Aunt Agatha (need help to get the day off work)

I know her, she is not very old! She is 36...

x=1.5(1.5((x-4)/2)))

I'm still here at work :-(

Apparently she is 72

When I tried to work it out I got to 36 too, I got pretty much the same formulae as you, how did we get it wrong?

The correct formula is
x = 1.5 * 1.5 * ((x / 2) - 4)

We can count on Mark!

Cheers Mark, That's what I got finally. but seeing as I got it at 17:00 it seemed pretty pointless going home 15 mins early.

Cheers for your help guys

I wish my boss would give me the day off for solving a riddle.

Have you got a Maths based riddle I can give to my boss?

Check the "worlds first riddle" thread...

Here's one:
There is a row a soldiers that is 1 km in length. They walk with a constant speed in a straight line, in one direction.

At the end walks a messenger, he has to bring a message to the captain walking all the way back at the beginning of the row.

The messenger starts walking past the soldiers and immediately turns around when arriving at the captain and walks back to the end of the row. When the messenger is back at end, the group of soldiers traveled a distance of 1 km.

The soldiers and captain are walking at the same constant speed. The messenger (walking faster then the soldiers) is also walking a constant speed.

You don't know anything about time or speed.

How many km did the messenger travel from the end of the row to the beginning and back?

Just to check the answer is it Approx 2236m (2.4km)?

It's 1+sqrt(2) km. About 2.414km.

2236m would be 2.236km.

I worked it out as an isocellese triangle with a height of 1km and a base of 1km

I split that into 2 right angled triangles of Oposite size of 500m and Adjascent size of 1000m

therefore as a^2 = b^2 + c^2

the walk to the front of the queue was sqrt( 500^2 + 1000^2) which is approx. 1118m

times that by 2 (to include the walk back) then I got approx. 2236m

How did you get 2414m ?

Call the length of the row 1. Below is a diagram.
The top row is the starting position. The messenger starts at the left and the captain is to the far right.
The middle row shows the messenger reaching the captain.
The third row shows the ending position.
The bottom two rows indicate distances. x is how far past the captain's original position the messenger walks.


On the way to the captain, the messenger walks 1+x in the same time that the row moves x.
On the way back, the messenger walks x in the same time that the row moves 1-x.

Therefore, (1+x)/x = x/(1-x)
1-x^2 = x^2
1 = 2*x^2
x = sqrt(1/2) = sqrt(2)/2
The messenger walks 1+x+x = 1 + sqrt(2)/2 + sqrt(2)/2 = 1 + sqrt(2).

Ah, I though the row was walking the other way.

In your example the row is walking left to righ (as a column)

I worked it out as the row walked top to bottom


I thought about it this way, my manager thought about it the way you said (but still got the answer wrong hehehe)

Did you get the day off? Maybe your manager should have to come in on the weekend. :wink:

I'll suggest the weekend to him see what he says