# Stumped on Algebra cube problem-problem solving

• Jul 18th 2012, 01:20 AM
Atrey_G
Stumped on Algebra cube problem-problem solving
Hello everybody
I was doing revision on past problem solving papers and came across this problem:

The diagram shows the net of a cube. On each face there is an integer:1, 2011, 1207, x, y, z.
(The diagram is shown like this, it is a net for a cube but can't show a proper diagram cause of the limited actions;)
1207
X,y,2011,z
1
(in that order, just imagine lined boxes around them)
(continuing with the problem)
If each of the numbers 1207, x, y, z equals the average of the numbers written on the four faces of the cube adjacent to it, find the value of x.

Can anyone plz explain this to me fairly simply if you can get the answer.(BTW I currently don't have the answers for this question)
• Jul 18th 2012, 01:21 AM
Prove It
Re: Stumped on Algebra cube problem-problem solving
Quote:

Originally Posted by Atrey_G
Hello everybody
I was doing revision on past problem solving papers and came across this problem:

The diagram shows the net of a cube. On each face there is an integer:1, 2011, 1207, x, y, z.
(The diagram is shown like this, it is a net for a cube but can't show a proper diagram cause of the limited actions;)
1207
X,y,2011,z
1
(in that order, just imagine lined boxes around them)
(continuing with the problem)
If each of the numbers 1207, x, y, z equals the average of the numbers written on the four faces of the cube adjacent to it, find the value of x.

Can anyone plz explain this to me fairly simply if you can get the answer.(BTW I currently don't have the answers for this question)

It would help if we could see the diagram too...
• Jul 18th 2012, 07:15 AM
Soroban
Re: Stumped on Algebra cube problem-problem solving
Hello, Atrey_G!

Quote:

$\text{The diagram shows the net of a cube.}$
$\text{On each face there is an integer: }1, 2011, 1207, x, y, z.$

$\text{The unfolded cube looks like this:}$

Code:

            *- - - *             |      |             | 1207 |             |      |       * - - -*- - - * - - -*- - - *       |      |      |      |      |       |  x  | 2011 |  y  |  z  |       |      |      |      |      |       * - - -*- - - * - - -*- - - *             |      |             |  1  |             |      |             *- - - *
$\text{If each of the numbers }1207, x, y, z\text{ equals the average of the numbers}$
$\text{on the four faces of the cube adjacent to it, find the value of }x.$

We are told that:

$x \:=\:\frac{z+1207 + 2011 + 1}{4} \quad\Rightarrow\quad 4x - z \:=\:3219\;\;[1]$

$y \:=\:\frac{z+1207 + 2011 + 1}{4} \quad\Rightarrow\quad 4y - z \:=\:3219\;\;[2]$

$z \:=\:\frac{x+y+1207 + 1}{4} \quad\Rightarrow\quad 4z - x - y \:=\:1208\;\;[3]$

$1207 \:=\:\frac{x+y+z+2011}{4} \quad\Rightarrow\quad x + y + z \:=\:2817\;\;[4]$

$\text{Subtract [1] - [2]: }\:4x-4y \:=\:0 \quad\Rightarrow\quad x \,=\,y$

$\text{Substitute into [3]: }\:4z - 2x \:=\:1208 \;\;[5]$

$\text{Substitute into [4]: }\:2x + z \:=\:2817\;\;[6]$

$\text{Add [5] + [6]: }\:5z \:=\:4025 \quad\Rightarrow\quad z \:=\:805$

$\text{Substitute into [6]: }\:2x + 805 \:=\:2817 \quad\Rightarrow\quad \boxed{x \:=\:1006}$
• Jul 18th 2012, 10:10 PM
Atrey_G
Re: Stumped on Algebra cube problem-problem solving
Soroban you have got the method sussed, but there is a problem in your cube
My mistake I couldnt provide a better net and data
x is not adjacent to 2011, y and z are. X is on the opposite face of 2011
But i suppose it doesnt really matter as we could just solve for z.
So as you have said the answer is 805

Prove It, Soroban has provided a nice net, you could use that

BTW could you plz tell me how you made the cube

Thank you
• Jul 19th 2012, 01:42 AM
Atrey_G
Re: Stumped on Algebra cube problem-problem solving
Just wondering
When I use x as 805 and y and z as 1006
I can't get the average of y correct
It comes up as
1006=1+1027+2011+805 divided by 4 which is 1006=3844
Which is definitely incorrect could anyone plz tell me what I am doing wrong