MSC #10 – Using a Base

We have used bases, mostly powers of ten, in our earlier ”specials” – in getting the ten’s complements and in completing the whole. Here we will show how with the use of bases, we can mentally solve lengthy calculations.

  
The common solution to our sample problem is to get the total of the numbers and divide it by 40: [ (25 x 80) + (15 x 88)] ÷ 40

For this, we may need a calculator or at least paper and pencil to perform the calculations.

Using 80 as base, we can even compute for the answer mentally:

1. 15 are (88 – 80) or 8 over the base
2. That means an excess of 8 x 15 = 120 of the base
3. This 120 is to be shared equally by the 40 elements, so 120/40 = 3.
4. The average then is 80 + 3 or 83.

We may also use 88 as the base.

1. Here 25 are 8 below the base
2. 8 x 25 or 200 is to be shared by 40, so 200/40 is 5.
3. The average then is 88 – 5 or 83.

We will be posting about base multiplication and base division in the next two days. These two topics will be discussed in the Inspirational Maths from India(Year 4) webinars on Dec 4-5 and 11-12, 2021.

Using a base is very useful in grade calculations as shown in these two examples:

Example 1: In the local science high school, the final grade is computed using the ff. formula: 40% Final Exam, 30% long tests, 20% short quizzes and 10% class participation where Wilfred got grades of 92, 91, 89 and 93 respectively. Compute his final rating.

Our common solution using the weights of the grades is as follows:

            92 x 40% = 36.8

            91 x 30% = 27.3

            89 x 20% = 17.8

            93 x 10% =   9.3   

                               91.2

Using a base of 90 and the relative weights of the grades is definitely easier and can be done mentally.

             2 x 4 =  8

             1 x 3 =  3

            -1 x 2 = -2

             3 x 1 =  3    

                      12

and 12/ 10 = 1.2. The weighted average is 90 + 1.2 = 91.2

Example 2) Carla’s average for her four previous exams was 88. What score does she have to get if she wants to have an average of 90 to maintain her scholarship?

The usual solution to this is let S be the score needed and solve for S in the equation

(4 x 88 + S)/ 5 = 90

352 + S = 450

S = 450 – 352 = 98.

In the solution above 4 x 88 represents the total score for the 4 previous tests. It also means that we can consider her getting a score of 88 in each of the 4 tests.

Using the desired average, 90 as base, we can consider all 5 grades as 90 but the 5^th exam also needs to raise the score of each of the 4 previous exams by 2 (90 – 88). So 90 + (4 x 2) = 98.