The six faces of a dice have been marked with alphabets A, B, C, D, E and F respectively. This dice is rolled down three times. The three positions are shown as:

Find the alphabet opposite A.

From figures (i), (ii) and (iii), we conclude that 1, 3, 5 and 6 dots appear adjacent to the face with 2 dots. Therefore, 4 dots will appear opposite to 2 dots. Now, from figures (i) and (ii), we conclude that 2, 3 and 5 dots appear adjacent to 1 dot Therefore, either 4 or 6 dots will appear opposite to 1 dot. But since, 4 dots appear opposite to 2 dots it follows that 6 dots will appear opposite 1 dots.

From figures (i), (ii) and (iv) We conclude that 6, 4, 3 and 1 lie adjacent to 2. Hence, 5 must lie opposite 2.

From positions X and Y we conclude that 1, 5, 6 and 3 lie adjacent to 4. Therefore, 2 must lie opposite 4. From positions Y and Z we conclude that 4, 3, 2 and 5 lie adjacent to 6. Therefore, 1 must lie opposite 6. Thus, 2 lies opposite 4, 1 lies opposite 6 and consequently 5 lies opposite 3.

As analysed above, the number on the face opposite 4 is 2. In position Y, since 4 lies on the top, therefore 2 must lie at the bottom face.

Number 3 is common to the two positions of the block. We assume the block in fig. (ii) to be rotated so that 3 appears at the same position as in fig. (i) i.e. on RHS face (i.e. on face II as per activity 1) and the numbers 5 and 2 move to the faces hidden behind the numbers 4 and 6 respectively [in fig. (i)]. Thus, the combined figure will have 3 on RHS face (i.e. face II), 4 on the Front face (i.e. face I), 6 on the Top face (i.e. face V), 5 on the Rear face (i.e. face III) and 2 on the Bottom face (i.e. face VI). Clearly, when 2 is at the bottom; then 6 is at the top.

From figures (i) and (ii), we conclude that 2, 3, 5 and 6 he adjacent to 1. Therefore, 4 lies opposite 1. Hence, when 4 is at the bottom, then1 must be on the top.

Number 3 is common to both the figures (i) and (ii). The dice in fig. (ii) is assumed to be rotated so that 3 remains on the FR-RH face (i.e. face I as per activity 1) and the numbers 5 and 2 move to the faces hidden behind the numbers 6 and 1 respectively [in fig. (i)]. Thus, the combined figure will have 3 on FR-RH face (i.e. face I), 5 on RR-RH face (i.e. face II), 2 on Bottom face (i.e. face VI), 1 on the Top face (i.e. face V) and 6 on FR-LH face (i.e. face IV). Clearly, 2 lies opposite 1. Hence, when 2 is at the bottom, then 1 will be at the top.

From figures (i), (ii) and (iii), we conclude that 3, 4, 6 and 1 dots appear adjacent to 2 dots. Therefore, 5 dots appear opposite 2 dots. Now, from figures (i) and (iii), we conclude that 2, 4 and 1 dots appear adjacent to 3 dots. Therefore, either 5 or 6 dots appear opposite 3 dots. Since, 5 dots appear opposite 2 dots, it follows that 6 dots appear opposite 3 dots.

From positions X and Y we conclude that 1, 5, 6 and 3 lie adjacent to 4. Therefore, 2 must lie opposite 4. From positions Y and Z we conclude that 4, 3, 2 and 5 lie adjacent to 6. Therefore, 1 must lie opposite 6. Thus, 2 lies opposite 4, 1 lies opposite 6 and consequently 5 lies opposite 3.

As analysed above, the number opposite 6 is 1 and the number opposite 5 is 3. Therefore, the numbers hidden behind the numbers 6 and 5 in position Z (these are the numbers opposite 5 and 6 respectively) are 1 and 3.

From figures (i) and (ii), we conclude that 5, 4, 6 and 2 dots appear adjacent to 3 dots. Therefore, 1 dot must appear opposite 3 dots. Thus, if the face with 1 dot is at the bottom, then the face with 3 dots will appear on the top.

From figures (i) and (ii) we conclude that the number 1, 2, 3 and 4 appear adjacent to 6. Thus, the number 5 will appear opposite 6. Therefore, when six is at the bottom, then 5 will be at the top.