Work through the levels in order. Type your answer and press Check — open the worked solution if you get stuck. Answers like -6, 3/4 or 0.25 are typed exactly as shown. A ⭐ Bonus Challenge at the end is extra credit — skip it if the rest was hard work.
Level 1 · Warm-up
1 On an island of knights (always truthful) and knaves (always lie), A says, “We are both knaves.” What are A and B?
Show worked solution
A knight cannot say he is a knave, so A is a knave. Then his statement is false, so they are not both knaves, and B is a knight: A is a knave and B is a knight. “Both are knaves” takes the statement as true, but a knave cannot say something true.
2 Ken has only 20c and 50c coins: 11 coins worth $3.10 altogether. How many of each does he have? Work from the table.
Show worked solution
Each swap adds 30c. $3.10 − $2.20 = 90c = 3 swaps, so 3 fifty-cent coins and 11 − 3 = 8 twenty-cent coins. 20c coins 8, 50c coins 3. Check: 8 × 20 + 3 × 50 = 160 + 150 = 310c ✓.
3 A drawer holds the socks shown, all mixed up. In the dark, how many socks must you take to be sure of having two socks of different colours?
Show worked solution
The worst case is all 8 black socks in a row, still one colour. The next sock must be white or grey. 8 + 1 = 9.
Level 2 · Core
4 Amir, Bea, Chen, Dana and Eli live in houses 1 to 5, numbered left to right, one person per house.
• Dana lives next to both Amir and Eli.
• Bea lives in an end house.
• Chen lives directly to the right of Bea.
• Eli does not live in an end house.
Who lives in house 5?
• Dana lives next to both Amir and Eli.
• Bea lives in an end house.
• Chen lives directly to the right of Bea.
• Eli does not live in an end house.
Who lives in house 5?
Show worked solution
Nobody is to the right of house 5, so Bea is in house 1 and Chen in house 2. Dana needs neighbours on both sides among houses 3 to 5, so Dana is in house 4, with Amir and Eli in houses 3 and 5. Eli is not at an end, so Eli is in 3 and Amir is in house 5. Check: Dana (4) is next to Eli (3) and Amir (5) ✓.
5 You have plenty of 10c, 20c and 50c coins. Count the different ways to make exactly $1 (the order of the coins does not matter).
Show worked solution
Two 50c coins: 1 way. One 50c coin: 50c from 20c and 10c with 0, 1 or 2 twenty-cent coins: 3 ways. No 50c coin: $1 with 0 to 5 twenty-cent coins: 6 ways. Total 10 ways; with at least one 50c coin, 1 + 3 = 4 ways.
6 A tank loses one third of its water. Then 10 L is added. Then half of the water in the tank is used, leaving 25 L. How many litres were in the tank at the start?
Show worked solution
Before using half: 25 × 2 = 50 L. Before adding 10 L: 40 L. 40 L is two thirds of the start, so the start was 40 ÷ 2 × 3 = 60 L. Check: 60 − 20 = 40, + 10 = 50, half used leaves 25 ✓. Adding a third back (40 + 13⅓) is the wrong inverse.
Level 3 · Challenge
7 On an island of knights (always truthful) and knaves (always lie), X says, “Y is a knight.” Y says, “X and I are different types.” Three statements:
1 X is a knave.
2 Y is a knight.
3 X and Y are the same type.
Which statements are correct?
1 X is a knave.
2 Y is a knight.
3 X and Y are the same type.
Which statements are correct?
Show worked solution
If X is a knight, Y is a knight, but then they are the same type and Y’s statement is false, a clash. So X is a knave, his statement is false, and Y is a knave too. Y’s statement “different types” is then false, as a knave’s must be. X is a knave (1 ✓); Y is not a knight (2 ✗); both are knaves, the same type (3 ✓). 1 and 3.
8 Cards are drawn from a set numbered 1 to 20. What is the least number you must draw to be sure that two of your cards add to 21? Then answer the same question for cards numbered 1 to 25 and a sum of 26.
Show worked solution
1 to 20 splits into 10 pairs adding to 21: (1, 20), (2, 19), …, (10, 11). Ten cards, one from each pair, can avoid 21, so you need 11. 1 to 25 gives 12 pairs adding to 26, (1, 25) to (12, 14), and 13 has no partner. The worst case is 13 plus one from each pair, 13 cards, so you need 14. Giving 13 forgets that the card 13 cannot pair with itself.
9 Ten people sit in a room. Each one is a knight (always truthful) or a knave (always lies), and each one says, “At least 5 of the other 9 people in this room are knaves.” How many knights are in the room?
Show worked solution
A knight sees 10 − k knaves among the others, and must be right: 10 − k ≥ 5, so k ≤ 5. A knave sees 9 − k knaves among the others, and must be wrong: 9 − k ≤ 4, so k ≥ 5. With both types present, k = 5. All knights (k = 10) fails, since each would see 0 knaves; all knaves (k = 0) fails, since each would see 9 knaves and be telling the truth. So 5. Check: each knight sees 5 knaves ✓; each knave sees only 4 and is lying ✓.