3 Logic Puzzles With Fully Explained Solutions

How to use these puzzles
Read each problem, stop before its solution, and write your reasoning. The rules mean exactly what they say; no secret wordplay is waiting behind a curtain. Speed is irrelevant. A puzzle is entertainment, not a diagnostic test or a tiny courtroom deciding whether you are clever.
Puzzle one: the mislabeled fruit boxes
Three closed boxes contain apples, oranges, or a mixture of both. They are labeled Apples, Oranges, and Mixed, but every label is wrong. You may draw one piece of fruit from one box without looking inside. How can you label all three boxes correctly?
Solution
Draw from the box labeled Mixed. Because every label is wrong, this box cannot contain a mixture. If you draw an apple, that entire box must be apples; if you draw an orange, it must be oranges.
Suppose you drew an apple. Relabel that box Apples. The box currently labeled Oranges cannot be oranges, because its label is wrong, and it cannot be apples, because the apple box is known. Therefore it is Mixed. The remaining box, currently labeled Apples, must be Oranges.
If you draw an orange, the same reasoning swaps apple and orange. One draw works because the all-wrong rule turns the “Mixed” label into useful exclusion evidence.
Puzzle two: three arrival days
Ada, Ben, and Cora arrived on Monday, Tuesday, and Wednesday, one person per day.
- Ada did not arrive Monday.
- Ben arrived before Cora.
- Cora did not arrive Tuesday.
Who arrived on each day?
Solution
Cora cannot be Monday because Ben must arrive before her. Cora also cannot be Tuesday by the third clue, so Cora arrived Wednesday.
Ben must be before Wednesday. If Ben were Tuesday, Ada would have to be Monday, contradicting the first clue. Therefore Ben arrived Monday and Ada arrived Tuesday.
The final order is Monday: Ben; Tuesday: Ada; Wednesday: Cora. Check every clue: Ada is not Monday, Ben is before Cora, and Cora is not Tuesday.
Find more checked problems in Logic Puzzles.
Puzzle three: one heavier coin
Nine coins look identical, but exactly one is heavier. You have a balance scale and may use it only twice. How do you guarantee finding the heavier coin?
Solution
Divide the coins into three groups of three: A, B, and C. First, weigh A against B.
- If they balance, the heavier coin is in C.
- If one side drops, the heavier coin is in that lower group.
You now have a suspect group of exactly three coins. For the second weighing, place one suspect coin on each side and leave the third off the scale.
- If one side drops, that coin is heavier.
- If the two balance, the unweighed suspect is heavier.
This always succeeds in two weighings. The first weighing reduces nine candidates to three; the second reduces three to one. The method depends on exactly one coin being heavier and the scale comparing weight reliably. Change those rules and the solution may no longer hold.
Audit before celebrating
A complete solution should account for every possible outcome. The fruit solution handles either fruit drawn. The arrival solution checks all clues. The coin solution handles left heavy, right heavy, and balance at both stages.
Use the systematic solving guide for a reusable method. If your answer differs, compare assumptions before declaring either solver defeated. Sometimes the villain is simply a word interpreted two ways, wearing a very small cape.