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?
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:
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?
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?
Show 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.
Updated On: Jul 13, 2026
- Arjun votes for the tax bill.
- Karn votes for the recreation bill.
- Karn votes against the school bill.
- Bhim votes against the school bill.
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Force Bhim's recreation vote.
Arjun votes for the recreation bill (given), and exactly two members vote for it. Karn votes against the recreation bill (given), so Karn is not the second yes-vote. The only member left is Bhim, so Bhim must vote for the recreation bill. This is fixed in every valid case.
Step 2: Note who is still free.
So far: Arjun = for-recreation, against-school; Karn = against-recreation; Bhim = for-recreation, against-tax (given). Exactly one member votes for the school bill, and Arjun is against it, so the sole yes-vote on school is Karn or Bhim. Exactly one member votes for the tax bill, and Bhim is against it, so the sole yes-vote on tax is Arjun or Karn.
Step 3: Use the "at least one for, at least one against" rule to pin Karn.
Arjun already has a for-vote (recreation) and an against-vote (school), so Arjun's tax vote is unrestricted by this rule alone. Bhim already has a for-vote (recreation) and an against-vote (tax), so Bhim's school vote is unrestricted by this rule alone. Karn, though, is against on recreation only so far, so Karn needs at least one for-vote among school and tax, otherwise Karn would be against on all three bills, which breaks the rule.
Step 4: Branch on who is the sole yes-vote for school.
Branch 1, Karn is for-school: Karn's need for a for-vote is already met, so Karn's tax vote is free, giving two further options for who is the sole yes-vote on tax, Arjun or Karn.
Branch 1a: Arjun is for-tax. Then Karn is against-tax (tax has only one yes-voter) and Bhim is against-school (school has only one yes-voter). This gives: Arjun(for R, against S, for T), Karn(against R, for S, against T), Bhim(for R, against S, against T).
Branch 1b: Karn is for-tax. Then Arjun is against-tax and Bhim is against-school. This gives: Arjun(for R, against S, against T), Karn(against R, for S, for T), Bhim(for R, against S, against T).
Branch 2, Bhim is for-school: Then Karn is NOT the school yes-voter, so Karn's only way to satisfy "at least one for-vote" is to be the sole yes-voter on tax, forcing Arjun against-tax. This gives: Arjun(for R, against S, against T), Karn(against R, against S, for T), Bhim(for R, for S, against T).
Step 5: Confirm each branch is internally consistent, giving exactly three valid worlds.
World 1 (Branch 1a): Arjun for R,T / against S. Karn for S / against R,T. Bhim for R / against S,T.
World 2 (Branch 1b): Arjun for R / against S,T. Karn for S,T / against R. Bhim for R / against S,T.
World 3 (Branch 2): Arjun for R / against S,T. Karn for T / against R,S. Bhim for R,S / against T.
Every member has at least one for-vote and one against-vote in all three worlds, and every bill's for-count matches the rules, so all three are valid and no other world is possible.
Step 6: Apply this question's extra condition to the three worlds.
The condition says the against-school group and the against-tax group are exactly the same two people. Check each world:
World 1: against-school = {Arjun, Bhim} (Karn is the school yes-vote). against-tax = {Karn, Bhim} (Arjun is the tax yes-vote). These two groups differ (one has Arjun, the other has Karn), so World 1 fails the condition.
World 2: against-school = {Arjun, Bhim} (Karn is the school yes-vote). against-tax = {Arjun, Bhim} (Karn is the tax yes-vote too). Both groups are exactly {Arjun, Bhim}, a perfect match. World 2 satisfies the condition.
World 3: against-school = {Arjun, Karn} (Bhim is the school yes-vote). against-tax = {Arjun, Bhim} (Karn is the tax yes-vote). These differ (Karn versus Bhim), so World 3 fails the condition.
Only World 2 survives: Arjun votes for recreation only (against school and tax); Karn votes for school and tax (against recreation); Bhim votes for recreation only (against school and tax).
Step 7: Check each statement against this single surviving world.
"Arjun votes for the tax bill": false, Arjun is against tax in World 2.
"Karn votes for the recreation bill": false, Karn is against recreation in every world, including World 2.
"Karn votes against the school bill": false, Karn is actually FOR school in World 2.
"Bhim votes against the school bill": true, Bhim is against school in World 2.
Final Answer:
Since World 2 is the only world consistent with the extra condition, and Bhim is against the school bill in that world, this is the statement that must be true. \[ \boxed{\text{Bhim votes against the school bill}} \]
Arjun votes for the recreation bill (given), and exactly two members vote for it. Karn votes against the recreation bill (given), so Karn is not the second yes-vote. The only member left is Bhim, so Bhim must vote for the recreation bill. This is fixed in every valid case.
Step 2: Note who is still free.
So far: Arjun = for-recreation, against-school; Karn = against-recreation; Bhim = for-recreation, against-tax (given). Exactly one member votes for the school bill, and Arjun is against it, so the sole yes-vote on school is Karn or Bhim. Exactly one member votes for the tax bill, and Bhim is against it, so the sole yes-vote on tax is Arjun or Karn.
Step 3: Use the "at least one for, at least one against" rule to pin Karn.
Arjun already has a for-vote (recreation) and an against-vote (school), so Arjun's tax vote is unrestricted by this rule alone. Bhim already has a for-vote (recreation) and an against-vote (tax), so Bhim's school vote is unrestricted by this rule alone. Karn, though, is against on recreation only so far, so Karn needs at least one for-vote among school and tax, otherwise Karn would be against on all three bills, which breaks the rule.
Step 4: Branch on who is the sole yes-vote for school.
Branch 1, Karn is for-school: Karn's need for a for-vote is already met, so Karn's tax vote is free, giving two further options for who is the sole yes-vote on tax, Arjun or Karn.
Branch 1a: Arjun is for-tax. Then Karn is against-tax (tax has only one yes-voter) and Bhim is against-school (school has only one yes-voter). This gives: Arjun(for R, against S, for T), Karn(against R, for S, against T), Bhim(for R, against S, against T).
Branch 1b: Karn is for-tax. Then Arjun is against-tax and Bhim is against-school. This gives: Arjun(for R, against S, against T), Karn(against R, for S, for T), Bhim(for R, against S, against T).
Branch 2, Bhim is for-school: Then Karn is NOT the school yes-voter, so Karn's only way to satisfy "at least one for-vote" is to be the sole yes-voter on tax, forcing Arjun against-tax. This gives: Arjun(for R, against S, against T), Karn(against R, against S, for T), Bhim(for R, for S, against T).
Step 5: Confirm each branch is internally consistent, giving exactly three valid worlds.
World 1 (Branch 1a): Arjun for R,T / against S. Karn for S / against R,T. Bhim for R / against S,T.
World 2 (Branch 1b): Arjun for R / against S,T. Karn for S,T / against R. Bhim for R / against S,T.
World 3 (Branch 2): Arjun for R / against S,T. Karn for T / against R,S. Bhim for R,S / against T.
Every member has at least one for-vote and one against-vote in all three worlds, and every bill's for-count matches the rules, so all three are valid and no other world is possible.
Step 6: Apply this question's extra condition to the three worlds.
The condition says the against-school group and the against-tax group are exactly the same two people. Check each world:
World 1: against-school = {Arjun, Bhim} (Karn is the school yes-vote). against-tax = {Karn, Bhim} (Arjun is the tax yes-vote). These two groups differ (one has Arjun, the other has Karn), so World 1 fails the condition.
World 2: against-school = {Arjun, Bhim} (Karn is the school yes-vote). against-tax = {Arjun, Bhim} (Karn is the tax yes-vote too). Both groups are exactly {Arjun, Bhim}, a perfect match. World 2 satisfies the condition.
World 3: against-school = {Arjun, Karn} (Bhim is the school yes-vote). against-tax = {Arjun, Bhim} (Karn is the tax yes-vote). These differ (Karn versus Bhim), so World 3 fails the condition.
Only World 2 survives: Arjun votes for recreation only (against school and tax); Karn votes for school and tax (against recreation); Bhim votes for recreation only (against school and tax).
Step 7: Check each statement against this single surviving world.
"Arjun votes for the tax bill": false, Arjun is against tax in World 2.
"Karn votes for the recreation bill": false, Karn is against recreation in every world, including World 2.
"Karn votes against the school bill": false, Karn is actually FOR school in World 2.
"Bhim votes against the school bill": true, Bhim is against school in World 2.
Final Answer:
Since World 2 is the only world consistent with the extra condition, and Bhim is against the school bill in that world, this is the statement that must be true. \[ \boxed{\text{Bhim votes against the school bill}} \]
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