Well done to everybody who sent in a completed grid, or found the value of ?

Click to see methods using each of the images suggested, or read below to see the other methods we were sent.

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Many solutions included a lot of 'educated guessing' or 'trial and improvement'. This method from Prim, from Bangkok Patana School in Thailand, shows a systematic trial approach:
 


Philip from Abingdon School in the UK came up with this neat solution:
   


If the top row is worth 28 and has 2 squares and 2 triangles and the second row is worth 30 and has 2 squares and 2 hexagons, we can deduce that 2 hexagons are worth 2 less than 2 triangles, or that a triangle is 1 less than a hexagon.




By repeating this process with the bottom two rows, we can work out that a square is worth 2 more than a triangle.
We can suppose that 3circles+1square=20. We then substitute the shapes by variables; the circle=$c$, and the square=$s$. Thus, the equation is now $3c+s=20$. For the second equation, the triangle=$t$. Therefore, the equation would be: $3c+t=18$.

In the first equation, we know that $s=t+2.$ Therefore, we can substitute the value of $s$ in terms of $t$ and a constant into the equation. $3c+s=20$ becomes: $3c+t+2=20.$

When we simplify the equation, we want to have like terms on the right side of the equal sign, so $t=18-3c.$

Now, by using the second column from the left side of the table, we can find the solution. We know that $s=t+2.$ In this column, however, there are 3 squares and 1 triangle. [To express 3 squares, we] multiply $3(t+2)$ and get $3t+6$
[So 3 squares + 1 triangle = 30 becomes] $30=3t+6+t,$ we then simplify the equation.
$4t+6=30$
Subtract 6 from the left and right side, we get $4t+6-6=30-6$ OR $4t=24$
Finally, to isolate $t,$ we divide $4t$ by $4$ and $24$ by $4.$ Thus, the triangle=6.

Now, to find the value of the circle, we use our previous equation, $t=18-3c$, and substitute the the $t$ by $6.$ Just by simplifying as we were doing, $6=18-3c$, and the circle equals $4.$

Because we know that the $s=t+2$, $s=6+2.$ Therefore, the square=8. Finally, for the hexagon, we can find the 2nd row form the top to have 2 hexagons and 2 squares. If the hexagon=$h,$ then $2h+2s=30$ so $2h+2(8)=30.$
$2h-16=30.$ Thus, the hexagon=7.

For the first column from the left, we have 1triangle+1hexagon+2circles=?
$t=6, h=7, c=4$ so $6+7+2(4)=?\Rightarrow 6+7+8=?$ Thus, the solution to this problem is:
$?=21$

Siddhant from Singapore International School of Mumbai in India, Lauren from Skipton Girls' High School, Eliya, Lauren and Hannah from North London Collegiate Junior in and Oliver from Abingdon School in the UK, Yerin, Prinn and Utkarsh from Bangkok Patana School and Anh Pham and Anh Minh from British Vietnamese International School used algebra starting from image 1. This is Lauren's work (Skipton Girls'):
To calculate this solution, you must first think of the second column and the first row. As you can see, they consist of squares ($s$) and triangles ($t$). To calculate this I did:

Square                                     Triangle (do this after square)
$2s+2t=28$                             $2s+2t=28$
$3s+t=30$                              $(2\times8)+2t=28$
$s+t=14$                                $16+2t=28$
$(3s+t)-(s+t)= 2s$             $2t=12$
$30-14=16$                            $t=6$
$2s=16$
$s=8$

After calculating this you can now move on to the bottom two rows. To do this you use substitution (like you did to calculate the triangle). The circles($c$) are the ones we want to find out. You do the steps as follows (there are several ways to do this so I will just show 2 of them):

$3c+t=18$                                                                   $3c+s=20$
$3c+6=18$                                     Or                          $3c+8=20$
$3c=12$                                                                         $3c=12$
$c=4$                                                                             $c=4$

Now that we have calculated the circles, squares and triangles, we can now move on to calculating the hexagons ($h$). There are a few ways to do this and I will show a few possibilities:

$2h+2s=30$                                                              $s+h+2c=23$
$2h+ (2\times8) =30$                                                       $8+h+(2\times4)=23$
$2h+ 16=30$                                 Or                          $8+h+8=23$
$2h=14$                                                                      $h+16=23$
$h=7$                                                                          $h=7$

Finally, we can find out what that mystery number is. We simply use substitution again to calculate it. We do as follows:
$t+h+2c=?$
$6+7+(2\times4)=?$                     
$6+7+8=?$
$21=?$