AMC8 2000

AMC8 2000 · Q8

AMC8 2000 · Q8. It mainly tests Arithmetic misc.

Three dice with faces numbered 1 through 6 are stacked as shown. Seven of the eighteen faces are visible, leaving eleven faces hidden (back, bottom, between). The total number of dots NOT visible in this view is

三个骰子如图所示堆叠。十八个面中七个可见，留下十一个隐藏的面（背面、底面、之间）。在这个视图中不可见的点数总数是

(A) 21 21

(B) 22 22

(D) 41 41

(E) 53 53

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The numbers on one die total $1+2+3+4+5+6=21$, so the numbers on the three dice total 63. Numbers 1, 1, 2, 3, 4, 5, 6 are visible, and these total 22. This leaves $63-22=41$ not seen.

答案（D）：一个骰子的点数总和为 $1+2+3+4+5+6=21$，所以三个骰子的点数总和为 63。可见的点数是 1、1、2、3、4、5、6，它们的和为 22。因此看不见的点数之和为 $63-22=41$。

Topics

AR.015 Arithmetic misc