CSAT (Paper II): previous-year questions
1280 real UPSC Prelims questions (2011–2026) with the official answer and a clear explanation. 400 of them came in the last five papers.
How often UPSC asks this
Questions per year, counted from the official papers.
Subtopics
Arrangements, ranking, ordering and puzzles
142 questions
Clocks and calendars
5 questions
Coding-decoding, series, patterns, cubes, dice and visual reasoning
110 questions
Communication, interpersonal skills and decision-making
10 questions
Counting, permutations, combinations and probability
1 questions
Directions, blood relations and Venn diagrams
60 questions
General English: grammar, vocabulary and comprehension
20 questions
General Hindi: grammar, vocabulary, dialects and comprehension
25 questions
Geometry and mensuration
26 questions
Number system and basic algebra (divisibility, HCF/LCM, remainders, digits, equations)
196 questions
Percentages, profit and loss, discount, simple and compound interest
47 questions
Ratio, proportion, averages, mixtures and alligation
47 questions
Syllogisms, statements and conclusions, assumptions and arguments
89 questions
Time and work; pipes and cisterns
2 questions
Time, speed and distance; trains, boats and streams; races
48 questions
Attempt the questions
Newest papers first. Tap an option; the answer and the reason appear instantly.
P and Q walk along a circular track. They start at 5:00 a.m. from the same point in opposite directions. P walks at an average speed of 5 rounds per hour and Q walks at an average speed of 3 rounds per hour. How many times will they cross each other between 5:20 a.m. and 7:00 a.m.?
Answer (b) & explanation
They cross 13 times (b). Step 1: walking in opposite directions on a circle, their speeds add: 5 + 3 = 8 rounds per hour. So they cross every 1/8 hour = 7.5 minutes. Step 2: crossings happen at 7.5, 15, 22.5, 30, … minutes after 5:00. The first one after 5:20 is at 22.5 minutes (the 3rd crossing). Step 3: 7:00 is 120 minutes, which is the 16th crossing. Between 5:20 and 7:00 (not counting 7:00), crossings 3 to 15 occur: 13 times. Options (a), (c) and (d) miscount the ends.
Tip: opposite directions add speeds; same direction subtracts them. Count crossings carefully at the start and end of the time window.
Answer as per the official UPSC answer key.
A question is given followed by two Statements I and II. Consider the Question and the Statements and mark the correct option.
Question : Let P, Q, R, S be distinct non-zero digits. If PP × PQ = RRSS, where P ≤ 3 and Q ≤ 4, then what is Q equal to?
Statement I : R = 1.
Statement II : S = 2.
Which one of the following is correct in respect of the above Question and the Statements?
Answer (d) & explanation
Q can be found without either statement (d), as per the key. Step 1: P can be 1, 2 or 3, and Q at most 4. Step 2: if P = 1 or 2, then PP × PQ is at most 22 × 24 = 528, too small for a 4-digit number RRSS. So P = 3. Step 3: try 33 × 31 = 1023, 33 × 32 = 1056, 33 × 34 = 1122 (Q cannot be 3, as digits are distinct). Only 1122 has the RRSS shape. So Q = 4, found from the question alone. The statements (R = 1, S = 2) only confirm it.
Tip: when the conditions leave very few cases, check them all before reading the statements. If the answer comes out, neither statement is needed.
Do not jump to a statement before testing the question on its own. Small limits like P ≤ 3 often settle it.
Answer as per the official UPSC answer key.
If 7 * 24 = 25 and 12 * 16 = 20, then what is 16 * 63 equal to?
Answer (c) & explanation
1. Look at the first two: 7 × 7 + 24 × 24 = 49 + 576 = 625 = 25 × 25. And 12 × 12 + 16 × 16 = 144 + 256 = 400 = 20 × 20.
2. So the rule is: square both numbers, add, and find the number whose square is the total.
3. 16 × 16 + 63 × 63 = 256 + 3969 = 4225 = 65 × 65. Answer 65.
Answer as per the official UPSC answer key.
What is the 489th digit in the number 123456789101112...?
Answer (d) & explanation
The 489th digit is 9 (d). Step 1: the numbers 1 to 9 use 9 digits. The numbers 10 to 99 use 90 × 2 = 180 digits. Total so far: 189. Step 2: 489 − 189 = 300 digits remain. Each three-digit number uses 3 digits, so that is exactly 100 numbers: 100 to 199. Step 3: the 489th digit is the last digit of 199: 9. Options (a), (b) and (c) come from miscounting the blocks.
Tip: count digits block by block: 1-digit (9), 2-digit (180), 3-digit (2700). If the leftover divides exactly, the answer is the last digit of that number.
Answer as per the official UPSC answer key.
The difference between any two natural numbers is 10. What can be said about the natural numbers which are divisible by 5 and lie between these two numbers?
Answer (c) & explanation
There can be more than one such number (c). Step 1: take two numbers 10 apart, such as 5 and 15. Strictly between them, only 10 is divisible by 5. That is one number. Step 2: now take 1 and 11. Between them, 5 and 10 are divisible by 5. That is two numbers. Step 3: the count can be one or two depending on the pair. So "only one" (a) and "only two" (b) are not always true, and (d) is false. The safe statement is (c): there can be more than one.
Tip: test both an "edge" case (ends are multiples of 5) and a general case. When answers vary, pick the option that allows for it.
Answer as per the official UPSC answer key.
Let both p and k be prime numbers such that (p2 + k) is also a prime number less than 30. What is the number of possible values of k?
Answer (b) & explanation
Five values of k are possible (b). Step 1: if p = 2, then p² = 4, and 4 + k must be a prime below 30. k = 3, 7, 13, 19 work (sums 7, 11, 17, 23). k = 2 gives 6, not prime. Step 2: if p is odd, p² is odd, so k must be 2 (the even prime) to keep the sum odd. p = 3 gives 9 + 2 = 11, prime. p = 5 gives 27, not prime. So k = 2 also works. Step 3: possible k: 2, 3, 7, 13, 19 = 5 values. Options (a), (c) and (d) miscount.
Tip: in prime questions, split into "2" and "odd" cases. Odd + odd is even, so the even prime 2 often decides things.
Answer as per the official UPSC answer key.
X can complete one-third of a certain work in 6 days, Y can complete one-third of the same work in 8 days and Z can complete three-fourth of the same work in 12 days. All of them work together for n days and then X and Z quit and Y alone finishes the remaining work in 8 2/3 days. What is n equal to?
Answer (b) & explanation
n is 4 days (b). Step 1: find each person's full time. X does 1/3 in 6 days, so the whole work in 18 days. Y: 1/3 in 8 days, so 24 days. Z: 3/4 in 12 days, so 16 days. Step 2: together per day: 1/18 + 1/24 + 1/16 = 8/144 + 6/144 + 9/144 = 23/144. Step 3: Y alone works 8⅔ = 26/3 days: (26/3) × (1/24) = 26/72 = 52/144. The rest, 92/144, was done together: 23n/144 = 92/144, so n = 4. Options (a), (c) and (d) fail.
Tip: turn part-work into full-work days first. Use the LCM (here 144) as the total work to avoid fractions.
Answer as per the official UPSC answer key.
Three prime numbers p, q and r, each less than 20, are such that p – q = q – r. How many distinct possible values can we get for (p + q + r)?
Answer (d) & explanation
More than 6 sums are possible (d). Step 1: p − q = q − r means p + r = 2q, so the sum p + q + r = 3q. Each different q gives a different sum. Step 2: the question does not say the primes must be different. So p = q = r is allowed (for example, 7 − 7 = 7 − 7). Then every prime below 20 can be q: 2, 3, 5, 7, 11, 13, 17, 19. Step 3: that gives 8 sums, which is more than 6. Options (a), (b) and (c) assume the primes must be distinct.
Tip: read the conditions closely. If a question does not say "distinct", equal values are allowed, and that often changes the answer.
read what is not said. Without "distinct", equal primes are allowed. With distinct primes you would get only 4 (q = 5, 7, 11, 13).
Answer as per the official UPSC answer key.
If P = +, Q = –, R = ×, S = ÷, then insert the proper notations between the successive numbers in the equation 60 _ 15 _ 3 _ 20 _ 4 = 20 :
Answer (b) & explanation
The correct order is QRPS (b). Step 1: decode: P = +, Q = −, R = ×, S = ÷. Step 2: test (b) QRPS: 60 − 15 × 3 + 20 ÷ 4. Do × and ÷ first: 15 × 3 = 45 and 20 ÷ 4 = 5. Then 60 − 45 + 5 = 20 ✓. Step 3: test (a) SPRQ: 60 ÷ 15 + 3 × 20 − 4 = 4 + 60 − 4 = 60 ✗. Option (c) QRSP gives 60 − 15 × 3 ÷ 20 + 4 = 61.75, and (d) SPQR gives 4 + 3 − 80 = −73.
Tip: test the options rather than building the answer yourself. Always apply BODMAS: × and ÷ before + and −.
Answer as per the official UPSC answer key.
A question is given followed by two Statements I and II. Consider the Question and the Statements and mark the correct option.
Question: How is Q related to P?
Statement I: P has two sisters, R and S.
Statement II: R's father is the brother of Q.
Which one of the following is correct in respect of the above Question and the Statements?
Answer (d) & explanation
Even both statements together are not enough (d). Step 1: statement I says P has two sisters, R and S. So R is P's sibling. Step 2: statement II says R's father is Q's brother. R's father is also P's father. So Q is a sibling of P's father. Step 3: Q could be P's uncle (if male) or aunt (if female). Q's gender is not given, so the exact relation cannot be fixed. Options (a), (b) and (c) fail.
Tip: in relation questions, check that the gender of the person asked about is known. "Brother of Q" tells us about the brother, not about Q.
Answer as per the official UPSC answer key.
The petrol price shot up by 10% as a result of the hike in crude oil prices. The price of petrol before the hike was ₹ 90 per litre. A person travels 2200 km every month and his car gives a mileage of 16 km per litre. By how many km should he reduce his travel if he wants to maintain his expenditure at the previous level?
Answer (b) & explanation
He must cut his travel by 200 km (b). Step 1: to keep spending the same, price × litres must stay fixed. The price rises by 10%, from ₹90 to ₹99, so the litres he buys must fall to 100/110 = 10/11 of before. Step 2: his travel falls in the same ratio: 2200 × 10/11 = 2000 km. Step 3: so he must reduce by 2200 − 2000 = 200 km. The mileage (16 km/l) is not needed. Option (c), 220 km, wrongly takes 10% of 2200.
Tip: a 10% price rise needs a cut of 10/110 = 1/11 (about 9.1%) in use, not 10%, to keep spending the same.
a 10% rise needs a 1/11 cut (about 9.1%), not a 10% cut of 220 km.
Answer as per the official UPSC answer key.
A mobile phone has been stolen. There are 3 suspects P, Q and R. They were questioned knowing that only one of them is guilty. Their responses are as follows :
P : I did not steal. Q stole it.
Q : R did not steal. I did not steal.
R : I did not steal. I do not know who did it.
Who stole the mobile phone?
Answer (d) & explanation
The thief cannot be concluded (d). Step 1: we know only one is guilty, but we are not told who speaks the truth or how many lie. Step 2: test each case. If P stole it, P lies, and Q and R can both be telling the truth. If Q stole it, Q lies, and P and R can be truthful. If R stole it, R lies, and Q's "R did not steal" is false, while P's "Q stole it" is false too. Step 3: without a rule like "only one lies", several cases fit. So no one can be named. Options (a), (b) and (c) fail.
Tip: in truth-teller puzzles, look for a rule about who lies. If none is given, the answer is often "cannot be concluded".
don't assume the innocent always tell the truth unless the question says so.
UPSC repeats traps like this year after year. The PYQ Book gathers every trap in one place, so you see them coming. Try the free sample
Answer as per the official UPSC answer key.
There are n sets of numbers each having only three positive integers with LCM equal to 1001 and HCF equal to 1. What is the value of n?
Answer (d) & explanation
There are more than 8 such sets (d). Step 1: 1001 = 7 × 11 × 13. Each number in a set must be made from these primes (or be 1). Step 2: LCM 1001 means every prime appears somewhere; HCF 1 means no prime is in all three. Step 3: list some sets: {7, 11, 13}, {1, 77, 91}, {1, 77, 143}, {1, 91, 143}, {7, 11, 1001}, {1, 7, 143}, {1, 11, 91}, {1, 13, 77}, {77, 91, 143}. That is already 9, so n is more than 8. Options (a), (b) and (c) are too small.
Tip: when an option says "more than", you only need to find enough cases to pass it; no full count is needed. 1001 = 7 × 11 × 13 is worth remembering.
Answer as per the official UPSC answer key.
What is X in the sequence 1, 3, 6, 11, 18, X, 42?
Answer (c) & explanation
X is 29 (c). Step 1: find the gaps: 3 − 1 = 2, 6 − 3 = 3, 11 − 6 = 5, 18 − 11 = 7. Step 2: the gaps are the prime numbers in order: 2, 3, 5, 7, so next come 11 and 13. Step 3: X = 18 + 11 = 29, and 29 + 13 = 42 ✓. Options (a) 26, (b) 27 and (d) 30 do not give 42 with a prime gap.
Tip: when gaps grow unevenly, check whether they are primes, squares or Fibonacci numbers. Always confirm with the last given term.
Answer as per the official UPSC answer key.
How many possible values of (p + q + r) are there satisfying 1/p + 1/q + 1/r = 1, where p, q and r are natural numbers (not necessarily distinct)?
Answer (c) & explanation
Three values are possible (c). Step 1: arrange p ≤ q ≤ r. The smallest, p, has the biggest fraction, so 1/p ≥ 1/3. Since 1/p must be less than 1, p is 2 or 3. Step 2: if p = 3, then q = r = 3. If p = 2, then 1/q + 1/r = 1/2, giving (4, 4) or (3, 6). Step 3: the solutions are (3, 3, 3), (2, 4, 4), (2, 3, 6) with sums 9, 10 and 11: three values. Options (a), (b) and (d) fail.
Tip: in unit-fraction equations, order the variables and bound the smallest one first. This keeps the cases few and complete.
Answer as per the official UPSC answer key.
A tram overtakes 2 persons X and Y walking at an average speed of 3 km/hr and 4 km/hr in the same direction and completely passes them in 8 seconds and 9 seconds respectively. What is the length of the tram?
Answer (c) & explanation
The tram is 20 m long (c). Step 1: the tram and walkers move the same way, so use relative speed = tram speed − walker speed. Length = (v − 3) × 8 = (v − 4) × 9, with speeds in km/hr and time in seconds. Step 2: 8v − 24 = 9v − 36, so v = 12 km/hr. Step 3: relative speed past X = 12 − 3 = 9 km/hr = 9 × 5/18 = 2.5 m/s. Length = 2.5 × 8 = 20 m. Check with Y: 8 km/hr = 20/9 m/s × 9 s = 20 m ✓. Options (a), (b) and (d) fail.
Tip: to change km/hr to m/s, multiply by 5/18. Passing a person (a point) means covering only the tram's own length.
Answer as per the official UPSC answer key.
Read the passage
A single number for inflation is aggregate across different commodities and Services the price rise differs for different items of consumption. So, the single number is arrived at by assigning weights to different commodities and services. For WPI, the weights in production are used; for CPI, the consumption basket is used. But people are not homogeneous.
The consumption basket is vastly different for the poor, the middle classes, and the rich. Hence, the CPI is different for each of these classes and a composite index requires averaging the baskets.
Which one of the following statements best reflects the most logical, rational and crucial message conveyed by the passage?
Answer (b) & explanation
The message is that the present single inflation number does not correctly show the price rise of individual items (b). Step 1: the passage says one inflation number is an average of many items, each given a weight. Prices of different items rise differently. Step 2: it adds that consumption baskets differ for the poor, middle class and rich, so one composite CPI averages them too. So the single number hides how individual prices move: (b). Step 3: option (a), scrap CPI and use only WPI, and (d), inflation data is useless, are extreme. Option (c), services under-represented, is not discussed.
Tip: WPI uses production weights; CPI uses consumption weights. Since 2014 the RBI targets CPI inflation at 4% (with a band of 2–6%).
Answer as per the official UPSC answer key.
A 4-digit number N is such that when divided by 3, 5, 6, 9 leaves a remainder 1, 3, 4, 7 respectively. What is the smallest value of N?
Answer (c) & explanation
The smallest value of N is 1078 (c). Step 1: notice the remainders: 3 leaves 1, 5 leaves 3, 6 leaves 4, 9 leaves 7. Each remainder is 2 less than its divisor. So N + 2 is divisible by all of 3, 5, 6 and 9. Step 2: LCM(3, 5, 6, 9) = 90. So N = 90k − 2. Step 3: the smallest 4-digit value: 90 × 12 = 1080, so N = 1078. (90 × 11 − 2 = 988 is only 3 digits.) Options (a), (b) and (d) fail the remainder checks.
Tip: when divisor − remainder is the same in every case, N + (that gap) is a multiple of the LCM. This is a classic shortcut.
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.
Let PQR be a 3-digit number, PPT be a 3-digit number and PS be a 2-digit number, where P, Q, R, S, T are distinct non-zero digits. Further, PQR − PS = PPT. If Q = 3 and T < 6, then what is the number of possible values of (R, S)?
Answer (b) & explanation
1. Write it as an ordinary subtraction: PQR − PS = PPT. The answer keeps P in front, and its middle digit is also P.
2. Middle column: 3 − P (minus 1 if we borrowed) must give P. Only P = 1 with a borrow works: 3 − 1 − 1 = 1.
3. Last column, after borrowing: (R + 10) − S = T. So S + T = R + 10.
4. T can be 2, 4 or 5 (not 1 or 3, which are taken, and less than 6). Try each, using only unused digits:
5. T = 2: S = R + 8 needs R = 1, which is taken. None.
6. T = 4: S = R + 6 gives R = 2, S = 8. One pair.
7. T = 5: S = R + 5 gives (R, S) = (2, 7) or (4, 9). Two pairs.
8. Total: 3 pairs. Check one: 132 − 18 = 114.
Count pairs (R, S), not separate digits: (2, 8), (2, 7) and (4, 9).
Answer as per the official UPSC answer key.