Number system and basic algebra (divisibility, HCF/LCM, remainders, digits, equations): previous-year questions
192 real UPSC Prelims questions (2011–2026) with the official answer and a clear explanation. 95 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.
What is the total number of digits printed, if a book containing 150 pages is to be numbered from 1 to 150?
Answer (b) & explanation
The total number of digits is 342 (b). Step 1: pages 1 to 9 use 1 digit each: 9 digits. Step 2: pages 10 to 99 are 90 pages with 2 digits each: 180 digits. Step 3: pages 100 to 150 are 51 pages with 3 digits each: 153 digits. Total = 9 + 180 + 153 = 342. Option (a), 262, and option (c), 360, are calculation slips. Option (d), 450, assumes 3 digits on every page.
Tip: split page numbers into 1-digit, 2-digit and 3-digit groups. Count pages in each group carefully: from 100 to 150 there are 51 pages, not 50.
Answer as per the official UPSC answer key.
If second and fourth Saturdays and all the Sundays are taken as only holidays for an office, what would be the minimum number of possible working days of any month of any year?
Answer (b) & explanation
The minimum is 22 working days (b). Step 1: every month has exactly two "second and fourth" Saturdays. Sundays number 4 or 5. Step 2: February in a normal year has 28 days and 4 Sundays: 28 − 4 − 2 = 22. A leap-year February with 5 Sundays: 29 − 5 − 2 = 22. Step 3: a 30-day month with 5 Sundays gives 30 − 5 − 2 = 23. A 31-day month gives at least 24. So the minimum is 22. Option (a), 23, is from a 30-day month. Options (c) and (d) would need more holidays than possible.
Tip: for "minimum in any month", test the shortest month first, February, with the most Sundays it can have.
a 30-day month with 5 Sundays gives 23. February is the shortest month.
Answer as per the official UPSC answer key.
The sum of income of A and B is more than that of C and D taken together. The sum of income of A and C is the same as that of B and D taken together. Moreover, A earns half as much as the sum of the income of B and D. Whose income is the highest?
Answer (b) & explanation
B has the highest income (b). Step 1: A earns half of B + D, so B + D = 2A. Also A + C = B + D, so A + C = 2A, which gives C = A. Step 2: A + B > C + D. Since C = A, this becomes B > D. Step 3: B + D = 2A and B > D, so B is above A and D is below A. So B > A = C > D, and B is highest. Option (a) A and option (c) C are equal and in the middle. Option (d) D is the lowest.
Tip: turn each sentence into an equation and substitute step by step. Finding that two people earn the same simplifies everything else.
Answer as per the official UPSC answer key.
A watch loses 2 minutes in every 24 hours while another watch gains 2 minutes in every 24 hours. At a particular instant, the two watches showed an identical time. Which of the following statements is correct if 24-hour clock is followed?
Answer (d) & explanation
None of the statements is correct (d). Step 1: one watch loses 2 minutes a day and the other gains 2 minutes a day. So they drift apart by 4 minutes a day. Step 2: on a 24-hour clock, they show the same time again only when the gap is a full 24 hours, which is 1440 minutes. Step 3: days needed = 1440 ÷ 4 = 360 days. This is not 30, 90 or 120. So options (a), (b) and (c) are wrong. Option (b), 90 days, would be the answer only for a 6-hour gap.
Tip: on a 12-hour clock the full gap is 720 minutes; on a 24-hour clock it is 1440. Read which clock the question uses.
on a 12-hour dial the answer would be 180 days. The question says 24-hour clock, and none of the options fit either way.
Answer as per the official UPSC answer key.
A clock strikes once at 1 o'clock, twice at 2 o'clock and thrice at 3 o'clock, and so on. If it takes 12 seconds to strike at 5 o'clock, what is the time taken by it to strike at 10 o'clock?
Answer (b) & explanation
The official answer is 24 seconds (b). Step 1: the official key treats the time as shared per strike. Five strikes take 12 seconds. Step 2: ten strikes are twice five strikes, so the time doubles: 12 × 2 = 24 seconds. Step 3: strictly, a clock's time runs between strikes. Five strikes have 4 gaps, so each gap is 3 seconds, and ten strikes have 9 gaps, which would give 27 seconds. Since 27 is not an option, UPSC's key follows the simple per-strike method. Options (a), (c) and (d) do not fit either method.
Tip: if the gap method gives no matching option, fall back on simple proportion, as UPSC did here.
Counting the gaps between strikes (4 gaps take 12 s, so 9 gaps take 27 s) gives 27 s, which is not an option; the official answer is (b).
Answer as per the official UPSC answer key.
There are some nectar-filled flowers on a tree and some bees are hovering on it. If one bee lands on each flower, one bee will be left out. If two bees land on each flower, one flower will be left out. The number of flowers and bees respectively are :
Answer (c) & explanation
There are 3 flowers and 4 bees (c). Step 1: one bee per flower leaves one bee out. So bees = flowers + 1. Step 2: two bees per flower leaves one flower empty. So bees = 2 × (flowers − 1). Step 3: put them together: f + 1 = 2f − 2, so f = 3 and bees = 4. Check: 4 bees on 3 flowers leaves 1 bee over, and 2 bees each on 2 flowers uses all 4 bees with 1 flower empty. Option (a), 2 and 4, fails the first rule. Options (b) and (d) give fewer bees than flowers, which breaks the first rule.
Tip: you can skip the algebra by testing each option against both rules. The one that passes both is the answer.
(d) swaps the order. The question asks for flowers first.
Answer as per the official UPSC answer key.
A class starts at 11:00 am and lasts till 2:27 pm. Four periods of equal duration are held during this interval. After every period, a rest of 5 minutes is given to the students. The exact duration of each period is :
Answer (a) & explanation
Each period lasts 48 minutes (a). Step 1: from 11:00 am to 2:27 pm is 3 hours 27 minutes = 207 minutes. Step 2: four periods have three breaks between them, because the class ends with the last period. Breaks = 3 × 5 = 15 minutes. Step 3: 4 periods = 207 − 15 = 192 minutes, so each period = 192 ÷ 4 = 48 minutes. Counting four breaks by mistake gives about 46.75 minutes, which matches no option. Options (b), (c) and (d) are all too long.
Tip: in "gaps between items" problems, n items have n − 1 gaps, like fence posts. Check: 4 × 48 + 15 = 207, which matches.
counting 4 breaks gives 46·75, which is not an option. That tells you the last break is not inside the class time.
Answer as per the official UPSC answer key.
In aid of charity, every student in a class contributes as many rupees as the number of students in that class. With the additional contribution of Rs. 2 by one student only, the total collection is Rs. 443. Then how many students are there in the class?
Answer (b) & explanation
There are 21 students (b). Step 1: each student gives as many rupees as there are students. With n students, the total is n × n = n². Step 2: one student gives Rs 2 extra, so n² + 2 = 443, which gives n² = 441. Step 3: 21 × 21 = 441, so n = 21. Option (a), 12, gives only 144. Options (c) 43 and (d) 45 are traps built from the digits of 443 and 441.
Tip: when a total is "n times n", look for a perfect square near the number. Learning squares up to 30 by heart saves a lot of time in CSAT. Check: 21 × 21 + 2 = 443, which matches.
Answer as per the official UPSC answer key.
A person is standing on the first step from the bottom of a ladder. If he has to climb 4 more steps to reach exactly the middle step, how many steps does the ladder have?
Answer (b) & explanation
The ladder has 9 steps (b). Step 1: he is on step 1 and climbs 4 more steps, so the middle step is step 5. Step 2: a middle step has the same number of steps below and above it. There are 4 steps below step 5. So there must be 4 above. Step 3: total = 4 + 1 + 4 = 9 steps. Option (a), 8, has no single middle step. Option (c), 10, also has no single middle step. Option (d), 11, would put the middle at step 6.
Tip: a "middle" item exists only when the total is odd. For an odd total n, the middle item is number (n + 1) ÷ 2. Here (9 + 1) ÷ 2 = 5, which checks out.
Answer as per the official UPSC answer key.
If R and S are different integers both divisible by 5, then which of the following is not necessarily true?
Answer (b) & explanation
"R + S is divisible by 10" is not always true (b). Step 1: R and S are multiples of 5, so write R = 5a and S = 5b. Step 2: R − S = 5(a − b), divisible by 5. R × S = 25ab, divisible by 25. R² + S² = 25(a² + b²), divisible by 5. So options (a), (c) and (d) always hold. Step 3: R + S = 5(a + b). This is divisible by 10 only if a + b is even. Take R = 5 and S = 10. Then R + S = 15, which is not divisible by 10. So (b) is not necessarily true.
Tip: for "not necessarily true", one counter-example is enough. Try the smallest numbers that fit the rules.
Answer as per the official UPSC answer key.
How many numbers are there between 100 and 300 which either begin with or end with 2?
Answer (a) & explanation
There are 110 such numbers (a). Step 1: between 100 and 300, the numbers that begin with 2 are 200 to 299. That is 100 numbers. Step 2: numbers that end with 2 and begin with 1 are 102, 112, 122 and so on up to 192. That is 10 numbers. The ones ending in 2 that begin with 2, like 202, are already counted. Step 3: total = 100 + 10 = 110. Option (b), 111, counts something twice. Options (c) and (d) do not fit.
Tip: for "either… or" counts, add the two groups and subtract the overlap. Here we simply skipped the overlap. This rule is called inclusion–exclusion.
adding 100 + 20 counts 202, 212, … twice.
Answer as per the official UPSC answer key.
There are five hobby clubs in a college — photography, yachting, chess, electronics and gardening. The gardening group meets every second day, the electronics group meets every third day, the chess group meets every fourth day, the yachting group meets every fifth day and the photography group meets every sixth day. How many times do all the five groups meet on the same day within 180 days?
Answer (d) & explanation
All five groups meet together 3 times (d). Step 1: the groups meet every 2, 3, 4, 5 and 6 days. They all meet on the same day only on days that are multiples of every one of these numbers. Step 2: find the LCM of 2, 3, 4, 5 and 6. 4 = 2 × 2 and 6 = 2 × 3, so LCM = 2 × 2 × 3 × 5 = 60. They all meet every 60 days. Step 3: within 180 days that is days 60, 120 and 180, so 180 ÷ 60 = 3 times. Option (b), 18, comes from dividing by 10 by mistake. Options (a) and (c) do not fit.
Tip: "when do things happen together again" is almost always an LCM question.
2 × 3 × 4 × 5 × 6 = 720 is not the LCM. Use the LCM.
Answer as per the official UPSC answer key.
A father is nine times as old as his son and the mother is eight times as old as the son. The sum of the father's and the mother's age is 51 years. What is the age of the son?
Answer (d) & explanation
The son is 3 years old (d). Step 1: call the son's age s. The father is 9s and the mother is 8s. Step 2: father + mother = 9s + 8s = 17s = 51. Step 3: s = 51 ÷ 17 = 3. Check: father 27, mother 24, total 51. Options (a), (b) and (c) give totals of 119, 85 and 68.
Tip: when everyone's age is a multiple of one person's age, add the multiples first (9 + 8 = 17) and divide the total by that. You can also test options: 3 × 17 = 51 works straight away.
Answer as per the official UPSC answer key.
Each of A, B, C and D has ₹ 100. A pays ₹ 20 to B, who pays ₹ 10 to C, who gets ₹ 30 from D. In this context, which one of the following statements is not correct?
Answer (c) & explanation
Statement (c) is not correct (c). Step 1: everyone starts with ₹100. Step 2: A pays ₹20 to B: A = 80, B = 120. B pays ₹10 to C: B = 110, C = 110. C gets ₹30 from D: C = 140, D = 70. Step 3: final amounts: A 80, B 110, C 140, D 70. Step 4: test each option. (a) C is the richest, with 140: correct. (b) D is the poorest, with 70: correct. (c) A + D = 80 + 70 = 150, which is more than C's 140. So (c) is not correct. (d) B (110) is richer than D (70): correct.
Tip: in money-transfer questions, track each person in a small table step by step. Then check every option against the final numbers.
the question asks for the statement that is not correct.
Answer as per the official UPSC answer key.
In a parking area, the total number of wheels of all the cars (four-wheelers) and scooters/motorbikes (two-wheelers) is 100 more than twice the number of parked vehicles. The number of cars parked is
Answer (c) & explanation
There are 50 cars (c). Step 1: let c be cars and s be two-wheelers. Total wheels = 4c + 2s. Twice the number of vehicles = 2c + 2s. Step 2: the wheels are 100 more: 4c + 2s = 2c + 2s + 100. Step 3: the 2s appears on both sides, so it cancels. That leaves 2c = 100, so c = 50. The number of two-wheelers does not matter. Options (a), (b) and (d) do not fit.
Tip: a simple way to see it: give every vehicle 2 wheels first. That uses "twice the vehicles". The extra 100 wheels are the 2 more wheels each car has. 100 ÷ 2 = 50 cars.
Answer as per the official UPSC answer key.
If ABC × DEED = ABCABC; where A, B, C, D and E are different digits, what are the values of D and E?
Answer (c) & explanation
D = 1 and E = 0 (c). Step 1: look at ABCABC. It is the number ABC written twice. Writing a three-digit number twice is the same as multiplying it by 1001. For example, 123 × 1001 = 123123. Step 2: so ABC × DEED = ABC × 1001, and DEED = 1001. Step 3: reading 1001 as D E E D gives D = 1 and E = 0. Options (a), (b) and (d) give 2002, 0110 and 1221, which do not work.
Tip: remember the 1001 trick: abc × 1001 = abcabc. Similarly, ab × 101 = abab. These patterns come up in CSAT digit puzzles.
Answer as per the official UPSC answer key.
Between 6 PM and 7 PM the minute hand of a clock will be ahead of the hour hand by 3 minutes at
Answer (c) & explanation
1. Between 6 and 7, the hour hand starts at the 30-minute mark and moves 1 minute-mark every 12 minutes.
2. Check (c) 6:36. The hour hand has moved 36 ÷ 12 = 3 marks, so it is at 33. The minute hand is at 36. It is 3 ahead. Works.
3. Quick check of the others: at 6:48 the hour hand is at 34 and the minute hand at 48 (14 ahead), so (d) fails; at 6:15 and 6:18 the minute hand is still behind.
4. Answer: 6:36 PM.
Answer as per the official UPSC answer key.
A cow costs more than 4 goats but less than 5 goats. If a goat costs between ₹ 600 and ₹ 800, which of the following is a most valid conclusion?
Answer (d) & explanation
A cow costs between ₹2,400 and ₹4,000 (d). Step 1: the cow costs more than 4 goats and less than 5 goats. A goat costs between ₹600 and ₹800. Step 2: the lowest a cow could cost is just over 4 × 600 = ₹2,400. Step 3: the highest is just under 5 × 800 = ₹4,000. Step 4: only option (d) covers every possible price. Option (a) is too narrow at the bottom (a cow might cost ₹2,450). Option (b) misses prices up to ₹4,000. Option (c) misses both ends.
Tip: for a "most valid" range, use the smallest values together for the minimum and the largest values together for the maximum.
(a), (b) and (c) can fail. A cow could cost ₹ 2,450 or ₹ 3,900. The most valid conclusion is the one that always holds.
Answer as per the official UPSC answer key.
In a box of marbles, there are three less white marbles than the red ones and five more white marbles than the green ones. If there are a total of 10 white marbles, how many marbles are there in the box?
Answer (b) & explanation
There are 28 marbles (b). Step 1: there are 10 white marbles. Step 2: white is 3 less than red, so red = 10 + 3 = 13. Step 3: white is 5 more than green, so green = 10 − 5 = 5. Step 4: total = 10 + 13 + 5 = 28. Options (a), (c) and (d) come from adding or subtracting the wrong way.
Tip: read "three less white marbles than the red ones" carefully: white = red − 3, so red = white + 3. Write each sentence as a simple equation before working it out to avoid flipping the direction.
Answer as per the official UPSC answer key.
Four persons, Alok, Bhupesh, Chander and Dinesh have a total of ₹ 100 among themselves. Alok and Bhupesh between them have as much money as Chander and Dinesh between them, but Alok has more money than Bhupesh; and Chander has only half the money that Dinesh has. Alok has in fact ₹ 5 more than Dinesh has. Who has the maximum amount of money?
Answer (a) & explanation
Alok has the most money (a). Step 1: Alok + Bhupesh = Chander + Dinesh, and all four total ₹100, so each pair has ₹50. Step 2: Chander has half of Dinesh's money. So Chander + Dinesh = 50 with Dinesh = 2 × Chander. That gives Chander = ₹16⅔ and Dinesh = ₹33⅓. Step 3: Alok has ₹5 more than Dinesh: 33⅓ + 5 = ₹38⅓. Then Bhupesh = 50 − 38⅓ = ₹11⅔. Step 4: the largest is Alok. Options (b), (c) and (d) have less.
Tip: split the total into the pairs first. "Half as much" means three equal parts: one for Chander and two for Dinesh, so 50 ÷ 3 per part.
Answer as per the official UPSC answer key.