28 logic riddles with answers

Behind door 2.

Test each door. If the prize were behind door 1, signs 1 and 2 would both be true. If it were behind door 3, signs 2 and 3 would both be true. With the prize behind door 2, only sign 3 is true.

The red-haired child.

One child is a boy and the other a girl, so if either statement were true, the other would be true as well. At least one is a lie, so both are lies: the black-haired child is a girl, and the red-haired child is the boy.

Ben.

Ben and Dana contradict each other, so one of them is the liar. That means Ada and Cam both tell the truth, and Ada says Ben did it. Check: Ben's claim is false and the other three are true, so there is exactly one liar.

Only Lu.

If Mo told the truth, Kit and Lu would both be liars, but then Kit's claim that Lu lies would be true, which cannot happen. So Mo lies. That makes Lu's claim true, so Lu tells the truth, and Kit's claim that Lu lies is false, so Kit lies. Mo's claim is false because Lu is truthful, which fits.

Quin lies and Rue tells the truth. Pip could be either.

Everyone asked that question answers yes: a truth-teller says yes honestly, and a liar lies and also says yes. So Pip cannot have said he always lies, which makes Quin a liar. Rue's claim that Quin lies is true, so Rue tells the truth. Pip's answer reveals nothing about Pip.

Both of them.

A liar could not say this, because the first part, "I am a liar," would make the whole sentence true. So Ada tells the truth, and her sentence is true. Its first part is false, so the second part must be true: Bea tells the truth.

Thursday.

Rowan can say this either on a truthful day when yesterday was a lying day (Thursday) or on a lying day when yesterday was not a lying day (Monday). For Sasha the same test gives Sunday or Thursday. The only day that works for both is Thursday.

The cat.

Whoever sits in the middle is next to both of the others. The dog and the mouse are not next to each other, so neither of them is in the middle, which leaves the cat.

Three: one dog, one cat and one parrot.

"All are dogs except two" means the cats and parrots together make 2. In the same way, the dogs and parrots make 2, and the dogs and cats make 2. Those three totals count every pet twice, so 2 x (all pets) = 6 and there are 3 pets, one of each kind.

All three of them.

Each friend knows only their own wish. If the first did not want dessert, she could answer no at once, so "I don't know" means she wants one. The second's answer means the same. The third now knows the first two want dessert, and she wants one too, so she can say yes for everyone.

Ms. Green wears white, Ms. Gray wears green and Ms. White wears gray.

The woman in white is talking to Ms. Gray, so she is not Ms. Gray, and she is not Ms. White, because nobody wears her own color. So Ms. Green wears white. Ms. Gray cannot wear gray, so she wears green, and Ms. White has the gray scarf.

Both children have mud on their foreheads.

If only one child were muddy, that child would see a clean forehead, know the mud must be on their own and step forward at the first clap. Nobody did, so each child learns that the other also saw mud. Each can see mud on the other, and now knows the other saw mud too, so by the second clap both know their own forehead is muddy.

Red.

If Cara had seen two blue hats, she would know hers was red, because there are only two blue hats. So Ben and Ana are not both wearing blue, and Ben can work that out. If Ana's hat were blue, Ben would know his own was red. He does not know, so Ana's hat is red.

Cy.

If green and white were the middle and right houses, the left house would be red, but Ana lives on the left and Ben lives in the red house. So the houses run green (Ana), white, red (Ben), and Cy lives in the white house and drinks tea. The dog's owner drinks milk, so Cy does not own the dog, and the cat is not in the middle, so Cy owns the fish. Ana drinks juice and has the cat; Ben drinks milk and has the dog.

June 5.

The days 7 and 8 appear only once, so Nia would know at once if she had been told either. Leo is sure she does not know, so his month contains neither: it is not March or April. That leaves June 3, June 5, September 3, September 4 and September 6. Nia now knows, so her day is not 3, which appears twice. That leaves June 5, September 4 and September 6. Leo now knows too, so his month has only one date left: June 5.

4.

Nobody can shake more than 8 hands, so the nine answers are 0 to 8. The person who shook 8 hands shook everyone except their partner, so everyone else shook at least one hand: the 8 and the 0 are partners. Leave that couple out (everyone left loses the one handshake with the 8) and the same reasoning pairs 7 with 1, 6 with 2 and 5 with 3. The person who shook 4 hands is left without a partner among the nine, so their partner is Ben, and that person is Ana.

Floor 1 Mona, floor 2 Joel, floor 3 Lars, floor 4 Kemi, floor 5 Ines.

Start with the top floor. It is not Joel or Kemi (clues), not Mona (she is below Kemi) and not Lars (Ines is above him). So Ines is on 5 and Lars on 3. Joel cannot be on 1, on 5, on 4 (next to Ines) or on 3 (Lars), so he is on 2. Mona and Kemi take 1 and 4, and Mona is lower.

Cal.

Number the seats 1 to 6 clockwise with Ana in seat 1. Ben is opposite her in seat 4, and Cal is in seat 2. Seats 3 and 5 are both next to Ben, so Dia takes seat 6. Seat 3 is next to Cal, so Eve takes seat 5 and Fin seat 3. The seat opposite 5 is seat 2, which is Cal's.

Ada, Cy, Bo, Dev.

Suppose Ada came second. Then Ada did not win and Bo was not second, so both of fan 1's claims are wrong, which breaks the rule. So Ada was not second, and fan 3's other claim is right: Dev came last. Dev was not third, so fan 2's other claim is right: Cy came second. Bo was not second, so fan 1's other claim is right: Ada won. That leaves Bo third.

3.8 cm.

When a book stands with its spine facing you, its front cover is on the right. So the first page of Volume 1 is on its right side, touching Volume 2, and the last page of Volume 3 is on its left side, also touching Volume 2. The worm goes through Volume 1's front cover, all of Volume 2 (two covers and 3 cm of pages) and Volume 3's back cover: 3 cm + 4 x 2 mm = 3.8 cm.

7 races.

Race five groups of five (5 races), then race the five group winners (race 6). The winner of race 6 is the fastest horse. Only five horses can still be second or third: the second and third horses of race 6, the second and third horses from the fastest horse's group, and the second horse from the group of race 6's runner-up. Race those five (race 7); its first two are second and third overall. After six races those five are still unranked against each other, so six races are not enough.

Every glim must be a tonk, but a tonk does not have to be a glim.

Picture circles inside circles: glims inside frells, and frells inside tonks. Every glim is then inside the tonk circle. The tonk circle can also hold things that are not frells or glims, so a tonk need not be a glim.

You cannot tell.

The rule says chess club members wear blue badges. It does not say that only chess club members do, so Mo could belong to another group that also wears them.

A and 7.

The A must have an even number on the back, so check it. The 7 must not have a vowel on the back, so check it too. The K does not matter, because the rule says nothing about consonants. The 4 does not matter either, because the rule does not say an even number needs a vowel.

Sentence 2.

The sentences contradict each other, so at most one of them is true. If none were true, all three would be false, which would make sentence 3 true. So exactly one is true and two are false, and the sentence that says so is sentence 2.

The amounts are the same.

Each glass ends with 200 ml, as it started. Whatever juice is missing from the juice glass has been replaced by exactly the same amount of water, and the missing juice is now in the water glass. So the juice in the water glass equals the water in the juice glass.

Yes.

Imagine both walks happening on the same day: one walker going up as Rin did on day one, and one coming down as she did on day two. They start at opposite ends of the same path at 7 a.m., so they must meet somewhere. At that spot and time of day, Rin was there on both days.

9.

The person at the back says "red" if they see an odd number of red hats and "blue" if the number is even. Now the other nine know whether their own red hats add up to odd or even. Each of them counts the red hats they see ahead plus the red hats already called by the others among the nine; if that total does not match the odd-or-even signal, their own hat is red, and otherwise it is blue. All nine are always right. The person at the back sees nothing that shows their own color, so they are the only one who can be wrong.