Deduction Choice Lab

A. No, all three boxes remain possible

The two signs are logical opposites, so exactly one is true wherever the token sits. The truth-count information adds no restriction. All three boxes remain possible; selecting A would require another independent clue.

Try next: Add one clue that really distinguishes a box from the other two.

B. It is not glow

If this token were glow, the rule would force it to be warm, contradicting the given observation. It therefore cannot be glow. This conclusion uses the stated implication without assuming its converse.

Try next: Explain why knowing only that a token is warm would not reverse the rule.

C. Impossible

An even pair needs two odd cards or two even cards. Here there are 3 odd cards and 1 even cards. An odd category count leaves an unmatched card, so it is impossible.

Try next: Try all three unordered ways to divide four cards into pairs.

A. Thursday

10 days contain 1 complete weeks and 3 extra days. Full weeks preserve the weekday; advancing the remainder from Monday reaches Thursday. Counting today as day one would shift the result incorrectly.

Try next: Check the answer by first advancing one full week.

B. Cal → Ari → Dee → Bea

The fixed clues place Ari in position 2 and Bea in 4. The remaining positions are 1 and 3. Cal must take the earlier one, leaving the later position for Dee.

Try next: Remove one fixed-position clue and see whether multiple orders become possible.

C. 4

Initially both pans weigh 11 units. Removing 4 from the left reduces that pan by 4, so removing the same amount from the right restores equality. The remaining left block does not determine the removal amount.

Try next: Write the original equality, then subtract the same value on both sides.

A. 2

Red and blue account for 5 of the 7 tokens. The remaining 2 must be green because the categories cover the whole box without overlap. This conclusion would fail if another color were allowed.

Try next: State which assumption makes subtraction identify the missing category.

B. No, the information is insufficient

The rule guarantees the second property for objects with the first property, but it does not exclude other objects from having that second property. The selected object may or may not be square.

Try next: Invent one collection containing a counterexample without breaking the given rule.

C. 2, 5

The qualifying numbers are 2, 5. Each makes one condition true and the other false. A number passing both is deliberately rejected because “exactly one” differs from an ordinary inclusive OR rule.

Try next: Make a two-column truth table for every candidate.

B. A and C

The opposite signs contribute no restriction because exactly one is always true. The independent observation does exclude B. A and C remain possible, so the observation narrows the set without identifying a unique box.

Try next: Add a new observation that distinguishes the two remaining possibilities.

C. Undetermined

Being round does not establish being striped. Both a striped round token and an unstriped round token satisfy the stated implication. The property remains undetermined because the rule does not say that its converse is true.

Try next: Construct one example for each permitted possibility.

C. Possible

An even pair needs two odd cards or two even cards. Here there are 2 odd cards and 2 even cards. Both category counts can be paired, so the task is possible.

Try next: Try all three unordered ways to divide four cards into pairs.

A. Sunday

17 days contain 2 complete weeks and 3 extra days. Full weeks preserve the weekday; advancing the remainder from Thursday reaches Sunday. Counting today as day one would shift the result incorrectly.

Try next: Check the answer by first advancing one full week.

B. Bea → Cal → Ari → Dee

The fixed clues place Ari in position 3 and Bea in 1. The remaining positions are 2 and 4. Cal must take the earlier one, leaving the later position for Dee.

Try next: Remove one fixed-position clue and see whether multiple orders become possible.

C. 3

Initially both pans weigh 11 units. Removing 3 from the left reduces that pan by 3, so removing the same amount from the right restores equality. The remaining left block does not determine the removal amount.

Try next: Write the original equality, then subtract the same value on both sides.

A. 4

Red and blue account for 6 of the 10 tokens. The remaining 4 must be green because the categories cover the whole box without overlap. This conclusion would fail if another color were allowed.

Try next: State which assumption makes subtraction identify the missing category.

A. Must be true

The second rule explicitly states that every small object is wooden. It applies directly to the selected small object, so the answer is definite. Unlike a one-way implication, the supplied pair includes the reverse direction.

Try next: Identify the exact rule needed without relying on the other one.

C. 4, 7

The qualifying numbers are 4, 7. Each makes one condition true and the other false. A number passing both is deliberately rejected because “exactly one” differs from an ordinary inclusive OR rule.

Try next: Make a two-column truth table for every candidate.

C. No consistent placement

The signs are exact logical opposites. For every possible placement, one is true and one is false; they cannot both be true. Therefore no token placement satisfies the narrator’s simultaneous-truth claim under the stated rules.

Try next: Test each box and record the truth value of both signs.

A. Must be true

The rule applies directly because the selected token is metal. It must therefore be heavy. This uses the implication in its stated direction, requiring neither a converse assumption nor information about tokens outside the metal category.

Try next: Explain why unrelated token categories do not affect this deduction.

C. Impossible

An even pair needs two odd cards or two even cards. Here there are 1 odd cards and 3 even cards. An odd category count leaves an unmatched card, so it is impossible.

Try next: Try all three unordered ways to divide four cards into pairs.

A. Monday

23 days contain 3 complete weeks and 2 extra days. Full weeks preserve the weekday; advancing the remainder from Saturday reaches Monday. Counting today as day one would shift the result incorrectly.

Try next: Check the answer by first advancing one full week.

B. Cal → Bea → Dee → Ari

The fixed clues place Ari in position 4 and Bea in 2. The remaining positions are 1 and 3. Cal must take the earlier one, leaving the later position for Dee.

Try next: Remove one fixed-position clue and see whether multiple orders become possible.

C. 5

Initially both pans weigh 14 units. Removing 5 from the left reduces that pan by 5, so removing the same amount from the right restores equality. The remaining left block does not determine the removal amount.

Try next: Write the original equality, then subtract the same value on both sides.

A. 4

Red and blue account for 9 of the 13 tokens. The remaining 4 must be green because the categories cover the whole box without overlap. This conclusion would fail if another color were allowed.

Try next: State which assumption makes subtraction identify the missing category.

C. Undetermined

The selected object lies outside the category governed by the rule. An unstriped object could be soft or not soft without violating the implication. Negating the first property does not establish the opposite of the second property.

Try next: Give two permitted examples with different softness outcomes.

C. 6, 9

The qualifying numbers are 6, 9. Each makes one condition true and the other false. A number passing both is deliberately rejected because “exactly one” differs from an ordinary inclusive OR rule.

Try next: Make a two-column truth table for every candidate.