This is a level 4-4+ activity from the Figure It Out series.
Ideally, the shunting problem should be solved using model railway track and the required rolling stock. If you don’t have this equipment, 8 wooden blocks with a cup hook screwed into each end would serve just as well. Mark the carriage blocks a1, a2, a3, and so on, and the engine blocks A and B. Mark out a track on a piece of cardboard, being careful to ensure that the siding can take only 2 carriages and an engine.
There are different possible solutions to this problem, but they all require the drivers of the two engines to co-operate. Diagrams and a possible solution are included in the Answers.
As an extension activity, the students could be asked to come up with two different solutions and then explain why one of them is better than the other. The criteria for such a judgment could include the number of separate forwards and backwards movements involved, the number of couplings and uncouplings required, and whether the locomotives and their carriages end up in the original order. There could be a class-wide search for the simplest solution.
The students will be familiar with games that involve challenges similar to that posed in mini-disaster two. If desired, they can easily model the pile of steel sheets using paper or cardboard rectangles. They will discover that, although the starting sequence is clear, there are different possibilities for the later sheets.
The domino problem is a perfect example of risk management. By placing a break between every 100 dominoes, Makil is reducing the risk of disaster. He is not removing it altogether, but the worst that can happen is that 100 dominoes can tumble unexpectedly. The removal strategy is critical. If all the breaks were to be taken out in sequence, the risk of untimely collapse would increase by 100 dominoes each time. Just before the last break is removed, the whole lot except the last 100 dominoes could collapse! So, by removing every second break, Makil increases the risk in a controlled fashion.
An opportunity exists here for discussing risk management and mathematical strategies that contain risk. For instance, if a sweet factory makes bags of jelly beans, can every bag be checked for quality and correct contents? How could the risk of poor product going to the shops be managed realistically?
For the links problem, there is a simple solution. By opening the links of the two 2-link chains, Merania has 4 loose links, which is exactly the number of joining links required to connect the 4 remaining chains.
Furthermore, it is an economical solution since it takes a minimal amount of opening and closing to achieve the desired result.
In the card problem, Yvette’s cards have to be placed in a strict sequence for the tricks to work. There are 10 cards in the pack, and every alternate card is placed down starting with the first, so we have the positions for 1, 2, 3, 4, and 5. Let the letters A, B, C, D, and E represent unknown cards for now. Then the sequence so far is 1, A, 2, B, 3, C, 4, D, 5, E. This means that 1 will be placed down and A moved to the back, 2 placed down and B moved to the back, and so on. After the 5 has been placed down and E moved to the back, the remaining cards will be in the order A, B, C, D, E. A will be the next card to place down, so it must be 6. B will go to the back of the pack, so C is 7 and E must be 8. The remaining order is B, D. E was placed down, so B goes to the back of pack. D is 9 and B is 10. Therefore, the original order of the cards was 1, 6, 2, 10, 3, 7, 4, 9, 5, 8.
Answers to Activity
1. A possible solution that minimises coupling and uncoupling is shown below. In each step, the diagram shows the position of the trains and carriages when the step is complete.
i. B reverses sufficiently far along on the right to allow A to go completely past the siding. A reverses into the siding, uncouples a3, and then returns to a position in front of B.
ii. Both trains move independently until they occupy the left section of the track. B reverses into the siding, links up with a3, and pulls it back out onto the left section of the track.
iii. Both trains move independently until they occupy the right section of the track. A then reverses into the siding with a1 and a2.
iv. B immediately takes its 4 carriages back into the left section of track. A pulls a1 and a2 out of the siding onto the right section of the track, then reverses and couples with a3, which B then uncouples.
v. Both trains head off in the desired directions with all their carriages in the original order.
2. One possible order is: i, c, d, k, a, h, f, j, g, e, b
3. One way to remove the break is to take out every second break each time. This limits the total number of dominoes that can fall by mistake at any one time. The arrows point to the breaks that are to be removed in each step.
4. The most economical solution is to undo the links on the 2-link pieces and use these four separate pieces to join the other pieces.
5. The order is: 1, 6, 2, 10, 3, 7, 4, 9, 5, 8