Two cubes are stacked as shown. The faces of each cube are labelled with $1,2,3,4,5$, and 6 dots. A total of five faces are shown. What is the total number of dots on the other seven faces of these two cubes?
(A) 13
(B) 14
(C) 18
(D) 21
(E) 24

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-057.jpg?height=437&width=268&top_left_y=1359&top_left_x=1319)
Solution 1

The faces not visible on the top cube are labelled with 2,3 and 6 dots.

The faces not visible on the bottom cube are labelled with $1,3,4$, and 5 dots.

Thus, the total number of dots on these other seven faces is $2+3+6+1+3+4+5=24$.

## Solution 2

Since each of the two cubes has faces labelled with 1, 2, 3, 4, 5, and 6 dots, then the total number of dots on the two cubes is $2 \times(1+2+3+4+5+6)=2 \times 21=42$.

The five visible faces have a total of $4+1+5+6+2=18$ dots.

Therefore, the seven other faces have a total of $42-18=24$ dots.