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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At which one of the following times, do the hour hand and the minute hand of the clock make an angle of 180° with each other ?
Answer (d) & explanation
The hands are opposite between 7:05 and 7:10 (d). Step 1: the angle between the hands is |30H − 5.5M|, where H is the hour and M the minutes. At 7:00 it is 30 × 7 = 210°, so (a) fails. Step 2: as minutes pass, the minute hand closes the gap at 5.5° per minute. We need 210 − 5.5M = 180, so M = 30 ÷ 5.5 ≈ 5.45 minutes. Step 3: so the hands are 180° apart at about 7:05:27. That is just after 7:05, so (b) and (c) fail.
Tip: the minute hand gains 5.5° per minute on the hour hand. This one fact solves most clock-angle questions.
at 7:00 the angle is 210°, not 180°. The minute hand has to catch up 30°.
Answer as per the official UPSC answer key.
In an objective type test of 90 questions, 5 marks are allotted for every correct answer and 2 marks are deducted for every wrong answer. After attempting all the 90 questions, a student got a total of 387 marks. What is the number of incorrect responses ?
Answer (a) & explanation
The student gave 9 wrong answers (a). Step 1: if all 90 were right, the score would be 90 × 5 = 450. Step 2: each wrong answer loses the 5 marks it would have earned and costs 2 more. So each wrong answer reduces the score by 7. Step 3: the score fell by 450 − 387 = 63. So wrong answers = 63 ÷ 7 = 9. Check: 81 × 5 − 9 × 2 = 405 − 18 = 387. Options (b), (c) and (d) fail this check.
Tip: in negative-marking questions, assume all are correct, then divide the shortfall by (marks for right + penalty for wrong).
Answer as per the official UPSC answer key.
Consider the following addition problem : 3P + 4P + PP + PP = RQ2; where P, Q and R are different digits. What is the arithmetic mean of all such possible sums ?
Answer (c) & explanation
The mean of the possible sums is 202 (c). Step 1: write out the sum: 3P + 4P + PP + PP = (30 + P) + (40 + P) + 11P + 11P = 70 + 24P. Step 2: the sum RQ2 ends in 2. So 24P must end in 2, which happens when P = 3 (sum 142) or P = 8 (sum 262). In both, P, Q and R are different digits. Step 3: the mean is (142 + 262) ÷ 2 = 202. Options (a), (b) and (d) do not match.
Tip: in digit puzzles, start with the units place. The last digit of a sum usually narrows the choices to one or two.
Answer as per the official UPSC answer key.
If 32019 is divided by 10, then what is the remainder ?
Answer (c) & explanation
The remainder is 7 (c). Step 1: the remainder on dividing by 10 is just the unit digit. Step 2: powers of 3 end in a repeating cycle: 3¹ = 3, 3² = 9, 3³ = 27, 3⁴ = 81. The cycle is 3, 9, 7, 1, of length 4. Step 3: 2019 ÷ 4 leaves a remainder of 3 (2019 = 4 × 504 + 3). So take the 3rd term of the cycle: 7. Option (b), 3, is the first term. Options (a) and (d) are the 4th and 2nd terms.
Tip: unit digits of powers repeat every 4 (or fewer). Divide the power by 4 and use the remainder; a remainder of 0 means the 4th term.
Answer as per the official UPSC answer key.
Consider the following multiplication problem :
(PQ) × 3 = RQQ, where P, Q and R are different digits and R ≠ 0.
What is the value of (P + R) ÷ Q ?
Answer (b) & explanation
(P + R) ÷ Q is 2 (b). Step 1: Q × 3 must end in Q. Only Q = 0 (0 × 3 = 0) and Q = 5 (5 × 3 = 15) work. Step 2: if Q = 0, then P0 × 3 = R00, which no single digit P can give. So Q = 5. Step 3: try P5 × 3 until the result ends in 55: 85 × 3 = 255. So P = 8 and R = 2. Then (P + R) ÷ Q = 10 ÷ 5 = 2. Option (d) is wrong because the data do fix the answer.
Tip: look for digits that repeat themselves when multiplied (0, 1, 5, 6). They often unlock such puzzles quickly.
Answer as per the official UPSC answer key.
The number 3798125P369 is divisible by 7. What is the value of the digit P ?
Answer (b) & explanation
P is 6 (b). Step 1: a test for 7 (it also works for 11 and 13): from the right, split the number into groups of three digits and add and subtract them in turn. 3798125P369 splits into 37 | 981 | 25P | 369. Step 2: 369 − 25P + 981 − 37 = 1313 − (250 + P) = 1063 − P. This must be divisible by 7. Step 3: 1063 = 7 × 151 + 6. So 1063 − P is divisible by 7 when P = 6. Options (a), (c) and (d) leave remainders.
Tip: the alternating three-digit sum works because 1001 = 7 × 11 × 13. It is the fastest check for large numbers.
Answer as per the official UPSC answer key.
Consider the following statements :
- 1. The sum of 5 consecutive integers can be 100.
- 2. The product of three consecutive natural numbers can be equal to their sum.
Which of the above statements is/are correct ?
Answer (c) & explanation
Both statements are correct (c). Step 1: the sum of 5 consecutive integers is 5 × the middle one. 100 = 5 × 20, so 18 + 19 + 20 + 21 + 22 = 100. Statement 1 is true. Step 2: for statement 2, try small natural numbers: 1 × 2 × 3 = 6, and 1 + 2 + 3 = 6. They are equal, so statement 2 is true. Step 3: so options (a), (b) and (d) fail.
Tip: for consecutive numbers, the sum = count × middle value (with an odd count). For "can be" statements, one example is enough to prove it true.
Answer as per the official UPSC answer key.
From January 1, 2021, the price of petrol (in Rupees per litre) on mth day of the year is 80 + 0·1m, where m = 1, 2, 3, ..., 100 and thereafter remains constant. On the other hand, the price of diesel (in Rupees per litre) on nth day of 2021 is 69 + 0·15n for any n. On which date in the year 2021 are the prices of these two fuels equal ?
Answer (b) & explanation
The prices are equal on 20th May (b). Step 1: in the first 100 days, petrol = 80 + 0·1m and diesel = 69 + 0·15n. Setting them equal gives 0·05n = 11, so n = 220. That is outside the first 100 days, so there is no match there. Step 2: after day 100, petrol stays at 80 + 10 = ₹ 90. Diesel = 90 when 69 + 0·15n = 90, so n = 140. Step 3: day 140 of 2021: January to April = 31 + 28 + 31 + 30 = 120 days. Day 140 is the 20th of May. Options (a), (c) and (d) are off by a day.
Tip: always check the range where a formula applies. 2021 is not a leap year, so February has 28 days.
the petrol formula stops at day 100. Using it beyond that gives a wrong day.
Answer as per the official UPSC answer key.
Joseph visits the club on every 5th day, Harsh visits on every 24th day, while Sumit visits on every 9th day. If all three of them met at the club on a Sunday, then on which day will all three of them meet again ?
Answer (b) & explanation
They meet again on a Wednesday (b). Step 1: they meet again after the LCM of 5, 24 and 9. 24 = 2³ × 3 and 9 = 3², so the LCM = 2³ × 3² × 5 = 360 days. Step 2: find the weekday shift: 360 ÷ 7 = 51 weeks and 3 days left over. Step 3: Sunday + 3 days = Wednesday. Option (d), Sunday, ignores the extra days. Options (a) and (c) miscount.
Tip: "every nth day" problems use the LCM. Then divide by 7 and use the remainder to move the weekday forward.
Answer as per the official UPSC answer key.
A boy plays with a ball and he drops it from a height of 1·5 m. Every time the ball hits the ground, it bounces back to attain a height 4/5th of the previous height. The ball does not bounce further if the previous height is less than 50 cm. What is the number of times the ball hits the ground before the ball stops bouncing ?
Answer (b) & explanation
The ball hits the ground 5 times (b). Step 1: each bounce reaches 4/5 of the previous height. It starts at 150 cm. Step 2: hit 1 → rises to 120 cm. Hit 2 → 96 cm. Hit 3 → about 77 cm. Hit 4 → about 61 cm. Hit 5 → about 49 cm. Step 3: after the fifth hit, the height is below 50 cm, so it does not bounce again. The ball hits the ground 5 times. Option (a), 4, stops too early. Options (c) and (d) go too far.
Tip: list each step in a small table. Converting metres to centimetres first (1·5 m = 150 cm) avoids decimal errors.
Some readers count one more landing after the last small bounce; the official answer is (b), as printed.
Answer as per the official UPSC answer key.
The difference between a 2-digit number and the number obtained by interchanging the positions of the digits is 54.
Consider the following statements :
- 1. The sum of the two digits of the number can be determined only if the product of the two digits is known.
- 2. The difference between the two digits of the number can be determined.
Which of the above statements is/are correct ?
Answer (b) & explanation
1. Swapping the two digits of a number always changes it by 9 × (the gap between the digits). Example: 71 − 17 = 54 = 9 × 6.
2. So 54 ÷ 9 = 6: the digits differ by 6. Statement 2 is correct.
3. The number could be 71, 82 or 93, with digit sums 8, 10 and 12. The sum is not fixed.
4. Statement 1 says the sum can be found 'only if' the product is known. Other facts could also fix it, so 'only if' is wrong. Statement 1 is incorrect.
Answer as per the official UPSC answer key.
Consider all 3-digit numbers (without repetition of digits) obtained using three non-zero digits which are multiples of 3. Let S be their sum. Which of the following is/are correct ?
- 1. S is always divisible by 74.
- 2. S is always divisible by 9.
Select the correct answer using the code given below :
Answer (c) & explanation
Both statements are correct (c). Step 1: the non-zero digits that are multiples of 3 are 3, 6 and 9. Without repeating digits, they make 3 × 2 × 1 = 6 numbers. Step 2: each digit appears in each place (hundreds, tens, units) twice. So S = 2 × (3 + 6 + 9) × 111 = 36 × 111 = 3996. Step 3: 3996 = 74 × 54, so it is divisible by 74 (statement 1). Its digits add up to 27, so it is divisible by 9 (statement 2). So options (a), (b) and (d) fail.
Tip: for the sum of all numbers made from given digits, use (number of times each digit sits in a place) × (sum of digits) × 111.
Answer as per the official UPSC answer key.
How many five-digit prime numbers can be obtained by using all the digits 1, 2, 3, 4 and 5 without repetition of digits?
Answer (a) & explanation
No such prime exists, so the answer is zero (a). Step 1: any number made from all of 1, 2, 3, 4 and 5 has a digit sum of 1 + 2 + 3 + 4 + 5 = 15. Step 2: 15 is divisible by 3, so every such number is divisible by 3. Step 3: a number divisible by 3 (and bigger than 3) is not prime. So there are zero such primes. Options (b), (c) and (d) fail.
Tip: before counting arrangements, check the digit sum. If it is divisible by 3, no arrangement can be prime. This trick saves you from listing 120 numbers.
Answer as per the official UPSC answer key.
What is the remainder when 51 × 27 × 35 × 62 × 75 is divided by 100?
Answer (a) & explanation
The remainder is 50 (a). Step 1: the remainder on dividing by 100 is just the last two digits. So keep only the last two digits at each step. Step 2: 35 × 62 = 2170 → 70. 70 × 75 = 5250 → 50. 50 × 27 = 1350 → 50. 50 × 51 = 2550 → 50. Step 3: the remainder is 50. Option (b), 25, and option (c), 5, come from multiplying only the 5-ending numbers. Option (d) fails.
Tip: an even number times a multiple of 25 gives a multiple of 50. Here 62 is even and 75 is a multiple of 25, and the other factors are odd, so the product ends in 50.
Answer as per the official UPSC answer key.
In the sum
⊗ + 1⊗ + 5⊗ + ⊗⊗ + ⊗1 = 1⊗⊗
for which digit does the symbol ⊗ stand?
Answer (b) & explanation
The symbol stands for 3 (b). Step 1: test the options. Put ⊗ = 3. Step 2: left side: 3 + 13 + 53 + 33 + 31 = 133. Step 3: right side: 1⊗⊗ = 133. Both sides match, so ⊗ = 3. Check another: ⊗ = 2 gives 2 + 12 + 52 + 22 + 21 = 109, but 1⊗⊗ = 122. No match. Options (a), (c) and (d) fail.
Tip: when an equation has one unknown digit and four options, plugging in is quickest. Algebra also works: the left side is 24⊗ + 61 and the right side is 100 + 11⊗, so 13⊗ = 39 and ⊗ = 3.
Trying the four options is quicker than any algebra here.
Answer as per the official UPSC answer key.
For what value of n, the sum of digits in the number (10n + 1) is 2?
Answer (b) & explanation
The digit sum is 2 for any whole number n (b). Step 1: for n ≥ 1, 10ⁿ + 1 looks like 100…01: one 1, some zeros, then another 1. Digit sum = 2. Step 2: for n = 0, 10⁰ = 1, so 10⁰ + 1 = 2. Digit sum is also 2. Step 3: so it works for every whole number (0, 1, 2, …). Option (c) leaves out 0. Option (a) is too narrow. Option (d) fails because a fractional n, like 0·5, gives a non-whole number.
Tip: whole numbers include 0; natural (positive) numbers start at 1. CSAT often tests this difference, so always check n = 0 separately.
don't forget n = 0. That is the difference between (b) and (c).
Answer as per the official UPSC answer key.
If you have two straight sticks of length 7·5 feet and 3·25 feet, what is the minimum length can you measure?
Answer (b) & explanation
The minimum length you can measure is 0·25 foot (b). Step 1: by adding and subtracting the two sticks, you can measure any multiple of their HCF, and nothing smaller. Step 2: write both in quarter-feet: 7·5 feet = 30 quarters; 3·25 feet = 13 quarters. Step 3: HCF(30, 13) = 1 quarter, which is 0·25 foot. For example, 30 − 2 × 13 = 4 quarters, and 13 − 3 × 4 = 1 quarter. Option (a), 0·05, is too small. Options (c) and (d) are not the minimum.
Tip: the smallest measurable length with two rods is their HCF. Convert decimals to whole units first to find it easily.
Answer as per the official UPSC answer key.
How many pairs of natural numbers are there such that the difference of whose squares is 63?
Answer (a) & explanation
There are 3 such pairs (a). Step 1: x² − y² = (x − y)(x + y) = 63. Both factors must be odd, because 63 is odd, and x + y is the larger factor. Step 2: factor pairs of 63: 1 × 63, 3 × 21, 7 × 9. Step 3: solve each. 1 and 63: x = 32, y = 31. 3 and 21: x = 12, y = 9. 7 and 9: x = 8, y = 1. That is 3 pairs. Check: 64 − 1 = 63. Options (b), (c) and (d) miscount.
Tip: for x² − y² = N, count the factor pairs of N with the same parity (both odd or both even). Each gives one solution.
Answer as per the official UPSC answer key.
One page is torn from a booklet whose pages are numbered in the usual manner starting from the first page as 1. The sum of the numbers on the remaining pages is 195. The torn page contains which of the following numbers?
Answer (b) & explanation
The torn page has 7 and 8 (b). Step 1: find the total before tearing. 1 + 2 + … + 20 = 20 × 21 ÷ 2 = 210. 1 to 19 gives only 190, which is less than 195, so the booklet had 20 pages. Step 2: missing sum = 210 − 195 = 15. Step 3: a torn leaf has two consecutive numbers that add to 15: 7 + 8. Option (a), 5 + 6 = 11. Option (c), 9 + 10 = 19. Option (d), 11 + 12 = 23. None of these is 15.
Tip: use n(n + 1) ÷ 2 for the sum of 1 to n. Find the smallest total just above the given sum; the difference is the torn page.
Answer as per the official UPSC answer key.
Which one of the following will have minimum change in its value if 5 is added to both numerator and the denominator of the fractions 2/3, 3/4, 4/5 and 5/6?
Answer (d) & explanation
5/6 changes the least (d). Step 1: adding the same number to the top and bottom pulls a fraction towards 1. The fraction already closest to 1 moves the least. Step 2: among 2/3, 3/4, 4/5 and 5/6, the closest to 1 is 5/6. Step 3: check: 5/6 → 10/11, a change of 5/66 (about 0·076). 2/3 → 7/8, a change of 5/24 (about 0·21). So 5/6 changes least. Options (a), (b) and (c) change more.
Tip: the change for a/b with k added is k(b − a) ÷ [b(b + k)]. A bigger b means a smaller change, which favours 5/6.
Answer as per the official UPSC answer key.