Every free question on Seating Arrangement, Ranking & Puzzles (Reasoning) from past WBCS Preliminary Examination and WBCS Main Examination papers, newest first, each with its answer and a short explanation. Free to read.
Answer: (D) 28th
Rank from the last = total − rank from the top + 1 = 35 − 8 + 1 = 28th. 27th forgets to add 1, since the student is counted in both rankings.
Answer: (A) D, C, A, B
C is older than D but younger than B, so B > C > D. A is "older but not eldest", i.e. A is older than C yet younger than B, giving B > A > C > D. In ascending order of age (youngest first) this is D, C, A, B. Check the other options: (B) puts C before D although C is older; (C) makes A the eldest, which is ruled out; and (D) starts with the eldest, B, so it is not ascending at all. In such questions always note whether ascending or descending order is asked.
Answer: (D) 44
B did better than A and exactly two students stand between them, so B's rank from the top is $10-3=7$. B is thus 7th from the top and 38th from the bottom, and the total number of students is $7+38-1=44$. The general rule is: total = (rank from top) + (rank from bottom) − 1, because the person concerned is counted twice. Had B been worse than A, B would have been 13th from the top and the total would have been 50, which is not among the options.
Answer: (C) 7
Total teaching periods per day = $3\times6$ (Classes VI, VII, VIII) $+\ 2\times7$ (Classes IX, X) $=18+14=32$. Each teacher can take at most 5 periods, so the number of teachers needed is at least $32\div5=6.4$, i.e. 7 after rounding up. Six teachers could cover only 30 periods, which is short by 2. Seven teachers can cover 35 periods, and since only five classes run at any one time, a workable timetable exists. In "minimum number" questions always round up, never down.
Answer: (C) (C, E)
Number the six seats 1 to 6 round the table. Put B at seat 1; A sits opposite at seat 4. B is between C and D, so C and D occupy seats 2 and 6; A is between E and F, so E and F occupy seats 3 and 5. E is next to D (to D's left), so if D is at seat 2, E is at seat 3, which leaves F at seat 5 and C at seat 6. Opposite pairs in a hexagon are 1–4, 2–5 and 3–6, i.e. A–B, D–F and C–E. Hence C and E face each other. The mirror arrangement (D at 6, E at 5) gives the same opposite pairs.
Answer: (D) 12
Follow the cards step by step, starting with 10 each. E borrows 3 from A: A = 7, E = 13. E gives 5 to C: E = 8, C = 15. C gives 4 to B: C = 11, B = 14. B passes 2 to D: B = 12, D = 12. So D has 12 cards. Final tally: A 7, B 12, C 11, D 12, E 8, which adds up to 50, the same as the starting total — a useful check. D takes part in only one transaction, receiving 2 cards, so $10+2=12$ can be read off directly.
Answer: (C) D
List each girl's games from the two statements in which she appears. A: football, volleyball, badminton. B: football, volleyball, cricket. C: cricket, football, badminton. D: badminton, cricket, volleyball. Only D plays badminton, cricket and volleyball, and D is option (C). A quicker way: the girl wanted does not play football, and the only two statements that leave out football are "C and D play badminton and cricket" and "D and A play volleyball and badminton"; D is common to both. (In the booklet the girls' names carry a bar over the letter.)
Answer: (B) C
Tall: A, D, E, so the short girls are B and C. Dark: B, D, E, so the fair girls are A and C. Fat: C and E, so the thin girls are A, B and D. The girl who is short and fair and fat must be in all three lists: short (B, C), fair (A, C) and fat (C, E). Only C is common, so the girl at the first desk is C, option (B). For reference, the full descriptions are: A – tall, fair, thin; B – short, dark, thin; C – short, fair, fat; D – tall, dark, thin; E – tall, dark, fat.
Answer: (D) 3
The middle of 15 chairs is the 8th chair, so P is 8th from either end. Q is 12th from the right, i.e. $15 - 12 + 1 = 4$th from the left. Between the 4th and the 8th chairs lie the 5th, 6th and 7th, i.e. $8 - 4 - 1 = 3$ chairs. Formula to remember: position from left = total - position from right + 1.
Answer: (C) D
From the clues: B > D > (C, E) > A. Since B is tallest and D is next, D is the second tallest person.
Answer: (B) 21
In a knock-out format, exactly one player is eliminated per match and all but the champion must be eliminated, so with 22 participants, 21 matches are needed.
Also asked as Prelims 2018
Answer: (C) Chirag and Ela
Placing all six around the hexagon per the given clues shows that Chirag and Ela end up directly opposite each other.
Answer: (B) 2nd
Overtaking the runner who was in 2nd place means you take over that position — you move into 2nd place, not 1st.
Answer: (C) 25
After the swap, A's new position (18th from left) equals B's original position, so B was originally 18th from left, meaning the row has (10 + 18 - 1) = 27 people in total, or 25 excluding A and B.
Answer: (C) Joseph
Since Aries is the oldest and John is not the youngest, by elimination Joseph must be the youngest.
Answer: (B) The 71st day
The caterpillar makes a net gain of 1 inch per full day-night cycle, reaching 70 inches after 70 days; on the 71st day, its daytime climb of 5 inches takes it from 70 to 75 inches, reaching the top before it can slide back.
Answer: (B) B
Let $A+C = B+D = k$. Since $A = \dfrac{1}{2}(B+D) = \dfrac{k}{2}$, we get $C = k - A = \dfrac{k}{2}$, so $A=C=\dfrac{k}{2}$. Also, since $A+B>C+D$ and $A=C$, it follows that $B>D$. Since $B+D=k$ and $B>D$, we must have $B>\dfrac{k}{2}$, so $D<\dfrac{k}{2}$. Putting it together: $D < A = C < B$, so B's income is the highest.
Answer: (C) Two
Scanning the sequence in overlapping triplets, exactly two starting numbers satisfy all three conditions (odd, divisible by 3 or 5, immediately followed by an odd number, which is in turn immediately followed by an even number): the triplet 3-25-18 and the triplet 21-45-46, giving two qualifying cases.
Answer: (B) Two
Checking each occurrence of the digit 3 in the sequence against the two exclusion conditions (immediately preceded by 6, or immediately followed by 9) shows that exactly two of the 3s satisfy neither condition and so qualify.
Answer: (B) E
For people facing the centre, 'left' means clockwise and 'right' means anticlockwise. Fix F at seat 0 and number the seats clockwise: D is second to the left of F (seat 2) and third to the right of H, so H is at seat 5; A is second to the right of F (seat 6) and is next to H. F is third to the right of B, so B is at seat 3, and C, second to the right of B, is at seat 1. G cannot sit next to F, so G takes seat 4 and E takes seat 7. Clockwise order: F, C, D, B, G, H, A, E - so E is to the immediate left of A.
