On Average(s)

One of the concepts that standardised tests use to trouble students is averages, ironically enough. Ironic, because calculating averages is one of the few things from maths classes we tend to retain over time – how could averages ever give us trouble?

While calculating averages is easy – simply add everything up and divide by how many things there are – understanding what they are often isn’t so straight forward. In fact, we tend to think of averages as being precisely what they aren’t – and the test is counting on that.

First off: averages aren’t the middle number in the distribution: that’s the median. And they’re also not the most likely number: that’s the mode.

So, what exactly are averages, then?

An average is best thought of as being the balancing point in the data: the place where all the deviations to one side are perfectly balanced by all the deviations to the other! Consider the following distribution:

1 2 3 4 5

It’s pretty obvious that the average here is 3 – and that’s in the middle, right? It is – but only because of where all the other numbers are compared to 3! Let’s look at the distance from ‘3’ to each of the numbers:

-2 -1 0 1 2
1 2 3 4 5

Those distances balance each other out around 3, so 3 is the average. (Also, (1+2+3+4+5)/5 totally equals 3, so the old method works too.) What’s interesting though is that, if we chose different numbers, we would still see the same thing! Consider:

1 5 6 9 19

Here, the average is 8: (1+5+6+9+19 = 40 and 40/5 = 8). So what happens with those distances?

-7 -3 -2 +1 +11
1 5 6 9 19

As you can see, the distances add up to zero! The numbers balance around the average, because that’s what an average does!

So How Does This Help?

By looking at the distances from the average instead of the numbers themselves, it makes it a lot easier to calculate an answer on the types of questions you are likely to get on a standardised exam. Basically – smaller numbers are simpler and easier to deal with, they take less time and you are less likely to make a mistake! Check out this example question:

A group of children had their heights measured for a project. The 18 boys had an average height of 47 inches, while the girls had an average height of 44. If the group as a whole had an average height of 46 inches, then how many girls are in the class?
a) 7
b) 8
c) 9
d) 10
e) 11

Answer and Explanation – The Hard Way

Attacking this question in the traditional way would involve a LOT of annoying calculation. You could try stating that the group’s average height would equal the sum of the kids individual heights divided by the number of kids – so:

46 = (47(18) + 44g)/(18+g)

where g is the number of girls in the group. We could then move that denominator to the other side to get

46(18+g) = 47(18) + 44g
828 + 46g = 846 + 44g
2g = 18
g = 9

All of that is true! And ‘9’ is the right answer! But, seriously – who has time for this? Not you – not on a standardised exam. So, what’s the easier way to do it? Let’s check out that balance.

Answer and Explanation – the Easy Way

First off: how far from the average height are the boys? Well, on average, they’re one inch taller and there are 18 boys – so, that’s 18 inches altogether! To balance that, the girls would have to be 18 inches shorter altogether – and since it’s 2 inches per girl, that gives us 9 girls.

The maths here are literally as simple as multiplying 18 times 1 and then dividing by 2: smaller numbers, easier calculations and it takes almost no time at all.

The Upshot

The point of this all is that maths questions on standardised tests are not simply about memorising formulae and working really quickly: they’re about testing your ability to think critically and use the proper tool for the proper job. Understanding what averages really are allows you to use methods that are simpler, easier and quicker – so you can get more questions right, feel less stress and achieve your high score!

Hercules and the Tortoise

One of the most frustrating maths questions you’re likely to get involves relative speeds. It’s not that the concepts are hard – it’s just that you probably never got taught the concepts at all! And relative speed is one area where conceptualisation makes all the difference.

To see what I mean, consider the ancient Greek philosopher, Zeno. Zeno was different to all the other ancient Greek philosophers, in that his goal wasn’t to show that logic and rationality could lead you to understand the universe – it was to prove that logic and rationality simply aren’t enough!

To show this, he developed some paradoxes – conclusions, logically arrived at, that are totally absurd. One of his most famous involved Hercules and the tortoise. The story goes like this:

Hercules is Big and Strong and Fast. The turtle is … not. So, if Hercules were to chase a turtle, you gotta figure he’s gonna catch him, right?

Hold on, says Zeno. Hercules would have to run up to the place where the turtle was to start with – and during that time, the turtle would have moved a bit. So, Hercules would have to run up to where the turtle is then – during which time, the turtle will have moved a bit. So, Hercules would have to run up to where the turtle is now – during which time, the turtle will have moved a bit. And that will keep happening … forever.

So, logically, Hercules never catches the turtle – even though he’s Bigger and Stronger and Faster!

Now, obviously, Hercules will absolutely catch that turtle in reality – the point is that, proceeding perfectly logically, we’ve ended up believing something crazy.

So, does this mean maths are insane and we should chuck it all in to live on a commune? Not necessarily (unless you’re into that kind of thing, anyway). Instead, recognise this: it doesn’t matter how fast Hercules is – it matter how much faster he is than the turtle!

Let’s say Hercules can run 12 mph, and the turtle can only run 3 mph. Hercules is currently 27 miles behind the turtle. How long will it take to catch him? Instead of trying to create simultaneous equations and thinking about two bodies in motion, let’s just say this: Hercules is 27 miles away from the turtle and he’s catching up at 12 – 3 = 9 miles every hour. So, at that rate, how long would it take Hercules to catch up? 27/9 = 3 hours. Simple as that.

After all – how fast are you moving right now? You might say zero miles per hour – but you are on the Earth and the Earth is rotating, no? And the Earth is circling the Sun, right And the Sun is circling the Galaxy, the Galaxy is circling the Universe and the Universe is accelerating in all directions at once! So how fast are you really going? Well, zero – compared to the Earth!

All speed is relative speed, just as all distance is relative distance. And there’s no reason to privilege Hercules’ speed compared to the Earth, when what matters is the turtle!

Simply conceptualising the problem this way makes an otherwise complicated issue incredibly easy. But then, that’s the GMAT’s way, isn’t it?

Prime Real Estate

One of the most important concepts to learn before taking a standardised exam is how to think about primes – unfortunately, it’s precisely this part of number theory that most students never get a firm grounding in! Getting to grips with primes allows you to find Least Common Multiples and Greatest Common Factors, add and subtract fractions with ease, and even take the roots of non-perfect-squares in your head!

To see why primes are so important, consider this: primes are the only integers divisible by only two different factors – themselves and 1. Which means every other integer can be broken down into factors that are either prime, or not prime – and the not-prime factors can be broken down the same way. What this means is that every integer in this world can be broken down into its prime factors – and that’s what the number really is! 18 is really 2 times 9, which itself is 3 times 3; so really, 18 is just one 2 and two 3s multiplied together! 18 = $2^1 3^2$

Here’s where the fun starts. If we want to take the square root of 18, all we have to do it look for pairs of prime factors – in this case, the two 3s. We can take the square root of $3^2$, since $3^2$ is 9 – and $\sqrt{9}$ is 3! We can’t take the root of 2, though – so we leave it alone. Consequently, the square root of 18 is simply $3\sqrt{2}$

Try it yourself! Break down the following numbers and see what roots you can take.

• 8
• 27
• 50
• 98
• 54
• 252
• 3,000,000,000,000 (for funzies…)