Let's solve a problem based on missing numbers from the hidden faces of dice.
Three standard dice are sitting next to each other as shown in the diagram. There are 7 faces visible. How many dots are hidden on the other 11 sides?
But try the problem first...
Australian Mathematics Competition - Middle Primary Division
Basic Arithmetic by Robert Moon
This one is very easy but if you really need any hints at first take a dice and try to understand the number of dots in each sides .
So for first dice : we can see 4,5 and 1. So the no of dots which are not seen 6,2 and 3
For Second Dice : The visible dots are 2 and 6 . So the dots in sides which are not visible are 4,3,1 and 5.
For Third Dice : The Visible dots are 3 and 1 .So the dots which are not visible 2,4,5 and 6.
I think you have already got the answer but if you really need this last step :
Now add all the number of dots from the invisible sides:
2+4+5+6+4+3+1+5+6+2+3 = 41 .