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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What is the remainder when 91 × 92 × 93 × 94 × 95 × 96 × 97 × 98 × 99 is divided by 1261?
Answer (d) & explanation
The remainder is 0 (d). Step 1: break the divisor into factors: 1261 = 13 × 97. Step 2: look for these factors in the product. 91 = 7 × 13, so 13 is there. 97 itself is one of the numbers. Step 3: so the product is divisible by 13 × 97 = 1261, and the remainder is 0. Options (a), (b) and (c) assume some remainder is left.
Tip: when dividing a long product, first factorise the divisor. If all its factors appear in the product, the remainder is zero.
Answer as per the official UPSC answer key.
Which date of June 2099 among the following is Sunday?
Answer (d) & explanation
1. Use a known day as an anchor: 1 January 2001 was a Monday. From 1 Jan 2001 to 1 Jan 2099 is 98 years, with 24 leap years (2004 to 2096). Odd days = 98 + 24 = 122, which leaves 3 on division by 7. So 1 January 2099 is Thursday.
2. January to May 2099 (not a leap year) = 31 + 28 + 31 + 30 + 31 = 151 days, which leaves 4. So 1 June 2099 is Thursday + 4 = Monday.
3. 1 June is Monday, so 7 June is Sunday.
Answer as per the official UPSC answer key.
What is the smallest number greater than 1000 that when divided by any one of the numbers 6, 9, 12, 15, 18 leaves a remainder of 3?
Answer (c) & explanation
The number is 1083 (c). Step 1: a number leaving remainder 3 for each divisor is 3 more than a common multiple. LCM(6, 9, 12, 15, 18) = 180. Step 2: so the number is 180k + 3. Step 3: 180 × 5 + 3 = 903 is below 1000. 180 × 6 + 3 = 1083 is the first above 1000. Options (a) 1063 and (b) 1073 are not 3 more than a multiple of 180. Option (d), 1183, is not either.
Tip: "same remainder r for all divisors" means the number = LCM × k + r. Find the LCM first.
Answer as per the official UPSC answer key.
A bill for ₹ 1,840 is paid in the denominations of ₹ 50, ₹ 20 and ₹ 10 notes. 50 notes in all are used. Consider the following statements :
- 1. 25 notes of ₹ 50 are used and the remaining are in the denominations of ₹ 20 and ₹ 10.
- 2. 35 notes of ₹ 20 are used and the remaining are in the denominations of ₹ 50 and ₹ 10.
- 3. 20 notes of ₹ 10 are used and the remaining are in the denominations of ₹ 50 and ₹ 20.
Which of the above statements are not correct?
Answer (d) & explanation
1. a + b + c = 50 and 50a + 20b + 10c = 1840. Subtract 10 × the first: 4a + b = 134.
2. 1: a = 25 gives b = 34, but then c = –9. Impossible. 2: b = 35 needs 4a = 99. Not a whole number.
3. 3: c = 20 gives a + b = 30 and 4a + b = 134, so a = 34·67. Impossible. All three are not correct.
the question asks which are not correct.
Answer as per the official UPSC answer key.
Let p be a two-digit number and q be the number consisting of same digits written in reverse order. If p × q = 2430, then what is the difference between p and q?
Answer (d) & explanation
The difference is 9 (d). Step 1: p and q use the same two digits in reverse. Their product 2430 ends in 0, so one digit must be 5 and the other even (or one digit 0, which is not possible here). Step 2: try 45 × 54 = 2430. It works. Step 3: the difference is 54 − 45 = 9. Options (a), (b) and (c) do not fit.
Tip: a two-digit number and its reverse always differ by a multiple of 9 (9 × the difference of digits). Here the digits differ by 1, so the gap is 9.
Answer as per the official UPSC answer key.
Which number amongst 240, 321, 418 and 812 is the smallest?
Answer (b) & explanation
3²¹ is the smallest (b). Step 1: write the powers of 2, 4 and 8 in base 2. 4¹⁸ = 2³⁶ and 8¹² = 2³⁶. So (c) and (d) are equal, and both are smaller than 2⁴⁰ (a). Step 2: compare 3²¹ with 2³⁶. Make the powers equal: 3²¹ = (3⁷)³ = 2187³, and 2³⁶ = (2¹²)³ = 4096³. Step 3: 2187 is less than 4096, so 3²¹ is the smallest. Options (a), (c) and (d) are larger.
Tip: to compare powers, change them to the same base or the same exponent. Also note that two equal options can never be "the smallest" alone.
Answer as per the official UPSC answer key.
Consider the following statements in respect of two natural numbers p and q such that p is a prime number and q is a composite number :
- 1. p × q can be an odd number.
- 2. q/p can be a prime number.
- 3. p + q can be a prime number.
Which of the above statements are correct?
Answer (d) & explanation
All three statements are correct (d). Step 1: "can be" needs just one example each. Step 2: statement 1: take p = 3 (prime) and q = 9 (composite). 3 × 9 = 27, odd. Statement 2: p = 2 and q = 6. 6 ÷ 2 = 3, a prime. Step 3: statement 3: p = 2 and q = 9. 2 + 9 = 11, a prime. So options (a), (b) and (c) fail.
Tip: for "can be" statements, look for one working example. Remember 2 is the only even prime, which often helps.
Answer as per the official UPSC answer key.
The digits 1 to 9 are arranged in three rows in such a way that each row contains three digits, and the number formed in the second row is twice the number formed in the first row; and the number formed in the third row is thrice the number formed in the first row. Repetition of digits is not allowed. If only three of the four digits 2, 3, 7 and 9 are allowed to use in the first row, how many such combinations are possible to be arranged in the three rows?
Answer (c) & explanation
Two arrangements are possible (c). Step 1: the third row is 3 × the first row and must stay a 3-digit number. So the first row is at most 329. Using three of 2, 3, 7, 9, the choices are 237, 239, 273, 279, 293, 297, 327 and 329. Step 2: check that all nine digits 1 to 9 appear once. 273 → 546 → 819 works. 327 → 654 → 981 works. Step 3: the others repeat a digit or produce a 0. So there are 2. Options (a), (b) and (d) miscount.
Tip: first use a size limit (×3 below 1000) to shrink the list. Then test the few candidates left.
Answer as per the official UPSC answer key.
Consider the following statements :
- 1. Between 3:16 p.m. and 3:17 p.m., both hour hand and minute hand coincide.
- 2. Between 4:58 p.m. and 4:59 p.m., both minute hand and second hand coincide.
Which of the above statements is/are correct?
Answer (c) & explanation
Both statements are correct (c). Step 1: after 3 o'clock, the hands meet at (60 ÷ 11) × 3 = 16 4/11 minutes. That is about 3:16:22, between 3:16 and 3:17 (1 is true). Step 2: at 4:58:00, the minute hand is at the 58-minute mark and the second hand at 0. The second hand moves about 59 marks per minute faster. Step 3: it catches up in 58/59 minute, about 59 seconds, near 4:58:59. That is between 4:58 and 4:59 (2 is true). So options (a), (b) and (d) fail.
Tip: the hour and minute hands meet every 65 5/11 minutes. After H o'clock, they meet at (60/11) × H minutes.
Answer as per the official UPSC answer key.
On one side of a 1·01 km long road, 101 plants are planted at equal distance from each other. What is the total distance between 5 consecutive plants?
Answer (b) & explanation
The distance is 40.4 m (b). Step 1: 101 plants in a line leave 100 gaps between them. The road is 1.01 km = 1010 m. Step 2: so each gap = 1010 ÷ 100 = 10.1 m. Step 3: 5 plants in a row have 4 gaps between them. 4 × 10.1 = 40.4 m. Option (d), 50.5 m, counts 5 gaps. Options (a) and (c) use 10 m per gap.
Tip: n posts in a line have n − 1 gaps. This "fence-post" rule is a common UPSC trap.
5 plants have 4 gaps, not 5.
Answer as per the official UPSC answer key.
If 15 × 14 × 13 × … × 3 × 2 × 1 = 3m × n where m and n are positive integers, then what is the maximum value of m?
Answer (b) & explanation
The maximum value of m is 6 (b). Step 1: count how many times 3 divides 15!. Multiples of 3 up to 15: 3, 6, 9, 12, 15. That gives 5 threes. Step 2: 9 = 3 × 3 has an extra 3. That adds 1 more. Step 3: m = 5 + 1 = 6. Option (c), 5, forgets the extra 3 in 9. Options (a) and (d) miscount.
Tip: to count a prime p in n!, add n/p + n/p² + ... (whole parts only). Here 15/3 + 15/9 = 5 + 1 = 6.
don't forget the extra 3 in 9.
Answer as per the official UPSC answer key.
A has some coins. He gives half of the coins and 2 more to B. B gives half of the coins and 2 more to C. C gives half of the coins and 2 more to D. The number of coins D has now, is the smallest two-digit number. How many coins does A have in the beginning?
Answer (d) & explanation
A had 52 coins at the start (d). Step 1: the smallest two-digit number is 10, so D has 10. Work backwards. Step 2: D got half of C's coins plus 2. So half of C's coins = 8, and C had 16. Step 3: C got half of B's plus 2, so half of B's = 14 and B had 28. B got half of A's plus 2, so half of A's = 26 and A had 52. Options (a), (b) and (c) fail this check.
Tip: when a chain of "half plus 2" ends at a known number, reverse each step: subtract 2, then double.
Answer as per the official UPSC answer key.
The sum of three consecutive integers is equal to their product. How many such possibilities are there?
Answer (c) & explanation
There are three such sets (c). Step 1: integers include 0 and negative numbers. Try small sets in a row. Step 2: 1, 2, 3: sum 6, product 6. Works. −1, 0, 1: sum 0, product 0. Works. −3, −2, −1: sum −6, product −6. Works. Step 3: bigger sets fail; for 2, 3, 4 the sum is 9 but the product is 24. The product grows much faster. So there are three. Option (a) counts only positive numbers.
Tip: "integers" means negatives and zero too. UPSC often hides extra answers there.
integers include zero and negatives. If you check only natural numbers, you find just one.
Answer as per the official UPSC answer key.
How many seconds in total are there in x weeks, x days, x hours, x minutes and x seconds?
Answer (d) & explanation
The total is 694861x seconds (d). Step 1: 1 week = 604800 s, 1 day = 86400 s, 1 hour = 3600 s, 1 minute = 60 s, plus 1 s. Step 2: add them: 604800 + 86400 + 3600 + 60 + 1 = 694861. So the total is 694861x. Step 3: options (a) and (b) leave out the week. Option (c), 694860x, forgets the last x seconds.
Tip: a quick check: "+ x seconds" makes the answer end in 1, which leaves (b) or (d). One week alone is over 6 lakh seconds, so (d).
Answer as per the official UPSC answer key.
Five friends P, Q, X, Y and Z purchased some notebooks. The relevant information is given below :
- 1. Z purchased 8 notebooks more than X did.
- 2. P and Q together purchased 21 notebooks.
- 3. Q purchased 5 notebooks less than P did.
- 4. X and Y together purchased 28 notebooks.
- 5. P purchased 5 notebooks more than X did.
If each notebook is priced ₹ 40, then what is the total cost of all the notebooks?
Answer (a) & explanation
The total cost is ₹ 2,600 (a). Step 1: P + Q = 21 and Q = P − 5. So 2P − 5 = 21, giving P = 13 and Q = 8. Step 2: P bought 5 more than X, so X = 8. Z bought 8 more than X, so Z = 16. X + Y = 28, so Y = 20. Step 3: total = 13 + 8 + 8 + 20 + 16 = 65 notebooks. 65 × ₹ 40 = ₹ 2,600. Options (b), (c) and (d) come from wrong values.
Tip: start with the pair of equations that has only two unknowns (P and Q). Each answer then unlocks the next person.
Answer as per the official UPSC answer key.
X said to Y, "At the time of your birth I was twice as old as you are at present." If the present age of X is 42 years, then consider the following statements :
- 1. 8 years ago, the age of X was five times the age of Y.
- 2. After 14 years, the age of X would be two times the age of Y.
Which of the above statements is/are correct ?
Answer (b) & explanation
Only statement 2 is correct (b). Step 1: when Y was born, X's age equalled the age gap. That gap is twice Y's present age. So X − Y = 2Y, which means X = 3Y. With X = 42, Y = 14. Step 2: statement 1: 8 years ago, X was 34 and Y was 6. 34 is not 5 × 6 = 30, so it is false. Step 3: statement 2: after 14 years, X will be 56 and Y 28. 56 = 2 × 28, so it is true. Options (a), (c) and (d) fail.
Tip: the age gap between two people never changes. "At your birth I was ..." gives you that gap directly.
Answer as per the official UPSC answer key.
Which day is 10th October, 2027 ?
Answer (a) & explanation
10 October 2027 is a Sunday (a). Step 1: use an anchor date: 10 October 2021 was a Sunday (the CSAT exam was held that day). Step 2: count odd days from 10 Oct 2021 to 10 Oct 2027. That is 6 years, and each normal year adds 1 odd day. Only one 29 February falls in between (2024), adding 1 more. Total = 6 + 1 = 7 odd days = 0 extra. Step 3: so the weekday stays the same: Sunday. Options (b), (c) and (d) come from missing or adding the leap day.
Tip: a normal year shifts the weekday by 1, a leap year by 2. Count only the 29 Februarys that fall inside your range.
Answer as per the official UPSC answer key.
When a certain number is multiplied by 7, the product entirely comprises ones only (1111...). What is the smallest such number ?
Answer (d) & explanation
The smallest number is 15873 (d). Step 1: multiply each option by 7 and look for an answer made only of ones. Step 2: 15873 × 7 = 111111. That works. Step 3: the others fail: 15713 × 7 = 109991, 15723 × 7 = 110061 and 15783 × 7 = 110481. Also, 111111 is the smallest all-ones number divisible by 7 (1, 11, 111, 1111 and 11111 are not). So (d).
Tip: 111111 = 3 × 7 × 11 × 13 × 37. It is a handy number: divisible by 7, 11 and 13 together.
Answer as per the official UPSC answer key.
Jay and Vijay spent an equal amount of money to buy some pens and special pencils of the same quality from the same store. If Jay bought 3 pens and 5 pencils, and Vijay bought 2 pens and 7 pencils, then which one of the following is correct ?
Answer (c) & explanation
A pen costs twice as much as a pencil (c). Step 1: both spent the same amount. So 3 pens + 5 pencils = 2 pens + 7 pencils. Step 2: take away 2 pens and 5 pencils from both sides. What is left: 1 pen = 2 pencils. Step 3: so a pen costs two times a pencil. Option (a) reverses it; option (b), equal prices, and option (d), three times, do not fit the equation.
Tip: when two purchases cost the same, cancel the common items. What remains is a direct price comparison, with no need for algebra.
Answer as per the official UPSC answer key.
A person P asks one of his three friends X as to how much money he had. X replied, "If Y gives me ₹ 40, then Y will have half of as much as Z, but if Z gives me ₹ 40, then three of us will have equal amount." What is the total amount of money that X, Y and Z have ?
Answer (b) & explanation
The total is ₹ 360 (b). Step 1: use the second condition first. If Z gives X ₹ 40, all three are equal. So X + 40 = Y = Z − 40. Step 2: the first condition: if Y gives X ₹ 40, Y has half of Z. So Y − 40 = Z ÷ 2. Put Y = Z − 40: Z − 80 = Z ÷ 2, so Z = 160. Step 3: then Y = 120 and X = 80. Total = 160 + 120 + 80 = ₹ 360. Options (a), (c) and (d) do not satisfy both conditions.
Tip: start with the condition that gives the simplest link ("all equal"). Then plug the answer options back in to check quickly.
Answer as per the official UPSC answer key.