A co-operative society has to transport 200 churns of milk. A pick-up and a tractor with a trailer are available. The pick-up can ferry 12 churns per trip and the tractor 20 per trip. The tractor can

Let $x$ be the number of trips made by the pick-up.
Let $y$ be the number of trips made by the tractor.

The pick-up ferries 12 churns per trip, and the tractor ferries 20 churns per trip. A total of 200 churns must be transported.

$$12x + 20y \ge 200$$

This inequality can be simplified by dividing by 4:

$$3x + 5y \ge 50$$

The tractor can make a maximum of 5 trips:

$$y \le 5$$

The total number of trips must not exceed 18:

$$x + y \le 18$$

Also, the number of trips cannot be negative:

$$x \ge 0$$ $$y \ge 0$$

The tractor costs Sh 60 per trip, and the pick-up costs Sh 100 per trip. The total cost $C$ is:

$$C = 100x + 60y$$

We need to find integer pairs $(x, y)$ such that:

1. $3x + 5y \ge 50$
2. $y \le 5$
3. $x + y \le 18$
4. $x \ge 0, y \ge 0$

Since $y \le 5$ and $y \ge 0$, we can test values for $y$ from 0 to 5.

• If $y = 0$: $3x \ge 50 \implies x \ge 16.67 \implies x \ge 17$ $x + 0 \le 18 \implies x \le 18$ Possible $(x, y)$ pairs: $(17, 0), (18, 0)$.

• If $y = 1$: $3x + 5 \ge 50 \implies 3x \ge 45 \implies x \ge 15$ $x + 1 \le 18 \implies x \le 17$ Possible $(x, y)$ pairs: $(15, 1), (16, 1), (17, 1)$.

• If $y = 2$: $3x + 10 \ge 50 \implies 3x \ge 40 \implies x \ge 13.33 \implies x \ge 14$ $x + 2 \le 18 \implies x \le 16$ Possible $(x, y)$ pairs: $(14, 2), (15, 2), (16, 2)$.

• If $y = 3$: $3x + 15 \ge 50 \implies 3x \ge 35 \implies x \ge 11.67 \implies x \ge 12$ $x + 3 \le 18 \implies x \le 15$ Possible $(x, y)$ pairs: $(12, 3), (13, 3), (14, 3), (15, 3)$.

• If $y = 4$: $3x + 20 \ge 50 \implies 3x \ge 30 \implies x \ge 10$ $x + 4 \le 18 \implies x \le 14$ Possible $(x, y)$ pairs: $(10, 4), (11, 4), (12, 4), (13, 4), (14, 4)$.

• If $y = 5$: $3x + 25 \ge 50 \implies 3x \ge 25 \implies x \ge 8.33 \implies x \ge 9$ $x + 5 \le 18 \implies x \le 13$ Possible $(x, y)$ pairs: $(9, 5), (10, 5), (11, 5), (12, 5), (13, 5)$.

The possible $(x, y)$ pairs and their corresponding costs $C = 100x + 60y$ are:

• $(17, 0): C = 100(17) + 60(0) = 1700$
• $(18, 0): C = 100(18) + 60(0) = 1800$
• $(15, 1): C = 100(15) + 60(1) = 1500 + 60 = 1560$
• $(16, 1): C = 100(16) + 60(1) = 1600 + 60 = 1660$
• $(17, 1): C = 100(17) + 60(1) = 1700 + 60 = 1760$
• $(14, 2): C = 100(14) + 60(2) = 1400 + 120 = 1520$
• $(15, 2): C = 100(15) + 60(2) = 1500 + 120 = 1620$
• $(16, 2): C = 100(16) + 60(2) = 1600 + 120 = 1720$
• $(12, 3): C = 100(12) + 60(3) = 1200 + 180 = 1380$
• $(13, 3): C = 100(13) + 60(3) = 1300 + 180 = 1480$
• $(14, 3): C = 100(14) + 60(3) = 1400 + 180 = 1580$
• $(15, 3): C = 100(15) + 60(3) = 1500 + 180 = 1680$
• $(10, 4): C = 100(10) + 60(4) = 1000 + 240 = 1240$
• $(11, 4): C = 100(11) + 60(4) = 1100 + 240 = 1340$
• $(12, 4): C = 100(12) + 60(4) = 1200 + 240 = 1440$
• $(13, 4): C = 100(13) + 60(4) = 1300 + 240 = 1540$
• $(14, 4): C = 100(14) + 60(4) = 1400 + 240 = 1640$
• $(9, 5): C = 100(9) + 60(5) = 900 + 300 = 1200$
• $(10, 5): C = 100(10) + 60(5) = 1000 + 300 = 1300$
• $(11, 5): C = 100(11) + 60(5) = 1100 + 300 = 1400$
• $(12, 5): C = 100(12) + 60(5) = 1200 + 300 = 1500$
• $(13, 5): C = 100(13) + 60(5) = 1300 + 300 = 1600$

The possible number of trips for the pick-up ($x$) and tractor ($y$) and their corresponding costs are:
\boxed{\begin{align*} &(17, 0) \text{ with cost Sh } 1700 \\ &(18, 0) \text{ with cost Sh } 1800 \\ &(15, 1) \text{ with cost Sh } 1560 \\ &(16, 1) \text{ with cost Sh } 1660 \\ &(17, 1) \text{ with cost Sh } 1760 \\ &(14, 2) \text{ with cost Sh } 1520 \\ &(15, 2) \text{ with cost Sh } 1620 \\ &(16, 2) \text{ with cost Sh } 1720 \\ &(12, 3) \text{ with cost Sh } 1380 \\ &(13, 3) \text{ with cost Sh } 1480 \\ &(14, 3) \text{ with cost Sh } 1580 \\ &(15, 3) \text{ with cost Sh } 1680 \\ &(10, 4) \text{ with cost Sh } 1240 \\ &(11, 4) \text{ with cost Sh } 1340 \\ &(12, 4) \text{ with cost Sh } 1440 \\ &(13, 4) \text{ with cost Sh } 1540 \\ &(14, 4) \text{ with cost Sh } 1640 \\ &(9, 5) \text{ with cost Sh } 1200 \\ &(10, 5) \text{ with cost Sh } 1300 \\ &(11, 5) \text{ with cost Sh } 1400 \\ &(12, 5) \text{ with cost Sh } 1500 \\ &(13, 5) \text{ with cost Sh } 1600 \end{align*}}