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.
How often UPSC asks this
Questions per year, counted from the official papers.
Attempt the questions
Newest papers first. Tap an option; the answer and the reason appear instantly.
Two identical straight rods are painted in five distinct colours so that each of them gets divided into five equal parts along the length. In one of them, the portions are marked P1, P2, P3, P4 and P5 (not necessarily in that order) whereas in the other, they are marked Q1, Q2, Q3, Q4 and Q5 (not necessarily in that order). When the rods are kept parallel to each other side by side, P1 and Q3 match, P4 matches Q1 or Q2, and Q4 matches P3 or P5. If Q3 and Q5 are adjacent, which of the following is/are possible?
- I. Q3 is marked at the middle portion of the straight rod.
- II. P2 is marked at one of the extreme portions of the straight rod.
Select the answer using the code given below.
Answer (c) & explanation
Both statements are possible (c). Step 1: "possible" needs just one arrangement that fits all rules. Number the portions 1 to 5 on each rod. Step 2: try rod Q: Q1, Q2, Q3, Q5, Q4 and rod P: P2, P4, P1, P5, P3. Step 3: check the rules: P1 and Q3 both sit at position 3 ✓; P4 is at 2, beside Q2 ✓; Q4 is at 5, beside P3 ✓; Q3 and Q5 are adjacent ✓. In this arrangement, Q3 is in the middle (I) and P2 is at an end (II). So both are possible, and options (a), (b) and (d) fail.
Tip: for "possible" questions, build one valid example. For "must be true", you would need to check every case.
'possible' asks for one example, not for something that is always true.
Answer as per the official UPSC answer key.
Seven persons A, B, C, D, E, F and G travel by three cars X, Y, Z. A and another two of them travel by X. Only E travels with G. C travels by Z, but B does not travel by Y. Besides, A and B do not travel by the same car. Then which of the following are correct?
- I. No one travels alone.
- II. Only D travels with F.
- III. Only C travels with B.
Select the answer using the code given below.
Answer (b) & explanation
1. B is not in Y and not with A (in X), so B is in Z with C.
2. E and G form a car of their own; X already has A and Z has B and C, so E and G are in Y. D and F join A in X.
3. Cars: X = A, D, F; Y = E, G; Z = B, C. I and III are true. II is false, because A also travels with F.
Answer as per the official UPSC answer key.
Seven cubes are identical in shape. Out of these, the weight of each of the six cubes is equal and the weight of the remaining cube is less than the weight of any other cube. A balance is used to identify the lightest cube. What is the minimum number of attempts required to distinguish the odd cube with certainty?
Answer (a) & explanation
Two weighings are enough (a). Step 1: put 3 cubes against 3, keeping one aside. If they balance, the cube kept aside is the light one: done in one weighing. Step 2: if one side is lighter, the odd cube is among those three. Weigh 1 against 1 of them. Step 3: the lighter pan holds the odd cube; if they balance, it is the third cube. So 2 weighings always work. One weighing (d) cannot cover all cases. Options (b) and (c) use more than needed.
Tip: each weighing has three outcomes (left, right, balance), so split into three groups. With n weighings you can find a light coin among up to 3ⁿ coins.
one weighing can only split the cubes into three groups (left, right, aside), so 7 cubes need at least two.
Answer as per the official UPSC answer key.
How many words can one form by shuffling the letters of the word QUEUE, if Q is always followed by U? The words thus formed need not necessarily have any meaning.
Answer (d) & explanation
12 words can be formed (d). Step 1: Q must always be followed by U, so tie them together as one block, QU. Step 2: the letters of QUEUE are Q, U, E, U, E. After making the block, we have QU, U, E, E: four items. Step 3: four items can be arranged in 4 × 3 × 2 × 1 = 24 ways. But the two Es are identical, so swapping them gives the same word: 24 ÷ 2 = 12. Options (a), (b) and (c) miss arrangements or divide wrongly.
Tip: "always together" means treat them as one block. Then divide by the factorial of each repeated letter's count.
don't halve again for the Us. One U is inside the block, so the two Us are no longer interchangeable items.
Answer as per the official UPSC answer key.
Eight persons P, Q, R, S, T, U, V and W sit around a round table in eight different seats placed with equal distance between any two consecutive seats. Both P and R are adjacent to Q. Both T and R are adjacent to S. Both U and W are adjacent to V. S and W are on opposite chairs. If while going in the clockwise direction around the table from P, one meets R before T, then how many persons shall Q cross while moving in the clockwise direction around the table before meeting W?
Answer (a) & explanation
1. P–Q–R and R–S–T join into one run P, Q, R, S, T. Clockwise from P, R comes before T, so number the seats 1 to 8 clockwise: P 1, Q 2, R 3, S 4, T 5.
2. U, V, W take seats 6, 7, 8 with V in the middle (seat 7). W must face S (seat 4), so W is at seat 8 and U at 6.
3. Clockwise from Q: R, S, T, U, V, then W. Q crosses 5 persons.
Answer as per the official UPSC answer key.
The top of a table is rectangular and its dimensions are 6′ × 10′. Two rectangular portions of the table top are painted in blue colour; both these portions have dimensions 2.5′ × 8′ and each of them has exactly two sides common with two edges of the table top. If the table is fixed to the ground and the remaining portion of the table top is painted in white, how many different patterns are possible when observed from above?
Answer (b) & explanation
Four patterns are possible (b). Step 1: each blue strip is 2.5′ × 8′. Its 8′ side cannot lie along the 6′ edge, so it lies along a 10′ edge, and it touches two edges, so it sits in a corner. Step 2: two strips on the same 10′ edge would need 8 + 8 = 16′, more than 10′, so they would overlap and merge into one shape. So one strip goes along each 10′ edge. Step 3: each strip can sit at the left or right end of its edge: 2 × 2 = 4 patterns. The table is fixed, so mirror images count as different. Options (a), (c) and (d) miscount.
Tip: in arrangement puzzles, first use the sizes to rule out positions. Then count the free choices and multiply.
Answer as per the official UPSC answer key.
There are four types of weights, namely 1 kg, 2 kg, 5 kg and 10 kg. What is the maximum number of different ways one can measure 20 kg, if at least eight but not more than eleven weights of 1 kg are to be used while measuring?
Answer (b) & explanation
There are 8 ways (b). Step 1: with eight 1 kg weights, we need 12 kg from 2, 5 and 10 kg weights (no more 1s): 10 + 2, 5 + 5 + 2, and six 2s: 3 ways. Step 2: with nine 1 kg, we need 11 kg: only 5 + 2 + 2 + 2: 1 way. Step 3: with ten 1 kg, we need 10 kg: 10, 5 + 5, five 2s: 3 ways. With eleven 1 kg, we need 9 kg: 5 + 2 + 2: 1 way. Total = 3 + 1 + 3 + 1 = 8. Options (a), (c) and (d) miscount.
Tip: fix one variable at a time and count the rest. Here, fixing the number of 1 kg weights keeps the cases clean.
an odd remainder needs exactly one 5 (two or more would be too much or leave an odd gap).
Fell for this one? Most aspirants do. The PYQ Book puts every UPSC trap side by side, so the pattern finally clicks. Try the free sample
Answer as per the official UPSC answer key.
P, Q, R, S and T are ranked 1 to 5 (not necessarily in that order). The rank of P is 4, the rank of Q is not 5, the rank of R is 1, the rank of S is not 2, the rank of T is not 3. Then which of the following is/are correct?
- I. If the rank of S is 3, then that of T is 2.
- II. If the rank of Q is 3, then that of T is 5.
Select the answer using the code given below.
Answer (d) & explanation
Neither statement is correct (d). Step 1: P = 4 and R = 1 are fixed. Ranks 2, 3 and 5 remain for Q, S and T, with rules: Q is not 5, S is not 2, T is not 3. Step 2: statement I: if S = 3, then Q and T share 2 and 5. Q cannot be 5, so Q = 2 and T = 5, not 2. I is wrong. Step 3: statement II: if Q = 3, then S and T share 2 and 5. S cannot be 2, so S = 5 and T = 2, not 5. II is wrong. Options (a), (b) and (c) fail.
Tip: fix the known ranks first, then apply each "if" condition and let the "not" rules decide the rest. Write the small table on paper.
Answer as per the official UPSC answer key.
Three teams P, Q, R participated in a tournament in which the teams play with one another exactly once. A win fetches a team 2 points and a draw 1 point. A team gets no point for a loss. Each team scored exactly one goal in the tournament. The team P got 3 points, Q got 2 points and R got 1 point. Which of the following statements is/are correct?
- I. The result of the match between P and Q is a draw with the score 0-0.
- II. The number of goals scored by R against Q is 1.
Which of the statements given above is/are correct?
Answer (c) & explanation
Both statements are correct (c). Step 1: points: P = 3 (a win and a draw), Q = 2, R = 1 (a draw and a loss). There are three matches: P–Q, P–R and Q–R. Step 2: suppose P beat Q and drew R. Then Q must beat R for 2 points. P has only one goal, so P–Q was 1–0 and P–R 0–0; R's one goal comes against Q, so Q needs 2 goals to win, but Q has only one. Impossible. So P drew Q, P beat R, Q drew R. Step 3: each team scored exactly one goal. P's single goal beat R 1–0, so P–Q was 0–0 (I true). Q's goal came against R in a 1–1 draw, so R scored 1 against Q (II true). Options (a), (b) and (d) fail.
Tip: in points tables, list each team's possible results first, then match goals.
Answer as per the official UPSC answer key.
What is the number of selections of 10 consecutive things out of 12 things in a circle taken in the clockwise direction ?
Answer (c) & explanation
There are 12 selections (c). Step 1: in a circle, a run of consecutive things is fixed by where it starts. Step 2: any of the 12 things can be the starting point. Going clockwise, each start gives a different run of 10. Step 3: so there are 12 selections. Option (a), 3, applies to a straight line of 12 (12 − 10 + 1). Option (d), 66, counts any 10 things, not consecutive ones.
Tip: in a line, runs of k from n = n − k + 1. In a circle (with k less than n), runs of k = n.
66 = 12C10 counts all selections, consecutive or not.
Answer as per the official UPSC answer key.
In an examination, the maximum marks for each of the four papers namely P, Q, R and S are 100. Marks scored by the students are in integers. A student can score 99% in n different ways. What is the value of n ?
Answer (d) & explanation
1. 99% of 400 = 396, so the student loses exactly 4 marks in total across the four papers.
2. Count the ways to share the 4 lost marks:
3. All 4 in one paper: 4 ways. 3 + 1: 4 × 3 = 12 ways. 2 + 2: 6 ways. 2 + 1 + 1: 12 ways. 1 + 1 + 1 + 1: 1 way.
4. Total 4 + 12 + 6 + 12 + 1 = 35.
'no paper above 100' does not cut any case here, because a loss can't be negative.
Answer as per the official UPSC answer key.
A flag has to be designed with 4 horizontal stripes using some or all of the colours red, green and yellow. What is the number of different ways in which this can be done so that no two adjacent stripes have the same colour ?
Answer (c) & explanation
There are 24 ways (c). Step 1: the top stripe can be any of 3 colours. Step 2: each stripe below only has to differ from the one above it, so it has 2 choices. Step 3: 3 × 2 × 2 × 2 = 24. "Some or all" colours means using just two colours (like red-green-red-green) is allowed. Option (d), 36, and the others miscount.
Tip: for "no two adjacent the same", the first item has n choices and each next one has n − 1.
Answer as per the official UPSC answer key.
Raj has ten pairs of red, nine pairs of white and eight pairs of black shoes in a box. If he randomly picks shoes one by one (without replacement) from the box to get a red pair of shoes to wear, what is the maximum number of attempts he has to make?
Answer (d) & explanation
The maximum is 45 attempts (d). Step 1: think of the unluckiest case. First he picks every non-red shoe: 9 + 8 = 17 pairs = 34 shoes. Step 2: still unlucky, he then picks all 10 red shoes of the same foot (say all left shoes): 34 + 10 = 44. Step 3: the next shoe must be a red right shoe, which makes a pair. So 45 picks at most. Option (c), 44, stops one short. Options (a) and (b) ignore the left-right problem.
Tip: shoes have a left and a right, unlike socks. In worst-case questions, always ask what "a pair" really needs.
(c) 44 forgets that ten left shoes still do not make a pair.
Answer as per the official UPSC answer key.
There are five persons P, Q, R, S and T each one of whom has to be assigned one task. Neither P nor Q can be assigned Task-1. Task-2 must be assigned to either R or S. In how many ways can the assignment be done ?
Answer (d) & explanation
There are 24 ways (d). Step 1: Task-2 must go to R or S: 2 ways. Step 2: Task-1 cannot go to P, Q or the person who got Task-2. So only 2 people are left for it: 2 ways. Step 3: the other 3 tasks go to the remaining 3 people in 3 × 2 × 1 = 6 ways. Total = 2 × 2 × 6 = 24. Options (a), (b) and (c) miss a factor.
Tip: fill the most restricted task first. Then each later choice becomes simple counting.
Answer as per the official UPSC answer key.
In how many ways can a batsman score exactly 25 runs by scoring single runs, fours and sixes only, irrespective of the sequence of scoring shots?
Answer (b) & explanation
1. Choose how many sixes and fours. The singles then fill up the rest of the 25, so they never cause a problem.
2. 0 sixes: fours can be 0 to 6 (up to 24 runs). That is 7 ways.
3. 1 six: 19 runs left, fours 0 to 4. 5 ways.
4. 2 sixes: 13 left, fours 0 to 3. 4 ways.
5. 3 sixes: 7 left, fours 0 or 1. 2 ways.
6. 4 sixes: 1 left, no four. 1 way.
7. Total: 7 + 5 + 4 + 2 + 1 = 19.
Answer as per the official UPSC answer key.
There are large number of silver coins weighing 2 gm, 5 gm, 10 gm, 25 gm, 50 gm each. Consider the following statements :
- 1. To buy 78 gm of coins one must buy at least 7 coins.
- 2. To weigh 78 gm using these coins one can use less than 7 coins.
Which of the statements given above is/are correct ?
Answer (c) & explanation
Both statements are correct (c). Step 1: buying 78 g: 50 + 10 + 10 + 2 + 2 + 2 + 2 = 78 with 7 coins. 50 + 25 leaves 3 g, which no coin can make. Other mixes need more coins. So at least 7 are needed (1 true). Step 2: weighing allows coins on both pans. Put 50 + 25 + 5 on one pan and 2 on the other. That measures 80 − 2 = 78 g with only 4 coins (2 true). Step 3: so options (a), (b) and (d) fail.
Tip: "buy" means adding coins only; "weigh" lets you subtract by using the other pan. The wording changes the answer.
'buy' and 'weigh' are different. On a balance, a coin can also subtract.
Answer as per the official UPSC answer key.
There are four letters and four envelopes and exactly one letter is to be put in exactly one envelope with the correct address. If the letters are randomly inserted into the envelopes, then consider the following statements :
- 1. It is possible that exactly one letter goes into an incorrect envelope.
- 2. There are only six ways in which only two letters can go into the correct envelopes.
Which of the statements given above is/are correct?
Answer (b) & explanation
Only statement 2 is correct (b). Step 1: if three letters are in the right envelopes, only one envelope is left, and it is the right one for the fourth letter. So exactly one wrong letter is impossible. Statement 1 is false. Step 2: for exactly two right letters, choose which two are right: AB, AC, AD, BC, BD, CD = 6 ways. Step 3: the other two must both be wrong, so they swap, in only 1 way. Total = 6 × 1 = 6. Statement 2 is true. Options (a), (c) and (d) fail.
Tip: in matching problems, "exactly one wrong" can never happen. A wrong letter always pushes at least one other letter out of place.
Answer as per the official UPSC answer key.
How many distinct 8-digit numbers can be formed by rearranging the digits of the number 11223344 such that odd digits occupy odd positions and even digits occupy even positions?
Answer (c) & explanation
1. The odd digits are 1, 1, 3, 3. They go into the 4 odd places. Two places get 1 and two get 3. Choosing which two places get the 1s can be done in 6 ways (list them: 12, 13, 14, 23, 24, 34).
2. The even digits 2, 2, 4, 4 go into the 4 even places in the same way: 6 ways.
3. Each choice on one side goes with each choice on the other: 6 × 6 = 36.
Answer as per the official UPSC answer key.
ABCD is a square. One point on each of AB and CD; and two distinct points on each of BC and DA are chosen. How many distinct triangles can be drawn using any three points as vertices out of these six points?
Answer (c) & explanation
20 triangles can be drawn (c). Step 1: any 3 of the 6 points make a triangle, unless all 3 lie on one straight line. Step 2: the most points on any one side is 2. So no three chosen points are in a line. Step 3: the number of ways to choose 3 from 6 = (6 × 5 × 4) ÷ (3 × 2 × 1) = 20. Options (a), (b) and (d) wrongly subtract or add cases.
Tip: number of triangles = C(n, 3) minus the groups of 3 points that are in a line. Here there are none to subtract.
Answer as per the official UPSC answer key.
A box contains 14 black balls, 20 blue balls, 26 green balls, 28 yellow balls, 38 red balls and 54 white balls. Consider the following statements :
- 1. The smallest number n such that any n balls drawn from the box randomly must contain one full group of at least one colour is 175.
- 2. The smallest number m such that any m balls drawn from the box randomly must contain at least one ball of each colour is 167.
Which of the above statements is/are correct ?
Answer (c) & explanation
Both statements are correct (c). Step 1: think of the worst case. To avoid completing any colour, take one less than each group: 13 + 19 + 25 + 27 + 37 + 53 = 174. The next ball must complete a group, so n = 175 (statement 1 true). Step 2: to miss a colour as long as possible, take every ball except the smallest group (14 black): 180 − 14 = 166. Step 3: the next ball must be black, so m = 167 (statement 2 true). Options (a), (b) and (d) fail.
Tip: worst-case questions: give the unlucky picker everything except what you need, then add one.
for 'one of each colour', the worst case leaves out the smallest group, not the largest.
Answer as per the official UPSC answer key.