Gavin is four times as old as Tom, who is six years younger than Xavier, who is a third of the age of John, who is a year younger than Gavin. How old is Xavier?

[pickledsiblings]

Please can someone give me some guidance on the methodology required to answer this type of question.

Thanks

NB I can work out the answer but I can't explain my reasoning easily.

G=4T
T+6=X
3X=J
J+1=G
You can then easily mix about the equations eliminating each letter in turn until you get an expression that is just numbers.

e.g. clearly 4T=J+1 and therefore
4T=3X+1

This can then be solved as a normal simultaneous equation with T+6=X

So that's how I did it but I'm finding it difficult explaining what to do 1st and why iykwim.

DS tried this, made a table expressing everything in terms of everything else and got himself in a right muddle.

Is there any more to the question?

No MustTidy that's it. It's amazing how it works out - a bit like magic.

Ok let tom be X.
Gavin 4x
Tom X
Xav X+6
J 3(X+6)
G 3(X+6)+1
So Gavin is
4x=3(X+6)+1 which solves to X=19
Xavier is 25

Better method than lots of letters and sim equations.

Why 'let tom be X'? Why not let someone else be X though? What's the 'reasoning' behind the approach?

Thanks Fish and Must btw.

I'm finding it difficult explaining what to do 1st and why

That's because there is no right order. You can start the way FishWithABicycle says or you could start by saying:

X=T+6

Therefore

3X=3T+18

Therefore

J=3T+18

Therefore

G=3T+19

Therefore

4T=3T+19

Therefore

T=19

You just need to manipulate the equations to end up with one which only contains one of the unknown ages. That allows you to determine that age from which you can determine all the others.

Why 'let tom be X'?

It doesn't matter who you start with. You can let any one of them be X. It really doesn't make any difference. Just choose one and go for it. You will always end up with the same answer.

manipulate the equations

DH and I can do this as we are old. How are young people taught to approach this kind of problem I wonder?

Thanks for giving me the confidence to tell DS to just jump in and give it a go. He has just given it another try and managed to solve it.

Only chose Tom because if Tom is X then Gavin is 4x rather than if you start with Gavin being X then Tom becomes X/4 so introducing division and better avoided as the final equation would be harder. This is the way they would be expected to approach it but making the right decision as to where to start is quite hard for them.

I just did it without looking at the rest of the thread.

My instinct was to make Xavier x, as that is the one we were looking for. (Plus his name starts with x which think is perhaps a hint)

X is x
T is x-6
G is 4(x-6)
J is 3x

3x + 1 = 4(x-6)
3x+1=4x-24
x=25

Xavier is 25.

I worked it out differently to all of you - I didn't use algebra, but more of a trial and error approach.

So first I worked out that, assuming everyone is an integer and the maximum age is about 100, Xavier must be at least 7 (so Tom can be 1) and no older than 33 (Gavin being 100, John being 99.)

I tried 7 to start. John is then 21 and Gavin 22. That makes Tom far too old. No. So I then started thinking about how John has to be a multiple of 3 and Gavin a multiple of 4 and they have to be one apart from each other. The next time this occurs is 27 and 28, which would make Xavier 9, but Tom older than 3.

I realised I had to make a larger jump forward, so I was going to try 15, but DH was going on about 27, so I tried that. John 81, Gavin 82. Not a multiple of 4 (I can't remember the proper word for this ) so fail. But I calculated tom anyway to see how close - 20.5.

Very close. So 26 - 78 - 79. Nope. 25 - 75 - 76, tom must be 19, and it all works, so Xavier is 25. :)

Probably the algebra is more efficient, I only took maths to GCSE, and that is the way that came to mind to solve it.

Interesting, anyway.

If we call Gavin's age G, Tom's age T etc then you can rearrange the information given (in English) in to equations (in Maths). So you get
G=4T
T=X-6
3X=J
G-J=1

There are four unknowns (G,T,J,X) and four equations so you know it's solvable.

It interesting to see the other approaches. Bertie, DS was tempted to go down the trial and error route as this has worked out well for him in the past but I think you are right, the algebra is more efficient provided you know how to make a start.

Thanks again to all of you for sharing your ideas.

Many ways to do this. One way (probably not the easiest, but the first way I did it):

G=4T ----> T=G/4
T=X-6 -->X=T+6 -> X= G/4 +6
X=J/3

J=G-1 --->X=(G-1)/3

(G-1)/3=G/4 + 6
G-1=3G/4 + 18
4G-4=3G=3G=72
G-4=72
G=76

J=75
X=75/3=25

It would be a lot easier for Tom, or any of them, to be X if one wasn't called bloody Xavier! what with X for X, x for 'multiplied by' and X for Xavier....

Brenda