UPSC CSAT 2026
CSAT (Civil Services Aptitude Test) Previous Year Questions (PYQs)
Explore 80 solved UPSC CSAT 2026 CSAT (Civil Services Aptitude Test) questions with detailed step-by-step bilingual solutions, option analysis, and answer keys.
Suppose $x$, $y$ and $z$ are variables taking positive real numbers as their possible values. It is given that $y$ is directly proportional to $x^2$ and $x$ is inversely proportional to $z$. For $z = \frac{7}{25}$, the values of $x$ and $y$ are $5$ and $50$, respectively. If $y = 98$, what is $z$ equal to?
Detailed Explanation:
Given: $y \propto x^2 \Rightarrow y = kx^2$ and $x \propto \frac{1}{z} \Rightarrow xz = c$ (constant).
Using $z = \frac{7}{25}, x = 5, y = 50$: From $y = kx^2$, we get $50 = k(25) \Rightarrow k = 2$. From $xz = c$, we get $c = 5 \times \frac{7}{25} = \frac{7}{5}$. So $y = 2x^2$ and $xz = \frac{7}{5}$.
When $y = 98$: $2x^2 = 98 \Rightarrow x^2 = 49 \Rightarrow x = 7$. Thus $z = \frac{c}{x} = \frac{7/5}{7} = \frac{1}{5}$.
Therefore, the value of z is $\frac{1}{5}$, making Option 2 the correct answer.
Three partners $A$, $B$ and $C$ entered into a business. $A$ invested one-third of the capital for one-third duration. $B$ invested one-fourth of the capital for one-fourth duration. $C$ invested the remaining capital for the whole duration. Out of a profit of ₹ $17,000$, how much profit will $C$ get?
Detailed Explanation:
Profit sharing in partnership = Capital × Time. Let total capital = K, total duration = T.
A's share: (K/3) × (T/3) = KT/9 | B's share: (K/4) × (T/4) = KT/16 | C's share: [K - K/3 - K/4] × T = (5K/12) × T = 5KT/12.
Ratio A : B : C = 1/9 : 1/16 : 5/12. Multiplying by LCM(9,16,12) = 144 → 16 : 9 : 60. Total parts = 85.
C's profit = (60/85) × 17,000 = 60 × 200 = ₹12,000. Therefore, Option 1 is the correct answer.
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 ${(x \times y) + z}$; then $M - N$ is
Detailed Explanation:
To maximize $(x \times y) + z$, choose the two largest values for the product: $x = 5, y = 4$ gives $5 \times 4 = 20$, and the largest remaining value $z = 3$ gives $M = 20 + 3 = 23$.
To minimize $(x \times y) + z$, choose the two smallest values for the product: $x = 2, y = 3$ gives $2 \times 3 = 6$, and the smallest remaining value $z = 4$ gives $N = 6 + 4 = 10$.
Therefore, $M - N = 23 - 10 = 13$, making Option 3 the correct answer.
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If $10^m \times 1000 \times n = 75^{25} \times 25^{32} \times 32^{75}$, where $n$ is not divisible by $10$, then the value of $m$ is
Detailed Explanation:
Prime factorize the RHS: $75^{25} = (3 \times 5^2)^{25} = 3^{25} \times 5^{50}$, $25^{32} = 5^{64}$, and $32^{75} = 2^{375}$. Combined: $2^{375} \times 3^{25} \times 5^{114}$.
Maximum power of 10: Since $10 = 2 \times 5$, the limiting factor is $\min(375, 114) = 114$. Thus, RHS = $10^{114} \times 2^{261} \times 3^{25}$, where $(2^{261} \times 3^{25})$ is not divisible by $10$ (no factor of $5$).
LHS comparison: $10^m \times 1000 \times n = 10^{m+3} \times n$. Since $n$ is not divisible by $10$, we equate: $m + 3 = 114 \Rightarrow m = 111$.
Therefore, m = 111, making Option 2 the correct answer.
A shopkeeper employs a delivery boy and gives him a motorcycle for home delivery. For every delivery, the boy is given ₹ $5$. At the end of the day, he also gets ₹ $2$ for every kilometre of the distance covered in the day. The boy wants to earn more than ₹ $500$ a day, but does not want to travel more than $100$ km. Which of the following numbers of deliveries would definitely meet his target?
Detailed Explanation:
Earnings formula: E = 5d + 2k, where d = deliveries and k = km travelled. Target: E > 500, with constraint k ≤ 100.
To definitely meet the target using d alone, the inequality must hold even in the worst case (k = 0), requiring 5d > 500 or d > 100. For the given options (d = 80, 85, 90), worst-case earnings are ₹400, ₹425, and ₹450 respectively—all fall short of ₹500. Since actual km travelled is unknown and could be anywhere from 0 to 100, none of the delivery counts guarantees the target.
Therefore, without knowing the distance covered, the question cannot be answered due to insufficient data, making Option 4 the correct answer.
The top of a table is rectangular and its dimensions are $6' \times 10'$. Two rectangular portions of the table top are painted in blue colour; both these portions have dimensions $2.5' \times 8'$ and each of them has exactly two sides common with two edges of the table top. If the table is fixed to the ground and the remaining portion of the table top is painted in white, how many different patterns are possible when observed from above?
Detailed Explanation:
The table is 6' × 10' (fixed orientation). Each blue rectangle is 2.5' × 8' with exactly two sides common with table edges—meaning each blue rectangle must occupy a corner. The 8' side must align with the 10' edge, and the 2.5' side with the 6' edge.
There are 4 corner positions. Since two blue rectangles are placed, we evaluate valid non-overlapping pairs: Adjacent corners on the same long/short side overlap (bottom-left + bottom-right, or top-left + top-right). Opposite or diagonal corner pairs do not overlap: (bottom-left + top-left), (bottom-right + top-right), (bottom-left + top-right), (bottom-right + top-left).
Therefore, 4 distinct non-overlapping patterns are possible, making Option 2 the correct answer.
For two distinct real numbers $x$ and $y$, which of them is bigger? Statement I : $x^2 < y < 1$ Statement II : $y < \sqrt{x} < 1$
Detailed Explanation:
Statement I alone: x² < y < 1 allows y to be anywhere between x² and 1. For x = 0.5, y = 0.3 gives x > y; for x = 0.5, y = 0.9 gives y > x. Insufficient.
Statement II alone: y < √x < 1 requires x ∈ (0,1) but doesn't fix the relationship. For x = 0.25, y = 0.4 gives y > x; for x = 0.81, y = 0.1 gives x > y. Insufficient.
Both together: x² < y < √x with x ∈ (0,1). Since x² < x < √x for x ∈ (0,1), y can still lie on either side of x. Examples: x = 0.25, y = 0.1 → x > y; x = 0.25, y = 0.4 → y > x. Both conditions are satisfied but the relationship remains ambiguous.
Therefore, the question cannot be answered even using both statements, making Option 4 the correct answer.
How many words can one form by shuffling the letters of the word QUEUE, if Q is always followed by U? The words thus formed need not necessarily have any meaning.
Detailed Explanation:
The word QUEUE has letters: Q, U, E, U, E. Since Q must always be followed by U, treat QU as a single block. The remaining letters are U, E, E.
We now arrange 4 objects: {QU, U, E, E}. Since the two E's are identical, the number of arrangements = 4!/2! = 24/2 = 12.
Therefore, 12 words can be formed, making Option 4 the correct answer.
If $x$, $y$ and $z$ are integers, each greater than $1$, then is $x$ a prime number? Statement I : $xy^2 = 116$ Statement II : $xz = 261$
Detailed Explanation:
Statement I: $xy^2 = 116 = 2^2 \times 29$. Since $y > 1$ is an integer, $y^2$ must be a perfect square divisor of $116$. The only such divisor is $4$, so $y^2 = 4$ and $y = 2$. Thus $x = 116/4 = 29$, which is prime. Statement I alone is sufficient.
Statement II: $xz = 261 = 3^2 \times 29$. Possible values for $x$ are $3, 9, 29, 87$. Since $x = 3$ or $29$ gives prime, while $x = 9$ or $87$ gives composite, we cannot determine definitively. Statement II alone is insufficient.
Therefore, the question can be answered using Statement I alone but not Statement II alone, making Option 1 the correct answer.
The speed of a train $T$ is $100$ km per hour and the speed of a person $P$ is $4$ km per hour. $T$ crosses $P$ in $15$ seconds, if $P$ travels along the direction of motion of $T$. If $P$ travels along the opposite direction of $T$, then in how much time does $T$ cross $P$, in seconds, approximately?
Detailed Explanation:
Same-direction crossing: Relative speed = 100 - 4 = 96 km/h = 80/3 m/s. Train length L = (80/3) × 15 = 400 m.
Opposite-direction crossing: Relative speed = 100 + 4 = 104 km/h = 260/9 m/s. Time = 400 ÷ (260/9) = 3600/260 ≈ 13.846 seconds.
Therefore, the train crosses in approximately 13.85 seconds, making Option 3 the correct answer.
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 : $$A + B + C + D = 17$$ $$A + C = 6$$ $$P + Q + S + D = 15$$ $$P + Q + R + B = 17$$ $$P = R \text{ and } Q = S$$
Which one of the following statements is correct?
Detailed Explanation:
From A + B + C + D = 17 and A + C = 6, we get B + D = 11.
Using P = R and Q = S, equations (3) and (4) become: P + 2Q + D = 15 and 2P + Q + B = 17. Adding these: 3P + 3Q + (B + D) = 32. Substituting B + D = 11 gives 3(P + Q) = 21, so P + Q = 7.
Since P + Q = 7 and A + C = 6, we have 7 > 6. Options 3 and 4 cannot be determined as individual values of P and Q are unknown.
Therefore, P and Q together weigh more than A and C, making Option 2 the correct answer.
If $x$ and $y$ are integers, then is $x$ even? Statement I : $x^2 y^2$ is even. Statement II : $1 + x^2 + y^2$ is odd.
Detailed Explanation:
We need to determine if x is even.
- Statement I (x²y² is even): At least one of x or y is even, but x could be odd (y even) or even (y odd/even) — insufficient alone.
- Statement II (1 + x² + y² is odd): x² + y² must be even, so x and y have the same parity (both even or both odd) — insufficient alone.
- Combining both: At least one is even (Statement I) and both have same parity (Statement II) → both x and y must be even.
- Therefore, both statements together uniquely determine that x is even, making Option 3 the correct answer.
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?
Detailed Explanation:
Let f = forward jumps and b = backward jumps. Total jumps: f + b = 31. Net displacement: 5f - 2b = 15.
Substitute b = 31 - f into the second equation: 5f - 2(31 - f) = 15 → 5f - 62 + 2f = 15 → 7f = 77 → f = 11. Therefore, b = 20.
Therefore, the difference is |11 - 20| = 9, making Option 4 the correct answer.
There are two chemicals which do not react with each other. A container contains $10$ litres of the chemical $A$. One litre of this chemical is removed from it and one litre of the chemical $B$ is poured. Then one litre of the mixture is removed from the container and one litre of $B$ is poured. If this process of replacing one litre of the mixture by one litre of $B$ is performed once more, then what is the volume of $B$ that is present in the container approximately (in percentage)?
Detailed Explanation:
After each replacement of 1 litre from 10 litres, the fraction of chemical A remaining is (9/10). After 3 replacements, the amount of A left is: 10 × (9/10)³ = 10 × 729/1000 = 7.29 litres.
Since the total volume remains 10 litres, chemical B = 10 – 7.29 = 2.71 litres, which is 27.1% ≈ 27%.
Therefore, the volume of B is approximately 27%, making Option 2 the correct answer.
In a recruitment process, the selection of candidates is based on their performance in three components. The weightages of the components 1, 2 and 3 are $0.2$, $0.3$ and $0.5$, respectively. Use the data given below and find the cutoff score if exactly three candidates are to be selected :
| Candidate | Score in component 1 | Score in component 2 | Score in component 3 |
|---|---|---|---|
| 1 | 5 | 4 | 6 |
| 2 | 4 | 6 | 5 |
| 3 | 3 | 2 | 8 |
| 4 | 9 | 4 | 3 |
| 5 | 8 | 8 | 2 |
Detailed Explanation:
Calculate weighted score for each candidate using formula: Score = (C1 × 0.2) + (C2 × 0.3) + (C3 × 0.5).
Candidate 1: 5(0.2) + 4(0.3) + 6(0.5) = 1.0 + 1.2 + 3.0 = 5.2
Candidate 2: 4(0.2) + 6(0.3) + 5(0.5) = 0.8 + 1.8 + 2.5 = 5.1
Candidate 3: 3(0.2) + 2(0.3) + 8(0.5) = 0.6 + 0.6 + 4.0 = 5.2
Candidate 4: 9(0.2) + 4(0.3) + 3(0.5) = 1.8 + 1.2 + 1.5 = 4.5
Candidate 5: 8(0.2) + 8(0.3) + 2(0.5) = 1.6 + 2.4 + 1.0 = 5.0
Ranking in descending order: 5.2, 5.2, 5.1, 5.0, 4.5. Top three candidates score 5.2, 5.2, and 5.1. The cutoff is the score of the third-highest candidate.
Therefore, the cutoff score is 5.1, making Option 1 the correct answer.
$X$, $Y$ 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?
Detailed Explanation:
At 8 AM, all three are at mark 199'. Since X jumps 4', Y jumps 6', and Z jumps 5', they will all land on the same mark again when the distance from 199' is a multiple of LCM(4, 6, 5) = 60'.
Common landing marks are 199 + 60k where k = 0, 1, 2, ... Within the range 195' < m < 1000', we get: 199, 259, 319, 379, 439, 499, 559, 619, 679, 739, 799, 859, 919, 979. The next mark (1039') exceeds 1000'.
Therefore, counting from k=0 to k=13 gives 14 common marks, making Option 4 the correct answer.
There are four statements X, Y, Z and W. Their relations are as follows:
If X is incorrect, then so is Z; if Y is incorrect, then W is correct; if W is correct, then X is incorrect.
Which of the following is/are correct?
I. If X is correct, then so is Y.
II. If Z is correct, then it is not necessary that Y is correct.
Select the answer using the code given below.
Detailed Explanation:
Given relations: ¬X ⇒ ¬Z; ¬Y ⇒ W; W ⇒ ¬X. Chaining these: ¬Y ⇒ W ⇒ ¬X ⇒ ¬Z.
Taking the contrapositive (reversing and negating): Z ⇒ X ⇒ ¬W ⇒ Y. This shows X ⇒ Y (Statement I is correct) and Z ⇒ Y (Statement II is incorrect, as Y must be correct if Z is correct).
Therefore, only Statement I is correct, making Option 1 the correct answer.
Seven cubes are identical in shape. Out of these, the weight of each of the six cubes is equal and the weight of the remaining cube is less than the weight of any other cube. A balance is used to identify the lightest cube. What is the minimum number of attempts required to distinguish the odd cube with certainty?
Detailed Explanation:
Divide the 7 cubes into groups of 3, 3, and 1. In the first weighing, place 3 cubes on each pan. If balanced, the lightest cube is the one left aside (identified in 1 attempt). If unbalanced, the lightest is among the 3 cubes on the lighter pan.
In the second weighing, take the 3 cubes from the lighter pan and weigh any 2 (one on each side), leaving the third aside. If balanced, the lightest is the one left aside. If unbalanced, the lightest is on the lighter pan.
Therefore, a maximum of 2 weighings guarantees identification of the lightest cube with certainty, making Option 1 the correct answer.
P, Q, R, S and T are ranked 1 to 5 (not necessarily in that order). The rank of P is 4, the rank of Q is not 5, the rank of R is 1, the rank of S is not 2, the rank of T is not
- Then which of the following is/are correct? I. If the rank of S is 3, then that of T is
- II. If the rank of Q is 3, then that of T is
- Select the answer using the
code given below.
Detailed Explanation:
Given: P=4, R=1, Q≠5, S≠2, T≠3. Remaining ranks {2,3,5} must be assigned to Q, S, T.
- Statement I: If S=3, then Q and T take {2,5}. Since Q≠5, Q=2 and T=5. Thus T=5, not 2. Statement I is incorrect.
- Statement II: If Q=3, then S and T take {2,5}. Since S≠2, S=5 and T=2. Thus T=2, not 5. Statement II is incorrect.
- Therefore, both statements are incorrect, making Option 4 the correct answer.
P has a son and a daughter. S is the mother of T. S is R's spouse. Q and R are children of P. Then how is Q related to S?
Detailed Explanation:
P has a son and a daughter: Q and R. S is the mother of T, making S female. S is R's spouse, so R is male (husband of S). Since P has one son and one daughter, R is the son and Q is the daughter.
Q is the sister of R, and R is the husband of S. This makes Q the sister of the husband of S (or sister-in-law of S).
Therefore, Q is the sister of the husband of S, making Option 4 the correct answer.
UPSC CSAT 2026 - CSAT (Civil Services Aptitude Test) Chapter-wise Distribution
Reading Comprehension
23 Qs (28.8%)Number System
19 Qs (23.8%)Arithmetic
11 Qs (13.8%)Analytical Ability
9 Qs (11.3%)Logical Reasoning
7 Qs (8.8%)Data Sufficiency
4 Qs (5%)Algebra
3 Qs (3.8%)Decision Making & Problem Solving
2 Qs (2.5%)General Mental Ability
1 Qs (1.3%)Data Interpretation
1 Qs (1.3%)UPSC CSAT 2026 - CSAT (Civil Services Aptitude Test) Questions FAQs
Q1 How many CSAT (Civil Services Aptitude Test) questions were asked in UPSC CSAT 2026?
Q2 What is the chapter-wise question distribution for CSAT (Civil Services Aptitude Test) in UPSC CSAT 2026?
- Reading Comprehension: 23 questions (28.8%)
- Number System: 19 questions (23.8%)
- Arithmetic: 11 questions (13.8%)
- Analytical Ability: 9 questions (11.3%)
- Logical Reasoning: 7 questions (8.8%)
- Data Sufficiency: 4 questions (5%)
- Algebra: 3 questions (3.8%)
- Decision Making & Problem Solving: 2 questions (2.5%)
- General Mental Ability: 1 questions (1.3%)
- Data Interpretation: 1 questions (1.3%)