UPSC CSAT 2026 Question Paper
Explore the complete solved question paper for UPSC CSAT 2026 featuring 80 solved questions with bilingual (English & Hindi) explanations, official answer key, and subject weightage breakdown.
X travels 6 km on a bicycle with average speeds of 5 km per hour, 10 km per hour and 4 km per hour during the first 1 km, the next 2 km and the remaining 3 km, respectively. Y travels the same distances with average speeds of 4 km per hour, 10 km per hour and 5 km per hour, respectively. How many minutes early will Y complete the journey if both X and Y start at the same time?
Detailed Explanation:
X's journey: Time = 1/5 + 2/10 + 3/4 = 0.2 + 0.2 + 0.75 = 1.15 hours = 69 minutes.
Y's journey: Time = 1/4 + 2/10 + 3/5 = 0.25 + 0.2 + 0.6 = 1.05 hours = 63 minutes.
Therefore, Y completes the journey 69 - 63 = 6 minutes earlier, making Option 4 the correct answer.
There are four types of weights, namely $1$ kg, $2$ kg, $5$ kg and $10$ kg. What is the maximum number of different ways one can measure $20$ kg, if at least eight but not more than eleven weights of $1$ kg are to be used while measuring?
Detailed Explanation:
We need a + 2b + 5c + 10d = 20 where a ∈ {8, 9, 10, 11} (1 kg weights) and b, c, d ≥ 0 (2 kg, 5 kg, 10 kg weights).
Case a = 8: 2b + 5c + 10d = 12 → (b,c,d) = (6,0,0), (1,2,0), (1,0,1) → 3 ways
Case a = 9: 2b + 5c + 10d = 11 (odd, so c odd) → (b,c,d) = (3,1,0) → 1 way
Case a = 10: 2b + 5c + 10d = 10 → (b,c,d) = (5,0,0), (0,2,0), (0,0,1) → 3 ways
Case a = 11: 2b + 5c + 10d = 9 (odd, so c odd) → (b,c,d) = (2,1,0) → 1 way
Therefore, total = 3 + 1 + 3 + 1 = 8, making Option 2 the correct answer.
The class average $x$ in a test increases by $4$ when the score of a student is rectified, whose corrected score is $100$ instead of $0$. Later, the score of another student was found to have been recorded as $81$ in place of $56$. If there are no other corrections and the final corrected average is $y$, then $y - x$ is
Detailed Explanation:
Let the number of students be n. When one student's score is corrected from 0 to 100, the total increases by 100 and the average increases by 4. Thus: 100/n = 4, giving n = 25 students.
The second correction changes a score from 81 to 56, decreasing the total by 25. This decreases the average by 25/25 = 1.
After the first correction, average = x + 4. After the second correction, y = x + 4 − 1 = x + 3. Hence y − x = 3, making Option 2 the correct answer.
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For $\frac{1}{3} < x < y < 2$, which of the following statements is/are always correct? I. $x + \frac{1}{x} < y + \frac{1}{y}$ II. $\frac{\sqrt{1 + y^2}}{y} < \frac{\sqrt{1 + x^2}}{x}$ Select the answer using the code given below.
Detailed Explanation:
Statement I: Define $f(t) = t + \frac{1}{t}$. Its derivative $f'(t) = 1 - \frac{1}{t^2}$ is negative on $(0,1)$ and positive on $(1,\infty)$. Thus $f(t)$ decreases on $(\frac{1}{3}, 1)$ and increases on $(1, 2)$—not monotonic throughout.
Counterexample: Take $x = \frac{1}{2}, y = \frac{3}{4}$. Then $x + \frac{1}{x} = 2.5$ but $y + \frac{1}{y} \approx 2.08$. So $x + \frac{1}{x} > y + \frac{1}{y}$, contradicting Statement I.
Statement II: Rewrite $\frac{\sqrt{1+t^2}}{t} = \sqrt{\frac{1}{t^2} + 1}$. For $t > 0$, as $t$ increases, $\frac{1}{t^2}$ decreases, making $\sqrt{\frac{1}{t^2}+1}$ strictly decreasing. Since $x < y$, we have $\frac{\sqrt{1+y^2}}{y} < \frac{\sqrt{1+x^2}}{x}$, confirming Statement II is always true.
Therefore, only Statement II is correct, making Option 2 the correct answer.
$A$ 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?
Detailed Explanation:
Let A = 10a + b and B = 10b + a, where a and b are digits with a ≠ b and b ≠ 0 (so B remains 2-digit).
A − B = 9(a − b). For this to be a multiple of 27, we need 9(a − b) divisible by 27, so (a − b) must be a multiple of 3. Since A > B, we have a > b, giving possible differences: 3, 6, or 9.
For a − b = 3: pairs (4,1), (5,2), (6,3), (7,4), (8,5), (9,6) = 6 numbers. For a − b = 6: pairs (7,1), (8,2), (9,3) = 3 numbers. For a − b = 9: pair (9,0) is invalid since B would not be 2-digit. Total = 6 + 3 = 9.
Therefore, 9 different values of A are possible, making Option 2 the correct answer.
There are three types of rectangular tiles : 3' * 3', 3' * 7' and 3' * 11'. An area of rectangular shape of dimensions 3' * 100' is to be covered using these tiles without breaking them. If x and y are the maximum and minimum numbers of tiles of various sizes, respectively, that can be used to cover the area exactly, then x - y is
Detailed Explanation:
The area 3' × 100' must be covered by tiles 3' × 3', 3' × 7', and 3' × 11'. Since all tiles have width 3', they fit along the length, giving the equation 3a + 7b + 11c = 100.
Maximum tiles (x): Use smallest tiles. 31 tiles of 3' give 93', leaving 7' for 1 tile of 7'. Total = 32 tiles. Minimum tiles (y): Use largest tiles. 8 tiles of 11' give 88', leaving 12' for 4 tiles of 3'. Total = 12 tiles.
Therefore, x − y = 32 − 12 = 20, making Option 1 the correct answer.
In a sequence of numbers, each number other than the first two is the sum of the two immediately preceding numbers from it. If the first two numbers in the sequence are 4 and 7, then the sixth number is
Detailed Explanation:
This is a Fibonacci-like sequence where each term equals the sum of the two preceding terms.
Starting with a₁ = 4 and a₂ = 7: a₃ = 4 + 7 = 11; a₄ = 7 + 11 = 18; a₅ = 11 + 18 = 29; a₆ = 18 + 29 = 47.
Therefore, the sixth number is 47, making Option 4 the correct answer.
A train has to complete a journey of $800$ km. If it meets a minor accident, its speed becomes half of the existing speed. If there is a mechanical defect, the speed becomes one-fourth of the existing speed. On its way, the train meets with a minor accident after $200$ km; and $400$ km thereafter, it develops a mechanical defect. Had the train developed the mechanical defect after $200$ km and met the minor accident $400$ km thereafter, it would have taken $4$ more hours to reach its destination. What was the original speed of the train in km per hour?
Detailed Explanation:
Let original speed be v km/h. The journey is 800 km divided into segments: 0–200 km, 200–600 km, 600–800 km. Speed reductions compound on the existing (current) speed.
Case 1: Minor accident at 200 km (speed → v/2), mechanical defect at 600 km (speed → v/2 × 1/4 = v/8). Time = 200/v + 400/(v/2) + 200/(v/8) = 200/v + 800/v + 1600/v = 2600/v hours.
Case 2: Mechanical defect at 200 km (speed → v/4), minor accident at 600 km (speed → v/4 × 1/2 = v/8). Time = 200/v + 400/(v/4) + 200/(v/8) = 200/v + 1600/v + 1600/v = 3400/v hours.
Given: Case 2 takes 4 hours more. So, 3400/v - 2600/v = 4 → 800/v = 4 → v = 200 km/h.
Therefore, the original speed was 200 km/h, making Option 1 the correct answer.
An alloy $P$ contains $20%$ copper and $80%$ zinc by weight. Another alloy $Q$ contains $60%$ copper and $40%$ zinc by weight. A third alloy $R$ is to be prepared from $P$ and $Q$ so that it contains equal amount of copper and zinc. In what ratio, amounts of $P$ and $Q$ be mixed in order to get $R$?
Detailed Explanation:
Use alligation method: Copper in P = 20%, in Q = 60%, desired in R = 50% (equal copper and zinc).
Ratio of P to Q = (60 - 50) : (50 - 20) = 10 : 30 = 1 : 3.
Therefore, P and Q must be mixed in the ratio 1:3, making Option 1 the correct answer.
A person saves 10% of his salary every month. If his salary increases by 12% and the expenditure increases by 10%, then what will be the change in his saving per month?
Detailed Explanation:
Assume initial salary = ₹100. Initial saving = 10% = ₹10; initial expenditure = ₹90.
New salary = 100 + 12% = ₹112. New expenditure = 90 + 10% = ₹99. New saving = 112 − 99 = ₹13.
Increase in saving = 13 − 10 = ₹3. Percentage increase = (3/10) × 100 = 30%.
Therefore, the saving increases by 30%, making Option 2 the correct answer.
The ratio of male to female workers in two companies $A$ and $B$ is $13 : 10$ and $7 : 5$, respectively. If both the companies have the same number of female workers, then what is the ratio of the total number of workers in $A$ to those in $B$?
Detailed Explanation:
Company A has male : female = 13 : 10, and company B has male : female = 7 : 5. Since both companies have the same number of female workers, equalize the female ratio parts by multiplying B's ratio by 2 → B becomes 14 : 10.
Now both companies have 10 units of female workers. Total workers in A = 13 + 10 = 23 units; total workers in B = 14 + 10 = 24 units.
Therefore, the ratio of total workers in A to B is 23 : 24, making Option 2 the correct answer.
How many times does $5$ appear in all two-digit positive integers?
Detailed Explanation:
Count the digit 5 in tens place: appears in 50, 51, 52, …, 59 — that's 10 occurrences.
Count the digit 5 in units place: appears in 15, 25, 35, 45, 55, 65, 75, 85, 95 — that's 9 occurrences.
Therefore, total = 10 + 9 = 19, making Option 2 the correct answer.
The digit in the unit place of the number $6^{129} \times 7^{307}$ is
Detailed Explanation:
Unit digit of 6^129: Any power of 6 always ends in 6 (since 6×6=36, 36×6=216, etc.). Thus, unit digit = 6.
Unit digit of 7^307: Powers of 7 cycle every 4: 7¹→7, 7²→9, 7³→3, 7⁴→1. Divide 307 by 4: remainder = 3. So unit digit matches 7³ = 3.
Final unit digit: Multiply 6 × 3 = 18, giving unit digit 8.
Therefore, the unit digit is 8, making Option 3 the correct answer.
How many three-digit numbers can be expressed as an integral power of $2$?
Detailed Explanation:
Three-digit numbers range from 100 to 999. We need to find integral powers of 2 in this range.
Calculate: 2⁷ = 128, 2⁸ = 256, 2⁹ = 512 (all three-digit). 2⁶ = 64 (two-digit) and 2¹⁰ = 1024 (four-digit) fall outside the range.
Therefore, exactly 3 three-digit numbers (128, 256, 512) are integral powers of 2, making Option 3 the correct answer.
$X$ and $Y$ are two runners who run for the same duration of time on the same circular track. They started running at the same time in the same direction with uniform speeds. When $X$ completed $7$ rounds, $Y$ did exactly $5$. After completing $5$ rounds, $Y$ changed his direction and started running in the opposite direction with speed which is double of his earlier speed. On the other hand, $X$ continued to run with the same speed. They stopped running when $X$ completed exactly $21$ rounds. How many times did $X$ and $Y$ meet after they had started and before they finally stopped?
Detailed Explanation:
Phase 1 (Same direction): When X completes 7 rounds, Y completes 5 rounds in time T. Speed ratio is 7:5. Relative speed = 7 − 5 = 2 rounds/T. In time T, they meet 2 times (at t = 0.5T and t = T).
Phase 2 (Opposite directions): After Y completes 5 rounds, he reverses and doubles his speed to 10 rounds/T. X continues at 7 rounds/T. Relative speed = 7 + 10 = 17 rounds/T. X runs 21 − 7 = 14 more rounds in time 2T. Meetings in Phase 2 = 17 × 2 = 34 times.
Therefore, total meetings after start and before final stop = 2 + 33 = 35 (excluding the meeting at final stop), making Option 1 the correct answer.
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?
Detailed Explanation:
Target = 298 litres. Optimal strategy: overshoot using large cylinders, then subtract excess.
Draw 3 × 100 = 300 litres (3 measurements). Excess = 300 − 298 = 2 litres. Pour back 2 × 1 litre = 2 litres (2 measurements). Total = 3 + 2 = 5 measurements.
Greedy approach (only adding) needs 13 measurements: 2×100 + 3×25 + 3×6 + 5×1. Using 4 measurements yields 299 or 294 litres, not exactly 298.
Therefore, the minimum is 5 measurements, making Option 2 the correct answer.
An explosion takes place at a certain distance from an army camp. As soon as the sensor in the camp receives the sound of the explosion, a drone starts flying towards the spot of explosion. The drone clicks a picture from the spot and the camp receives it at the same time. Immediately another drone starts flying to the spot and it also sends a picture as soon as it reaches the spot. The two pictures were received at 5:02 PM and 5:05 PM, respectively. If the speed of the drones is 30 m/s, at what time did the explosion take place? Assume that the speed of sound is 300 m/s.
Detailed Explanation:
The second drone leaves at 5:02 PM (when first picture arrives) and reaches at 5:05 PM, taking 180 seconds. At 30 m/s, distance d = 5400 m.
Working backward from 5:02 PM: Sound traveled 5400/300 = 18 s to reach camp, then first drone took 5400/30 = 180 s to reach spot. Total elapsed time = 198 s = 3 min 18 s.
Therefore, explosion occurred at 5:02:00 − 3:18 = 4:58:42 PM, making Option 3 the correct answer.
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.
Detailed Explanation:
Let two distinct numbers be a and b. Using HCF × LCM = a × b, and given this product equals the cube of one number (say a³): a × b = a³, so b = a².
Statement I: Difference = a² − a = a(a − 1), the product of two consecutive integers, which is always even. ✓
Statement II: One number is a², which is a perfect square. ✓
Therefore, both statements I and II are correct, making Option 3 the correct answer.
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?
Detailed Explanation:
Let correct answers = x, wrong answers = y. First condition: 5x - 2y = 69. Second condition: 4x - y = 84.
From second equation, y = 4x - 84. Substituting into first: 5x - 2(4x - 84) = 69 → 5x - 8x + 168 = 69 → -3x = -99 → x = 33. Thus y = 132 - 84 = 48.
Total questions = 33 + 48 = 81, making Option 2 the correct answer.
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 $11$?
Detailed Explanation:
For the number 4x5y790 to be divisible by 11, apply the alternating digit sum rule: (sum of odd-position digits from right) − (sum of even-position digits from right) must be divisible by 11.
Odd positions (from right): 0, 7, 5, 4 → sum = 16. Even positions: 9, y, x → sum = x + y + 9. Thus, 16 − (x + y + 9) = 7 − (x + y) must be ≡ 0 (mod 11).
Since x, y are digits (0–9), x + y ∈ [0, 18], so 7 − (x + y) ∈ [−11, 7]. The only multiples of 11 in this range are 0 and −11. If 7 − (x + y) = 0 → x + y = 7; if 7 − (x + y) = −11 → x + y = 18. In both cases, (x + y) mod 11 = 7.
Therefore, the remainder when x + y is divided by 11 is 7, making Option 4 the correct answer.
UPSC CSAT 2026 Questions Paper - Subject-wise Question Distribution
UPSC CSAT 2026 Question Paper - FAQs & Analysis
Q1 How many total questions were asked in UPSC CSAT 2026?
Q2 What is the subject-wise question breakdown for UPSC CSAT 2026?
- CSAT (Civil Services Aptitude Test): 80 questions (100%)