AMA Maths and Calculus Day

1. AMA Maths and Calculus Day November 27

3. Junior School Problem-Solving • Lowest Common Denominator • Graph Work and Ratios

4. If Janet can paint a house in 6 days while John takes 9 days to paint the same house, how long will it take them if they paint it together?

5. Now 18 is the LCM of 6 and 9 and is the smallest number of days in which they would each paint the house a whole number of times. In fact Janet paints it 3 times, while John paints it twice, so working together they would paint it 5 times in 18 days, and therefore they would complete it once in 3.6 days

6. The notion can be extended to 3 participants, or any number for that matter. The question can also be posed in reverse. If two people can set a table in 5 minutes and Jane can do it by herself in 8 minutes, then how long would it take Jacob?

7. Consider what would happen in 40 minutes. They could set the table 8 times between them, but Jane could do it 5 times by herself, leaving Jacob setting the table 3 times in 40 minutes i.e. once in 40/3 minutes.

8. Other contexts where this sort of thinking can apply are: • Bath filling with water from both hot and cold taps, with and without the plug in. • Two trains travelling simultaneously on the same line • Two swimmers simultaneously doing laps in a pool at different speeds

9. Let us look at the last case. Paul and James are swimming lengths of a pool. If Paul can swim a length in 20 secs and James can swim a length in 25 secs, and they start together, when will they next meet?

10. When they meet again, going in opposite directions of course, they have completed two lengths between them. In 100 secs they can complete 9 lengths between them, which is once in 100/9 secs and therefore twice in 200/9 secs which is 22.2 secs.

11. Now let us turn our attention to problem-solving with graphs. Note that I do not use a single equation, nor do I mention the words gradient and intercept. In other words this is basically drawing pictures of situations and employing a little intuition in places.

12. Jenny goes for a bike ride to the top of a hill, turns around and comes straight home again by the same route. On the trip out she travels at a speed of 15 km/h, while with the wind behind her and going down hill, she travels with a speed of 20km/h on the return trip. If she is away for a total of 3½ hrs, what is the total distance she travelled?

14. This graph shows that the journey each way was 30kms, and that the turn-around point occurred after 2 hrs. Therefore total distance travelled was 60kms.

15. Now let us look at a problem involving two candles. Two candles of the same length are lit simultaneously. If one candle takes 8 hours to burn out, and the other takes 10 hrs to burn out, when will the slower-burning candle be twice the length of the faster-burning one?

17. From these two graphs we derive a graph of the difference in length between the two. This is because the length of the slower-burning candle is twice the length of the faster-burning one when the difference in length equals the length of the faster-burning candle. I shall label this graph the Gap Graph.

19. The graphs show us the length of the slower-burning one is twice the length of the faster-burning one after about 6 and 2/3hrs. Other contexts include 2 runners on the 400m track, and when one is twice as far from finish as the other.

20. Now let us have a look at how those LCM problems can be solved graphically. If Johnny can paint a house in 10 days while Peter could paint the same house in 15 days, how long would it take them if they worked together?

21. We use ideas very similar to the previous one, but one of the graphs shows howmuch one of the men has done, while the other shows how much theother has left. When one has completed what the other still has to complete, then they have painted the house between them.

23. After 6 days, the work done by Johnny equals the work still to be done by Peter. Therefore they have done the paint job between them.

24. If we add in a third person who can do the job by himself in 12 days , then these graphs show the paint job can be done in 4 days

26. Let us now look at solving some mixture problems using graphs. In this work • all the combinations that satisfy a given ratio form a line • The actual amount represented by a point is the sum of the two components at that point

27. Let us look at one ratio of, say, vinegar and water, in the ratio 2:1.

29. The graph can be used to add the coordinates of a point by passing an ‘x + y’ line through the point. Through the point (4,2), this line cuts both axes at 6, thus giving us the total of 4 and 2.

31. Using these ideas let us solve the following mixture problem: If I have a 2:1 mixture of vinegar and water and a 4:5 mixture of vinegar and water, how should I combine these mixtures to obtain 18 mLs of a 1:1 mixture?

33. This slide shows the two available mixtures and the desired mixture. Note that mixtures outside the rays of the two available mixtures are not possible as they represent concentrations in the components stronger than the strongest ones available .

35. We have added the demand line for 18 mLs of 1:1 mixture, thus isolating the critical point (9,9)

36. Now we perform a ‘vector’ addition of the two available ‘vectors’, so that their sum is the ‘vector’ (9,9). We draw a line through (9,9) parallel to the 1:1 line until it reaches the 4:5 line.

38. This gives vector OH, representing the quantity of 4:5 mixture which is needed, and vector HG which represents the quantity of 2:1 mixture which should be used.

39. We now use the ‘x + y’ line through H to work out the actual number of mLs of 4:5 mixture to be used. The sum will be the intercept value on either axis, while the number of mLs of 2:1 mixture will be what is required to make up the 18 mLs.

41. The graph shows that to achieve 18mLs of 1:1 we need to mix 13.5 mLs of the 4:5 mixture with 4.5 mLsof the 2:1 mixture.

42. Here is another mixture problem which can be quickly done by graphs. After mixing 21 litres of cordial and water in the ratio 5:2, I realise that I should be mixing it in the ratio 2:5. How much of what do I have to add to rectify the situation?

44. My email address is: mcenpat@hotmail.com