Averages

jupiter00

Could ye help me with a sixth class maths question, I know its a bit basic here but..... If the average of 4 nos is 14 and the average of 3 nos is 15, what is the fourth number? Is there a formula for this? In future is there a site where I could get help for 6th class maths problems? Thanks in advance.

Eliot Rosewater

If the average of 4 numbers is 14, it means that the sum of all those numbers is 14x4=56.

If the average of 3 numbers is 15, it means that the sum of all those numbers is 15x3=45.

The difference between these is the fourth number, ie 11.

Permabear

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Doyler92

I want to answer but it's one of those things I know how to do but can't explain for some reason.

jupiter00

Thanks for the answers, we actually worked this out but we didn't feel confident about the answer! Sorry but I still don't really understand how this is the fourth number! Is there a formula to find the third number,second and first? This might help me to understand how to work backwards if you know what I mean. I'd just like to understand it. If I'm driving you all crazy here, sorry guys!:)

Permabear

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sponsoredwalk

Here is a nice way to think about this concept of averages.

If you have 6 sweets and want to give 3 people a fair number of sweets
each so that nobody is being greedy what is the average number of
sweets you're going to give each person? Well obviously you're going to
give each person two sweets each. Said another way, the average number of
sweets each person gets is 2.

I'm going to explain this more mathematically now,
[latex] \frac{number \ of \ sweets}{number \ of \ people \ who \ want \ sweets} \ = \ average \ number \ of \ sweets \ each \ person \ gets}[/latex]

[latex] \frac{ 6 \ sweets}{3 \ people} \ = \ 2 \ \frac{sweets}{person} [/latex]

See, in mathematical symbols we have "sweets" on top on the left hand
side, and "people on the bottom", well on the right hand side we can read
the writing as "sweets per person", as in the average number of sweets
shared between 6 people is 2 sweets per person.

We can extend this simple idea to the example you've written up here.

The average of 4 numbers is 14.

We can rewrite this more mathematically like this:

[latex] \frac{x_1 \ + \ x_2 \ + \ x_3 \ + \ x_4}{4} \ = \ 14[/latex]

[latex]x_1[/latex] and the others are just unknown numbers,
we have no idea what they are but we know that we have four numbers
as the question told us and that because we're taking the average of them
that we have to add them together. Remember in the sweets example
above I had 6 sweets, I had:
1 sweet + 1 sweet + 1 sweet + 1 sweet + 1 sweet + 1 sweet = 6 sweets.
The same idea goes for the fraction with the [latex]x_1[/latex] etc...

This is a very important idea, in averages you add up all the numbers and
divide by the amount of terms you have to get the average.

So, if we use basic laws of algebra we can multiply both sides of the
equation by 4. If we do this we get rid of the fraction on one side and
find a definite answer on the other.

[latex] 4 \ \cdot \ ( \frac{x_1 \ + \ x_2 \ + \ x_3 \ + \ x_4}{4} ) \ = \ 4 \cdot \ 14[/latex]

[latex] ( \frac{4}{4}) \cdot \ (x_1 \ + \ x_2 \ + \ x_3 \ + \ x_4) \ = \ 4 \cdot \ 14[/latex]

[latex] 1 \cdot \ (x_1 \ + \ x_2 \ + \ x_3 \ + \ x_4) \ = \ 4 \cdot \ 14[/latex]

[latex]x_1 \ + \ x_2 \ + \ x_3 \ + \ x_4 \ = \ 56[/latex]

Now, this just says that the four unknown numbers, when added together,
gives 56. the [latex] \cdot [/latex] just means multiplication and the
thing I did with the fraction, moving the "4"'s on top of each other was
a way to make them cancel i.e. [latex] \frac{4}{4} \ = \ 1 [/latex].
It's a smart trick to get rid of the fraction by turning it into 1
Remember, it has to balance out on the other side of the = sign in order to
be true!

So, the question also told us that the average of 3 numbers is 15.
We can use the same kind of thing we did in the first example!
Can you do it before reading on???

[latex] \frac{x_1 \ + \ x_2 \ + \ x_3}{3} \ = \ 15 [/latex]

Remember, we can do the same thing again only multiplying through
by 3 to cancel the 3 in the fraction!

[latex] 3 \cdot \ ( \frac{x_1 \ + \ x_2 \ + \ x_3}{3}) \ = \ 3 \ \cdot \ 15 [/latex]

[latex] \frac{3}{3} \ \cdot \ (x_1 \ + \ x_2 \ + \ x_3) \ = \ 3 \cdot \ 15 [/latex]

[latex] 1 \ \cdot \ (x_1 \ + \ x_2 \ + \ x_3) \ = \ 45 [/latex]

[latex] \ (x_1 \ + \ x_2 \ + \ x_3) \ = \ 45 [/latex]

So, now we know that the average of 4 numbers is 56, the average of
3 numbers is 45.

[latex] \ (x_1 \ + \ x_2 \ + \ x_3) \ = \ 45 [/latex]
[latex]x_1 \ + \ x_2 \ + \ x_3 \ + \ x_4 \ = \ 56[/latex]

If we add 1 number to the first equation, we'll get 56, what is that?

Look,

[latex] \frac{x_1 \ + \ x_2 \ + \ x_3}{3} \ = \ 15 [/latex]

is missing an [latex] x_4 [/latex] term, and we're only dividing by 3, not 4!
That is how you find the fourth number.

Is that clearer? As donegalfella said, you should watch some of the videos
at www.khanacademy.org to see a great explanation.
http://www.khanacademy.org/video/averages
This is a good one The other videos are really helpful too!

jupiter00

Thanks for taking the time to help esp to donegalfella and sponsored walk, it must have taken a time to type all that. I remember the algebra from LC and x is useful for unknowns, I think I was making it harder than it need be. I will check out the websites too for the future. Much appreciated.

Permabear

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