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.
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In 2002, Meenu's age was one-third of the age of Meera, whereas in 2010, Meenu's age was half the age of Meera. What is Meenu's year of birth?
Answer (b) & explanation
Meenu was born in 1994 (b). Step 1: test the options. Step 2: born 1994: in 2002 Meenu is 8, so Meera is 3 × 8 = 24. In 2010, Meenu is 16 and Meera is 32. 16 is half of 32. It works. Step 3: born 1992: in 2002 Meenu is 10 and Meera 30. In 2010 they are 18 and 38; 18 is not half of 38. So (a) fails. Options (c) and (d) also fail the check.
Tip: algebra works too. Let Meenu be x in 2002, so Meera is 3x. In 2010: 2(x + 8) = 3x + 8, so x = 8. Born 2002 − 8 = 1994. But testing options is often faster in CSAT.
Answer as per the official UPSC answer key.
Sunita cuts a sheet of paper into three pieces. Length of first piece is equal to the average of the three single digit odd prime numbers. Length of the second piece is equal to that of the first plus one-third the length of the third. The third piece is as long as the other two pieces together. The length of the original sheet of paper is
Answer (d) & explanation
The sheet is 30 units long (d). Step 1: single-digit odd primes are 3, 5 and 7. Their average is 15 ÷ 3 = 5. So the first piece = 5. Step 2: let the third piece be t. The second piece = 5 + t/3. The third equals the other two together: t = 5 + (5 + t/3), so t − t/3 = 10, giving t = 15. The second piece = 5 + 5 = 10. Step 3: sheet = 5 + 10 + 15 = 30 units. Option (b), 15, is only the third piece. Options (a) and (c) do not fit.
Tip: 2 is the only even prime, so "odd primes" leaves out 2. Shortcut: the third piece is half the sheet, so the sheet is twice 15.
15 is the third piece. The question asks for the whole sheet.
Answer as per the official UPSC answer key.
Rakesh and Rajesh together bought 10 balls and 10 rackets. Rakesh spent ₹ 1300 and Rajesh spent ₹ 1500. If each racket costs three times a ball does, then what is the price of a racket?
Answer (c) & explanation
A racket costs ₹ 210 (c). Step 1: together they spent 1300 + 1500 = ₹ 2800 on 10 balls and 10 rackets. Step 2: let a ball cost b. A racket costs 3b. So 10b + 10 × 3b = 40b = 2800, and b = ₹ 70. Step 3: a racket = 3 × 70 = ₹ 210. Option (a), ₹ 70, is the price of a ball. Options (b) and (d) do not fit. How the cost was split between Rakesh and Rajesh does not matter.
Tip: CSAT often adds extra data to confuse you. Here the separate spends of ₹ 1300 and ₹ 1500 are not needed; only the total matters.
₹ 70 is the ball, not the racket.
Answer as per the official UPSC answer key.
An 8-digit number 4252746B leaves remainder 0 when divided by 3. How many values of B are possible?
Answer (c) & explanation
B can take 4 values (c). Step 1: a number is divisible by 3 if the sum of its digits is divisible by 3. Step 2: digits without B: 4 + 2 + 5 + 2 + 7 + 4 + 6 = 30. That is already a multiple of 3. Step 3: so B itself must be a multiple of 3: 0, 3, 6 or 9. That is 4 values. Option (b), 3, forgets 0. Options (a) and (d) miscount.
Tip: remember divisibility rules: by 3 or 9, use the digit sum; by 4, the last two digits; by 8, the last three; by 11, the difference of alternate digit sums. And 0 is a valid digit, so do not skip it.
don't forget B = 0.
Answer as per the official UPSC answer key.
The ratio of a two-digit natural number to a number formed by reversing its digits is 4 : 7. The number of such pairs is
Answer (b) & explanation
There are 4 such pairs (b). Step 1: let the number be 10a + b; reversed, it is 10b + a. Given (10a + b) : (10b + a) = 4 : 7, so 7(10a + b) = 4(10b + a). Step 2: 70a + 7b = 40b + 4a, so 66a = 33b, which gives b = 2a. Step 3: b must be a single digit, so a can be 1, 2, 3 or 4. The numbers are 12, 24, 36 and 48 (12 : 21 = 4 : 7, and so on). a = 5 would give b = 10, which is not a digit. Options (a), (c) and (d) fail.
Tip: write a two-digit number as 10a + b. Always check that the digits stay between 0 and 9.
Answer as per the official UPSC answer key.
Seeta and Geeta go for a swim after a gap of every 2 days and every 3 days respectively. If on 1st January both of them went for a swim together, when will they go together next?
Answer (d) & explanation
They swim together next on 13th January (d). Step 1: a gap of 2 days means Seeta swims on day 1, rests 2 days, then swims on day 4, so every 3rd day. A gap of 3 days means Geeta swims every 4th day. Step 2: they meet again after LCM(3, 4) = 12 days. Step 3: 1st January + 12 days = 13th January. Option (a), 7th January, uses LCM(2, 3) = 6. Options (b) and (c) also misread the gap.
Tip: "after a gap of n days" means the event repeats every n + 1 days. This wording trap appears often in CSAT calendar sums.
LCM(2, 3) = 6 gives 7 January. The gap is not the cycle.
Answer as per the official UPSC answer key.
If x is greater than or equal to 25 and y is less than or equal to 40, then which one of the following is always correct?
Answer (c) & explanation
(y − x) ≤ 15 is always correct (c). Step 1: y − x is largest when y is as big as possible and x as small as possible: 40 − 25 = 15. So y − x can never be more than 15. Statement (c) always holds. Step 2: check the others. x = 25, y = 40 makes y bigger than x, so (a) fails. It also gives y − x = 15, not greater than 15, so (b) fails. Step 3: a very small y, like y = −100, gives x + y below 65, so (d) fails.
Tip: for "always true" questions with limits, test the extreme values. The edge case (here 25 and 40) usually breaks the wrong options.
Answer as per the official UPSC answer key.
Ena was born 4 years after her parents' marriage. Her mother is three years younger than her father and 24 years older than Ena, who is 13 years old. At what age did Ena's father get married?
Answer (b) & explanation
Ena's father married at 23 (b). Step 1: Ena is 13. Her mother is 24 years older, so 37. Her father is 3 years older than her mother, so 40. Step 2: Ena was born 4 years after the marriage. So the marriage was 13 + 4 = 17 years ago. Step 3: father's age then = 40 − 17 = 23 years. Option (d), 25, uses the mother's age plus a wrong step. Option (c), 24, confuses the age gap. Option (a) forgets the 3-year difference.
Tip: in age puzzles, first find everyone's present age, then move back in time. Write each step on a line to avoid mixing up who is older.
Answer as per the official UPSC answer key.
Consider the sequence given below :
4/12/95, 1/1/96, 29/1/96, 26/2/96, ....
What is the next term of the series?
Answer (b) & explanation
The next date is 25/3/96 (b). Step 1: find the gap. From 4 Dec 1995 to 1 Jan 1996: December has 31 days, so 31 − 4 + 1 = 28 days. From 1 Jan to 29 Jan is also 28 days. Step 2: from 29 Jan to 26 Feb: January has 31 days, so 2 + 26 = 28 days again. So each term adds 28 days. Step 3: 1996 is a leap year, so February has 29 days. From 26 Feb: 3 days finish February, then 25 more days give 25 March. Option (c), 26/3, forgets the leap day. Options (a) and (d) also fail.
Tip: in date series, always check for a leap year. A year divisible by 4 (like 1996) has 29 days in February.
forgetting the leap year gives 26/3/96.
Answer as per the official UPSC answer key.
If x – y = 8, then which of the following must be true?
- 1. Both x and y must be positive for any value of x and y.
- 2. If x is positive, y must be negative for any value of x and y.
- 3. If x is negative, y must be positive for any value of x and y.
Select the correct answer using the code given below.
Answer (d) & explanation
None of the three statements must be true (d). Step 1: statement 1 says both must be positive. But x = 2, y = −6 gives x − y = 8, with y negative. So 1 fails. Step 2: statement 2 says if x is positive, y must be negative. But x = 10, y = 2 works, with both positive. So 2 fails. Step 3: statement 3 says if x is negative, y must be positive. But y = x − 8, so if x is negative, y is even more negative. For example, x = −1, y = −9. So 3 fails too. Options (a), (b) and (c) fail.
Tip: to test "must be true", find one example that breaks it. Try a mix of positive, negative and zero values.
Answer as per the official UPSC answer key.
Consider the following sum :
• + 1• + 2• + •3 + •1 = 21•
In the above sum, • stands for
Answer (d) & explanation
1. Test the options by putting the digit in place of •.
2. Try 8: 8 + 18 + 28 + 83 + 81 = 218. The answer has the form 21• with • = 8. Works.
3. Try 4: 4 + 14 + 24 + 43 + 41 = 126, not 214. Try 5: 5 + 15 + 25 + 53 + 51 = 149. Try 6: 6 + 16 + 26 + 63 + 61 = 172. All fail.
4. • = 8.
Answer as per the official UPSC answer key.
If X is between –3 and –1, and Y is between –1 and 1, then X2 – Y2 is in between which of the following?
Answer (d) & explanation
X² − Y² lies between 0 and 9 (d). Step 1: squaring removes the minus sign. X is between −3 and −1, so X² is between 1 and 9. Y is between −1 and 1, so Y² is between 0 and 1. Step 2: the largest value of X² − Y² is just under 9 − 0 = 9. Step 3: the smallest is just over 1 − 1 = 0. So the answer is between 0 and 9. Option (a), −9 and 1, and option (b), −9 and −1, forget that squares are never negative. Option (c), 0 and 8, gets the upper end wrong.
Tip: for the range of a difference, make the first part as big as possible and the second as small, then the reverse.
multiplying the end points without squaring first gives the negative options.
Answer as per the official UPSC answer key.
X and Y are natural numbers other than 1, and Y is greater than X. Which of the following represents the largest number?
Answer (a) & explanation
XY is the largest (a). Step 1: try simple numbers that fit: X = 2 and Y = 3. Then XY = 6, X/Y = 0.67, Y/X = 1.5 and (X + Y)/XY = 5/6. Step 2: XY is the biggest here. Step 3: check that it always holds. XY ÷ (Y/X) = X², and since X is more than 1, X² > 1. So XY is always bigger than Y/X. X/Y is less than 1, and (X + Y)/XY is at most 1. So options (b), (c) and (d) are smaller.
Tip: testing small numbers that obey the conditions is a quick way to pick the answer. Then a one-line check confirms it works in general.
Answer as per the official UPSC answer key.
A number consists of three digits of which the middle one is zero and their sum is 4. If the number formed by interchanging the first and last digits is greater than the number itself by 198, then the difference between the first and last digits is
Answer (b) & explanation
The difference between the first and last digits is 2 (b). Step 1: the number has the form a0c, with a zero in the middle. Swapping the end digits changes the number by 99 × (c − a). Step 2: the new number is 198 more, so 99(c − a) = 198, which gives c − a = 2. Step 3: check: the digits add to 4, so a + c = 4. With c − a = 2, we get a = 1 and c = 3. The number is 103, and 301 − 103 = 198. Options (a), (c) and (d) fail.
Tip: for any 3-digit number, swapping the first and last digits changes it by 99 times their difference. This saves solving long equations.
Answer as per the official UPSC answer key.
A bookseller sold 'a' number of Geography textbooks at the rate of ₹ x per book, 'a + 2' number of History textbooks at the rate of ₹ (x + 2) per book and 'a – 2' number of Mathematics textbooks at the rate of ₹ (x – 2) per book. What is his total sale in ₹?
Answer (b) & explanation
The total sale is 3ax + 8 (b). Step 1: add the three sales. Geography: a × x = ax. History: (a + 2)(x + 2) = ax + 2a + 2x + 4. Mathematics: (a − 2)(x − 2) = ax − 2a − 2x + 4. Step 2: add them. The 2a and 2x terms cancel out. Total = 3ax + 8. Step 3: check with numbers: a = 3, x = 3 gives 9 + 25 + 1 = 35, and 3 × 3 × 3 + 8 = 35. Option (a), 3x + 3a, gives 18. Option (c), 9ax, gives 81. Option (d) gives 729.
Tip: putting small numbers into the question and the options is a fast, safe check in algebra questions.
Answer as per the official UPSC answer key.
Six boys A, B, C, D, E and F play a game of cards. Each has a pack of 10 cards. F borrows 2 cards from A and gives away 5 to C who in turn gives 3 to B while B gives 6 to D who passes on 1 to E. Then the number of cards possessed by D and E is equal to the number of cards possessed by
Answer (b) & explanation
D and E together have as many cards as B, C and F (b). Step 1: everyone starts with 10. F borrows 2 from A: A 8, F 12. F gives 5 to C: F 7, C 15. Step 2: C gives 3 to B: C 12, B 13. B gives 6 to D: B 7, D 16. D gives 1 to E: D 15, E 11. Step 3: final count: A 8, B 7, C 12, D 15, E 11, F 7. D + E = 26. B + C + F = 7 + 12 + 7 = 26. Option (a) gives 27. Option (c) gives 22. Option (d) gives 27.
Tip: track each person's count after every move in a small table. Check that the total stays 60; if not, you have made a slip.
Answer as per the official UPSC answer key.
Certain 3-digit numbers have the following characteristics :
- 1. All the three digits are different.
- 2. The number is divisible by 7.
- 3. The number on reversing the digits is also divisible by 7.
How many such 3-digit numbers are there?
Answer (b) & explanation
There are 4 such numbers (b). Step 1: write the number as 100a + 10b + c. Number − reverse = 99(a − c). Both are divisible by 7, so 7 must divide 99(a − c). Since 7 does not divide 99, it must divide a − c. Step 2: the digits are all different, so a ≠ c. So a − c = ±7, giving end digits (1, 8), (8, 1), (2, 9), (9, 2). Step 3: test middle digits for each: 168, 861, 259 and 952 all work (168 = 7 × 24, 259 = 7 × 37). That is 4 numbers. Options (a), (c) and (d) do not fit.
Tip: number − reverse is always a multiple of 99 for 3-digit numbers. This fact unlocks many reversal puzzles.
Answer as per the official UPSC answer key.
The age of Mr. X last year was the square of a number and it would be the cube of a number next year. What is the least number of years he must wait for his age to become the cube of a number again?
Answer (b) & explanation
He must wait 38 years (b). Step 1: his age last year was a square, and next year will be a cube. These two ages are 2 years apart. Look for a square and a cube that differ by 2: 25 and 27. Step 2: so last year he was 25, next year he will be 27, and now he is 26. Step 3: the next cube after 27 is 64. He must wait 64 − 26 = 38 years. Option (a), 42, does not fit. Option (c), 25, and option (d), 16, are squares from the puzzle, not the wait.
Tip: learn small squares and cubes by heart: cubes are 1, 8, 27, 64, 125. Checking pairs from such lists solves these puzzles fast.
64 – 27 = 37 counts from next year, not from now.
Answer as per the official UPSC answer key.
A 2-digit number is reversed. The larger of the two numbers is divided by the smaller one. What is the largest possible remainder?
Answer (d) & explanation
The largest possible remainder is 45 (d). Step 1: when the quotient is 1, the remainder is just the difference of the two numbers. That difference is always 9 × (difference of the digits). Step 2: to make the difference large while the quotient stays 1, the bigger number must be less than twice the smaller. Take digits 9 and 4: 94 ÷ 49 = 1, remainder 45. Step 3: try 95 and 59: 95 ÷ 59 = 1, remainder 36. Bigger digit gaps make the quotient 2 or more and the remainder smaller. Options (a), (b) and (c) are possible remainders but not the largest.
Tip: test the highest option first. If it works and nothing beats it, you are done.
Answer as per the official UPSC answer key.
There are certain 2-digit numbers. The difference between the number and the one obtained on reversing it is always 27. How many such maximum 2-digit numbers are there?
Answer (d) & explanation
None of the options is right (d). Step 1: number − reverse = 9 × (difference of the digits). For a difference of 27, the digits must differ by 3. Step 2: numbers where the tens digit is 3 more than the units digit: 41, 52, 63, 74, 85 and 96. That is 6. Their reverses (14, 25, 36, 47, 58, 69) also differ by 27, if we allow the difference either way, giving 12. Step 3: 30 has reverse 03, which is not a 2-digit number. So the count is 6 or 12, never 3, 4 or 5. Options (a), (b) and (c) fail.
Tip: learn that any 2-digit number minus its reverse is a multiple of 9.
Answer as per the official UPSC answer key.