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 3rd day of a month is Monday, which one of the following will be the fifth day from 21st of this month?
Answer (c) & explanation
The fifth day from the 21st is a Wednesday (c). Step 1: the 3rd is a Monday. Add 7 at a time to find more Mondays: 3, 10, 17, 24. Step 2: "the fifth day from the 21st" means count five days after the 21st: 21 + 5 = the 26th. Step 3: the 24th is a Monday, so the 25th is Tuesday and the 26th is Wednesday. Options (a), (b) and (d) fall on the wrong day.
Tip: in calendar questions, jump in steps of 7 from a known day, because the weekday repeats every 7 days. Then count only the small gap that is left.
counting the 21st itself as day 1 gives the 25th (Tuesday). Fifth day from the 21st is the 26th.
Answer as per the official UPSC answer key.
For a charity show, the total tickets sold were 420. Half of these tickets were sold at the rate of ₹ 5 each, one-third at the rate of ₹ 3 each and the rest for ₹ 2 each. What was the total amount received?
Answer (c) & explanation
The total collected is ₹1,610 (c). Step 1: split the 420 tickets. Half = 210 tickets at ₹5 = ₹1,050. One-third = 140 tickets at ₹3 = ₹420. Step 2: the rest = 420 − 210 − 140 = 70 tickets at ₹2 = ₹140. Step 3: total = 1,050 + 420 + 140 = ₹1,610. Options (a), (b) and (d) come from miscounting "the rest".
Tip: always work out "the rest" by subtracting from the total, not by guessing a fraction. Here half plus one-third is five-sixths, so the rest is one-sixth: 420 ÷ 6 = 70.
Answer as per the official UPSC answer key.
Five persons fire bullets at a target at an interval of 6, 7, 8, 9 and 12 seconds respectively. The number of times they would fire the bullets together at the target in an hour is
Answer (b) & explanation
UPSC's final key accepts both 7 (b) and 8 (c). Step 1: find when all five fire together again. That is the smallest number that 6, 7, 8, 9 and 12 all divide into, the LCM. 8 covers 2 and 4; 9 covers 3; 7 is separate; 6 and 12 are already covered by 8 × 9. So LCM = 7 × 8 × 9 = 504 seconds. Step 2: one hour is 3,600 seconds. 3,600 ÷ 504 = 7 and a bit. So after the first shot they fire together 7 more times, option (b). If you also count the starting shot, it is 8, option (c). The question does not say which, so both were accepted. Options (a) and (d) fit neither count.
Tip: for "together again" questions, always use the LCM, and check whether the start is counted.
The question does not say whether to count the first shot, so UPSC's final key accepts both (b) 7 and (c) 8.
Answer as per the official UPSC answer key.
A group of 630 children is seated in rows for a group photo session. Each row contains three less children than the row in front of it. Which one of the following number of rows is not possible ?
Answer (d) & explanation
1. Try each option by writing the rows out (each row has 3 fewer than the one in front).
2. 3 rows: 213 + 210 + 207 = 630. Works.
3. 4 rows: 162 + 159 + 156 + 153 = 630. Works.
4. 5 rows: 132 + 129 + 126 + 123 + 120 = 630. Works.
5. 6 rows: the average row would be 630 ÷ 6 = 105, so the two middle rows would be 106·5 and 103·5. Half a child is impossible, so 6 rows cannot be done.
Answer as per the official UPSC answer key.
A straight line segment is 36 cm long. Points are to be marked on the line from both the end points. From each end, the first point is at a distance of 1 cm from the end, the second point is at a distance of 2 cm from the first point and the third point is at a distance of 3 cm from the second point and so on. If the points on the ends are not counted and the common points are counted as one, what is the number of points?
Answer (b) & explanation
There are 12 points (b). Step 1: from the left end, gaps are 1, 2, 3, 4 and so on. Points fall at running totals: 1, 3, 6, 10, 15, 21, 28. The next total would be 36, the far end, which is not counted. That is 7 points. Step 2: from the right end, measured from the left, points fall at 36 − 1 = 35, then 33, 30, 26, 21, 15, 8. Another 7 points. Step 3: the points at 15 and 21 appear in both lists, so count them once. Total = 7 + 7 − 2 = 12. Options (a), (c) and (d) do not match.
Tip: convert both sets into distances from one end so that shared points are easy to spot.
Answer as per the official UPSC answer key.
A bell rings every 18 minutes. A second bell rings every 24 minutes. A third bell rings every 32 minutes. If all the three bells ring at the same time at 8 o'clock in the morning, at what other time will they all ring together?
Answer (b) & explanation
They ring together again at 12:48 (b). Step 1: find the LCM of 18, 24 and 32. Break them into factors: 18 = 2 × 3 × 3; 24 = 2 × 2 × 2 × 3; 32 = 2 × 2 × 2 × 2 × 2. Step 2: take the highest power of each: five 2s (32) and two 3s (9). LCM = 32 × 9 = 288 minutes. Step 3: 288 minutes = 4 hours 48 minutes. 8:00 + 4:48 = 12:48. Options (a), (c) and (d) are a few minutes off.
Tip: "ring together again" always means LCM. Turn the minutes into hours and minutes before adding to the start time.
Answer as per the official UPSC answer key.
There are five hobby clubs in a college viz, photography, yachting, chess, electronics and gardening. The gardening group meets every second day, the electronics group meets every third day, the chess group meets every fourth day, the yachting group meets every fifth day and the photography group meets every sixth day. How many times do all the five groups meet on the same day within 180 days?
Answer (a) & explanation
3 times is correct (a). Step 1: all five groups meet on the same day only when the day number is a multiple of 2, 3, 4, 5 and 6 together. Step 2: the LCM of 2, 3, 4, 5 and 6 is 60. (4 = 2², 6 = 2 × 3, so LCM = 2² × 3 × 5 = 60.) Step 3: within 180 days, the common days are 60, 120 and 180, which is 3 times. So (b), (c) and (d) fail.
Tip: 'meet together every…' questions are LCM questions. Then divide the period by the LCM (180 ÷ 60 = 3). Check the wording: if day 0 were counted as a meeting day, the answer would change, but here the count starts after day 0.
Answer as per the official UPSC answer key.
A gardener has 1000 plants. He wants to plant them in such a way that the number of rows and the number of columns remains the same. What is the minimum number of plants that he needs more for this purpose?
Answer (b) & explanation
24 is correct (b). Step 1: equal rows and columns means the plants form a square, so the total must be a perfect square. Step 2: 31² = 961, which is less than 1,000, so too small. The next square is 32² = 1,024. Step 3: he needs 1,024 − 1,000 = 24 more plants. (a) 14 gives 1,014, (c) 32 gives 1,032 and (d) 34 gives 1,034, none of which is a perfect square. So (a), (c) and (d) fail.
Tip: learn squares up to 35 or so; they save time in CSAT. 'Minimum number to add' means go up to the next perfect square; 'minimum to remove' means go down to the one before it (here 961, removing 39).
Answer as per the official UPSC answer key.
A sum of Rs. 700 has to be used to give seven cash prizes to the students of a school for their overall academic performance. If each prize is Rs. 20 less than its preceding prize, then what is the least value of the prize?
Answer (b) & explanation
Rs. 40 is correct (b). Step 1: the seven prizes fall in equal steps of Rs. 20, so they form an arithmetic series. In such a series, the middle (4th) prize equals the average: 700 ÷ 7 = Rs. 100. Step 2: the smallest prize is three steps below the middle: 100 − 3 × 20 = Rs. 40. Step 3: check: 160 + 140 + 120 + 100 + 80 + 60 + 40 = 700. So (a), (c) and (d) fail.
Tip: for an odd number of equally spaced values, the middle value is the average. This avoids writing equations. With an even count, the average lies halfway between the two middle values.
Answer as per the official UPSC answer key.
A contract on construction job specifies a penalty for delay in completion of the work beyond a certain date is as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day etc., the penalty for each succeeding day being Rs. 50 more than that of the preceding day. How much penalty should the contractor pay if he delays the work by 10 days?
Answer (b) & explanation
Rs. 4,250 is correct (b). Step 1: the penalties form a series rising by Rs. 50 each day: 200, 250, 300 and so on. The 10th day's penalty = 200 + 9 × 50 = Rs. 650. Step 2: the total of an evenly rising series = number of days × (first + last) ÷ 2 = 10 × (200 + 650) ÷ 2 = 10 × 425 = Rs. 4,250. Step 3: (d) Rs. 650 is only the last day's penalty, a trap. (a) and (c) come from wrong sums. So (a), (c) and (d) fail.
Tip: for any steady increase, the average of the first and last term, times the number of terms, gives the total. No need to add ten numbers.
Rs. 650 is only the 10th day, not the total.
Answer as per the official UPSC answer key.
A person has only Rs. 1 and Rs. 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs. 75, then the number of Rs. 1 and Rs. 2 coins are, respectively
Answer (d) & explanation
25 and 25 is correct (d). Step 1: suppose all 50 coins were Rs. 1 coins. That gives only Rs. 50. Step 2: she has Rs. 75, which is Rs. 25 more. Each time a Rs. 1 coin is swapped for a Rs. 2 coin, the total rises by Rs. 1. So there must be 25 Rs. 2 coins. Step 3: the rest, 50 − 25 = 25, are Rs. 1 coins. Check: 25 × 1 + 25 × 2 = 75. So (a), (b) and (c) fail.
Tip: the 'assume all are the cheaper kind' trick solves most two-item coin and ticket problems in seconds. Or just test each option quickly.
Answer as per the official UPSC answer key.
Three persons start walking together and their steps measure 40 cm, 42 cm and 45 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?
Answer (a) & explanation
25 m 20 cm is correct (a). Step 1: each person must cover the same distance in whole steps, so the distance must be a multiple of 40, 42 and 45. The minimum is their LCM. Step 2: break into primes: 40 = 2³ × 5; 42 = 2 × 3 × 7; 45 = 3² × 5. LCM = 2³ × 3² × 5 × 7 = 2,520 cm. Step 3: 2,520 cm = 25 m 20 cm. The other options are 2, 3 and 4 times this, so they are not the minimum. So (b), (c) and (d) fail.
Tip: 'minimum common distance' or 'first time together' means LCM. Take the highest power of each prime factor.
Answer as per the official UPSC answer key.