Question

Directions: Each group of questions is based on a set of conditions. Read the passage below and answer the question that follows.

Krishnapuram's town council has exactly three members: Arjun, Karn and Bhim. During one week the council votes on exactly three bills: a recreation bill, a school bill and a tax bill. Each member votes either for or against each bill. The following is known:

• Each member votes for at least one of the bills and against at least one of the bills.
• Exactly two members vote for the recreation bill.
• Exactly one member votes for the school bill.
• Exactly one member votes for the tax bill.
• Arjun votes for the recreation bill and against the school bill.
• Karn votes against the recreation bill.
• Bhim votes against the tax bill.

If the set of members of the council who vote against the school bill are the only ones who also vote against the tax bill, then which one of the following statements must be true?

• A. Arjun votes for the tax bill.
• B. Karn votes for the recreation bill.
• C. Karn votes against the school bill.
• D. Bhim votes against the school bill.

Hint:

Test the extra clue (against-school set equals against-tax set) against each of the three base worlds from questions 108/110/111; only one world survives, and you can read the answer straight off it.

Answer 1

The Correct Option is D

Rather than re-deriving the three worlds from scratch with plain sentences, build the same 3-by-3 grid used for this whole caselet, fill it in from the base rules, and simply filter the finished grids by this question's extra clue: the people who are "no" on school must be the exact same people who are "no" on tax, not a superset, not a subset, the identical pair.

Draw a 3-by-3 grid, one row per member (Arjun, Karn, Bhim) and one column per bill (recreation, school, tax), and fill in only what the passage states directly: Arjun is yes on recreation and no on school; Karn is no on recreation; Bhim is no on tax.

Recreation needs exactly two yes-votes overall. Arjun already supplies one of those two. Karn is a confirmed no, so the second yes on recreation cannot be Karn, which leaves only Bhim. Fill in Bhim as yes on recreation right away, before touching anything else.

School needs exactly one yes-vote overall, and Arjun's cell there is already a no, so the single yes on school belongs to either Karn or Bhim, not both. Tax needs exactly one yes-vote overall, and Bhim's cell there is already a no, so the single yes on tax belongs to either Arjun or Karn, not both.

Every row of the grid needs at least one yes and at least one no. Arjun's row already has a yes (recreation) and a no (school), so Arjun's tax cell can go either way without breaking this rule. Bhim's row already has a yes (recreation) and a no (tax), so Bhim's school cell can go either way too. Karn's row so far has only a no (recreation), so Karn's row needs at least one yes somewhere among school and tax, or the row rule breaks.

1. Grid A: put the school yes on Karn, and the tax yes on Arjun. This forces Karn to no on tax (tax already has its one yes, from Arjun) and Bhim to no on school (school already has its one yes, from Karn). Row totals: Arjun yes on recreation and tax; Karn yes on school only; Bhim yes on recreation only.
2. Grid B: put the school yes on Karn, and the tax yes on Karn as well. This forces Arjun to no on tax and Bhim to no on school. Row totals: Arjun yes on recreation only; Karn yes on school and tax; Bhim yes on recreation only.
3. Grid C: put the school yes on Bhim instead of Karn. Since Karn's row still needs a yes somewhere, and school is now taken by Bhim, Karn's yes has to be on tax, which also forces Arjun to no on tax. Row totals: Arjun yes on recreation only; Karn yes on tax only; Bhim yes on recreation and school.

Every row in all three grids carries at least one yes and one no, and every column total matches the passage's counts, so all three grids are valid, and since the school/tax assignment choices above cover every possibility, no fourth grid exists.

Now apply the extra filter to each grid:
Grid A: no-on-school = {Arjun, Bhim} (Karn is the school yes). no-on-tax = {Karn, Bhim} (Arjun is the tax yes). These two sets are not identical, Arjun sits in one but not the other, so Grid A is filtered out by the new clue.
Grid B: no-on-school = {Arjun, Bhim} (Karn is the school yes). no-on-tax = {Arjun, Bhim} (Karn is the tax yes too). Identical sets, so Grid B passes the filter.
Grid C: no-on-school = {Arjun, Karn} (Bhim is the school yes). no-on-tax = {Arjun, Bhim} (Karn is the tax yes). Not identical, Karn versus Bhim differ, so Grid C is filtered out too.

Only Grid B survives the filter, so it is the single world we now read the answer from: Arjun is yes only on recreation, Karn is yes on school and tax, Bhim is yes only on recreation. Reading the five original choices off this row: Arjun is a no on tax, so "Arjun votes for the tax bill" is false. Karn is a no on recreation, so "Karn votes for the recreation bill" is false. Karn is actually a yes on school, so "Karn votes against the school bill" is false. Bhim is a confirmed no on school, so "Bhim votes against the school bill" is true.

Since Grid B is the only surviving world, whatever is true in it is true in every case that satisfies this question's condition. That makes "Bhim votes against the school bill" the statement that must be true, since it holds in the one and only remaining grid.