Arrangements, ranking, ordering and puzzles: previous-year questions
141 real UPSC Prelims questions (2011–2026) with the official answer and a clear explanation. 35 of them came in the last five papers.
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Questions per year, counted from the official papers.
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Consider the following statements :
- 1. A is older than B.
- 2. C and D are of the same age.
- 3. E is the youngest.
- 4. F is younger than D.
- 5. F is older than A.
How many statements given above are required to determine the oldest person/persons ?
Answer (d) & explanation
All five statements are needed (d). Step 1: put them in a chain: C = D (2), D > F (4), F > A (5), A > B (1), and E is the youngest (3). So C = D > F > A > B, and E is last. Step 2: to call C and D the oldest, we must show they beat everyone. B is beaten only through statement 1. E is placed only by statement 3. Step 3: C is linked to D only through statement 2. Drop any one and someone's position is unknown. So all five are needed. Options (a), (b) and (c) fail.
Tip: to name the oldest, every other person must be placed below. Check each person is covered by at least one statement.
Answer as per the official UPSC answer key.
What is the sum of all 4-digit numbers less than 2000 formed by the digits 1, 2, 3 and 4, where none of the digits is repeated ?
Answer (a) & explanation
The sum is 7998 (a). Step 1: below 2000, the first digit must be 1. The other digits 2, 3 and 4 can be arranged in 3 × 2 × 1 = 6 ways. Step 2: thousands place: 6 numbers × 1000 = 6000. Step 3: each of 2, 3, 4 sits twice in each of the other three places: (2 + 3 + 4) × 2 × 111 = 1998. Total = 6000 + 1998 = 7998. Options (b), (c) and (d) miscount.
Tip: for sums of arranged numbers, handle the fixed digit separately, then use (sum of digits) × (times each appears) × 111.
Answer as per the official UPSC answer key.
In a tournament of Chess having 150 entrants, a player is eliminated whenever he loses a match. It is given that no match results in a tie/draw. How many matches are played in the entire tournament?
Answer (c) & explanation
149 matches are played (c). Step 1: every match has a loser, and every loser is knocked out. No match is drawn. Step 2: in the end only one champion is left. So 150 − 1 = 149 players must be knocked out. Step 3: one match per knocked-out player means 149 matches. Option (b), 150, counts the champion too. Options (a) and (d) are off by one in other ways.
Tip: in any knockout tournament with n players, the number of matches is n − 1, whatever the number of rounds or byes.
Answer as per the official UPSC answer key.
There are 9 cups placed on a table arranged in equal number of rows and columns out of which 6 cups contain coffee and 3 cups contain tea. In how many ways can they be arranged so that each row should contain at least one cup of coffee?
Answer (d) & explanation
There are 81 ways (d). Step 1: the cups form a 3 × 3 grid. Placing the 3 tea cups decides everything. The number of ways to pick 3 places from 9 = 9 × 8 × 7 ÷ (3 × 2 × 1) = 84. Step 2: a row has no coffee only if all 3 tea cups fill that one row. This can happen in 3 ways (one per row). Step 3: 84 − 3 = 81. Options (a), (b) and (c) miscount.
Tip: for "at least one" conditions, count all cases and subtract the bad ones. It is usually easier than counting the good ones directly.
Answer as per the official UPSC answer key.
How many 3-digit natural numbers (without repetition of digits) are there such that each digit is odd and the number is divisible by 5 ?
Answer (b) & explanation
There are 12 such numbers (b). Step 1: a number divisible by 5 ends in 0 or 5. Since every digit must be odd, the last digit must be 5. Step 2: the first two digits come from the other odd digits: 1, 3, 7, 9. No repeats allowed. Step 3: the first digit has 4 choices and the second has 3. So 4 × 3 = 12. Option (a), 8, and the larger options (c) 16 and (d) 24 miscount.
Tip: fill the most restricted place first (here, the last digit). Then count the choices for the rest.
Answer as per the official UPSC answer key.
What is the number of numbers of the form 0.XY, where X and Y are distinct non-zero digits?
Answer (a) & explanation
There are 72 such numbers (a). Step 1: X can be any non-zero digit: 9 choices (1 to 9). Step 2: Y must be non-zero and different from X: 8 choices. Step 3: 9 × 8 = 72. Option (b), 81, allows X and Y to be the same. Options (c) and (d) include zero, which the question rules out. Note that X and Y are only the two decimal places, so 0.XY always has exactly two digits after the point.
Tip: "distinct" removes one choice for the second place. "Non-zero" removes 0 from both places.
Answer as per the official UPSC answer key.
The letters A, B, C, D and E are arranged in such a way that there are exactly two letters between A and E. How many such arrangements are possible?
Answer (c) & explanation
There are 24 arrangements (c). Step 1: five places in a row. With exactly two letters between them, A and E can be at places (1, 4) or (2, 5). Step 2: in each pair, A and E can swap. So 2 pairs × 2 = 4 ways to place A and E. Step 3: the other three letters fill the three free places in 3 × 2 × 1 = 6 ways. Total = 4 × 6 = 24. Options (a), (b) and (d) miss the swap or the extra pair.
Tip: place the restricted letters first, remember they can swap, then arrange the rest freely.
Answer as per the official UPSC answer key.
A, B and C are three places such that there are three different roads from A to B, four different roads from B to C and three different roads from A to C. In how many different ways can one travel from A to C using these roads?
Answer (c) & explanation
There are 15 ways (c). Step 1: going A to B and then B to C: 3 × 4 = 12 ways. For "and then", multiply. Step 2: going directly from A to C: 3 ways. Step 3: the two routes are alternatives, so add: 12 + 3 = 15. Option (a), 10, adds all the numbers. Option (b), 13, misses some routes. Option (d), 36, multiplies everything.
Tip: "and" (one step after another) means multiply; "or" (choose one route) means add.
Answer as per the official UPSC answer key.
Three persons A, B and C are standing in a queue not necessarily in the same order. There are 4 persons between A and B, and 7 persons between B and C. If there are 11 persons ahead of C and 13 behind A, what could be the minimum number of persons in the queue?
Answer (a) & explanation
The minimum is 22 persons (a). Step 1: 11 people are ahead of C, so C is 12th. Step 2: 7 people stand between B and C. So B is 4th or 20th. To keep the queue short, take B = 4th. Step 3: 4 people stand between A and B. A cannot be before 1st, so A is 9th. 13 people are behind A, so the queue has 9 + 13 = 22. C at 12th lies inside it. Options (b), (c) and (d) come from other, longer placements.
Tip: for a minimum, place people as close to the front as the rules allow, then check every condition.
Taking B = 20th pushes A to 15th or 25th and gives 28 or more.
Answer as per the official UPSC answer key.
Eight students A, B, C, D, E, F, G and H sit around a circular table, equidistant from each other, facing the centre of the table, not necessarily in the same order. B and D sit neither adjacent to C nor opposite to C. A sits in between E and D, and F sits in between B and H. Which one of the following is definitely correct?
Answer (d) & explanation
None of the options is definitely correct (d). Step 1: the blocks are E–A–D and B–F–H. C must not sit next to or opposite B or D. Step 2: several seatings meet all the rules. For example, going round: A, D, G, C, H, F, B, E. Step 3: in this seating, B is not between A and G (a fails), C is not opposite G (b fails) and E is not opposite F (c fails). Since one valid seating breaks each option, none is definite.
Tip: for "definitely correct", find one valid arrangement that breaks the option. One counter-example is enough to reject it.
'definitely correct' means true in every possible seating. One valid seating where it fails is enough to reject it.
Answer as per the official UPSC answer key.
Six persons A, B, C, D, E and F are sitting equidistant from each other around a circular table (facing the centre of the table).
Consider the Question and two Statements given below :
Question : Who is sitting on the immediate left of A?
Statement-1 : B is sitting opposite to C and D is sitting opposite to E.
Statement-2 : F is sitting on the immediate left of B.
Which one of the following is correct in respect of the Question and the Statements?
Answer (d) & explanation
Even both statements are not enough (d). Step 1: Statement-1 says B faces C and D faces E. With six seats, A must face F. Step 2: Statement-2 says F is on B's immediate left. This fixes B and F side by side, and C ends up on A's right. Step 3: A's immediate left is then a seat held by D or E, and D and E can swap. So it cannot be fixed. Options (a), (b) and (c) fail.
Tip: in circular seating, check if any two people can swap without breaking a rule. If yes, the answer is not fixed.
Answer as per the official UPSC answer key.
Six lectures A, B, C, D, E and F, each of one hour duration, are scheduled between 8:00 a.m. and 2:00 p.m.
Consider the Question and two Statements given below :
Question : Which lecture is in the third period?
Statement-1 : Lecture F is preceded by A and followed by C.
Statement-2 : There is no lecture after lecture B.
Which one of the following is correct in respect of the Question and the Statements?
Answer (d) & explanation
Even both statements are not enough (d). Step 1: Statement-2 says no lecture comes after B, so B is in the 6th period. Step 2: Statement-1 says A, F and C come one after another as a block A–F–C. This block sits somewhere in periods 1 to 5, along with D and E. Step 3: the block can start in period 1, 2 or 3. So the third period could hold C, F or A. It cannot be fixed. Options (a), (b) and (c) fail.
Tip: "preceded by A and followed by C" makes a fixed block. Count how many places the block can slide into.
Answer as per the official UPSC answer key.
There is a numeric lock which has a 3-digit PIN. The PIN contains digits 1 to 7. There is no repetition of digits. The digits in the PIN from left to right are in decreasing order. Any two digits in the PIN differ by at least 2. How many maximum attempts does one need to find out the PIN with certainty?
Answer (c) & explanation
At most 10 attempts are needed (c). Step 1: the digits go from big to small. So once three digits are chosen, the PIN is fixed. We only need to count the choices. Step 2: list sets from 1 to 7 where any two digits differ by at least 2: 531, 631, 641, 642, 731, 741, 742, 751, 752, 753. Step 3: that is 10 PINs, so 10 attempts make it certain. Options (a), (b) and (d) miscount.
Tip: when order is forced (decreasing), arrangements become combinations. A shortcut: choose 3 from 7 with gaps of 2 = C(5, 3) = 10.
Answer as per the official UPSC answer key.
There are eight equidistant points on a circle. How many right-angled triangles can be drawn using these points as vertices and taking the diameter as one side of the triangle?
Answer (a) & explanation
24 right-angled triangles can be drawn (a). Step 1: an angle in a semicircle is always a right angle. So any triangle with a diameter as one side is right-angled. Step 2: eight equally spaced points form 4 diameters (each point has one opposite point). Step 3: each diameter can join any of the other 6 points to make a triangle. 4 × 6 = 24. Options (b), (c) and (d) miscount the diameters or the third points.
Tip: remember the circle rule: the angle in a semicircle is 90°. It turns this geometry question into simple counting.
Answer as per the official UPSC answer key.
One non-zero digit, one vowel and one consonant from English alphabet (in capital) are to be used in forming passwords, such that each password has to start with a vowel and end with a consonant. How many such passwords can be generated?
Answer (c) & explanation
945 passwords can be made (c). Step 1: the password starts with a vowel and ends with a consonant. So the order is fixed: vowel, digit, consonant. Step 2: there are 5 vowels, 9 non-zero digits and 21 consonants. Step 3: 5 × 9 × 21 = 945. Option (d), 1050, uses 10 digits (including 0). Options (a) and (b) leave out a factor.
Tip: when positions are fixed, simply multiply the choices for each place. Check "non-zero" carefully: it removes one digit.
Answer as per the official UPSC answer key.
Integers are listed from 700 to 1000. In how many integers is the sum of the digits 10 ?
Answer (d) & explanation
9 integers have a digit sum of 10 (d). Step 1: numbers from 700 to 799 are written 7ab. We need a + b = 3: 703, 712, 721, 730. That is 4. Step 2: 8ab needs a + b = 2: 802, 811, 820. That is 3. 9ab needs a + b = 1: 901, 910. That is 2. Step 3: 1000 has digit sum 1, so it does not count. Total = 4 + 3 + 2 = 9. Options (a), (b) and (c) miss some.
Tip: two digits adding to k (with k ≤ 9) can be arranged in k + 1 ways. So 3 gives 4, 2 gives 3, 1 gives 2.
Answer as per the official UPSC answer key.
Seven books P, Q, R, S, T, U and V are placed side by side. R, Q and T have blue covers and other books have red covers. Only S and U are new books and the rest are old. P, R and S are law reports; the rest are Gazetteers. Books of old Gazetteers with blue covers are
Answer (c) & explanation
The old Gazetteers with blue covers are Q and T (c). Step 1: list each group. Gazetteers: all except the law reports P, R and S, so Q, T, U, V. Step 2: old books: all except the new ones S and U, so P, Q, R, T, V. Blue covers: Q, R, T. Step 3: books in all three groups: Q and T. Option (a) includes R, a law report. Options (b) and (d) include U, a new book.
Tip: in grouping puzzles, write each list and find the common members. Using "all except" lists helps avoid errors when a group is described by what it is not.
Answer as per the official UPSC answer key.
Consider two Statements and a Question :
Statement-1 : Priya is 4 ranks below Seema and is 31st from the bottom.
Statement-2 : Ena is 2 ranks above Seema and is 37th from the bottom.
Question : What is Seema's rank from the top in the class of 40 students ?
Which one of the following is correct in respect of the Statements and the Question ?
Answer (c) & explanation
Either statement alone is enough (c). Step 1: in a class of 40, rank from top = 40 − rank from bottom + 1. Step 2: Statement-1: Priya is 31st from the bottom, so she is 40 − 31 + 1 = 10th from the top. Seema is 4 ranks above her, so 6th. Step 3: Statement-2: Ena is 37th from the bottom, so 4th from the top. Seema is 2 ranks below her, so again 6th. Each statement alone gives the answer. Options (a) and (b) wrongly say one is not enough. Option (d) wrongly says both are needed.
Tip: in data sufficiency, test each statement on its own first. Only combine them if neither works alone.
Answer as per the official UPSC answer key.
Consider two Statements and a Question :
Statement-1 : Each of A and D is heavier than each of B, E and F, but none of them is the heaviest.
Statement-2 : A is heavier than D, but is lighter than C.
Question : Who is the heaviest among A, B, C, D and E ?
Which one of the following is correct in respect of the Statements and the Question ?
Answer (c) & explanation
1. Statement-1 alone: A and D beat B, E and F, but we are not told who beats A and D. C is never mentioned, so we cannot name the heaviest.
2. Statement-2 alone: C > A > D. We know nothing about B and E, so again we cannot say.
3. Together: C > A > D, and A, D are heavier than B and E. So C is the heaviest, and both statements are needed (c).
4. A different reading of Statement-1 makes the data unclear, which is why (d) is also accepted.
The words 'none of them' can be read two ways, so UPSC's final key accepts both (c) and (d).
Answer as per the official UPSC answer key.
On a chess board, in how many different ways can 6 consecutive squares be chosen on the diagonals along a straight path ?
Answer (b) & explanation
There are 6 ways (b). Step 1: a chess board has two main diagonals with 8 squares each. Step 2: a straight run of 6 squares on a diagonal of 8 can start at square 1, 2 or 3. That gives 3 ways on each diagonal. Step 3: two diagonals × 3 = 6. Option (a), 4, and the bigger options (c) 8 and (d) 12 do not match this count. UPSC's key counts only the two main diagonals here.
Tip: the number of runs of length k in a line of n squares is n − k + 1. Here 8 − 6 + 1 = 3.
Answer as per the official UPSC answer key.