Question 1

How do you solve logic puzzles systematically?

Accepted Answer

Write down every constraint explicitly, look for the one that eliminates the most possibilities, and check your answer against all constraints before committing. Most wrong answers satisfy some of the conditions and not all of them.

Question 2

Are logic puzzles useful for aptitude tests?

Accepted Answer

Yes — campus and graduate aptitude papers draw heavily on the same families: weighings, work rates, relative motion and conditional probability. Recognising the family is most of the speed.

Question 3

A server room has one bulb inside and three switches outside. Exactly one switch controls the bulb. The door is closed, and you may open it only once. How do you identify the right switch?

Accepted Answer

Turn on switch 1 and leave it for several minutes. Turn it off, turn on switch 2, and go in. If the bulb is lit it is switch 2. If it is dark but warm, it is switch 1. If it is dark and cold, it is switch 3.. The puzzle looks like it needs three observations from one look. Heat is a second channel of information that persists after the state changes, which turns one observation into two.

Question 4

You have two lengths of fuse. Each burns for exactly one hour end to end, but neither burns evenly — half a fuse does not mean half an hour. With only a lighter, measure exactly forty-five minutes.

Accepted Answer

Light fuse A at both ends and fuse B at one end simultaneously. When A is consumed, thirty minutes have passed. At that instant light B's other end; its remaining thirty minutes now burn in fifteen. Total: forty-five minutes.. Uneven burning defeats any attempt to measure by length, but not by time: burning from both ends always halves the remaining duration regardless of distribution. The trick is to reason about time rather than distance.

Question 5

Eight identical servers sit in a rack. One has a faulty fan and runs measurably hotter; the rest are identical. You have a balance scale that compares the total heat of any two groups. Find the faulty one in two weighings.

Accepted Answer

Split into 3, 3 and 2. Weigh the two threes. If one is heavier, the faulty server is in it — weigh two of those three against each other to find it. If they balance, the faulty one is in the pair, and one comparison identifies it.. A balance gives a ternary result — left, right or equal — so each weighing distinguishes three cases, not two. Splitting into thirds rather than halves is why eight (and up to nine) items are solvable in two weighings.

Question 6

Two doors, one to the exit and one to a locked room. Two guards, one who always tells the truth and one who always lies — and you cannot tell which is which. You may ask one guard one question. What do you ask?

Accepted Answer

Ask either guard: "Which door would the other guard tell me is the exit?" — then take the other door. The truthful guard reports the liar's false answer; the liar misreports the truthful guard's correct one. Both point at the wrong door.. Routing the question through the other guard guarantees exactly one lie in the chain regardless of who you asked, so the answer is reliably inverted. Building a known-wrong answer is often easier than building a known-right one.

Question 7

Three students stand in a line facing forward, each wearing a cap that is either black or white. The one at the back sees both caps ahead; the middle sees one; the front sees none. They are told at least one cap is white. The back student says nothing. The middle student then names her own colour correctly. What is it, and how did she know?

Accepted Answer

White. If both caps ahead were black, the back student would have known hers was white and spoken. Her silence proves at least one of the two ahead is white. The middle student sees the front cap — if it were white she could not conclude, so it must be black, making hers white.. The absence of a statement carries information when everyone knows the others are reasoning too. This is common knowledge reasoning, and it is why the puzzle collapses if any student is not known to be logical.

Question 8

One build in a batch of one thousand is corrupted. You have ten test machines. A machine given any set of builds reports, after one hour, whether the corrupted build was among them — but is then unusable. Find the bad build in a single hour.

Accepted Answer

Number the builds 1 to 1000 in binary. Machine k tests every build whose k-th bit is 1. After an hour, read the machines as bits: the pattern of failures is the binary number of the corrupted build.. Each machine answers one yes/no question, so ten of them distinguish 2^10 = 1024 possibilities. Encoding the search space in binary is the general technique whenever you have parallel one-bit tests.

Question 9

Four students must cross a dark footbridge that holds only two at a time, sharing one torch that must travel with every crossing. They walk at 1, 2, 5 and 10 minutes; a pair moves at the slower speed. Everyone is across in 17 minutes. How?

Accepted Answer

1 and 2 cross (2). 1 returns (1). 5 and 10 cross together (10). 2 returns (2). 1 and 2 cross (2). Total 17.. Pairing the two slowest makes them cost one crossing instead of two, and the extra return trips by the fast pair cost less than the time saved. The general lesson is that a locally greedy plan — always send the fastest — is not globally optimal.

Question 10

Three boxes hold cables: one all HDMI, one all USB, one mixed. Every label is wrong. You may draw one cable, without looking, from one box. Can you relabel all three correctly?

Accepted Answer

Draw from the box labelled "mixed". Since every label is wrong, that box is pure — whatever you draw is its true content. The remaining two then follow: the box labelled with what you drew cannot be that or its own label, so it is the mixed one, and the last box takes the remaining type.. The constraint that *every* label is wrong is what makes one draw sufficient — it turns the "mixed" box into the only one whose single sample is conclusive. Puzzles often hinge on a constraint that reads as flavour text.

Question 11

You have two identical phones and a hundred-floor tower, and you want the highest floor a phone survives being dropped from. Broken phones are gone for good. What strategy guarantees the answer in the fewest worst-case drops?

Accepted Answer

Fourteen drops. Start at floor 14, then 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100 — each gap one smaller than the last. If the first phone breaks at the n-th drop, the second walks up the remaining gap, and every path totals fourteen.. Fixed intervals make later cases more expensive than earlier ones. Shrinking the interval by one each time equalises the worst case across all outcomes — the smallest n where n(n+1)/2 ≥ 100 is 14. Balancing the worst case is a general optimisation move.

Question 12

A lecturer tells a class of thirty that it is more likely than not two of them share a birthday. The students object — thirty is nowhere near 365. Who is right, and why?

Accepted Answer

The lecturer. Thirty students form 435 pairs, and each pair is a chance of a match. The probability of at least one shared birthday passes 50% at 23 people and is about 70% at 30.. The intuition fails because people compare against one fixed birthday rather than counting pairs, which grow quadratically. The same error underlies underestimating hash collisions — the birthday bound is why a 128-bit hash needs about 2^64 items before collisions become likely.

Question 13

Nine laptops look identical; one is slightly heavier. With a balance scale, find it in two weighings.

Accepted Answer

Weigh 3 against 3. If they balance, the heavy one is in the untouched three; otherwise it is in the heavier pan. Weigh one against another from that group of three — if they balance it is the third.. Each weighing distinguishes three cases, so two weighings separate nine possibilities and three separate twenty-seven. Whenever a test has k outcomes, divide into k groups, not two — the halving instinct wastes information.

Question 14

Twelve components, one defective — but you are not told whether it is heavier or lighter. Three weighings on a balance. Find it, and say which it is.

Accepted Answer

Weigh 4 v 4. If balanced, the odd one is in the last four and two weighings against known-good items find it. If unbalanced, relabel: the heavy pan's four are "possibly heavy", the light pan's four "possibly light", and the remaining four are known good — then weigh a mixed group against known-good items, which narrows to at most three candidates with a known direction.. Three weighings give 3³ = 27 outcomes and twelve items with two directions need 24 — so it fits, but only just. That tightness is why items must be moved between pans rather than eliminated: every weighing has to carry near its full information capacity.

Question 15

You have a five-litre jug and a three-litre jug, neither marked. Measure exactly four litres.

Accepted Answer

Fill the five. Pour into the three, leaving two in the five. Empty the three, move the two litres into it. Fill the five again and top up the three, which takes one litre — leaving exactly four in the five-litre jug.. Every reachable amount is a combination of the two capacities, and because 5 and 3 share no common factor you can reach any whole number up to five. Two jugs whose sizes share a factor — say 4 and 6 — can only ever measure multiples of that factor.

Question 16

A technician must ferry a laptop, a magnet and a backup drive across a river. The boat holds her and one item. The magnet ruins the laptop if left with it; the laptop is fine but the magnet wipes the drive if those two are alone. How does everything cross intact?

Accepted Answer

Take the magnet over. Return empty. Take the laptop over, bring the magnet back. Leave the magnet, take the drive over. Return empty and finally take the magnet.. The item that conflicts with everything must move first and be temporarily returned. The insight most people miss is that a crossing may go backwards — the search space only opens up once you stop assuming progress must be monotonic.

Question 17

A hundred lockers, all shut. A hundred students file past. The first flips every locker, the second every second one, the third every third, and so on. Which lockers end open?

Accepted Answer

The perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100.. Each locker is toggled once per divisor, and divisors pair up (2 and 18 both divide 36) giving an even count — except for perfect squares, where the square root pairs with itself. An odd number of flips leaves the locker open.

Question 18

Two trains, 200 km apart, approach each other at 50 km/h each. A bird flies at 80 km/h from one to the other, turning instantly each time they meet, until the trains collide. How far does the bird fly?

Accepted Answer

160 km. The trains close at 100 km/h over 200 km, so they meet after 2 hours; the bird flies at 80 km/h for those 2 hours.. The tempting approach is summing an infinite series of shrinking legs, which converges to the same answer with far more effort. Asking "how long does this last?" instead of "what happens at each step?" is a reframing that collapses many iterative problems.

Question 19

Three doors, one prize. You pick one. The host, who knows where the prize is, opens a different door revealing nothing, then offers you the swap. Take it?

Accepted Answer

Yes. Switching wins two times in three; staying wins one time in three.. Your first choice is right 1/3 of the time, so the prize is behind the other two doors 2/3 of the time — and the host's knowing removal concentrates all of that 2/3 onto the single remaining door. The host's knowledge is what makes it work; a random opening would give a genuine 50/50.

Question 20

A colleague has two children. You know at least one is a girl. What is the chance both are girls?

Accepted Answer

One in three.. The equally likely pairs are BB, BG, GB, GG. Knowing at least one is a girl eliminates BB, leaving three cases of which one is GG. The intuition fails because "at least one" is information about the pair, not about a specific child — being told the elder is a girl really does give 1/2.

Question 21

A textbook and a highlighter cost ₹1,100 together. The textbook costs ₹1,000 more than the highlighter. What does the highlighter cost?

Accepted Answer

₹50. The textbook is ₹1,050, which is ₹1,000 more, and together they are ₹1,100.. The intuitive ₹100 makes the difference ₹900, not ₹1,000. This is the classic cognitive-reflection item: the mind substitutes an easier question and answers that instead. The defence is mechanical — always substitute your answer back into every constraint.

Question 22

Algae on a lake doubles in area every day and covers the whole lake on day 48. On which day is the lake half covered?

Accepted Answer

Day 47.. Exponential growth does almost all its visible work at the end: the lake is a quarter covered on day 46 and 1/1000th covered around day 38. This is why exponential problems — capacity, outbreaks, compounding — look harmless until one step before they are not.

Question 23

Five servers process five jobs in five minutes. How long do a hundred servers take to process a hundred jobs?

Accepted Answer

Five minutes — exactly as long as before.. One server takes five minutes per job, so a hundred servers working in parallel finish a hundred jobs in the same five minutes. The trap is treating the numbers as a ratio to scale rather than deriving the per-unit rate — and it is exactly the error behind "add more servers, halve the time" claims.

Question 24

Ten bottles of identical tablets; one bottle's tablets each weigh a gram more. You have a digital scale and may use it exactly once. Which bottle?

Accepted Answer

Take one tablet from bottle 1, two from bottle 2, and so on up to ten from bottle 10. Weigh the lot. The excess over the expected weight, in grams, is the bottle number.. Encoding identity into quantity turns one measurement into ten distinguishable outcomes. It is the same idea as the binary group-testing trick, and it generalises: when you get one reading, make each candidate contribute a unique, separable amount.

Question 25

A hundred people are held separately and taken to a room with one lamp in random order, indefinitely often. They may agree a strategy beforehand and never communicate again. How can one of them eventually be certain everyone has visited?

Accepted Answer

Appoint a counter. Everyone else turns the lamp on the first time they find it off, and never again. The counter turns it off each time they find it on, counting. At 99, everyone has visited.. The lamp is a one-bit channel, so the strategy must make each bit unambiguous. Restricting others to signalling once and centralising the count in a single person removes double-counting — the general technique of turning a shared, noisy channel into a reliable one by convention.

Question 26

In a dark room are twenty coins, exactly five showing heads. You may move and flip coins but never see them. Split them into two groups with equal numbers of heads.

Accepted Answer

Take any five coins as group A and flip all five. If A originally held h heads, it now holds 5 − h; the other group holds 5 − h too.. If group A takes h of the five heads, the remainder holds 5 − h. Flipping A's five coins turns its h heads into 5 − h. You never needed to know h — the algebra does the work that observation could not.

Question 27

Four people stand in a line facing forward, a wall between the third and fourth. Two black hats, two white. The one behind the wall sees nobody; the front person sees nobody; the others see everyone ahead of them. Who speaks first, and why?

Accepted Answer

The second person. The third sees two hats: if they matched, they would know theirs is the other colour and would speak. Their silence proves the two ahead differ — so the second, seeing the first, knows their own is the opposite.. Again the information is in the silence. A puzzle where everyone is known to be logical makes non-statements as informative as statements — and it collapses entirely if anyone might simply be slow.

Question 28

Three students pay ₹300 for a room, ₹100 each. The manager realises it costs ₹250 and sends ₹50 back; the porter pockets ₹20 and returns ₹10. So each paid ₹90, totalling ₹270, plus the porter's ₹20 is ₹290. Where is the missing ₹10?

Accepted Answer

Nowhere — the sum is nonsense. The students paid ₹270, which is ₹250 to the hotel plus ₹20 to the porter. Adding the ₹20 again double-counts it.. The puzzle works by inviting an addition that has no meaning: money paid and money received are different sides of a ledger and must balance, not sum. The lesson generalises to any accounting or metrics discrepancy — check the sum is meaningful before hunting the difference.

Question 29

Between noon and midnight, how many times do a clock's hour and minute hands overlap exactly?

Accepted Answer

Eleven times.. In twelve hours the minute hand completes twelve laps and the hour hand one, so it gains exactly eleven laps — and each gained lap is one overlap. Counting relative motion rather than absolute positions is the technique, and it is the same reasoning used for orbital and gear problems.

Question 30

Ten people meet and everyone shakes everyone else's hand exactly once. How many handshakes?

Accepted Answer

Forty-five.. Ten people × nine handshakes each = 90, but every handshake involves two people, so divide by two. The n(n−1)/2 formula is the same quantity behind the birthday paradox and behind why communication overhead grows quadratically as a team grows.

Question 31

Remove two opposite corner squares from a chessboard. Can the remaining 62 squares be tiled with 31 dominoes, each covering two adjacent squares?

Accepted Answer

No.. Opposite corners share a colour, so removing both leaves 32 of one colour and 30 of the other. Each domino covers exactly one of each, so 31 dominoes need 31 and 31. A colouring argument proves impossibility without any search — the same trick that settles many tiling and parity problems.

Question 32

Two fuses, each burning for exactly one hour end to end, both unevenly. Measure fifteen minutes.

Accepted Answer

Light fuse A at both ends and fuse B at one end. When A is gone, 30 minutes have passed; immediately light B's second end, and B's remaining 30 minutes burn in 15.. Uneven burning makes length useless but leaves time exact: two ends always halve whatever duration remains. Reasoning about the quantity that stays reliable — here, time — rather than the one you can see is the transferable move.

Question 33

On an island, everyone knows everyone else's eye colour but never their own, and anyone who deduces theirs must leave that night. A visitor says aloud: "At least one of you has blue eyes." If exactly five do, what happens?

Accepted Answer

On the fifth night, all five leave together.. The visitor adds no new fact but makes it common knowledge — everyone now knows that everyone knows, without limit. With one blue-eyed person they leave night one; with two, each waits one night and infers on night two; the induction gives night five for five.

Question 34

A drawer holds black and blue socks in the dark. How many must you take to be certain of a matching pair?

Accepted Answer

Three.. Two socks might be one of each; a third must repeat a colour. This is the pigeonhole principle — with n categories, n+1 items force a repeat — and it is the same argument that guarantees hash collisions once you exceed the output space.

Question 35

A spoonful of milk is stirred into a cup of tea, then a spoonful of that mixture is returned to the milk. Is there more milk in the tea, or tea in the milk?

Accepted Answer

Exactly equal.. Both cups return to their original volume, so whatever quantity of milk is absent from the milk cup must have been replaced by precisely that volume of tea. Conservation settles it in one line; tracking the mixing ratios gives the same answer after a page of algebra.

Question 36

The product of three children's ages is 36 and their sum is the number on the door opposite. The visitor says that is not enough. The parent adds, "the eldest plays chess." Now it is solvable. What are the ages?

Accepted Answer

2, 2 and 9.. Two triples sum to 13 — (1,6,6) and (2,2,9) — so ambiguity identifies the door number as 13. "The eldest" implies a unique oldest child, which rules out (1,6,6). The information is carried by the *failure* to solve, which is the puzzle's real move.

Question 37

Twenty-five candidates, five interview rooms, no timing — each round only ranks the five in it. How many rounds to find the top three overall?

Accepted Answer

Seven.. Five rounds rank each group; a sixth of the five winners fixes the overall first and eliminates whole groups — anyone in a group whose leader placed fourth or fifth cannot be top three. Six candidates remain, one is already known first, and a seventh round settles second and third.

Question 38

Two candles of different thickness both burn for exactly four hours, unevenly. Using only these, measure three hours.

Accepted Answer

Light candle A at both ends and candle B at one end. A is gone after 2 hours; immediately light B's other end, and B's remaining 2 hours burn in 1. Total: 3 hours.. Identical in structure to the fuse puzzle, which is the point — once you recognise "both ends halves the remainder" as a primitive, a whole family of timing puzzles becomes routine. Recognising the shape is faster than re-deriving it.

Question 39

One of a thousand bottles is contaminated, and a test takes a full day to show. You have ten test strips and one day. Find it.

Accepted Answer

Number the bottles 0–999 in binary and let strip k test every bottle whose k-th bit is 1. After a day, the pattern of positive strips read as binary is the bottle number.. Ten parallel one-bit tests span 2¹⁰ = 1024 outcomes, comfortably covering a thousand. Whenever tests are parallel and binary, encode the search space in binary — the same principle behind error-correcting codes and group testing.

Question 40

Three colleagues: one always truthful, one always lying, one answering at random. You may ask three yes/no questions, each to one person, to identify all three. Where do you start?

Accepted Answer

Ask A a self-referencing question — such as "If I asked you whether B is random, would you say yes?" — which forces truth-teller and liar to answer identically. Use the answer to identify someone who is definitely not random, then question that person directly.. Nesting a question makes the liar lie twice, so the double negation cancels and both reliable people answer alike. Constructing a channel that behaves identically regardless of which reliable source you hit is the general technique — the same idea as the two-guards puzzle, generalised.

Question 41

Two people, one cake, no trust and no ruler. How do you divide it so neither can complain?

Accepted Answer

One cuts, the other chooses.. The cutter is motivated to divide evenly because they get whichever piece is left. The mechanism removes the need for trust or measurement entirely — the same design principle behind escrow, sealed bids and separation of duties.

Question 42

Five ranked colleagues split a hundred units of budget. The most senior proposes a split; if at least half agree it stands, otherwise they are removed and the next proposes. Everyone is perfectly rational and prefers more. What does the most senior propose?

Accepted Answer

98 for themselves, 0 to the second, 1 each to the third and fifth (or an equivalent minimal bribe of two others).. Backward induction: with two left the senior always wins, so the third-from-last would get nothing and will accept a single unit. Bribing the cheapest voters with one unit each secures the majority. Reasoning from the terminal state outward is the standard method for sequential games.

Question 43

Everyone in a large group picks a number from 0 to 100. The winner is whoever is closest to two-thirds of the group average. What do you pick?

Accepted Answer

In theory 0; in practice something around 20–30, depending on how sophisticated you judge the group to be.. Level-0 players pick 50, level-1 picks 33, level-2 picks 22, and infinite iteration converges to 0 — the Nash equilibrium. Real groups stop after one or two levels, so the winning answer depends on modelling others' depth of reasoning rather than on the maths.

Question 44

Two envelopes; one holds twice the other. You open one and find ₹1,000. Switching seems to offer an expected ₹1,250, which suggests you should always switch — even before opening. What is wrong?

Accepted Answer

The reasoning assumes a uniform prior over all amounts, which cannot exist. With any real distribution, seeing ₹1,000 tells you something about which envelope you hold, and the symmetric always-switch conclusion evaporates.. No probability distribution over the positive reals is uniform, so "equally likely to be double or half" cannot hold everywhere. The paradox is a reminder that an expected-value calculation is only as sound as the prior it quietly assumed.

Question 45

Twelve identical-looking batches, one entirely counterfeit and lighter per unit. A digital scale, one weighing. Which batch?

Accepted Answer

Take 1 from the first batch, 2 from the second, and so on. Weigh together; the shortfall from the expected total, divided by the per-unit difference, gives the batch number.. Same encoding trick as the tablets puzzle, in subtraction rather than addition. Once identity is encoded in quantity, a single scalar reading carries as much information as many separate tests.

Question 46

A vehicle carries fuel for 500 km but must cross 800 km of desert. Fuel may be cached along the way, and only this vehicle exists. Can it cross?

Accepted Answer

Yes. Make repeated shuttle runs to build a fuel dump partway out — for instance ferrying fuel to the 100 km mark over several trips — then make the final run from the dump with a full tank.. Caching converts range into trips: each unit of fuel moved forward costs fuel to carry, so effective range grows sub-linearly with effort. It is the same accounting as a rocket's staging problem, and the reason exploration logistics are dominated by supply rather than distance.

Question 47

A machine in a sealed room is controlled by one of four switches outside. You may enter once. Identify the switch.

Accepted Answer

Leave switch 1 on for ten minutes then off. Turn on switch 2 and enter. Running = 2; stopped but warm = 1; a third switch can be left on only briefly to leave it barely warm, and cold and stopped = 4.. One observation seems to give one bit, but temperature adds a graded dimension — hot, warm, cold — turning a binary look into a multi-valued reading. Finding a second, slower-changing channel is the recurring trick in single-observation puzzles.

Question 48

Every student in a lecture theatre reports that their row is more crowded than average. All of them are telling the truth. How?

Accepted Answer

Averaging over students is not averaging over rows. Crowded rows hold more students, so more people experience crowding than the per-row average suggests.. This is the inspection paradox, and it explains why your bus always seems late and your class always seems large: sampling weighted by experience over-represents the crowded cases. Whenever an average is disputed, ask what population it was taken over.

Question 49

A number is doubled, increased by ten, halved, and the original number subtracted. What is left?

Accepted Answer

Five, always.. (2x + 10)/2 − x = x + 5 − x = 5. The starting number cancels, which is why every "think of a number" trick works: the operations are chosen so the unknown eliminates itself. Writing the algebra reveals in one line what testing examples only suggests.

Question 50

A test for a condition affecting 1 in 1,000 people is 99% accurate. You test positive. What is the chance you have it?

Accepted Answer

About 9%.. In 100,000 people, 100 have it (99 detected) and 99,900 do not (about 999 false positives). So roughly 99 of 1,098 positives are real. Base rates dominate when a condition is rare, which is why screening a low-prevalence population produces mostly false alarms.

Question 51

You reach a fork. One road leads home, one does not. A local stands there who either always lies or always tells the truth, and you cannot tell which. One question.

Accepted Answer

Ask: "If I asked you whether the left road leads home, would you say yes?" Both answer truthfully about the destination, because the liar's two lies cancel.. Self-referencing the question forces double negation, so both types produce the same reliable answer. It is the single-person version of the two-guards puzzle, and the general lesson is that you can build a trustworthy channel out of an untrustworthy one if its unreliability is consistent.

Question 52

You must pay a contractor one link of a seven-link chain each day for a week, settling daily. What is the fewest links you may cut?

Accepted Answer

One — cut the third link. That gives a single link, a two-link piece and a four-link piece, and every daily total from 1 to 7 can be made by handing over and taking back.. Cutting one link produces three usable pieces of sizes 1, 2 and 4 — the binary weights that combine to any total up to 7. Allowing exchange rather than only giving is what makes one cut sufficient, and it is the same reasoning as making change with binary denominations.

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