DIRECTIONS: Krishnapuram's town council has exactly three members: Arjun,…

Concept: In a for/against constraint puzzle, each stated count ("exactly two members vote for X") is an invariant — once every member but one is already fixed on that bill, the invariant forces the last member's vote. Solve by locking in every directly given vote first, then let each count invariant force the remaining unknowns in turn, and finally verify the completed grid against every original constraint.

1. Arjun votes for the recreation bill and against the school bill (given directly); Karn votes against the recreation bill (given); Bhim votes against the tax bill (given).

2. Exactly two members vote for the recreation bill. Arjun already supplies one "for" vote and Karn is against it, so the second supporter of the recreation bill must be Bhim.

3. The question adds that Karn votes for exactly two of the three bills. Since Karn is already against the recreation bill, both of his "for" votes must fall on the remaining two bills — the school bill and the tax bill.

4. Exactly one member votes for the school bill. Karn already supplies that single "for" vote, and Arjun is against the school bill (given), so Bhim must also be against the school bill.

5. Exactly one member votes for the tax bill. Karn already supplies that single "for" vote, and Bhim is against the tax bill (given), so Arjun must also be against the tax bill.

Cross-check: Every member now has at least one "for" and at least one "against" vote (Arjun: 1 for / 2 against, Karn: 2 for / 1 against, Bhim: 1 for / 2 against), and all three count constraints — two for the recreation bill, one for the school bill, one for the tax bill — hold exactly in the completed grid, confirming a unique, consistent solution.

Karn's votes are therefore for the school bill and for the tax bill.