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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Let A3BC and DE2F be four-digit numbers where each letter represents a different digit greater than 3. If the sum of the numbers is 15902, then what is the difference between the values of A and D?
Answer (c) & explanation
1. Units: C + F ends in 2 and both digits are at least 4, so C + F = 12, carry 1. Tens: B + 2 + 1 ends in 0, so B = 7, carry 1.
2. Hundreds: 3 + E + 1 ends in 9, so E = 5. Thousands: A + D = 15.
3. Digits left (greater than 3, not 5 or 7): C, F = 4, 8 and A, D = 6, 9. Difference = 3.
Answer as per the official UPSC answer key.
A digit n > 3 is divisible by 3 but not divisible by 6. Which one of the following is divisible by 4?
Answer (d) & explanation
3n + 1 is divisible by 4 (d). Step 1: n is a single digit above 3, divisible by 3 but not by 6. Digits divisible by 3 above 3 are 6 and 9. 6 is divisible by 6, so n = 9. Step 2: test each option. 2n = 18; 3n = 27; 2n + 4 = 22; 3n + 1 = 28. Step 3: only 28 is divisible by 4. So the answer is (d). Options (a), (b) and (c) fail.
Tip: "digit" means 0 to 9. That small range lets you find n directly, then test options. Divisible by 3 but not 6 means an odd multiple of 3.
Answer as per the official UPSC answer key.
If in a particular year 12th January is a Sunday, then which one of the following is correct?
Answer (c) & explanation
12th July is a Sunday if the year is a leap year (c). Step 1: days from 12 January to 12 July = 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 31 (May) + 30 (Jun) = 181 days. In a leap year, February has 29, so it is 182. Step 2: 182 = 7 × 26, an exact number of weeks. So in a leap year, 12 July falls on the same day, Sunday. Step 3: option (d) says the opposite, so it fails. Options (a) and (b) about 15 July do not work: 15 July would be Wednesday (leap) or Tuesday (non-leap).
Tip: count days between dates and divide by 7. The remainder tells how many days the weekday moves forward.
Answer as per the official UPSC answer key.
A vessel full of water weighs 40 kg. If it is one-third filled, its weight becomes 20 kg. What is the weight of the empty vessel?
Answer (a) & explanation
The empty vessel weighs 10 kg (a). Step 1: full weighs 40 kg; one-third full weighs 20 kg. Going from full to one-third full removes 2/3 of the water, and the weight drops by 40 − 20 = 20 kg. Step 2: so 2/3 of the water = 20 kg, and all the water = 30 kg. Step 3: vessel = 40 − 30 = 10 kg. Check: one-third water = 10 kg, plus vessel 10 kg = 20 kg. Options (b), (c) and (d) fail.
Tip: the difference between two states removes the vessel's weight. Use it to find the contents first, then subtract from the total.
Answer as per the official UPSC answer key.
How many integers are there between 1 and 100 which have 4 as a digit but are not divisible by 4?
Answer (c) & explanation
12 such integers exist (c). Step 1: list the numbers from 1 to 100 with a 4 as a digit: 4, 14, 24, 34, 40 to 49 (ten numbers), 54, 64, 74, 84, 94. That is 19 numbers. Step 2: among them, those divisible by 4: 4, 24, 40, 44, 48, 64, 84. That is 7. Step 3: 19 − 7 = 12. Option (b), 11, and option (d), 13, miscount by one. Option (a) misses the 40s.
Tip: count the 40s block separately; it is easy to miss that 44 has two 4s but counts once. A number is divisible by 4 if its last two digits are divisible by 4.
Answer as per the official UPSC answer key.
What is the largest number among the following?
Answer (c) & explanation
(1/3)⁻⁴ is the largest (c). Step 1: a negative power flips the fraction: (1/a)⁻ⁿ = aⁿ. Step 2: so (1/2)⁻⁶ = 2⁶ = 64; (1/4)⁻³ = 4³ = 64; (1/3)⁻⁴ = 3⁴ = 81; (1/6)⁻² = 6² = 36. Step 3: the largest is 81, option (c). Options (a) and (b) tie at 64. Option (d) is the smallest.
Tip: remember x⁻ⁿ = 1/xⁿ. Flipping first turns the question into simple powers. Know powers of 2 up to 2¹⁰ = 1024 and of 3 up to 3⁶ = 729.
Answer as per the official UPSC answer key.
How many zeroes are there at the end of the following product? 1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60
Answer (a) & explanation
The product ends in 10 zeros (a). Step 1: each zero needs one 2 and one 5. Step 2: count the 5s. The 12 numbers from 5 to 60 each give one 5. 25 and 50 give one extra each. Total = 14 fives. Step 3: count the 2s. Only the even numbers give 2s: 10 (one), 20 (two), 30 (one), 40 (three), 50 (one), 60 (two). Total = 10 twos. Zeros = the smaller count = 10. Option (c), 14, counts only the 5s. Options (b) and (d) are also wrong.
Tip: usually 2s are plentiful and 5s are scarce. Here most numbers are odd multiples of 5, so the 2s run out first. Always count both.
here the 2s run out first. Counting only the 5s gives 14, which is a trap option.
Answer as per the official UPSC answer key.
What is the greatest length x such that 3½ m and 8¾ m are integral multiples of x?
Answer (d) & explanation
The greatest length is 1¾ m (d). Step 1: write both in quarter-metres: 3½ m = 14/4 and 8¾ m = 35/4. Step 2: HCF(14, 35) = 7. So x = 7/4 = 1¾ m. Step 3: check: 3½ = 2 × 1¾ and 8¾ = 5 × 1¾. Both are whole multiples. Option (c), 1¼, divides 8¾ (7 times) but not 3½. Options (a) and (b) fail too.
Tip: for the HCF of fractions, put them over a common denominator and take the HCF of the numerators. Or use HCF of numerators ÷ LCM of denominators.
Answer as per the official UPSC answer key.
Let XYZ be a three-digit number, where (X + Y + Z) is not a multiple of 3. Then (XYZ + YZX + ZXY) is not divisible by
Answer (b) & explanation
The sum is not divisible by 9 (b). Step 1: in XYZ + YZX + ZXY, each digit appears once in every place (hundreds, tens, units). So the sum = 111X + 111Y + 111Z = 111(X + Y + Z). Step 2: 111 = 3 × 37. So the sum is divisible by 3 (a), by 37 (c) and by (X + Y + Z) (d). Step 3: to be divisible by 9, we need a second factor of 3. That would have to come from X + Y + Z, but the question says it is not a multiple of 3. So the sum is not divisible by 9.
Tip: check with numbers: 124 + 241 + 412 = 777 = 111 × 7. 777 ÷ 9 is not whole.
Answer as per the official UPSC answer key.
The recurring decimal representation 1·272727... is equivalent to
Answer (b) & explanation
1·272727... equals 14/11 (b). Step 1: split it: 1 + 0·2727... Step 2: a two-digit repeating block over 99: 0·2727... = 27/99 = 3/11. Step 3: 1 + 3/11 = 14/11. Check: 14 ÷ 11 = 1·2727... Option (a), 13/11 = 1·1818... Option (c), 127/99 = 1·2828... Option (d), 137/99 = 1·3838... All fail.
Tip: for pure recurring decimals, put the repeating digits over as many 9s as there are digits: 0·ab ab... = ab/99. Then simplify. If one digit does not repeat, like 1·2333..., use 9s for repeating digits and 0s for the rest.
Answer as per the official UPSC answer key.
Let p, q, r and s be natural numbers such that p – 2016 = q + 2017 = r – 2018 = s + 2019. Which one of the following is the largest natural number?
Answer (c) & explanation
r is the largest (c). Step 1: call the common value k. Then p − 2016 = k gives p = k + 2016. q + 2017 = k gives q = k − 2017. Step 2: r − 2018 = k gives r = k + 2018. s + 2019 = k gives s = k − 2019. Step 3: compare: r has the biggest number added, so r is the largest. p is second. q and s are smaller. Options (a), (b) and (d) fail.
Tip: setting a chain of equal expressions to one letter, k, makes comparison easy. Watch the signs: "p minus 2016" means p is bigger than k, while "q plus 2017" means q is smaller than k.
Answer as per the official UPSC answer key.
What is the least four-digit number when divided by 3, 4, 5 and 6 leaves a remainder 2 in each case?
Answer (b) & explanation
The least such number is 1022 (b). Step 1: a number leaving remainder 2 when divided by 3, 4, 5 and 6 has the form LCM × k + 2. LCM(3, 4, 5, 6) = 60. Step 2: the smallest four-digit multiple of 60: 60 × 17 = 1020. Step 3: add 2: 1022. Option (a), 1012, is 60 × 16·8 + 2, not valid. Option (c), 1122 = 60 × 18·67 + 2, fails. Option (d) also fails.
Tip: same remainder for all divisors → number = LCM × k + remainder. Find the LCM, then the first suitable multiple.
Answer as per the official UPSC answer key.
A wall clock moves 10 minutes fast in every 24 hours. The clock was set right to show the correct time at 8:00 a.m. on Monday. When the clock shows the time 6:00 p.m. on Wednesday, what is the correct time?
Answer (a) & explanation
The correct time is 5:36 p.m. (a). Step 1: the clock gains 10 minutes every 24 hours. So it shows 1450 minutes for every 1440 true minutes. Step 2: from 8:00 a.m. Monday to 6:00 p.m. Wednesday, the clock shows 58 hours = 3480 minutes. True time = 3480 × 1440/1450 = 3456 minutes = 57 hours 36 minutes. Step 3: 8:00 a.m. Monday + 57 h 36 min = 5:36 p.m. Wednesday. Option (b), 5:30 p.m., and the others use the wrong base.
Tip: a fast clock's shown time includes its gain. Convert using the ratio true : shown = 1440 : 1450, not by subtracting a fixed amount.
the clock is fast, so the true time is behind the clock. Adding the 24 minutes instead gives a time not in the options.
Answer as per the official UPSC answer key.
If the numerator and denominator of a proper fraction are increased by the same positive quantity which is greater than zero, the resulting fraction is
Answer (b) & explanation
The new fraction is always greater than the original (b). Step 1: take a proper fraction a/b, where a < b. Add k to both: (a + k)/(b + k). Step 2: the difference = (a + k)/(b + k) − a/b = k(b − a) / [b(b + k)]. Since b > a and k > 0, this is positive. Step 3: check: 1/2 becomes 2/3 when 1 is added. 2/3 is bigger. So (a), (c) and (d) fail.
Tip: adding the same amount to top and bottom pulls any fraction towards 1. A proper fraction (below 1) rises; an improper fraction (above 1) falls. This one rule answers many CSAT fraction questions instantly.
Answer as per the official UPSC answer key.
The number of times the digit 5 will appear while writing the integers from 1 to 1000 is
Answer (c) & explanation
The digit 5 appears 300 times (c). Step 1: write the numbers 000 to 999 as three-digit strings. There are 1000 strings with 3 places each, so 3000 digits in all. Step 2: each digit 0 to 9 appears equally often: 3000 ÷ 10 = 300. Leading zeros do not affect the count of 5s. Step 3: the number 1000 has no 5, so the answer stays 300. Check another way: 5 appears 100 times in the units place, 100 in the tens and 100 in the hundreds. Options (a), (b) and (d) are wrong.
Tip: from 1 to 999, every non-zero digit appears 300 times. Counting place by place (units, tens, hundreds) is the safest method.
269 and 271 come from counting only the numbers that contain a 5. A number like 55 has two 5s.
Answer as per the official UPSC answer key.
A printer numbers the pages of a book starting with 1 and uses 3089 digits in all. How many pages does the book have?
Answer (c) & explanation
The book has 1049 pages (c). Step 1: count digits used. Pages 1 to 9: 9 digits. Pages 10 to 99: 90 pages × 2 = 180. Pages 100 to 999: 900 × 3 = 2700. Total so far = 2889. Step 2: digits left = 3089 − 2889 = 200. Each page from 1000 on uses 4 digits, so 200 ÷ 4 = 50 pages. Step 3: last page = 999 + 50 = 1049. Option (d), 1050, adds one too many. Options (a) and (b) undercount.
Tip: remember the block totals: 9, 189 (to 99) and 2889 (to 999). Then divide the remainder by the digits per page.
Answer as per the official UPSC answer key.
Mr 'X' has three children. The birthday of the first child falls on the 5th Monday of April, that of the second one falls on the 5th Thursday of November. On which day is the birthday of his third child, which falls on 20th December?
Answer (b) & explanation
1. April has 30 days. A 5th Monday needs Monday on 29 or 30 April, so 1 April is a Monday or a Sunday.
2. 1 April to 1 November = 214 days = 30 weeks + 4 days. So 1 November is Friday or Thursday. A 5th Thursday in November (30 days) needs 1 November to be Wednesday or Thursday, so 1 November = Thursday.
3. 20 December is 49 days later = exactly 7 weeks. Thursday.
Answer as per the official UPSC answer key.
Which year has the same calendar as that of 2009?
Answer (d) & explanation
2015 has the same calendar as 2009 (d). Step 1: two years share a calendar if they start on the same day and are both leap or both non-leap. Step 2: add odd days (days beyond full weeks). A normal year has 1 odd day; a leap year has 2. 2009, 2010 and 2011 give 1 each. 2012 (leap) gives 2. 2013 and 2014 give 1 each. Total = 7 = 0 odd days. Step 3: so 2015 starts on the same day as 2009, and both are non-leap. Options (a), (b) and (c) do not total a multiple of 7.
Tip: a non-leap year's calendar usually repeats after 6 or 11 years. Counting odd days settles it quickly.
Answer as per the official UPSC answer key.
In a school every student is assigned a unique identification number. A student is a football player if and only if the identification number is divisible by 4, whereas a student is a cricketer if and only if the identification number is divisible by 6. If every number from 1 to 100 is assigned to a student, then how many of them play cricket as well as football?
Answer (b) & explanation
8 students play both cricket and football (b). Step 1: a number divisible by both 4 and 6 must be divisible by their LCM, which is 12. It is not 24, because 4 and 6 share the factor 2. Step 2: multiples of 12 up to 100 are 12, 24, 36, 48, 60, 72, 84 and 96. Step 3: that is 8 numbers (100 ÷ 12 = 8, ignoring the remainder). Option (a), 4, uses 24 instead of 12. Options (c) and (d) are wrong.
Tip: "divisible by both" means divisible by the LCM, not the product. LCM of 4 and 6 is 12; multiplying them gives 24, which misses numbers like 12, 36 and 60.
multiplying 4 × 6 = 24 gives 4, a trap option. Use the LCM.
Answer as per the official UPSC answer key.
Number 136 is added to 5B7 and the sum obtained is 7A3, where A and B are integers. It is given that 7A3 is exactly divisible by 3. The only possible value of B is
Answer (d) & explanation
1. Try each value of B in the sum 5B7 + 136.
2. B = 2: 527 + 136 = 663. It must start with 7. Fails. B = 5: 557 + 136 = 693. Fails.
3. B = 7: 577 + 136 = 713. Digit sum 7 + 1 + 3 = 11 is not divisible by 3. Fails.
4. B = 8: 587 + 136 = 723. Digit sum 12 is divisible by 3. Works, so B = 8.
Answer as per the official UPSC answer key.