# MSC #23 – Proportionately: Partitive solution

Our **23 ^{rd} 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**.

If however, 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.

Imagine now a **5**-meter-long weightless beam unevenly placed in a fulcrum with **2** meter 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 1

^{st}quarter of 2020.