The mathematical section of a standardized test includes various types of questions. Some of them (a minority) are “conventional” questions, which can be relatively easily dealt with by anyone familiar with school algebra. A vast majority of the questions, however, tend to test our ability to “think” and analyze rather than checking our accumulated knowledge.
Those problems that are “regular” (or seem like ones) can be solved more effectively if the examinee is adding his “rational” to his “technical luggage”. One of such examples is a question that requires calculation of averages.
A note has to be stated that modern mathematics recognizes several different averages. Those are, for example, the geometric average, the harmonic average, the quadratic average – each of those has a meaning and, of course, an exact formula. Most people, however, mean the arithmetic average (or arithmetic mean) when they use such terms as “average value”, “average number”, “average speed” etc.
The arithmetical average is defined by the simple math formula of , where n is the number of numbers and x is their respective values.
The formula may not be that familiar, but you are probably familiar with the basic calculation of average – summing all the values and dividing the result by the total number of figures.
For example, when you are required to calculate the average of 68 and 72, all you have to do is to summarize the numbers (68+72 = 140) and divide by 2, getting 70 as the mean.
When you have more numbers – 52,56,64,76 – you summarize them all (52+56+64+76 = 248) and divide by 4, getting 62. There are, of course, many places for a mistake, as you are working with relatively large numbers. And although I hope you have calculated via 52+56+64+76 = (56+64)+(52+76) = 120+128 = 248; and divided via 248/4 = 240/4+8/4 = 60+2 = 62, the calculation process is still error-prone.
Let’s try and make the average calculation procedure easier, limiting the possibility of an error.
In the first example, there were two numbers, 68 and 72. What one should always remember, is that the average is located BETWEEN the numbers. And when we are talking about two numbers – it is located EXCATLY in the middle (when looking on the numbers line). This means, that the average is “distanced” equally from the bigger number (in our case – 72) and the smaller one (68). It is easy to see that the number is 70. I think it is much easier than doing the (68+72)/2 computations.
Of course, we can’t always “see” that easily (for example, when we are talking about 98 and 48). But there is a mathematical way to locate the middle: subtract the smaller number from the bigger one: 98-48 = 50. This is the “total distance”. As our average is located in the “middle of the way”, let’s divide it by two: 50/2 = 25. This is the “distance” form each of the numbers (48 and 98) to the mean number. Of course, when calculating form the smaller number we should ADD the distance (48+25 = 73) and when going from the bigger number – we SUBTRACT it (98-25=73).
When we have more than two numbers, the average is also located somewhere in between. Thus, there is no need to sum up the numbers and reach (sometimes) “astronomical” values. Many times, we can divide the number to pairs:
To calculate the average for 52,56,64 and 76 the pairs can vary. For example:
The average of 52 and 56 is 54 (easy to see, and: 56-52 = 4, 4/2 = 2, 52+2 = 54).
The average of 64 and 76 is 70 (similar)
Now, the average if 54 and 70 (remember, we can do this as we had EQUAL number of members (two) in each of the mid-average sections) is: 70-54 = 16; 16/2 = 8; 70-8 =62. That’s it!
Another technique used is “intelligent guessing”. Suppose, you have 5 numbers to calculate the average: 1010, 1027, 1015, 1006, 1007. Adding them up and dividing will definitely produce some mistakes. As we already know, the average is located somewhere “in between”. Let’s assume it is 1010. Now, let’s calculate, how much each number “diverts” from average:
1010 – 1010 = 0
1027 – 1010 = 17
1015 – 1010 = 5
1006 – 1010 = -4
1007 – 1010 = -3
Now, let’s sum up the “diversions”: 0+17+5-4-3 = 15. In total, we have mistaken by 15 points. This number need to be divided by the total number of members (five in our case) and then added to the “guessed” average: 15/5 + 1010 = 1013. This is the correct answer.
What would have happened, had we guessed better? Let’s see: suppose, our initial guess was “1013”. Then:
1010 – 1013 = -3
1027 – 1013 = 14
1015 – 1013 = 2
1006 – 1013 = -7
1007 – 1013 = -6
-3+14+2-7-6 = 0. When we have 0 – our “guessed” average is indeed the correct average.
So, there are two of the techniques to calculate averages (namely, the arithmetic means): locating the middle (for a pair) and “intelligent guessing” – recommended for larger numbers.