# 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:

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

We may also use 88 as the base.

- Here 25 are 8 below the base
- 8 x 25 or 200 is to be shared by 40, so 200/40 is 5.
- 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

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.**