Puzzle Test Questions and Answers

Answer: C

Take the conditions one at a time. A's floor is immediately higher than F's, and C's floor is immediately higher than A's, so F, A and C occupy three consecutive floors in that order going up. B is immediately below F, so B comes just under F. E is lower than B, so E is at the bottom. From the bottom the order is E, B, F, A, C, and since C is lower than D, D must be above C on the sixth floor. D is therefore on the top floor, option (C). F, E and C all lie below D, so they cannot be on the top floor.

Answer: B

The box fourth from the bottom is 1.

Box 3 is second from the bottom and above box 5, so 5 is at the bottom and 3 is second. Box 7 is just below box 4, and 7 is above 2. Box 6 is above 2 with exactly one box between them. The only arrangement that works from the bottom is 5, 3, 2, 1, 6, 7, 4: 6 is two places above 2, 7 is above 2 and 4 is directly above 7.

The fourth place from the bottom therefore holds box 1. Boxes 5, 2 and 7 sit at the first, third and sixth places.

Placing the fixed facts first makes it quick: 5 at the bottom, 3 second, then the pair 4 over 7 as a block, with 2 below 7, and 6 two places above 2. Trying 2 in the fourth place from the bottom leaves no room for 6, so 2 must be in the third place and box 1 fills the fourth.

Answer: B

The true statement is B likes Nokia.

A lives on an odd floor above 2, so on floor 3, and likes Redmi (not iPhone, Samsung or Nokia). One person lives above A, so one person is below D, which puts D on floor 2. B is then on floor 1 and C, who lives above B and likes iPhone, is on floor 4. D does not like Nokia and the iPhone and Redmi are taken, so D likes Samsung, leaving Nokia for B.

D likes Samsung, C lives on the 4th floor and A lives on the 3rd floor, so the other statements are false.

Check the answer against the other statements: D likes Samsung, not Redmi, so the statement 'D likes Redmi' is false. C lives on the top floor, the 4th, and not the 2nd. A lives on the 3rd floor, not the 1st. Only 'B likes Nokia' is true.

Answer: D

The house number is 76.

Each statement is an 'if ... then' rule, so a number passes a rule automatically when the 'if' part is false. Test the options:

• 76: not a multiple of 3, so rule i does not apply; a multiple of 4, so rule ii does not apply; not a multiple of 6, so rule iii requires 70 to 79, and 76 is. All three hold.
• 54: not a multiple of 4, so rule ii demands 60 to 69. It fails.
• 65: not a multiple of 6, so rule iii demands 70 to 79. It fails.
• 66: a multiple of 3, so rule i demands 50 to 59. It fails.

Answer: C

The southernmost hill is H3.

By (i) and (ii), neither H2 nor H3 is at either end of the west-east order, so H1 and H4 occupy the two ends and H2 and H3 the middle two places. By (iv), two hills lie west of H2, so H2 is third from the west and H3 is second from the west.

By (iii), the southernmost hill is not at either end, so it is H2 or H3. By (v), it has at least two hills to its east. H2, third from the west, has only one hill to its east, while H3 has two. So H3 is the southernmost.

H1 and H4 are ruled out by (iii), and H2 by (v).

Answer: C

The youngest person is R.

Write ages with \(\lt\) meaning 'younger than'. P's statement gives \(P \lt S\). R's statement, that P is older than R, gives \(R \lt P\). Together, \(R \lt P \lt S\). Q is neither the youngest nor the oldest, so Q falls somewhere between R and S, and the order from youngest is R, then P and Q in some order, then S.

Either way R is the youngest. P cannot be youngest because R is younger than P; Q's own statement rules Q out; and S is older than P, so S is if anything the oldest.

Answer: D

Neither can be inferred, so the answer is neither (i) nor (ii).

From the statements: P always beats Q, Q always beats R, and R always beats S. P beats S only sometimes, so S sometimes beats P.

(i) is not certain: P always beats Q, but nothing says P ever plays R or how often P beats R, and P loses to S part of the time, so 'likely to beat all three' is not established. (ii) is not certain either: S always loses to R, but S sometimes beats P, who always beats Q, so S is not simply the worst player. The chain of 'always' results is broken by statement III, which is the point of the question.

Answer: B

The correct statement is S cannot be placed at a corner.

Among the six points, two lines are parallel only when one joins the mid-points of two sides and the other lies along the third side. Suppose S were a corner. Then QS lies along a side through S, so P and R must be the mid-points of the other two sides, one of which is the side opposite S. U must not share a side with S, and the only point not on a side through S is the mid-point of the side opposite S, which is already P or R. So there is no place for U, and S cannot be a corner.

The other options are possible. Take corners X, Y, Z with T = X, Q and S the mid-points of XY and XZ, and P = Y, R = Z, U the mid-point of YZ. Every condition holds, yet P and R sit at corners and U at a mid-point, so options A, C and D are false.

Answer: A

The correct statement is The line segment joining R and T is parallel to the line segment joining Q and S.

In a regular hexagon there are three lengths: side, short diagonal and long diagonal. A long diagonal is perpendicular only to short diagonals, and a short diagonal is perpendicular to sides and long diagonals. Trying the possibilities, RS longer than and perpendicular to PQ, together with RU parallel to TQ, can only be met with RS a long diagonal and PQ a short diagonal.

Up to rotation and reflection the order round the hexagon is then forced: P, S, Q, U, R, T. Check: R and S are opposite; P and Q sit either side of S, so PQ is a short diagonal perpendicular to RS; RU is a side and TQ is the long diagonal parallel to it.

In this order RT and QS are opposite sides, so they are parallel. TQ and PU are two different long diagonals, so they meet at the centre and are not parallel. RP is a short diagonal, and UQ and QS are sides that each meet its direction at 30 degrees, not 90, so the two perpendicularity claims fail.

Answer: B

The hardest pair is learnt by Q.

Chain the comparisons from easiest to hardest: Spanish is easier than French, French easier than Dutch, Dutch easier than Japanese, and Japanese easier than Chinese:

Spanish \(\lt\) French \(\lt\) Dutch \(\lt\) Japanese \(\lt\) Chinese

Q studies Chinese and Japanese, the two hardest languages, so Q's pair is the most difficult.

S studies Dutch and Japanese, the next-hardest pair; P has French and Dutch; R has Spanish and French, the easiest pair of all.