ISRO EC 2023 Q.87: solution and answer

Official answer: ASource: ISRO's revised and final answer key of 16 February 2024

1. Write each domino as $[\text{top}|\text{bottom}]$, counting an empty half as $0$. The first group is $[3|2],[2|1],[1|0]$.

2. In this group the bottom of each domino equals the top of the next one ($2$, then $1$), and each domino has its top one more than its bottom. Read as a chain, the top halves go $3,2,1$ and the bottom halves go $2,1,0$. The group counts down.

3. The second group is $[0|1],[1|2],[2|3]$. Here the bottom of each domino equals the top of the next one ($1$, then $2$), and each bottom is one more than its top. The group counts up, from $0$ to $3$.

4. The groups therefore alternate: down, up, then down again. The third group is $?,[1|0]$, with three dominoes in all, so it must be the counting-down group $[3|2],[2|1],[1|0]$. It ends with the printed $[1|0]$, as the first group does.

5. So the missing pair is $[3|2],[2|1]$, which is option 1 (letter (a)).

6. Check the other options. Option 2 gives $[2|1],[1|2]$: the second domino counts up, so the chain breaks. Option 3 gives $[2|1],[3|2]$: the bottom $1$ is followed by a top of $3$, so the chain breaks. Option 4 gives $[2|1],[1|0]$, and followed by the printed $[1|0]$ it repeats $[1|0]$ and has no domino with a top of $3$, so it does not form the counting-down group.