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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A is a 2-digit number with different digits. B is also a 2-digit number and is obtained by reversing the digits of A. If A – B is a multiple of 27, where A > B, how many such different A's are possible?
Answer (b) & explanation
Nine such numbers are possible (b). Step 1: let A = 10a + b and B = 10b + a, with a > b. Then A − B = 9(a − b). Step 2: for this to be a multiple of 27, a − b must be a multiple of 3: 3, 6 or 9. Step 3: a − b = 3: b = 1 to 6 (A = 41, 52, 63, 74, 85, 96): 6 numbers. a − b = 6: b = 1 to 3 (A = 71, 82, 93): 3. a − b = 9 needs b = 0, making B a one-digit number (09), so it is not allowed. Total = 9. Options (a), (c) and (d) fail.
Tip: A number minus its reverse is always 9 × (digit difference). B must also be two-digit, so the units digit of A cannot be 0.
B must also be a 2-digit number, so 90 is out.
Answer as per the official UPSC answer key.
If 10m × 1000 × n = 7525 × 2532 × 3275, where n is not divisible by 10, then the value of m is
Answer (b) & explanation
m is 111 (b). Step 1: each 10 needs one 2 and one 5. Count the 5s: 75²⁵ = (3 × 5²)²⁵ = 3²⁵ × 5⁵⁰, and 25³² = 5⁶⁴. Total 5s = 50 + 64 = 114. Step 2: count the 2s: 32⁷⁵ = (2⁵)⁷⁵ = 2³⁷⁵. Far more than 114. Step 3: so the right side has exactly 114 factors of 10 (n has no more, since n is not divisible by 10). The left side has 10ᵐ × 1000 = 10^(m + 3). So m + 3 = 114 and m = 111. Options (a), (c) and (d) fail.
Tip: the number of 10s equals the smaller count of 2s and 5s. Do not forget the extra 10s hidden in 1000.
remember the 1000 on the left: it already holds 103.
Answer as per the official UPSC answer key.
For 1/3 < x < y < 2, which of the following statements is/are always correct?
- I. x + 1/x < y + 1/y
- II. √(1 + y2)/y < √(1 + x2)/x
Select the answer using the code given below.
Answer (b) & explanation
Only statement II is always correct (b). Step 1: statement I: t + 1/t falls until t = 1, then rises. So it is not always increasing in this range. Test x = 1/2, y = 1: x + 1/x = 2.5, y + 1/y = 2. Here 2.5 is not less than 2, so I fails. Step 2: statement II: √(1 + y²)/y = √(1/y² + 1). As y grows, 1/y² shrinks, so this value gets smaller. Since x < y, the y-value is always smaller than the x-value. II always holds. Step 3: options (a), (c) and (d) fail.
Tip: "always correct" needs to hold for every case; one counter-example kills it. Rewrite expressions to see if they rise or fall.
the range crosses 1, where t + 1/t turns around. "Always" fails with one counter-example.
Answer as per the official UPSC answer key.
What is the minimum number of times one needs to measure to get 298 litres of water from a tank, if the measuring cylinders have capacities 1 litre, 6 litres, 25 litres and 100 litres?
Answer (b) & explanation
The minimum is 5 measures (b). Step 1: adding only, the best is 2 × 100 + 2 × 25 + 8 × 6 = 298, which needs 12 measures. That is not an option, so think about taking water out too. Step 2: 298 = 3 × 100 − 2 × 1. Pour in 100 litres three times, then remove 1 litre twice: 5 measures. Step 3: could 4 work? 3 × 100 leaves 2 litres to remove, and no single cylinder holds 2 litres, so 4 is impossible. Options (a), (c) and (d) fail.
Tip: when adding gives too many steps, try overshooting and subtracting. Measuring questions often allow both pouring in and taking out.
Answer as per the official UPSC answer key.
Three variables x, y and z take values 2, 3, 4 or 5 such that their values are always distinct. If M and N denote the largest possible value and the smallest possible value, respectively, for the expression {(x × y) + z}; then M – N is
Answer (c) & explanation
M − N is 13 (c). Step 1: to make (x × y) + z largest, multiply the two biggest numbers and add the next one: M = 5 × 4 + 3 = 23. (Check: 5 × 3 + 4 = 19 and 4 × 3 + 5 = 17 are smaller.) Step 2: to make it smallest, multiply the two smallest and add the next smallest: N = 2 × 3 + 4 = 10. (Check: 2 × 4 + 3 = 11 is bigger.) Step 3: M − N = 23 − 10 = 13. Options (a), (b) and (d) come from other pairings.
Tip: multiplication has a bigger effect than addition, so give the product the biggest (or smallest) numbers. Quickly check one or two other pairings to be safe.
Answer as per the official UPSC answer key.
In an objective type question paper, 5 marks are awarded for a correct answer and 2 marks are deducted for a wrong answer. A student attempted all the questions and got a score of 69. Had he been awarded 4 marks for a correct answer and 1 mark deducted for a wrong answer, he would have scored 84. How many questions were there in the question paper?
Answer (b) & explanation
There were 81 questions (b). Step 1: let c be correct and w wrong answers. First scheme: 5c − 2w = 69. Second scheme: 4c − w = 84. Step 2: from the second, w = 4c − 84. Put it in the first: 5c − 2(4c − 84) = 69, so 5c − 8c + 168 = 69, giving 3c = 99 and c = 33. Then w = 132 − 84 = 48. Step 3: he attempted all questions, so total = 33 + 48 = 81. Check: 165 − 96 = 69 ✓; 132 − 48 = 84 ✓. Options (a), (c) and (d) fail.
Tip: two marking schemes give two equations. Solve, then check both before answering.
Answer as per the official UPSC answer key.
The digit in the unit place of the number 6129 × 7307 is
Answer (c) & explanation
The unit digit is 8 (c). Step 1: every power of 6 ends in 6 (6, 36, 216, …). Step 2: powers of 7 end in a cycle of four: 7, 9, 3, 1. Divide 307 by 4: remainder 3. So 7³⁰⁷ ends in the third digit of the cycle: 3. Step 3: 6 × 3 = 18, so the unit digit is 8. Options (a), (b) and (d) use the wrong place in the cycle.
Tip: digits 0, 1, 5 and 6 keep the same unit digit for every power. For others, find the cycle of four and use the remainder of the power divided by 4.
Answer as per the official UPSC answer key.
How many three-digit numbers can be expressed as an integral power of 2?
Answer (c) & explanation
Three three-digit numbers are powers of 2 (c). Step 1: keep doubling from 1: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024. Step 2: only 128 (2⁷), 256 (2⁸) and 512 (2⁹) have three digits. 1024 (2¹⁰) has four. Step 3: so the count is 3. Options (a), (b) and (d) miscount.
Tip: learn powers of 2 up to 2¹⁰ = 1024. Between any two powers of 10 there are always three or four powers of 2, since 2¹⁰ is about 10³.
Answer as per the official UPSC answer key.
How many times does 5 appear in all two-digit positive integers?
Answer (b) & explanation
The digit 5 appears 19 times (b). Step 1: 5 in the tens place: 50, 51, …, 59 = 10 times. Step 2: 5 in the units place: 15, 25, 35, 45, 55, 65, 75, 85, 95 = 9 times (05 is not a two-digit number). Step 3: total = 10 + 9 = 19. The number 55 has two 5s, and both are counted. Option (a), 18, counts 55 once. Options (c) and (d) wrongly add 05 or 5.
Tip: count place by place. The question asks how many times the digit appears, not how many numbers contain it (that would be 18).
Answer as per the official UPSC answer key.
The weight of X, in kg, is denoted by X. The weights of A, B, C, D, P, Q, R and S are measured. Given :
A + B + C + D = 17
A + C = 6
P + Q + S + D = 15
P + Q + R + B = 17
P = R and Q = S
Which one of the following statements is correct?
Answer (b) & explanation
1. From the first two equations: B + D = 11.
2. Put R = P and S = Q, then add the last two equations: 3P + 3Q + B + D = 32, so 3(P + Q) = 21 and P + Q = 7.
3. 7 > 6 = A + C, so (b) is true. (a) fails because 11 > 7. P and Q separately cannot be fixed, so (c) and (d) cannot be claimed.
Answer as per the official UPSC answer key.
If the product of the HCF and LCM of two distinct numbers is the cube of one of the numbers, then which of the following statements is/are correct?
- I. The difference of the numbers is an even number.
- II. One of the numbers is a perfect square.
Select the answer using the code given below.
Answer (c) & explanation
Both statements are correct (c). Step 1: for any two numbers, HCF × LCM = product of the numbers. So a × b = a³ (the cube of one of them), giving b = a². Step 2: statement II: b = a² is a perfect square. True. Step 3: statement I: the difference is a² − a = a(a − 1), the product of two consecutive numbers. One of any two consecutive numbers is even, so the product is always even. True. Example: 3 and 9: HCF 3 × LCM 9 = 27 = 3³; difference 6 is even, and 9 is a square. Options (a), (b) and (d) fail.
Tip: remember HCF × LCM = a × b for two numbers. It turns many HCF-LCM puzzles into simple algebra.
Answer as per the official UPSC answer key.
X, Y and Z jump forward 4', 6' and 5', respectively. At 8 AM, they all land on mark 199'. How many times will they all land on the same mark (need not be at the same moment) between mark 195' and 1000', if all of them cross mark 1000' by 9 AM?
Answer (d) & explanation
They land on a common mark 14 times (d). Step 1: all three are at 199' together. After that, they share a mark every LCM(4, 6, 5) = 60'. So common marks are 199, 259, 319, … Step 2: the largest common mark up to 1000' is 199 + 60k ≤ 1000, so k ≤ 13 (199 + 780 = 979). Step 3: k = 0 to 13 gives 14 marks. The mark before 199 would be 139', which is below 195', so it does not count. Options (a), (b) and (c) miss the starting mark or the end.
Tip: common points repeat at the LCM. Count from the first shared point, and include both ends when they fall inside the range.
199' itself counts. It lies between 195' and 1000'.
Answer as per the official UPSC answer key.
If x and y are two digits and the number 4x5y790 is divisible by 11, then what is the remainder, if x + y is divided by 11?
Answer (d) & explanation
1. Test for 11: add every second digit, add the others, and the two totals must differ by 0 or 11.
2. Digits 4, x, 5, y, 7, 9, 0. First group: 4 + 5 + 7 + 0 = 16. Second group: x + y + 9.
3. So 16 − (x + y + 9) = 7 − (x + y) must be 0 or −11. That gives x + y = 7 or x + y = 18.
4. 7 ÷ 11 leaves 7, and 18 ÷ 11 also leaves 7. Either way the remainder is 7.
Answer as per the official UPSC answer key.
A toy T jumps forward or backward. In each forward jump, it moves 5' forward whereas in each backward jump, it moves 2' backward. If in 31 jumps, T moves exactly 15' forward, then what is the difference of the number of forward and backward jumps?
Answer (d) & explanation
The difference is 9 jumps (d). Step 1: let f be forward jumps and b backward jumps. Total jumps: f + b = 31. Net movement: 5f − 2b = 15. Step 2: put b = 31 − f: 5f − 2(31 − f) = 15, so 5f − 62 + 2f = 15, giving 7f = 77 and f = 11. Then b = 20. Step 3: difference = 20 − 11 = 9. Check: 5 × 11 − 2 × 20 = 55 − 40 = 15 ✓. Options (a), (b) and (c) fail.
Tip: two unknowns need two equations: one for the count and one for the total effect. Always check your answer in both.
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.
Let P = QQQ be a 3-digit number. What is the HCF of P and 481?
Answer (c) & explanation
The HCF is 37 (c). Step 1: a number like QQQ (for example 555) equals Q × 111, and 111 = 3 × 37. So every QQQ has 37 as a factor. Step 2: 481 = 13 × 37. Step 3: the common part is 37. Could the HCF be bigger? Only if QQQ also had 13, but Q is a single digit (1–9) and 3 × Q cannot contain 13. So the HCF is always 37. Options (a), (b) and (d) fail.
Tip: remember 111 = 3 × 37, so every three-same-digit number (111, 222, …, 999) is divisible by 37. Also 1001 = 7 × 11 × 13.
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.
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.
What is the unit digit in the multiplication of 1 × 3 × 5 × 7 × 9 × . . . × 999?
Answer (c) & explanation
The unit digit is 5 (c). Step 1: the product 1 × 3 × 5 × … × 999 has only odd numbers, and one of them is 5. Step 2: any odd number × 5 ends in 5 (for example 3 × 5 = 15, 7 × 5 = 35). Step 3: to end in 0, the product would need a factor 2, and there is none. So the unit digit stays 5. Options (a), (b) and (d) are wrong.
Tip: odd × 5 always ends in 5; even × 5 always ends in 0. Check for 2s and 5s first in unit-digit questions; they decide the answer quickly.
Answer as per the official UPSC answer key.
The 5-digit number PQRST (all distinct digits) is such that T ≠ 0, P is thrice T. S is greater than Q by 4, while Q is greater than R by 3. How many such 5-digit numbers are possible?
Answer (b) & explanation
Four such numbers are possible (b). Step 1: P = 3T with T not 0: (T, P) can be (1, 3), (2, 6), (3, 9). Step 2: Q = R + 3 and S = Q + 4 = R + 7. S is a digit, so R can be 0, 1 or 2: (R, Q, S) = (0, 3, 7), (1, 4, 8), (2, 5, 9). Step 3: combine and keep only numbers with all digits distinct: 35291 ✓, 63072 ✓, 64182 ✓, 94183 ✓. Others repeat a digit (for example 93073 repeats 3; 65292 repeats 2). So four. Options (a), (c) and (d) miscount.
Tip: list the few cases for each condition, then combine and strike out repeats. "All distinct" is the condition people forget.
Answer as per the official UPSC answer key.