 # 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 a base, we can mentally solve problems which usually require lengthy calculations.

Consider this problem which was asked in a nationwide Math contest:

The average of a set of 25 numbers is 80, and the average of another set of 15 numbers is 88. What is the arithmetic mean of all 40 numbers?

The common solution is to get the total of the numbers and divide it by (25 + 15), that is,

A         = [ (25 x 80) + (15 x 88)] ÷ (25 + 15)

= [2000 + 1320] ÷ 40

= 3320 ÷ 40

= 83

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 using the following mental steps:

1. 15 are (88 – 80) or 8 over the base
2. That means an excess of 8 x 15 = 120 over 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.

Let us see how we can apply a  base in our illustrated example: 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 to let S be the score needed and solve for S in the equation

[(4 x 880) + 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 5th exam also needs to raise the score of each of the 4 previous exams by 2 (90 – 88). So 90 + (4 x 2) = 98.