Puzzle Test Questions and Answers

Answer: B

Two boxes lie between F and D. Number the positions 1 to 6 from the top. D is second last, so D is 5. Three boxes are between D and C, so C is 1. Four boxes are between C and G, so G is 6. One box is between D and E, so E is 3. The remaining positions 2 and 4 go to F and H, and F is above H, so F is 2 and H is 4.

Between F (2) and D (5) are positions 3 and 4, so two boxes. One and three miscount that gap, and zero would need F and D to be adjacent.

Answer: A

Yashika was born in 2000. Nitu and Monika are 14 years apart and Monika was born before 1993, so Monika is 1990 and Nitu is 2004. Saanvi is 3 years younger than Monika, so Saanvi is 1993. The years left are 1997, 2000 and 2002. Divya is 5 years older than Tania, which fits only 1997 and 2002, so Divya is 1997 and Tania is 2002. That leaves 2000 for Yashika.

1993 is Saanvi's year, 2002 is Tania's and 1997 is Divya's.

A check on the years: Nitu (2004) is 14 years younger than Monika (1990), Saanvi (1993) is 3 years younger than Monika, and Divya (1997) is 5 years older than Tania (2002).

Answer: D

Only four persons live between Cheema and Deepanshu. Bindu is just below Cheema and both are above floor 4, so Bindu is on 5 and Cheema on 6. Cheema is three floors above Aarush, so Aarush is on 3. Gopal is two floors below Harish; of the free floors 1, 2 and 4 the only pair two apart is Gopal on 2 and Harish on 4, which leaves Deepanshu on 1, not adjacent to Aarush as required. Floors 2, 3, 4 and 5 lie between 1 and 6, so four persons live between Cheema and Deepanshu. Harish is two floors, not one, above Gopal. Aarush (3) is not just above Bindu (5), and only one person (on floor 4) lives between Aarush and Bindu.

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.