ISRO CS 2025 Q.7: solution and answer

Official answer: CSource: ISRO's final answer key of 15 May 2026

1. The array after three iterations is $\{2,5,10,16,8,9\}$. The first four elements are sorted and the last two, $8,9$, are in their original positions.

2. Bubble sort, pass 1: compare adjacent pairs and swap when out of order, which gives $\{5,10,16,2,8,9\}$, then $\{5,10,2,16,8,9\}$, then $\{5,10,2,8,16,9\}$, then $\{5,10,2,8,9,16\}$. The largest element $16$ is now at the last position and stays there. The target has $9$ in the last position, so bubble sort can never produce it. Not bubble sort.

3. Selection sort: iteration 1 swaps the minimum $2$ with $10$, giving $\{2,5,16,10,8,9\}$. Iteration 2 finds the minimum $5$ already in place. Iteration 3 swaps the minimum of $\{16,10,8,9\}$, which is $8$, with $16$, giving $\{2,5,8,10,16,9\}$. This differs from the target. Not selection sort.

4. Insertion sort treats the first element as a sorted prefix and inserts one new element per iteration. Iteration 1 inserts $5$: $\{5,10,16,2,8,9\}$. Iteration 2 inserts $16$, already in place: $\{5,10,16,2,8,9\}$. Iteration 3 inserts $2$ at the front, shifting $5,10,16$ one place right: $\{2,5,10,16,8,9\}$.

5. This matches the target exactly, with $8,9$ untouched. Only insertion sort fits, so option (d), all of the above, is wrong.