Deductive reasoning

Lay them out on a line in order of speed. E is faster than C, C than A, A than B, and B than D, so the single line E-C-A-B-D is fixed. The fourth is B. Choosing C or A comes from reading the direction of a condition the wrong way round: C is second and A is third. D is last.

Lining them up from the left in order of height: R is taller than Q, Q than P, and P than S, giving R-Q-P-S. The second is Q. R is the tallest, P is third and S is the shortest. Choosing P comes from misreading 'S is shorter than P' as making P second.

In order from the top the chain is C-B-A-D-E. C is above B, B above A, A above D and D above E, so the lowest is E. Choosing D comes from overlooking 'D scored higher than E'. A is third and B is second.

D is fixed at 5th. The pair of places for A and C is 1st and 3rd, 2nd and 4th, or 3rd and 5th, but 5th is D, so 3rd and 5th is out. With 1st and 3rd, A would be 1st and there would be no room for B above, so A is 2nd, C is 4th, B is 1st, and E, who is left, is 3rd. Choosing 4th comes from mixing up E and C.

In order of weight the chain is W-X-Y-Z. X is heavier than Y and Y than Z, so X is second. Choosing W comes from answering with the heaviest box. Y is third and Z is the lightest.

The run A-B-C-D is fixed, but E may be slotted in anywhere among those four, so five orderings satisfy the conditions. Since B is above C and C is above D, transitivity puts B above D in all five. 'A is 1st' is contradicted by the ordering in which E is 1st, 'E is 5th' by that same ordering, and 'C is 4th' by any ordering that puts E after C.

The order A-B-C-D is fixed, and the remaining person E can go into any of five positions, giving five orderings in all. Only E can come above A, so A is either 1st or 2nd, which is 2 possibilities. Answering 1 forgets the case where E slots in ahead of A. Answering 3 or more overlooks that the order A-B-C-D is fixed.

With the original conditions alone there are 6 orderings. Checking them all, only adding 'A is above D and B is above C' narrows it to the single ordering A-B-C-D. 'A is above C' (from which A above D already follows), 'B is 4th' and 'C is 1st' each still leave 3 orderings, so none of them settles it. The trap is assuming that adding any one condition must settle the matter.

A scored 9, so B scored 6. C and D together make 30 minus 15, that is 15, and C is 1 more than D, so D scored 7 and C scored 8. Trying every value from 0 to 10 gives no other combination. Answering 7 gives D's score and answering 6 gives B's.

A is 2nd. D is above B and above C, and B is above E, so D is above every one of B, C and E; since 2nd is A, D is fixed in 1st. B, C and E then fill 3rd to 5th under the single condition that B is above E, which allows 3 orderings, so counter-examples can be built against 'B is 3rd', 'C is 5th' and 'E is 5th' alike.

In order of height the chain is S-P-Q-R-T. S is taller than P, and P is taller than Q and R, so S is the tallest. Choosing P comes from overlooking 'S is taller than P'. T is the shortest.

Two runners in between means the gap in placings between A and B is 3. D is 1st, so A and B can only be the pair 2nd and 5th. If A were 2nd there would be no room for C ahead of A, so B is 2nd and A is 5th. C and E fill 3rd and 4th, and since C is ahead of E, C is 3rd and E is 4th. Counting the gap as 2 makes this pairing impossible and loses the answer.

B has 1000 yen, so A has 2000 yen, C has 500 yen less than A, that is 1500 yen, and D has 300 yen more than B, that is 1300 yen. In descending order that is A, C, D, B, so the second is C. Choosing A answers with the first; choosing D comes from subtracting 500 yen from B instead of from A when working out C.

C is 3rd and D is 2nd, and A is above C, so A is fixed in 1st. B, E and F fill 4th to 6th, with E above F. The pair E and F can be 4th and 5th, 4th and 6th, or 5th and 6th, so F is 5th or 6th. If F were 4th there would be nowhere to put E, so any option including 4th is wrong. '5th only' overlooks the ordering with E 4th and F 6th.

Put a circle at B and tennis, then cross out the rest of the tennis column and the rest of B's row. A plays neither baseball nor tennis, so A plays soccer. C, who is left, is fixed on baseball. Drawing the grid settles it uniquely, so the 'cannot be settled' option is wrong.

P is from Tokyo, so Tokyo, Osaka and Fukuoka are all ruled out for Q, who must be from Hokkaido. R is not from Fukuoka, and Hokkaido and Tokyo are taken, so R is from Osaka. S, who is left, is from Fukuoka. Answering Osaka comes from mixing up R and S.

Number the positions 1 to 5 from the left: C is at 3, A at 4 and B at 5. D and E fill 1 and 2, and since D is to the left of E, D is at 1 and E at 2. The left end is D. Choosing E comes from getting left and right the wrong way round.

A likes red, so blue belongs to B or C. Since C does not like mandarin oranges, C cannot be the blue one without breaking the rule that the blue one likes mandarin oranges, so B is blue and likes mandarin oranges. C is therefore green. A does not like grapes, so A likes the apple, and the remaining grapes go to C. Blue and mandarin orange is B's combination, and red is A's colour, so both are wrong.

Y keeps the fish, so W keeps neither dog nor cat nor fish and must keep the bird. X keeps either the cat or the bird, and the bird is W's, so X keeps the cat. Z, who is left, keeps the dog, which also fits the condition that Z does not keep a bird. Answering bird comes from mixing up W and X.

A is in 4, B in 3, F in 5 and D in 7. C is to the left of B, so C is in 1 or 2, and E is further left still, so E is fixed in 1 and C in 2. The remaining seat 6 is G's. Answering 5 gives F's seat and answering 2 gives C's.

Monday is B and Friday is A, so C, D and E fill Tuesday, Wednesday and Thursday. C coming the day after D allows Tuesday-Wednesday or Wednesday-Thursday, but D is not on Tuesday, so D is Wednesday and C is Thursday. The remaining Tuesday is E's. Answering D comes from judging on the day-after condition alone and forgetting the one that rules Tuesday out.

Each plays 3 matches. A is 3 wins 0 losses and D is 0 wins 3 losses. B loses to A and beats C and D, so B is 2 wins 1 loss. C loses to A and B and beats D, so C is 1 win 2 losses. Laying out the results of all 6 matches gives no other possibility. '0 wins 3 losses' confuses C with D.

The person from Akita is in the swimming club, so Akita is ruled out for C (judo), for D (not swimming) and for B (Yamagata), leaving A from Akita in the swimming club. D is not from Aomori either, so D is from Iwate, and the remaining Aomori goes to C. D, from Iwate, is not in the soccer club, and swimming and judo are taken, so D is in the baseball club, and the remaining soccer goes to B. 'Iwate and soccer' skips over the Iwate condition.

A is on 5, D on 1 and E on 4, so B and C are on 2 and 3. B is one floor above C, so C is on 2 and B on 3. Choosing B comes from getting above and below the wrong way round; B is on 3.

A is top left, so B directly below is bottom left; C is bottom right; D immediately to B's right is bottom middle; and E, in the same right-hand column as C but in the top row, is top right. The remaining top middle is F's. The middle of the bottom row is D's place and the right end of the top row is E's, so both are wrong.

C wears green, so A wears neither red nor blue nor green and is fixed on yellow. B and D split the remaining red and blue, but which takes which is not settled, so there are 2 possibilities. Counter-examples can therefore be built against 'B wears red', 'B wears blue' and 'D wears blue' alike. The only certainty is that A wears yellow.

If B is good at Japanese, then A is good at neither mathematics nor Japanese and so at English, and the remaining mathematics goes to C, settling it uniquely. Adding 'C is good at mathematics', 'B is not good at mathematics' or 'A is good at Japanese' each leaves the other two subjects interchangeable, so 2 possibilities remain. Counting the solutions exhaustively makes the difference plain.

Assume each in turn to be the culprit and count the truthful speakers. If A did it, B and D speak the truth, giving 2. If B did it, 3 speak the truth. If C did it, 2 do. If D did it, only A among the four speaks the truth, giving 1. Only D fits the condition. Choosing A comes from believing B's statement alone, and in that case 2 people speak the truth, which breaks the condition.

Try all 8 combinations of honest and liar for the three. Only one fits: A a liar, B honest, C a liar. Then A's statement is false, B's statement is true because C is a liar, and C's statement is false because B is honest, so everything is consistent. If A and C were both honest, C's statement 'both A and B are liars' would make A a liar, contradicting A being honest.

Let white be x; then red is x+3 and blue is 2x, so the total is 4x+3. Setting that equal to 19 gives x=4. There are 4 white, 7 red and 8 blue. Answering 4 gives the white count and 8 the blue count. Trying every value exhaustively gives no other set that totals 19.

Let R be x; then Q is x+3 and P is x+5, so the total is 3x+8. Setting that equal to 20 gives x=4. R has 4, Q has 7 and P has 9, which totals 20 as required. Answering 7 gives Q's count and 9 would be P's. Answering 5 comes from adding one of the differences the wrong way.

The contrapositive of 'if you are in the tennis section then you like sport' is 'if you do not like sport then you are not in the tennis section', which always holds. 'Everyone who likes reading likes sport' is the converse, and cannot be claimed, since there may be a reader who dislikes sport. 'Every member likes reading' and 'there is a member who likes reading' are both refuted by the case where the readers are all outside the tennis section.

Chaining the two conditions, coming by train means holding a season ticket, and holding a season ticket means not cycling in, so 'a pupil who comes by train does not cycle in' always holds. 'Every pupil holding a season ticket comes by train' is the converse, and so is 'every pupil without a season ticket cycles in'. Thinking of a single pupil who walks in gives a counter-example to the option that not cycling implies coming by train.

The five together total 72 times 5, that is 360 marks, and the four excluding E total 70 times 4, that is 280 marks. The difference, 80 marks, is E's score. Answering 74 comes from adding the 2-mark gap in the averages just once instead of once per person. The rule is to convert to totals first and then subtract.

C scored 3, so A, B and D total 27. B is 5 more than A, so twice A plus 5 plus D is 27, that is, D is 22 minus twice A. D reaches 10 or more only when A is 6 or less; if A were 4 then D would be 14, breaking the highest score of 12, and if A were 6 then the highest would be 11, breaking the condition. That leaves A at 5, with B at 10 and D at 12, which satisfies the total of 30, the highest of 12 and the lowest of 3. Answering 7 comes from forgetting the condition that D scored 10 or more.

Assume each in turn to be the honest one. If A, then A's statement makes B honest too, giving 2, a contradiction. If B, then the liar A's statement that B is honest would be true, a contradiction. If D, then the liar B's statement that C is a liar would be true, a contradiction. Only with C does everything fit: A's, B's and D's statements are all false and only C's is true.

D is below C, so C is always somewhere above D, and D can therefore never be 1st. 'C is 1st' is refuted by an ordering with A above C. 'E is 5th' is refuted by an ordering in which B is last. A and B are not linked by any condition, so orderings exist with either above the other. Listing every ordering that satisfies the conditions confirms that the only thing always true is that D is not 1st.

With twice as many red as blue and 3 yellow, three times the blue makes 12, giving the single solution of 4 blue, 8 red and 3 yellow. With 5 yellow there are 4 ways to split the remaining 10 between red and blue; with only '2 more red than blue' there are 6; and with blue and yellow equal there are 4. Counting the solutions exhaustively shows only one option narrows it to a single case.

A is above E and E is above D, so the run A-E-D is the backbone. C is above D, so there are 3 places to slot C into that run of three: at the front, between A and E, or between E and D. For each of those, B goes into a gap after A: with C at the front there are 3 ways, and in the other two shapes 4 ways each, making 11 in all. Counting exhaustively also gives 11.

Let red be x; then black is 3x and blue is x+4, so the total is 5x+4. Setting that equal to 24 gives x=4. There are 4 red, 12 black and 8 blue, which totals 24. Answering 6 comes from forgetting to add the 4 extra blue and solving 3x + x = 24. Answering 5 comes from misreading black as twice red and solving 2x + x + (x + 4) = 24.