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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If the sum of the two-digit numbers AB and CD is the three-digit number 1CE, where the letters A, B, C, D, E denote distinct digits, then what is the value of A?
Answer (a) & explanation
A must be 9 (a). Step 1: AB + CD = 1CE. Two two-digit numbers add to at most 198, so the hundreds digit 1 comes from a carry. Step 2: in the tens column, A + C + (carry from units) gives a result ending in C, with a carry of 1. So A + carry = 10. Step 3: A is a single digit, so the carry from units must be 1 and A = 9. (Example: 97 + 25 = 122 fails distinctness, but 98 + 27 = 125 works: A = 9, C = 2, E = 5.) Option (d) is wrong, since A is fixed.
Tip: in letter-sum puzzles, start with the carries. A sum of two two-digit numbers can carry only 1 into the hundreds.
Answer as per the official UPSC answer key.
421 and 427, when divided by the same number, leave the same remainder 1. How many numbers can be used as the divisor in order to get the same remainder 1?
Answer (c) & explanation
Three numbers work as the divisor (c). Step 1: the divisor must leave remainder 1 for both, so it must divide 421 − 1 = 420 and 427 − 1 = 426. Step 2: it must divide their HCF. 426 − 420 = 6, and 6 divides both, so HCF = 6. Divisors of 6: 1, 2, 3, 6. Step 3: a remainder of 1 needs a divisor greater than 1 (dividing by 1 leaves 0). So only 2, 3 and 6 remain: three numbers. Option (d), 4, wrongly counts 1.
Tip: the remainder must always be smaller than the divisor. That rule quietly removes small divisors in such questions.
counting 1 as a divisor gives (d). Dividing by 1 leaves remainder 0, not 1.
Answer as per the official UPSC answer key.
Three numbers x, y, z are selected from the set of the first seven natural numbers such that x > 2y > 3z. How many such distinct triplets (x, y, z) are possible?
Answer (d) & explanation
1. Start with z. If z = 2, then 2y > 6 means y ≥ 4 and x > 8: impossible. So z = 1.
2. With z = 1, 2y > 3 gives y ≥ 2. y = 2 needs x > 4: x = 5, 6, 7. y = 3 needs x > 6: x = 7. y = 4 needs x > 8: none.
3. Four triplets: (5, 2, 1), (6, 2, 1), (7, 2, 1), (7, 3, 1).
Answer as per the official UPSC answer key.
A can X contains 399 litres of petrol and a can Y contains 532 litres of diesel. They are to be bottled in bottles of equal size so that whole of petrol and diesel would be separately bottled. The bottle capacity in terms of litres is an integer. How many different bottle sizes are possible?
Answer (b) & explanation
Four bottle sizes are possible (b). Step 1: the bottle size must divide both 399 and 532 exactly, so it must be a divisor of their HCF. Step 2: 399 = 3 × 7 × 19 and 532 = 4 × 7 × 19. HCF = 7 × 19 = 133. Step 3: divisors of 133 are 1, 7, 19 and 133: four sizes. Options (a), (c) and (d) miscount.
Tip: "how many sizes" means count the divisors of the HCF; "the largest size" means the HCF itself. To count divisors quickly, write the number as primes: 7¹ × 19¹ gives (1 + 1)(1 + 1) = 4.
Answer as per the official UPSC answer key.
The total cost of 4 oranges, 6 mangoes and 8 apples is equal to twice the total cost of 1 orange, 2 mangoes and 5 apples. Consider the following statements :
- 1. The total cost of 3 oranges, 5 mangoes and 9 apples is equal to the total cost of 4 oranges, 6 mangoes and 8 apples.
- 2. The total cost of one orange and one mango is equal to the cost of one apple.
Which of the statements given above is/are correct?
Answer (c) & explanation
Both statements are correct (c). Step 1: write o, m, a for the prices. 4o + 6m + 8a = 2(o + 2m + 5a) = 2o + 4m + 10a. Step 2: simplify: 2o + 2m = 2a, so o + m = a. One orange and one mango cost the same as one apple: statement 2 is right. Step 3: statement 1: (3o + 5m + 9a) − (4o + 6m + 8a) = a − o − m = 0. The two totals are equal, so statement 1 is right too. Options (a), (b) and (d) fail.
Tip: form one equation and simplify it first. Then test each statement by subtracting the two sides and checking for zero.
Answer as per the official UPSC answer key.
Consider the following statements in respect of the sum S = x + y + z, where x, y and z are distinct prime numbers each less than 10:
- 1. The unit digit of S can be 0.
- 2. The unit digit of S can be 9.
- 3. The unit digit of S can be 5.
Which of the statements given above are correct?
Answer (c) & explanation
Statements 1 and 3 are correct (c). Step 1: the primes below 10 are 2, 3, 5 and 7. We must pick three different ones, so there are only four possible sets. Step 2: list the sums: 2 + 3 + 5 = 10, 2 + 3 + 7 = 12, 2 + 5 + 7 = 14, 3 + 5 + 7 = 15. Step 3: unit digit 0 occurs (10), so 1 is true. Unit digit 5 occurs (15), so 3 is true. Unit digit 9 never occurs, so 2 is false. Options (a), (b) and (d) fail.
Tip: when the cases are few, simply list them all. Remember 1 is not a prime and 2 is the only even prime.
Answer as per the official UPSC answer key.
325 + 227 is divisible by
Answer (c) & explanation
The number is divisible by 10 (c). Step 1: write everything as powers of 2. 32 = 2⁵, so 32⁵ = 2²⁵. Step 2: 2²⁵ + 2²⁷ = 2²⁵ × (1 + 2²) = 2²⁵ × (1 + 4) = 5 × 2²⁵. Step 3: it has a factor 5 and a factor 2, so it is divisible by 10. It has no factor 3, 7 or 11, so options (a), (b) and (d) fail.
Tip: when the bases are powers of the same number, convert them to one base. Then take out the smaller power as a common factor. This turns a huge sum into a simple product.
when the bases differ, check if they are powers of the same number.
Answer as per the official UPSC answer key.
A father said to his son, “n years back I was as old as you are now. My present age is four times your age n years back”. If the sum of the present ages of the father and the son is 130 years, what is the difference of their ages?
Answer (a) & explanation
The age difference is 30 years (a). Step 1: let the father be F and the son S. "n years back I was as old as you are now" means F − n = S, so n = F − S. Step 2: "my age is four times your age n years back": F = 4(S − n) = 4(S − F + S) = 4(2S − F). So 5F = 8S, giving F : S = 8 : 5. Step 3: 8 + 5 = 13 parts = 130 years, so one part = 10. F = 80, S = 50, difference = 30 years. Check: n = 30; the son 30 years back was 20, and 4 × 20 = 80. Options (b), (c) and (d) fail.
Tip: in age puzzles, note that the age difference never changes. Turn each sentence into one equation.
Answer as per the official UPSC answer key.
Let p and q be positive integers satisfying p < q and p + q = k. What is the smallest value of k that does not determine p and q uniquely?
Answer (c) & explanation
The smallest such k is 5 (c). Step 1: p and q are positive whole numbers with p < q. Step 2: k = 3: only 1 + 2, so unique. k = 4: only 1 + 3, because 2 + 2 is not allowed (p must be less than q). Unique again. Step 3: k = 5: both 1 + 4 and 2 + 3 work. Two answers, so k = 5 is the first value that does not fix p and q. Option (b), 4, forgets the p < q rule. Options (a) and (d) are wrong.
Tip: test the smallest cases one by one. Read the conditions carefully: "p < q" rules out equal numbers.
Answer as per the official UPSC answer key.
Let X be a two-digit number and Y be another two-digit number formed by interchanging the digits of X. If (X + Y) is the greatest two-digit number, then what is the number of possible values of X?
Answer (d) & explanation
There are 8 possible values of X (d). Step 1: let X = 10a + b, so Y = 10b + a. Then X + Y = 11(a + b). The greatest two-digit number is 99, so 11(a + b) = 99, giving a + b = 9. Step 2: Y must also be a two-digit number, so b cannot be 0. That removes 90. Step 3: X can be 18, 27, 36, 45, 54, 63, 72, 81: 8 values. Option (c), 6, and the others miscount.
Tip: a number plus its reverse is always a multiple of 11; the difference is always a multiple of 9. Watch for hidden conditions like "Y is also two-digit", which rule out a zero digit.
90 looks valid, but its reverse, 09, is not a two-digit number.
Answer as per the official UPSC answer key.
In some code, letters P, Q, R, S, T represent numbers 4, 5, 10, 12, 15. It is not known which letter represents which number. If Q − S = 2S and T = R + S + 3, then what is the value of P + R − T?
Answer (b) & explanation
1. Q − S = 2S means Q = 3S. From the list, (S, Q) is (4, 12) or (5, 15).
2. If S = 4, Q = 12: T = R + 7 must come from 5, 10, 15. No pair works. So S = 5, Q = 15.
3. Then T = R + 8 from 4, 10, 12: R = 4, T = 12, and P = 10. P + R − T = 10 + 4 − 12 = 2.
Answer as per the official UPSC answer key.
How many times the hour hand and the minute hand coincide in a clock between 10:00 a.m. and 2:00 p.m. (same day)?
Answer (a) & explanation
The hands coincide 3 times (a). Step 1: the hands meet every 12/11 hours, about 1 hour 5 minutes. In 12 hours they meet only 11 times, because between 11 and 1 they meet just once, at 12:00. Step 2: after 10:00 a.m., they meet at about 10:54, then at 12:00, then at about 1:05. Step 3: the next meeting is about 2:11, after the window. So 3 times. Option (b), 4, wrongly assumes once every hour.
Tip: hands meet 11 times in 12 hours and 22 times in a day. The 11-to-1 gap holds only one meeting, at 12 o'clock.
four hours does not mean four meetings. The 11 to 1 stretch has only one.
Traps like this come back in new clothes. The PYQ Book collects all of them, topic by topic, so none catch you twice. Try the free sample
Answer as per the official UPSC answer key.
In the expression 5 * 4 * 3 * 2 * 1, * is chosen from +, −, × each at most two times. What is the smallest non-negative value of the expression?
Answer (d) & explanation
The smallest non-negative value is 0 (d). Step 1: zero is the smallest non-negative number, so first check whether 0 is possible. Step 2: try 5 − 4 − 3 + 2 × 1. Multiplication comes first (BODMAS): 2 × 1 = 2. Step 3: then 5 − 4 − 3 + 2 = 0. We used minus twice, plus once and times once, which the rule ("each at most two times") allows. Since 0 is reachable, options (a) 3, (b) 2 and (c) 1 cannot be the smallest.
Tip: when asked for the smallest or largest value, test the extreme option first. If it works, you are done. Always remember that × and ÷ are done before + and −.
always test the extreme option first. If 0 works, nothing smaller can be the answer.
Answer as per the official UPSC answer key.
The calendar for the year 2025 is same for
Answer (c) & explanation
2031 has the same calendar as 2025 (c). Step 1: an ordinary year moves the weekday forward by 1 day; a leap year by 2. Step 2: add the shifts from 2025: 2025 (1) + 2026 (1) + 2027 (1) + 2028 (2, leap) + 2029 (1) + 2030 (1) = 7, which is 0 extra days. So 2031 starts on the same weekday as 2025. Step 3: 2031 is also not a leap year, like 2025, so the whole calendar matches. Options (a) 2029, (b) 2030 and (d) 2033 do not add up to a multiple of 7.
Tip: keep adding odd days until you reach a multiple of 7, and check that both years are the same type (leap or ordinary).
Answer as per the official UPSC answer key.
How many consecutive zeros are there at the end of the integer obtained in the product 12 × 24 × 36 × 48 × … × 2550?
Answer (d) & explanation
There are 200 zeros at the end (d). Step 1: each trailing zero comes from one 2 and one 5. There are far more 2s than 5s, so just count the 5s. Step 2: the term k^(2k) gives 2k × (number of 5s in k). Only multiples of 5 matter: k = 5, 10, 15, 20 give 10 + 20 + 30 + 40 = 100. Step 3: k = 25 = 5 × 5 has two 5s, so it gives 2 × 25 × 2 = 100. Total = 200 fives, so 200 zeros. Options (a), (b) and (c) miss the extra 5s.
Tip: in zero-counting, numbers like 25, 125 hide extra 5s. Forgetting them is the most common trap.
(c) 100 is what you get if you forget that 25 carries two 5s.
Answer as per the official UPSC answer key.
Let p, q, r and s be distinct positive integers. Let p, q be odd and r, s be even. Consider the following statements :
- 1. (p − r)2(qs) is even.
- 2. (q − s)q2s is even.
- 3. (q + r)2(p + s) is odd.
Which of the statements given above are correct?
Answer (d) & explanation
All three statements are correct (d). Step 1: rule: a product is even if any factor is even; it is odd only if all factors are odd. p, q are odd; r, s are even. Step 2: statement 1: (p − r)² × q × s. s is even, so the product is even. Statement 2: (q − s) × q² × s: again s makes it even. Step 3: statement 3: q + r = odd + even = odd, so its square is odd. p + s = odd + even = odd. Odd × odd = odd. So all three are true, and options (a), (b) and (c) fail.
Tip: odd ± even = odd; odd ± odd = even. One even factor makes a whole product even.
Answer as per the official UPSC answer key.
What is the sum of all digits which appear in all the integers from 10 to 100?
Answer (b) & explanation
The sum is 856 (b). Step 1: from 10 to 99, the tens place shows each digit 1 to 9 exactly 10 times. That gives 45 × 10 = 450. Step 2: the units place shows each digit 0 to 9 exactly 9 times (once per ten). That gives 45 × 9 = 405. Step 3: add the digits of 100, which is 1. Total = 450 + 405 + 1 = 856. Option (a), 855, forgets 100. Options (c) and (d) miscount.
Tip: count digit sums place by place. Each place cycles through the digits evenly.
don't drop the 100 at the end. 855 is the trap option.
Answer as per the official UPSC answer key.
If ABC and DEF are both 3-digit numbers such that A, B, C, D, E and F are distinct non-zero digits such that ABC + DEF = 1111, then what is the value of A + B + C + D + E + F ?
Answer (d) & explanation
The sum of the six digits is 31 (d). Step 1: units place: C + F must end in 1. Two non-zero digits cannot add to 1, so C + F = 11, carrying 1. Step 2: tens place: B + E + 1 must end in 1, so B + E = 10, carrying 1. Step 3: hundreds place: A + D + 1 = 11, so A + D = 10. Total = 11 + 10 + 10 = 31. Options (a), (b) and (c) miss the carries.
Tip: in addition puzzles, work from the units place and track the carry at each step.
don't forget the carries. Without them you get 1 + 1 + 1 per column and land on a wrong option.
Answer as per the official UPSC answer key.
Choose the group which is different from the others :
Answer (d) & explanation
Group (d) is the odd one out (d). Step 1: check whether each number is prime. Groups (a) 17, 37, 47, 97; (b) 31, 41, 53, 67; and (c) 71, 73, 79, 83 contain only primes. Step 2: in group (d), 91 = 7 × 13, which is not prime. Step 3: so (d) is different. Options (a), (b) and (c) all fit the pattern.
Tip: 91 looks prime but is 7 × 13. Other common traps are 51 (3 × 17), 57 (3 × 19) and 87 (3 × 29).
91 looks prime but is 7 × 13. Learn this one by heart.
Answer as per the official UPSC answer key.
D is a 3-digit number such that the ratio of the number to the sum of its digits is least. What is the difference between the digit at the hundred's place and the digit at the unit's place of D?
Answer (c) & explanation
The difference is 8 (c). Step 1: to make (number ÷ digit sum) small, keep the hundreds digit small and the other two digits large. Step 2: with hundreds digit 1, try the best options: 199 ÷ 19 ≈ 10.47; 189 ÷ 18 = 10.5; 198 ÷ 18 = 11. So 199 is best. With hundreds digit 2, the ratio is at least 299 ÷ 20 ≈ 15. Step 3: D = 199. The difference between hundreds and units digits = 9 − 1 = 8. Options (a), (b) and (d) fail.
Tip: the hundreds digit adds 100 per unit but only 1 to the digit sum. So keep it as small as possible.
Answer as per the official UPSC answer key.