MSC #23 – Proportionately: Partitive solution
Our 23rd MATH-Inic Special for Christmas is my personal favorite and I think will draw the most comments. I originally titled it the “See Saw Solution” the reason for which you will realize later. I use it to mentally solve not only average speed problems, but also combination and mixture problems and even in engineering mechanics(statics). I just “feel’ the answer.
Let us consider this problem first: A car travels at a uniform rate of 40 kph for 3 hours, then speeds up to a uniform rate of 60 kph for 2 hours. What was his average speed for the entire trip?
We will normally compute the total distance traveled and divide it by the total time to get the average speed.
[(3 x 40) + (2 x 60)]/(3 + 2) = (120 + 120)/ 5 = 240/5 = 48 kph.
This is simply solving for the weighted average of the two speeds.
Now if we consider using the 40 kph speed as the base, the computation is much easier. (see Chapter 10, Vol. 1 of the book, 30 Master Strategies in Computing) Here we will assume that the car is at rest for 3 hours and then travels for 2 hours at the speed of 20 kph. The total distance traveled in 5 hours is only 2 x 20 kilometers. Thus the average speed is 40/5= 8 kph. We can the add the value of the base, 40 kph to get the real average speed of 48 kph.
However, if the car traveled 7 hours instead of 2 hours at the faster speed of 60 kph while the time for the 40 kph speed remains at 3 hours. What is the average speed? Again using 40 kph as the base, the total distance traveled would be 20 x 7 or 140 kilometers in a total time of 10 hours for an average of 14 kph. This would give us 40 + 14 = 54 kph as the average for the whole trip.
But this is still the conventional solution.
Now imagine a 5-meter-long weightless beam unevenly placed in a fulcrum with 2 meters on one side and 3 meters on the other. How can we place 20 1-newton weights on the two ends of this “see-saw” so that they will balance?
We can use “trial and error” here or we could simply partition the 20 into two groups in the ratio of 2 : 3 which is 8 : 12 and place 8 pieces at the end of the 3 meter side while placing 12 pieces at the end of the 2 meter side. We will have a balanced see-saw with 24 newton-meters per side.
The most common solution to our featured problem is to get the least common multiple of the speeds and ASSUME that this is the distance between the destinations. Then using this assumed distance, we compute the assumed times of travel. But this is time consuming and totally unnecessary. Given the same distance the speed is inversely proportional to the time. So if the ratio of the speed is 40 : 60 or 2 : 3, then their time to travel the same distance would be in the ratio of 3 : 2.
Thus the lower speed will contribute to 3/5 of the average speed while the faster speed will account for the other 2/5. If we use 40 kph as the base then we need to compute only 2/5 of the difference in their speed which is (60 -40) or 20. This is also the same as partitioning 20 in to a ratio of 2 : 3. For 20 it is 8:12. So The average speed is 40 + 8 = 48 kph.
This same strategy can be used this problem from the 2019 MTAP Grade 8 competitions: The mean of a set of 40 numbers is 85, and the mean of another set of 50 numbers is 76. What is the mean when the two sets are combined?
The numbers in the two sets are the ratio 40:50 or 4:5 while the difference in the group averages is 85 -76 or 9. This 9 can be exactly partitioned into 4:5. So the average of the two groups is exactly 76 + 4 = 80.
This see-saw strategy can also be used to solve a problem like this: How many liters of a 15% salt solution must be mixed with 5 liters of a 40% solution to produce a 30% solution.
The 40% solution is 10% away from the 30% final mixture while the 15% solution is 15% away. 10% : 15% can be reduced to 2:3. The amount of each solution then must be in the ratio of 3:2.
But it is given that 5 liters of the 40% solution is used, then 5:x = 3:2; X = 10/3 liters.
More examples and applications are discussed in Chapter 23 of 30 Master Strategies in Computing which will be released in the 1st quarter of 2020.
