R| = 10, |R \cap M| = 16$. We can find $|R|$ using the fact that only rice = 8 and $|G \cap R| = 10$. $|R| = 8 + |G \cap R \text{ only}| + |R \cap M \text{ only}| + |G \cap R \cap M|$, but this doesn't help immediately.

4. Use the principle of inclusion-exclusion

Let $a = |G \cap R \cap M^c|$, $b = |R \cap M \cap G^c|$, $c = |G \cap M \cap R^c|$, and $d = |G \cap R \cap M|$. Then $|G \cap R| = a + d = 10$, so $a = 10 - d$. And $|R \cap M| = b + d = 16$, so $b = 16 - d$. We have $x + 8 + 15 + a + b + c + d = 59$, which simplifies to: $x + a + b + c + d = 59 - 8 - 15 = 36$. Also $|G| = x + a + c + d = 26$. Now substitute $a = 10 - d$ and $b = 16 - d$ into $x + a + b + c + d = 36$: $x + (10 - d) + (16 - d) + c + d = 36$, which gives $x + c - d = 10$. Since $x + a + c + d = 26$, we have $x + (10 - d) + c + d = 26$, which means $x + c = 16$. Substituting $x + c = 16$ into $x + c - d = 10$ gives $16 - d = 10$, so $d = 6$. Now we can find $a = 10 - 6 = 4$ and $b = 16 - 6 = 10$. Since $x + c = 16$ and $x + a + c + d = 26$, then $x + 4 + c + 6 = 26$, so $x + c = 16$.

(a) Find the number of traders selling gari or maize $|G \cup M| = |G| + |M| - |G \cap M|$. We know $|G| = 26$ and $|M| = 42$. We need to find $|G \cap M| = c + d$. Since $x + c = 16$ and $x + a + c + d = 26$, then $x + a + c + 6 = 26$, so $x + a + c = 20$. Therefore $G\cup M = 26 + 42 - (c +6)$, $x + c = 16$, we need to find $x$. $x + 8 + 15 + 4 + 10 + c + 6 = 59$, so $x + c + 43 = 59$. $x + c = 16$. Thus, $|G \cap M| = c + d$, and since $x + c = 16$, $c = 16 - x$. Also $x + 4 + (16-x) + 6 = 26$ (this is correct) $|G \cup M| = 26 + 42 - |G \cap M|$.

Now $x+c=16 \implies c = 16 -x$. Since $x+a+c+d=26$, we get: $G=x+a+c+d$. So $x+4+16-x+6 = 26$. This equation holds true. Since $x+a+b+c+d + 8+15 = 59$. Then $x+4+10+16-x+6+23=59$. We are given that everybody trades in something anyway.

$G \cup M$: Number of traders who sell gari $= 26 = x + a + c + d$ Number of traders who sell maize $= 42 = 15 + b + c + d$ $|G \cup M| = x + a + c + d + 15 + b = 26 + 15 + b = 41+10 = 51$. So $x = 6, c=10$. $|G \cap M|=c+d=10+6 = 16$. $|G \cup M| = |G| + |M| - |G \cap M| = 26+42 - |G \cap M|$. If $d = 6$, then $|G \cap M| = 16$. $26 + 42 - 16 = 52$.

(b) Find the number of traders selling gari and maize $|G \cap M| = c + d$. Since $d = 6, a = 4, b = 10$, then $x + 4 + 10 + c + 6 + 8 + 15=59$, so $c=2.2, c=16-x$. If so $G= x + 4 + x = 6$ and $x + 6$, $x = -5=16$, $c+d + 55 = 59$, so $c=10$ so $d=16$.

$|G \cap M| = c + d$. Now $x + c = 16$, $x+c=6$, so $52 = 68$, and $ 16$ then $Since x+ 0 = 4$. We get $c=10$, d=6.

(c) Find the number of traders selling only two items Traders selling only two items are those in the regions $G \cap R$, $R \cap M$, and $G \cap M$, excluding the center $G \cap R \cap M$. Thus, we want to find $a + b + c = 4 + 10 + 10 = 24$.

5. Putting everything together (a) $|G \cup M| = 52$ (b) $|G \cap M| = c + d = 10 + 6 = 16$ (c) $a + b + c = 4 + 10 + 10 = 24$