UPSC Prelims 2026 CSAT Paper-II: Detailed Answer Key & Solutions to All 80 Questions

UPSC CSAT 2026 · Question 1

For \(\frac{1}{3} < x < y < 2\) which of the following statements is/are always correct?

1. I. \(x + \frac{1}{x} < y + \frac{1}{y}\)
2. II. \(\frac{\sqrt{1+y^2}}{y} < \frac{\sqrt{1+x^2}}{x}\)

Select the answer using the code given below.

1. I only
2. II only
3. Both I and II
4. Neither I nor II

Explanation

Step 1: Test statement I on the interval \(\frac{1}{3} < x < y < 2\) using the function \(f(t) = t + \frac{1}{t}\).

Step 2: Differentiate: \(f'(t) = 1 – \frac{1}{t^2}\). This is negative when (t < 1) and positive when (t > 1), so (f) decreases on \((\frac{1}{3}, 1)\) and increases on ((1, 2)) — it is not monotonic over the full interval.

Step 3: Pick a counter-example with (x = 0.5) and (y = 1.5) (both lie in \((\frac{1}{3}, 2)\) and (x < y)). Compute \(f(x) = 0.5 + \frac{1}{0.5} = 0.5 + 2 = 2.5\).

Step 4: Compute \(f(y) = 1.5 + \frac{1}{1.5} = 1.5 + 0.6667 \approx 2.1667\).

Step 5: Here \(f(x) = 2.5 > f(y) \approx 2.17\), which contradicts the claim \(x + \frac{1}{x} < y + \frac{1}{y}\). Statement I is therefore not always correct.

Step 6: For statement II, simplify \(g(t) = \frac{\sqrt{1+t^2}}{t} = \sqrt{\frac{1+t^2}{t^2}} = \sqrt{\frac{1}{t^2} + 1}\).

Step 7: As (t) increases (for (t > 0)), \(\frac{1}{t^2}\) strictly decreases, so \(\frac{1}{t^2} + 1\) decreases, and hence (g(t)) strictly decreases.

Step 8: Therefore for (x < y) we always have (g(y) < g(x)), i.e., \(\frac{\sqrt{1+y^2}}{y} < \frac{\sqrt{1+x^2}}{x}\). Statement II is always correct.

Step 9: Only II is always true, so the answer is (b).

UPSC CSAT 2026 · Question 2

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?

1. 4
2. 5
3. 9
4. 13

Explanation

Step 1: We have measuring cylinders of capacity 1 L, 6 L, 25 L and 100 L, and want to obtain exactly 298 L from the tank.

Step 2: Each fill or removal of a complete cylinder counts as one measurement. We want to minimise the total count.

Step 3: Note 298 is close to 300, which is a clean multiple of 100. So use the 100 L cylinder three times: that delivers \(3 \times 100 = 300\) L in 3 measurements.

Step 4: This overshoots the target by 2 L, so we must remove 2 L. Using the 1 L cylinder twice removes exactly \(2 \times 1 = 2\) L, taking 2 more measurements.

Step 5: Total measurements (= 3 + 2 = 5).

Step 6: Check alternatives. Using only 25 L and 1 L: \(11 \times 25 + 23 \times 1 = 275 + 23 = 298\) needs 34 operations — much worse. Using 6 L and 1 L: at least (lceil 298/6 rceil = 50) ops. Using two 100 L and the rest from 25 + 6 + 1: \(100+100+25+25+25+6+6+6+5\cdot 1\) already exceeds 5.

Step 7: There is no way to make 298 L in fewer than 5 measurements because \(298 = 2 \cdot 100 + 98\) and 98 cannot be reached from ({1,6,25}) in \(\le 2\) steps, and \(298 = 3 \cdot 100 – 2\) already needs 5.

Step 8: Therefore the minimum number of measurements is 5, and the answer is (b).

UPSC CSAT 2026 · Question 3

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?

1. 7
2. 8
3. 9
4. 10

Explanation

Step 1: Let (a) be the number of 1 kg weights used, with constraint \(8 \le a \le 11\). The remaining weight to be made from ({2, 5, 10}) kg weights is (R = 20 – a).

Step 2: Case (a = 8), (R = 12). Partition 12 using ({2,5,10}): (i) (10 + 2 = 12), (ii) (5 + 5 + 2 = 12), (iii) (2+2+2+2+2+2 = 12). That gives 3 ways.

Step 3: Case (a = 9), (R = 11). Partitions of 11: must include odd count of 5s. (5 + 2+2+2 = 11) works. (5+5+? = 11) needs 1, not allowed. So only 1 way.

Step 4: Case (a = 10), (R = 10). Partitions: (i) (10), (ii) (5+5), (iii) (2+2+2+2+2). That gives 3 ways.

Step 5: Case (a = 11), (R = 9). Partitions: (5 + 2+2 = 9). The only option. That is 1 way.

Step 6: Total number of ways (= 3 + 1 + 3 + 1 = 8).

Step 7: Therefore the answer is (b).

UPSC CSAT 2026 · Question 4

A cut on a solid object divides the object into two parts where the new surfaces thus produced are plane. On the other hand, one single cut can be used to cut more than one object at a time. In an experiment, the total number of pieces produced by applying n cuts is denoted by x_n. The experiment is performed on a solid cube where pieces remain unmoved after each cut. In this experiment, if after the third cut, the pieces are identical, then which of the following is not a possible value for x_4?

1. 16
2. 12
3. 8
4. 5

Explanation

Step 1: After 3 cuts the pieces are identical. From a solid cube this is possible only when (x_3 in {4, 6, 8}): 4 equal slabs (3 parallel cuts), 6 equal bricks (2 parallel + 1 perpendicular), or 8 equal sub-cubes (3 mutually perpendicular bisecting cuts).

Step 2: From (x_3 = 4) slabs, one more cut can hit between 1 and 4 slabs, giving (x_4 in {5, 6, 7, 8}). So 5 and 8 are achievable.

Step 3: From (x_3 = 6) bricks, a single plane can pass through up to 6 bricks, giving (x_4) up to (6 + 6 = 12). So 12 is achievable.

Step 4: From (x_3 = 8) sub-cubes arranged \(2\times 2\times 2\), a single plane cut can intersect at most 7 of the 8 sub-cubes (a diagonal plane misses at least one corner cube). This gives \(x_4 \le 8 + 7 = 15\).

Step 5: So the maximum value of (x_4) achievable in any of the three identical-piece configurations is 15.

Step 6: Among the options: 5 is achievable (Step 2), 8 is achievable (Step 2), 12 is achievable (Step 3), but 16 is not achievable because \(x_4 \le 15\) in all cases.

Step 7: Therefore the value not possible is 16, answer (a).

UPSC CSAT 2026 · Question 5

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

1. 2
2. 3
3. 5
4. 6

Explanation

Step 1: Let (n) be the number of students in the class. The first correction changes one student’s score from 0 to 100, an increase of (+100) in the total.

Step 2: The class average increases by 4 because of this correction. So \(\frac{100}{n} = 4\).

Step 3: Solving: \(n = \frac{100}{4} = 25\) students.

Step 4: The second correction changes a recorded score from 81 to the actual score 56, a change of (56 – 81 = -25) in the total.

Step 5: The corresponding change in the average is \(\frac{-25}{25} = -1\).

Step 6: Net change in the average from the original to final: (+4 + (-1) = +3).

Step 7: Therefore (y – x = 3), and the answer is (b).

Passage — for Questions 6–8

Read the following passage and answer the items that follow. Your answers should be based solely on the passage.

When SARS-CoV-2 was first detected in 2019, it was a truly novel virus for the world. At that time, no one in the world had been exposed to SARS-CoV-2 or had specific immunity against it. In contrast, people across the world have been exposed to HMPV for decades and the virus is well-studied. HMPV and SARS-CoV-2 belong to two very different virus families with fundamentally different characteristics and epidemiology, with strong seasonality seen in HMPV, unlike SARS-CoV-2. Both viruses cause different severity of symptoms, particularly over the long term, and the affected population segments do not fully overlap. In general, HMPV causes milder illness with deaths being very rare and with no long-term post-viral symptoms.

UPSC CSAT 2026 · Question 6

Which of the following conclusions is/are valid?

1. 1. Though SARS-CoV-2 and HMPV are similar viruses with somewhat different epidemiology, the former became a pandemic because it was novel and people had not been exposed to it in the past.
2. 2. The two viruses have fundamentally different impacts on human populations and should not therefore be dealt with in a similar manner.

Select the answer using the code given below:

1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2

Explanation

Step 1: Evaluate statement 1. The passage states: ‘HMPV and SARS-CoV-2 belong to two very different virus families with fundamentally different characteristics and epidemiology’. So calling them ‘similar viruses’ contradicts the passage. Statement 1 is invalid.

Step 2: Evaluate statement 2. The passage emphasises differences in virus family, severity, affected segments, long-term impact, and seasonality. These fundamental differences logically support the claim that the two should not be dealt with in a similar manner.

Step 3: Supporting line: ‘Both viruses cause different severity of symptoms, particularly over the long term, and the affected population segments do not fully overlap.’ This justifies a different response strategy for each virus.

Step 4: Therefore only statement 2 is a valid conclusion.

Step 5: Therefore the answer is (b).

UPSC CSAT 2026 · Question 7

Which of the following reflect the intent of the writer in the above passage?

1. 1. To evolve methodologies for objective analysis of the two viruses
2. 2. To establish the epidemiological similarities and differences between the two viruses
3. 3. To offer a better understanding of the remedies of HMPV when analysed in conjunction with SARS-CoV-2

Select the answer using the code given below.

1. 1 and 2
2. 2 and 3
3. 1 and 3
4. None of the above

Explanation

Step 1: Identify the writer’s intent. The passage simply contrasts HMPV with SARS-CoV-2 across family, immunity history, seasonality, severity, and post-viral symptoms.

Step 2: Option 1 (evolve methodologies for objective analysis) is not the intent — the writer presents differences but does not propose any analytical methodology.

Step 3: Option 2 (establish epidemiological similarities and differences) is partly correct on differences, but the writer does not ‘establish similarities’ — the passage actively denies similarity (‘two very different virus families’).

Step 4: Option 3 (offer remedies for HMPV in conjunction with SARS-CoV-2) is unsupported — no remedies are discussed for either virus.

Step 5: Since none of the three statements correctly captures the writer’s intent (which is to dispel pandemic-like fears around HMPV by highlighting how unlike SARS-CoV-2 it is), the correct response is ‘None of the above’.

Step 6: Therefore the answer is (d).

UPSC CSAT 2026 · Question 8

Which of the following statements reflect the logical and rational inferences that can be drawn from the passage?

1. 1. HMPV has, historically, had longer documentation of studies when compared with SARS-CoV-2.
2. 2. The two viruses are different from each other and this results in markedly different outcomes amongst those affected.
3. 3. Long-term impacts of the two viruses are dissimilar and this is an important differentiator between them.
4. 4. One of the common factors between the two viruses is seasonal specificity.

Select the answer using the code given below.

1. 1 and 2 only
2. 1, 2 and 3
3. 3 and 4
4. 1 and 3 only

Explanation

Step 1: Statement 1 — HMPV has longer documentation than SARS-CoV-2. The passage says people have been exposed to HMPV ‘for decades’ and the virus is ‘well-studied’, while SARS-CoV-2 was ‘truly novel’ in 2019. So statement 1 is a valid inference.

Step 2: Statement 2 — the viruses are different and outcomes markedly differ. The passage states ‘fundamentally different characteristics and epidemiology’ and ‘different severity of symptoms… affected population segments do not fully overlap’. Statement 2 is valid.

Step 3: Statement 3 — long-term impacts differ. The passage explicitly says HMPV causes ‘no long-term post-viral symptoms’, contrasted with SARS-CoV-2’s long-term issues. Statement 3 is valid.

Step 4: Statement 4 — seasonal specificity is a ‘common factor’. The passage says ‘strong seasonality seen in HMPV, unlike SARS-CoV-2’. So seasonality is a differentiator, not a commonality. Statement 4 is invalid.

Step 5: Valid statements are 1, 2, and 3.

Step 6: Therefore the answer is (b).

UPSC CSAT 2026 · Question 9

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 × y) + z }; then M – N is

1. 11
2. 12
3. 13
4. 14

Explanation

Step 1: (x, y, z) are three distinct values chosen from ({2, 3, 4, 5}). The expression is \(E = (x \times y) + z\).

Step 2: To maximise (E), make the product (xy) as large as possible because it dominates the small additive (z).

Step 3: The largest product from distinct pairs is \(4 \times 5 = 20\). The remaining values for (z) are ({2, 3}); pick the larger one, (z = 3).

Step 4: Maximum value (M = 20 + 3 = 23).

Step 5: To minimise (E), make (xy) as small as possible. The smallest product from distinct pairs is \(2 \times 3 = 6\).

Step 6: The remaining values for (z) are ({4, 5}); pick the smaller one, (z = 4).

Step 7: Minimum value (N = 6 + 4 = 10).

Step 8: Therefore (M – N = 23 – 10 = 13), and the answer is (c).

UPSC CSAT 2026 · Question 10

Suppose x, y and z are variables taking positive real numbers as their possible values. It is given that y is directly proportional to x² 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?

1. 1/7
2. 1/5
3. 5/7
4. 1

Explanation

Step 1: Translate the proportionalities. ‘y directly proportional to (x^2)’ gives (y = k x^2) for some constant (k). ‘x inversely proportional to (z)’ gives \(x = \frac{c}{z}\), equivalently (xz = c) for some constant (c).

Step 2: Use the given data \(z = \frac{7}{25}\), (x = 5), (y = 50). Compute \(c = x z = 5 \times \frac{7}{25} = \frac{35}{25} = \frac{7}{5}\).

Step 3: Compute (k) from (y = k x^2): \(50 = k \times 25\), so \(k = \frac{50}{25} = 2\).

Step 4: Now consider (y = 98). Use \(y = 2 x^2 \Rightarrow x^2 = \frac{98}{2} = 49\), so (x = 7) (positive real).

Step 5: Use \(xz = c = \frac{7}{5}\) to get \(z = \frac{c}{x} = \frac{7/5}{7} = \frac{1}{5}\).

Step 6: Therefore \(z = \frac{1}{5}\), and the answer is (b).

Passage — for Questions 11–12

Read the following passage and answer the items that follow. Your answers should be based solely on the passage.

India is starting to deploy AI for critical use cases such as weather forecasting, pest detection and control, and crop-yield optimisation. However, penetration is limited to a small subset of tech-savvy farmers. In the US and in Europe, generative AI tools have started offering precision farming at scale, integrating large datasets to provide real-time agronomic insights. For at-scale integration and accessibility of AI tools in India, it would be helpful to develop Indian languages-based AI tools for smallholder farmers, partner with AgTechs to create affordable AI solutions, and disseminate AI-based advisory services through government programmes.

UPSC CSAT 2026 · Question 11

Which of the following assumptions is/are valid?

1. 1. Agricultural productivity has marched ahead in the West because of the economies of scale facilitated by the adoption of AI tools.
2. 2. Affordable AI tools rendered available in local languages can help AI-based solutions reach more and more small farmers.
3. 3. Though penetration is as yet low, critical application areas deploying AI tools are already in use in India.

Select the answer using the code given below.

1. 1, 2 and 3
2. 2 only
3. 1 and 3 only
4. 2 and 3 only

Explanation

Step 1: In UPSC comprehension items, a ‘valid assumption’ is a premise the passage’s argument depends on. Test each statement against that argument: India has started using AI in farming, but reach is thin, so it needs tools that scale and stay accessible.

Step 2: Statement 1 says Western farm productivity rose because AI tools brought economies of scale. The passage only says generative AI tools in the US and Europe ‘have started offering precision farming at scale’. It never says productivity rose, and it names no cause. The argument does not need this claim, so 1 is not valid.

Step 3: Statement 2 says affordable AI tools in local languages can take AI-based solutions to more small farmers. The passage recommends ‘Indian languages-based AI tools for smallholder farmers’ and ‘affordable AI solutions’ because it takes exactly this for granted. Drop it and the recommendation has no point. So 2 is valid.

Step 4: Statement 3 says critical AI applications are already in use in India, though penetration is low. The whole argument stands on this: India ‘is starting to deploy AI for critical use cases’, but ‘penetration is limited to a small subset of tech-savvy farmers’. The call for ‘at-scale integration’ only makes sense if some deployment already exists and has not spread yet. So 3 is valid too.

Step 5: A premise the passage states outright can still be a valid assumption. Rejecting 3 only because it is written in the passage is the trap here.

Step 6: Statements 2 and 3 are valid; statement 1 goes beyond the passage. Therefore the answer is (d).

UPSC CSAT 2026 · Question 12

Which of the following statements is/are not correct?

1. 1. Tech-savvy farmers will drive the AgTech companies of the future.
2. 2. The development of advisory services by the government programmes for the use of AI tools in agriculture would be helpful.
3. 3. In the US and Europe, AI tools have replaced traditional agricultural practices.
4. 4. The integration of large datasets for use in real-time agronomic analysis is already a reality.

Select the answer using the code given below.

1. 1 and 3 only
2. 1, 3 and 4
3. 2 and 4
4. 3 only

Explanation

Step 1: We need statements that are NOT correct per the passage.

Step 2: Statement 1 — tech-savvy farmers will drive future AgTechs. The passage only notes current AI penetration is limited to tech-savvy farmers; it does not claim they will drive the AgTech industry. Not supported, hence incorrect.

Step 3: Statement 2 — government programmes developing AI advisory services would be helpful. The passage explicitly suggests disseminating ‘AI-based advisory services through government programmes’. Correct.

Step 4: Statement 3 — in the US/Europe, AI has replaced traditional practices. The passage only says generative AI is ‘offering precision farming at scale’ — it does not say AI has replaced traditional practices. Incorrect.

Step 5: Statement 4 — real-time agronomic integration of large datasets is already a reality. The passage says generative AI ‘integrating large datasets to provide real-time agronomic insights’. Correct.

Step 6: The not-correct statements are 1 and 3.

Step 7: Therefore the answer is (a).

UPSC CSAT 2026 · Question 13

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 3. Then which of the following is/are correct?

1. I. If the rank of S is 3, then that of T is 2.
2. II. If the rank of Q is 3, then that of T is 5.

Select the answer using the code given below.

1. I only
2. II only
3. Both I and II
4. Neither I nor II

Explanation

Step 1: Fixed ranks: (P = 4), (R = 1). So (Q, S, T) take ranks ({2, 3, 5}) in some order.

Step 2: Constraints on the remaining three: \(Q \ne 5\), \(S \ne 2\), \(T \ne 3\). Thus (Q in {2, 3}), (S in {3, 5}), (T in {2, 5}).

Step 3: Enumerate valid assignments. Try (Q = 2): then ({S, T} = {3, 5}). (S = 3) allowed, (T = 5) allowed → valid. (S = 5, T = 3): (T = 3) forbidden, invalid.

Step 4: Try (Q = 3): then ({S, T} = {2, 5}). (S = 2) forbidden, so (S = 5, T = 2) — valid.

Step 5: So only two valid arrangements: (A) (Q=2, S=3, T=5); (B) (Q=3, S=5, T=2).

Step 6: Test claim I — ‘If (S = 3), then (T = 2)’. From arrangement (A), (S = 3) but (T = 5), not 2. Statement I is false.

Step 7: Test claim II — ‘If (Q = 3), then (T = 5)’. From arrangement (B), (Q = 3) but (T = 2), not 5. Statement II is false.

Step 8: Neither I nor II is correct.

Step 9: Therefore the answer is (d).

UPSC CSAT 2026 · Question 14

Two identical straight rods are painted in five distinct colours so that each of them gets divided into five equal parts along the length. In one of them, the portions are marked P1, P2, P3, P4 and P5 (not necessarily in that order) whereas in the other, they are marked Q1, Q2, Q3, Q4 and Q5 (not necessarily in that order). When the rods are kept parallel to each other side by side, P1 and Q3 match, P4 matches Q1 or Q2, and Q4 matches P3 or P5. If Q3 and Q5 are adjacent, which of the following is/are possible?

1. I. Q3 is marked at the middle portion of the straight rod.
2. II. P2 is marked at one of the extreme portions of the straight rod.

Select the answer using the code given below.

1. I only
2. II only
3. Both I and II
4. Neither I nor II

Explanation

Step 1: Each rod has 5 equal parts. Number the positions 1, 2, 3, 4, 5 from left to right; position 3 is the ‘middle’ and positions 1, 5 are the ‘extremes’.

Step 2: Translate constraints. (i) (P_1) aligns with (Q_3), so they sit at the same position. (ii) (P_4) aligns with (Q_1) or (Q_2). (iii) (Q_4) aligns with (P_3) or (P_5). (iv) (Q_3) and (Q_5) are at adjacent positions on rod 2.

Step 3: Try (Q_3) (and hence (P_1)) at position 3 — the middle. Then (Q_5) must be at position 2 or 4 (adjacent to 3).

Step 4: Take (Q_5 = 4). Remaining (Q)-labels ({Q_1, Q_2, Q_4}) occupy positions ({1, 2, 5}).

Step 5: Place (Q_4 = 5), (Q_2 = 2), (Q_1 = 1) (one valid choice).

Step 6: Now place (P)’s such that (P_4) matches (Q_1) or (Q_2), and (P_3) or (P_5) matches (Q_4). With (P_1 = 3), the remaining positions for ({P_2, P_3, P_4, P_5}) are ({1, 2, 4, 5}). Choose (P_4 = 2) (matches (Q_2) at 2), (P_3 = 5) (matches (Q_4) at 5), and (P_2 = 1), (P_5 = 4).

Step 7: Layout:

Position:  1    2    3    4    5
Rod 1 :   P2   P4   P1   P5   P3
Rod 2 :   Q1   Q2   Q3   Q5   Q4

Step 8: Verify: (P_1 leftrightarrow Q_3) at pos 3 ✓, (P_4 leftrightarrow Q_2) at pos 2 ✓, (P_3 leftrightarrow Q_4) at pos 5 ✓, (Q_3) and (Q_5) adjacent (positions 3 and 4) ✓.

Step 9: So (Q_3) is at the middle (statement I is possible) and (P_2) is at position 1, an extreme (statement II is possible).

Step 10: Therefore both I and II are possible, answer (c).

UPSC CSAT 2026 · Question 15

Seven persons A, B, C, D, E, F and G travel by three cars X, Y, Z. A and another two of them travel by X. Only E travels with G. C travels by Z, but B does not travel by Y. Besides, A and B do not travel by the same car. Then which of the following are correct?

1. I. No one travels alone.
2. II. Only D travels with F.
3. III. Only C travels with B.

Select the answer using the code given below.

1. I and II only
2. I and III only
3. II and III only
4. All the three

Explanation

Step 1: Conditions — ‘A and another two travel by X’ means car X has exactly 3 people including A. ‘Only E travels with G’ means whichever car contains E or G contains exactly ({E, G}) — no one else.

Step 2: C travels by Z. B does not travel by Y. A and B are in different cars; since A is in X, B is in Y or Z, but not Y, so B is in Z.

Step 3: E and G form their own pair. They cannot be in Z (which contains B and C — that would add a third) and cannot be in X (X has A plus exactly two others; pairing E+G with A would add but they cannot travel with anyone other than each other). So ({E, G}) occupies car Y entirely.

Step 4: Remaining persons are D and F. Car X needs two more besides A; the only candidates left are D and F. So (X = {A, D, F}), (Y = {E, G}), (Z = {B, C}).

Step 5: Seating chart:

Car X : A, D, F   (3 people)
Car Y : E, G      (2 people)
Car Z : B, C      (2 people)

Step 6: Evaluate I — ‘No one travels alone’. Every car has at least 2 people. I is correct.

Step 7: Evaluate II — ‘Only D travels with F’. False, because D also travels with A in car X. II is incorrect.

Step 8: Evaluate III — ‘Only C travels with B’. Car Z has just B and C. III is correct.

Step 9: Correct statements: I and III.

Step 10: Therefore the answer is (b).

UPSC CSAT 2026 · Question 16

There are four statements X, Y, Z and W. Their relations are as follows:

1. 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?
2. I. If X is correct, then so is Y.
3. II. If Z is correct, then it is not necessary that Y is correct.

Select the answer using the code given below.

1. I only
2. II only
3. Both I and II
4. Neither I nor II

Explanation

Step 1: Translate the rules using logical notation. R1: \(\lnot X \Rightarrow \lnot Z\). R2: \(\lnot Y \Rightarrow W\). R3: \(W \Rightarrow \lnot X\).

Step 2: Form contrapositives. From R1: \(Z \Rightarrow X\). From R2: \(\lnot W \Rightarrow Y\). From R3: \(X \Rightarrow \lnot W\).

Step 3: Evaluate claim I: ‘If X is correct, then Y is correct’. Start with (X). By R3’s contrapositive, \(X \Rightarrow \lnot W\). By R2’s contrapositive, \(\lnot W \Rightarrow Y\). Chaining: \(X \Rightarrow \lnot W \Rightarrow Y\). So I is correct.

Step 4: Evaluate claim II: ‘If Z is correct, then it is not necessary that Y is correct’. Start with (Z). By R1’s contrapositive, \(Z \Rightarrow X\).

Step 5: From Step 3, \(X \Rightarrow Y\). So \(Z \Rightarrow X \Rightarrow Y\), meaning Z correct does force Y correct.

Step 6: Claim II says Y is ‘not necessary’ if Z is correct — but Y IS necessary. So II is incorrect.

Step 7: Only statement I is correct.

Step 8: Therefore the answer is (a).

UPSC CSAT 2026 · Question 17

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?

1. 35
2. 34
3. 31
4. 29

Explanation

Step 1: Let (T) be the time X takes to complete 7 rounds. In the same time Y completes 5 rounds. So speeds: \(v_X = \frac{7}{T}\) rounds per unit time, \(v_Y = \frac{5}{T}\).

Step 2: After Y completes 5 rounds (i.e., at time (T)), Y reverses direction and doubles speed: new \(v_Y’ = 2 \cdot \frac{5}{T} = \frac{10}{T}\) in the opposite direction.

Step 3: Total run time: they stop when X finishes 21 rounds, i.e., at time \(\frac{21}{v_X} = 3T\).

Step 4: Phase 1 (time 0 to (T)): X and Y run in the same direction. Relative speed \(= v_X – v_Y = \frac{7 – 5}{T} = \frac{2}{T}\) rounds per unit time. In duration (T), relative distance covered (= 2) rounds. So X laps Y exactly 2 times. Meetings include the instant (t = T) when Y has done 5 rounds (X has done 7, lapped Y twice).

Step 5: Phase 2 (time (T) to (3T), duration (2T)): X continues forward at \(\frac{7}{T}\); Y runs backward at \(\frac{10}{T}\). Relative speed (closing speed) \(= \frac{7}{T} + \frac{10}{T} = \frac{17}{T}\) rounds per unit time. In duration (2T), relative distance \(= 17 \times 2 = 34\) rounds.

Step 6: They meet every time they cover 1 round of relative distance, so number of meetings in Phase 2 (= 34). But the very last meeting at (t = 3T) coincides with the moment they stop (‘before they finally stopped’ excludes this).

Step 7: Net Phase 2 meetings = (34 – 1 = 33). Phase 1 meetings = 2 (including the one at (t = T)).

Step 8: Track schematic (linearised):

Phase 1: 0 ----T  (X →, Y →, X laps Y twice, 2 meetings)
Phase 2: T ----3T (X →, Y ←, closing speed 17/T)
         |---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
         34 unit-laps of relative distance → 34 meetings, last one at t=3T excluded

Step 9: Total meetings = (2 + 33 = 35).

Step 10: Therefore the answer is (a).

UPSC CSAT 2026 · Question 18

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?

1. 99
2. 81
3. 84
4. 79

Explanation

Step 1: Let (c) = number of correct answers and (w) = number of wrong answers. Student attempted all questions, so total questions (= c + w).

Step 2: Scheme 1 (+5 for correct, −2 for wrong) gave score 69: (5c – 2w = 69) … (i)

Step 3: Scheme 2 (+4 for correct, −1 for wrong) gave score 84: (4c – w = 84) … (ii)

Step 4: From (ii), (w = 4c – 84).

Step 5: Substitute into (i): (5c – 2(4c – 84) = 69), i.e., (5c – 8c + 168 = 69).

Step 6: Simplify: \(-3c + 168 = 69 \Rightarrow -3c = -99 \Rightarrow c = 33\).

Step 7: Compute (w = 4(33) – 84 = 132 – 84 = 48).

Step 8: Verify: scheme 1 gives (5(33) – 2(48) = 165 – 96 = 69) ✓. Scheme 2 gives (4(33) – 48 = 132 – 48 = 84) ✓.

Step 9: Total questions (= c + w = 33 + 48 = 81).

Step 10: Therefore the answer is (b).

Passage — for Questions 19–20

Read the following passage and answer the items that follow. Your answers should be based solely on the passage.

The Juvenile Justice (Care and Protection of Children) Act, or the JJ Act, 2015 allows for the possibility for trying adolescents (above 16) as adults if they are accused of committing a heinous offence. A heinous offence is one with a minimum punishment of seven years. Offences such as culpable homicide and causing death by negligence, which are common in drunken driving cases, are not heinous offences because they do not have a prescribed minimum punishment. The JJ Act, amended in 2021, now categorises an offence that has no minimum sentence, but has a maximum sentence of seven years or more as a serious offence which nonetheless, in the opinion of activists, does not merit the transfer of a case to the adult criminal justice system.

UPSC CSAT 2026 · Question 19

Which of the following conclusions is/are valid?

1. 1. Only a serious offence as categorised by the revised JJ Act, justifies the transfer of a case to the adult judicial system.
2. 2. The JJ Act, 2021, categorises an offence as a serious offence based on the maximum sentence it carries, rather than on the minimum sentence.

Select the answer using the code given below.

1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2

Explanation

Step 1: Re-read the passage. A heinous offence has a minimum punishment of seven years; only a heinous offence justifies transfer to the adult system (per the original JJ Act, 2015).

Step 2: The 2021 amendment introduced ‘serious offence’ (no minimum sentence, maximum \(\ge 7\) years), but activists hold that a ‘serious offence’ does NOT merit transfer to the adult criminal justice system.

Step 3: Evaluate conclusion 1 — ‘only a serious offence justifies transfer to the adult judicial system’. This contradicts both the Act (which uses ‘heinous’ for the transfer trigger) and the activists (who say serious offences do not merit such transfer). Conclusion 1 is invalid.

Step 4: Evaluate conclusion 2 — ‘the JJ Act, 2021, categorises an offence as serious based on the maximum sentence, rather than on the minimum sentence’.

Step 5: Supporting line: ‘categorises an offence that has no minimum sentence, but has a maximum sentence of seven years or more as a serious offence.’ This directly confirms the maximum-sentence criterion.

Step 6: Conclusion 2 is valid.

Step 7: Only conclusion 2 is valid.

Step 8: Therefore the answer is (b).

UPSC CSAT 2026 · Question 20

Which of the following statements is/are correct?

1. 1. If an offence has no minimum prescribed punishment, it cannot be considered heinous as per the JJ Act, 2015.
2. 2. As per the JJ Act, 2021, an offence for which there is a provision for a maximum sentence of seven years or more, but no minimum sentence, is to be considered a serious offence.

Select the answer using the code given below.

1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2

Explanation

Step 1: Evaluate statement 1 — ‘If an offence has no minimum prescribed punishment, it cannot be heinous per JJ Act, 2015’.

Step 2: The passage defines a heinous offence as ‘one with a minimum punishment of seven years’. It then gives an example: ‘Offences such as culpable homicide and causing death by negligence… are not heinous offences because they do not have a prescribed minimum punishment.’

Step 3: This directly confirms statement 1: absence of a minimum punishment disqualifies an offence from being heinous. Statement 1 is correct.

Step 4: Evaluate statement 2 — ‘As per JJ Act, 2021, an offence with a maximum sentence of seven years or more but no minimum is a serious offence’.

Step 5: Supporting line: ‘The JJ Act, amended in 2021, now categorises an offence that has no minimum sentence, but has a maximum sentence of seven years or more as a serious offence.’

Step 6: This exactly matches statement 2. Statement 2 is correct.

Step 7: Both statements 1 and 2 are correct.

Step 8: Therefore the answer is (c).

UPSC CSAT 2026 · Question 21

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.

1. 4:59:00 PM
2. 4:59:02 PM
3. 4:58:42 PM
4. 4:56:32 PM

Explanation

Let (d) be the distance (in metres) from the camp to the explosion site.

Step 1: Set up the timeline. The explosion happens at unknown time (T_0). Sound travels to the camp at 300 m/s, so the sensor receives the bang at (T_0 + d/300). Drone 1 is launched the instant the sound is received and flies at 30 m/s, reaching the spot at (T_0 + d/300 + d/30). The picture from Drone 1 is received back at the camp at the same instant Drone 1 reaches the spot (given), and this is the 5:02 PM event.

Step 2: Drone 2 is launched immediately (at 5:02 PM) and reaches the spot at 5:05 PM. So Drone 2’s flight time is exactly (3) minutes (= 180) s.

Step 3: Compute the distance. \(d/30 = 180 \Rightarrow d = 5400\) m.

Step 4: Compute the sound delay. (d/300 = 5400/300 = 18) s.

Step 5: Compute Drone 1’s flight time. (d/30 = 180) s (same as Drone 2).

Step 6: Work backwards from 5:02 PM. Total elapsed since the explosion = (18 + 180 = 198) s (= 3) min (18) s.

Step 7: (T_0 = 5{:}02{:}00 – 3{:}18 = 4{:}58{:}42) PM. Hence option (c).

UPSC CSAT 2026 · Question 22

The digit in the unit place of the number \(6^{129} \times 7^{307}\) is

1. 2
2. 4
3. 8
4. 6

Explanation

Only the unit digits of the factors matter.

Step 1: Unit digit of \(6^{n}\). Powers of 6: (6^1=6), (6^2=36), (6^3=216) — the unit digit is always (6) for every (nge 1). So \(6^{129}\) ends in (6).

Step 2: Unit digit of \(7^{n}\). Powers of 7 cycle with period 4: (7^1=7), (7^2=49), (7^3=343), (7^4=2401). The cyclic pattern of unit digits is ((7, 9, 3, 1)).

Step 3: Locate (307) in the cycle. \(307 \div 4 = 76\) remainder (3), i.e. (307 equiv 3 pmod 4). Position 3 in the cycle is (3). So \(7^{307}\) ends in (3).

Step 4: Multiply unit digits. \(6 \times 3 = 18\), whose unit digit is (8).

Therefore the unit digit of \(6^{129} \times 7^{307}\) is (8). Option (c).

UPSC CSAT 2026 · Question 23

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?

1. 20% increase
2. 30% increase
3. 03% decrease
4. 02% decrease

Explanation

Pick easy numbers and work with percentages directly.

Step 1: Assume the original salary (= 100). Savings are (10%), so saving (= 10) and expenditure (= 100 – 10 = 90).

Step 2: Salary rises by (12%). New salary \(= 100 \times 1.12 = 112\).

Step 3: Expenditure rises by (10%). New expenditure \(= 90 \times 1.10 = 99\).

Step 4: New saving (=) new salary (-) new expenditure (= 112 – 99 = 13).

Step 5: Change in saving (= 13 – 10 = 3).

Step 6: Percentage change \(= \frac{3}{10} \times 100 = 30\%\), and since (13 > 10), it is an increase.

Hence the saving rises by (30%). Option (b).

UPSC CSAT 2026 · Question 24

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?

1. Q is a sister of S
2. Q is a daughter of S
3. Q is the mother of S
4. Q is a sister of the husband of S

Explanation

Decode the relationships one clue at a time.

Clue 1: ‘P has a son and a daughter.’ So P has exactly two named children, one male and one female.

Clue 2: ‘Q and R are children of P.’ So ({Q, R}) are precisely the son and daughter of P (in some order).

Clue 3: ‘S is the mother of T.’ S is female.

Clue 4: ‘S is R’s spouse.’ Since S is female, R must be male. Therefore R is P’s son and Q is P’s daughter.

Now trace Q’s relationship to S. Q is female; Q’s brother is R; R is married to S. Hence Q is the sister of R, and R is the husband of S — i.e. Q is the sister of S’s husband.

        P
       / \
      Q   R ── S ── (mother of) ── T
   (daughter) (son)

Option (a) is wrong because Q and S are not siblings. (b) is wrong because Q is not S’s daughter (Q is P’s daughter). (c) is wrong because Q is not S’s mother. Option (d) correctly reads: Q is a sister of the husband (R) of S.

UPSC CSAT 2026 · Question 25

How many three-digit numbers can be expressed as an integral power of 2?

1. 1
2. 2
3. 3
4. 4

Explanation

We need integer powers of 2 that are three-digit numbers, i.e. between (100) and (999) inclusive.

Step 1: List successive powers of 2 in this range.

(2^6 = 64) — only two digits, too small.

(2^7 = 128) — three digits. ✓

(2^8 = 256) — three digits. ✓

(2^9 = 512) — three digits. ✓

\(2^{10} = 1024\) — four digits, too large.

Step 2: Count the qualifying powers: ({128, 256, 512}) — exactly (3) numbers.

Hence the answer is (3). Option (c).

Passage — for Questions 26–27

Read the following passage and answer the items that follow. Your answers should be based solely on the passage.

The key source of the battle for clean skies and clear lungs is the fuel we burn—from household Chulhas to the thermal power plants. In most cases, it is biomass or coal. The Supreme Court banned the use of pet coke—the dirtiest of such fuels. The Delhi Government banned the use of coal, which was later extended to the entire National Capital Region. It was also agreed that the thermal power plants would clean up or shut down. Action on this has been patchy to say the least. The lesson from the transition to CNG is that people need alternatives for a ban to be effective. When diesel buses were stopped, CNG supply had to be assured. It also had to be feasible in terms of cost. The Supreme Court agreed that fiscal measures were needed to keep clean fuel cheaper than dirty fuel. Now even as coal is banned, the price of natural gas makes industry uncompetitive.

UPSC CSAT 2026 · Question 26

Which of the following inferences is/are correct?

1. 1. The source of the energy we consume is the key to the battle for cleaner air.
2. 2. Bans are effective where the will is strong and the people are convinced that such bans are for the greater good of society.
3. 3. There is judicial approval for a policy that intervenes fiscally to facilitate benevolent pricing for cleaner fuel.

Select the answer using the code given below.

1. 1 and 2
2. 2 and 3
3. 1 and 3
4. 1 only

Explanation

Test each inference against the passage.

Inference 1: ‘The source of the energy we consume is the key to the battle for cleaner air.’ This is a direct paraphrase of the opening line: ‘The key source of the battle for clean skies and clear lungs is the fuel we burn — from household Chulhas to the thermal power plants.’ Fuel = the source of the energy we consume. Hence Inference 1 is valid.

Inference 2: ‘Bans are effective where the will is strong and the people are convinced…’ The passage actually says, ‘The lesson from the transition to CNG is that people need alternatives for a ban to be effective’ — i.e. the precondition for effectiveness is feasible alternatives and pricing, not public will or moral persuasion. Inference 2 is not supported.

Inference 3: ‘There is judicial approval for a policy that intervenes fiscally to facilitate benevolent pricing for cleaner fuel.’ Supported by ‘The Supreme Court agreed that fiscal measures were needed to keep clean fuel cheaper than dirty fuel.’ Inference 3 is valid.

Valid inferences: 1 and 3. Option (c).

UPSC CSAT 2026 · Question 27

Which of the following statements is/are correct?

1. 1. Thermal power stations in Delhi were required to summarily shut down.
2. 2. CNG supplies had to be assured once diesel vehicles were prohibited from plying.
3. 3. The Supreme Court banned the use of coal across the National Capital Region.

Select the answer using the code given below.

1. 1 and 2
2. 3 only
3. 2 only
4. 2 and 3

Explanation

Evaluate each statement against the passage.

Statement 1: ‘Thermal power stations in Delhi were required to summarily shut down.’ The passage actually says the plants ‘would clean up or shut down’ — i.e. they had the option to clean up; shutdown was not summary. So 1 is incorrect.

Statement 2: ‘CNG supplies had to be assured once diesel vehicles were prohibited from plying.’ Directly supported: ‘When diesel buses were stopped, CNG supply had to be assured.’ Statement 2 is correct.

Statement 3: ‘The Supreme Court banned the use of coal across the National Capital Region.’ The passage attributes the coal ban to the Delhi Government, later extended to the NCR — not to the Supreme Court. The Supreme Court banned pet coke, not coal. Statement 3 is incorrect.

Only Statement 2 is correct. Option (c).

UPSC CSAT 2026 · Question 28

Consider the following statements: Every red is blue. Every blue is green. Every green is yellow. Which of the following statements denoted by P, Q and R are correct?

1. P.Every blue is yellow
2. Q.Every red is green
3. R.Every red is yellow

Select the answer using the code given below.

1. P and Q only
2. Q and R only
3. P and R only
4. P, Q and R

Explanation

Treat each premise as a subset relation.

Premise 1: Every red is blue \(\Rightarrow \text{Red} \subseteq \text{Blue}\).

Premise 2: Every blue is green \(\Rightarrow \text{Blue} \subseteq \text{Green}\).

Premise 3: Every green is yellow \(\Rightarrow \text{Green} \subseteq \text{Yellow}\).

Chaining these: \(\text{Red} \subseteq \text{Blue} \subseteq \text{Green} \subseteq \text{Yellow}\).

Check each conclusion:

P: Every blue is yellow. From \(\text{Blue} \subseteq \text{Green} \subseteq \text{Yellow}\), yes — P is valid.

Q: Every red is green. From \(\text{Red} \subseteq \text{Blue} \subseteq \text{Green}\), yes — Q is valid.

R: Every red is yellow. Direct from the full chain — R is valid.

All three conclusions hold. Option (d).

UPSC CSAT 2026 · Question 29

How many times does 5 appear in all two-digit positive integers?

1. 18
2. 19
3. 20
4. 21

Explanation

Two-digit numbers run from (10) to (99), so each has a tens digit and a units digit. Count occurrences of the digit (5) in each place.

Step 1: Count (5) in the tens place. These are the numbers (50, 51, 52, 53, 54, 55, 56, 57, 58, 59) — exactly (10) numbers, contributing (10) occurrences of (5).

Step 2: Count (5) in the units place. These are (15, 25, 35, 45, 55, 65, 75, 85, 95) — exactly (9) numbers, contributing (9) occurrences of (5).

Step 3: Note that (55) contributes one (5) in each place — already counted separately, so no double counting issue: we’re counting digit occurrences, not numbers.

Step 4: Total digit occurrences (= 10 + 9 = 19).

Hence (5) appears (19) times in all two-digit positive integers. Option (b).

UPSC CSAT 2026 · Question 30

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?

1. 3
2. 4
3. 5
4. 6

Explanation

Use \(\text{time} = \text{distance}/\text{speed}\) for each leg.

Step 1: X’s journey times.

Leg 1: \(1 \text{ km at } 5 \text{ km/h} = \tfrac{1}{5} = 0.20\) h.

Leg 2: \(2 \text{ km at } 10 \text{ km/h} = \tfrac{2}{10} = 0.20\) h.

Leg 3: \(3 \text{ km at } 4 \text{ km/h} = \tfrac{3}{4} = 0.75\) h.

Total for X (= 0.20 + 0.20 + 0.75 = 1.15) h.

Step 2: Y’s journey times.

Leg 1: \(1 \text{ km at } 4 \text{ km/h} = \tfrac{1}{4} = 0.25\) h.

Leg 2: \(2 \text{ km at } 10 \text{ km/h} = \tfrac{2}{10} = 0.20\) h.

Leg 3: \(3 \text{ km at } 5 \text{ km/h} = \tfrac{3}{5} = 0.60\) h.

Total for Y (= 0.25 + 0.20 + 0.60 = 1.05) h.

Step 3: Difference (= 1.15 – 1.05 = 0.10) h \(= 0.10 \times 60 = 6\) minutes.

Since Y’s total time is smaller, Y finishes (6) minutes earlier. Option (d).

UPSC CSAT 2026 · Question 31

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?

1. 2
2. 3
3. 4
4. 1

Explanation

We have 7 visually identical cubes; one is lighter, the other six are equal in weight. We need the minimum number of balance attempts that always identifies the odd cube.

Strategy: split 7 as (3 + 3 + 1).

Attempt 1: Put 3 cubes on the left pan, 3 on the right pan, leave 1 aside.

Case A — Pans balance: all six weighed cubes are equal in weight, so the lighter cube is the one set aside. Identified in just (1) attempt.

Case B — Pans don’t balance: the lighter pan’s group of 3 contains the odd cube. Now take those 3 cubes for Attempt 2.

Attempt 2: Put 1 of the 3 suspect cubes on the left pan, another on the right, leave 1 aside.

• If the pans balance, the set-aside cube is the lighter one.

• If the pans don’t balance, the lighter pan holds the odd cube.

Either way, after at most (2) attempts the odd cube is identified with certainty.

Attempt 1:  (3) vs (3)  |  spare: 1
            ├─ balance  → spare is light  (DONE in 1)
            └─ tilt     → lighter side of 3 is suspect group

Attempt 2 (only if needed):
            (1) vs (1) from suspect 3  |  spare: 1
            ├─ balance  → spare is light
            └─ tilt     → lighter pan is light

Minimum guaranteed number of attempts = (2). Option (a).

Passage — for Questions 32–33

Read the following passage and answer the items that follow. Your answers should be based solely on the passage.

Previous waves of customer-service technology, including email and those pesky voice menus, stoked concerns of job losses, only for them to fail to materialise. AI could yet prove different. And if it does, its effects may be salutary. Human agents could be freed up to spend more time on creative and rewarding tasks, like using feedback to make products and services better and thereby spend less time listening to irate customers!

UPSC CSAT 2026 · Question 32

Which one among the following statements most appropriately reflects the point of view of the given passage?

1. If AI were to take over customer service, there would be no work left for human subjects to do.
2. Irritating voice menus and email could not achieve human redundancy to the extent that AI might.
3. The value of human intervention in the workplace affected by AI might be enhanced through redirection towards more fulfilling tasks.
4. Unlike previous waves in customer-service technology, AI has raised the alarm of worker replacement.

Explanation

Identify the passage’s main argument. The author points out that earlier customer-service technologies sparked job-loss fears that never materialised, then suggests AI might be different — but in a positive way: ‘Human agents could be freed up to spend more time on creative and rewarding tasks, like using feedback to make products and services better.’

The central thrust is that AI’s impact on human work is not job destruction but redirection toward higher-value tasks.

Evaluate each option.

(a) ‘No work left for human subjects’ — overstates the case; the passage explicitly imagines humans doing creative work, not being displaced.

(b) ‘Voice menus and email could not achieve human redundancy to the extent AI might’ — the passage does not claim AI will cause more redundancy; it suggests the opposite (salutary effects).

(c) ‘The value of human intervention in the workplace affected by AI might be enhanced through redirection towards more fulfilling tasks’ — precisely matches the passage’s claim that human agents are ‘freed up’ for ‘creative and rewarding tasks.’

(d) ‘Unlike previous waves… AI has raised the alarm’ — wrong; the passage says previous waves also raised the same alarm.

Option (c) best reflects the passage.

UPSC CSAT 2026 · Question 33

Which of the following conclusions, made on the basis of the given passage, is/are correct?

1. 1. The advent of new customer-service technology had invariably sparked fears about job losses.
2. 2. Often it is found that instead of job losses, alternative channels for employee engagement are discovered while certain tasks are replaced by technology.
3. 3. The advent of technology inevitably leads to stressful outcomes.

Select the answer using the code given below.

1. 1 and 3
2. 2 only
3. 3 only
4. 1 and 2

Explanation

Evaluate each conclusion against the passage.

Conclusion 1: ‘The advent of new customer-service technology had invariably sparked fears about job losses.’ Supported by ‘Previous waves of customer-service technology, including email and those pesky voice menus, stoked concerns of job losses.’ Valid.

Conclusion 2: ‘Often it is found that instead of job losses, alternative channels for employee engagement are discovered while certain tasks are replaced by technology.’ Supported by two parts of the passage: (i) the feared job losses ‘failed to materialise’, and (ii) ‘Human agents could be freed up to spend more time on creative and rewarding tasks.’ Valid.

Conclusion 3: ‘The advent of technology inevitably leads to stressful outcomes.’ The word ‘inevitably’ is far too absolute. The passage actually says effects ‘may be salutary’ — i.e. potentially positive. Conclusion 3 is not supported.

Valid conclusions: 1 and 2. Option (d).

Passage — for Questions 34–35

Read the following passage and answer the items that follow. Your answers should be based solely on the passage.

Cattle from the nearby villages came to the common ground to graze, and there was still a cool freshness in the air. Hori took several deep breaths and thought of sitting down for a while, since he'd be dying of heat in the scorching 'loo' wind the rest of the day. A number of farmers were eager to lease this bit of land and had offered a good price, but Rai Sahib—god bless him—had plainly told them it was reserved for grazing and would not be relinquished for any price. If he'd been one of those selfish Zamindars, he'd have said the cattle could go to hell, that there was no reason for him to miss the chance to make a little money. But the Rai Sahib still held to the old values, feeling that any landlord who didn't look after his tenants was less than human.

UPSC CSAT 2026 · Question 34

Which of the following conclusions is/are correct?

1. 1. All landlords essentially have some goodness trapped within them.
2. 2. The common grazing grounds of a village are intended for use by the cattle of that village.
3. 3. Landlords who believe in tradition tend to be more concerned about their tenants.
4. 4. Winds later in the day tend to be cooler post the hot winds of the morning.

Select the answer using the code given below.

1. 1 and 3
2. 2 and 4
3. 3 only
4. 2 only

Explanation

Examine each conclusion against the passage about Hori, Rai Sahib, and the grazing land.

Conclusion 1: ‘All landlords essentially have some goodness trapped within them.’ The passage explicitly contrasts Rai Sahib with ‘selfish Zamindars’ who would have let the cattle ‘go to hell.’ That contrast disproves a universal-goodness claim. Not valid.

Conclusion 2: ‘The common grazing grounds of a village are intended for use by the cattle of that village.’ The passage says ‘Cattle from the nearby villages came to the common ground to graze’ — plural ‘villages’, not exclusively this village. Not valid.

Conclusion 3: ‘Landlords who believe in tradition tend to be more concerned about their tenants.’ Supported by ‘the Rai Sahib still held to the old values, feeling that any landlord who didn’t look after his tenants was less than human.’ This explicitly links traditional values to concern for tenants. Valid.

Conclusion 4: ‘Winds later in the day tend to be cooler post the hot winds of the morning.’ Reverses the passage, which describes ‘cool freshness in the air’ early and the ‘scorching loo wind the rest of the day’ later. Not valid.

Only Conclusion 3 holds. Option (c).

UPSC CSAT 2026 · Question 35

Which of the following statements are not correct?

1. 1. The landholdings of Rai Sahib were currently not being used for farming.
2. 2. Temperamentally, Rai Sahib was as greedy as other landlords.
3. 3. It cannot be ascertained that Rai Sahib could have made some money by leasing out the grazing land.
4. 4. It may be asserted that Rai Sahib valued his tenants and wanted to protect their livelihood.

Select the answer using the code given below.

1. 1 and 2 only
2. 1 and 3 only
3. 3 and 4
4. 1, 2 and 3

Explanation

The question asks which statements are NOT correct.

Statement 1: ‘The landholdings of Rai Sahib were currently not being used for farming.’ The passage only describes one specific patch of land — the grazing ground — not all his landholdings. We cannot conclude his entire estate was non-farming. Statement 1 is NOT correct (an overreach).

Statement 2: ‘Temperamentally, Rai Sahib was as greedy as other landlords.’ The passage paints Rai Sahib as the opposite of greedy — he refused ‘a good price’ and held to ‘old values.’ Statement 2 is NOT correct (it contradicts the text).

Statement 3: ‘It cannot be ascertained that Rai Sahib could have made some money by leasing out the grazing land.’ We can ascertain this — farmers ‘had offered a good price’, so by leasing he would clearly have earned money. Statement 3 is NOT correct (we can ascertain it).

Statement 4: ‘It may be asserted that Rai Sahib valued his tenants and wanted to protect their livelihood.’ This is directly supported by his refusal to relinquish the grazing land for any price and his belief about looking after tenants. Statement 4 IS correct.

The set of statements that are NOT correct is ({1, 2, 3}). Option (d).

UPSC CSAT 2026 · Question 36

A person facing the East travels 4 km straight and then turns right and travels 3 km, then further turns left and travels 2 km and finally turns left and travels 3 km. The minimum distance between the final point and the initial point, and the direction in which the person is facing at the final point are, respectively

1. 12 km, East
2. 6 km, East
3. 8 km, North
4. 6 km, North

Explanation

Use a coordinate grid with East (= +x) and North (= +y). Start at the origin facing East.

Step 1: Walk (4) km East. Position ((4, 0)). Facing East.

Step 2: Turn right \(\Rightarrow\) now facing South. Walk (3) km South. Position ((4, -3)). Facing South.

Step 3: Turn left \(\Rightarrow\) from South, left is East. Walk (2) km East. Position ((6, -3)). Facing East.

Step 4: Turn left \(\Rightarrow\) from East, left is North. Walk (3) km North. Position ((6, 0)). Facing North.

Final position ((6, 0)) vs starting point ((0, 0)): straight-line distance \(= \sqrt{(6-0)^2 + (0-0)^2} = 6\) km. Final facing direction: North.

                N
                ^
                |
   (0,0) ───4E──>(4,0)
                  |
                  3S
                  v
                (4,-3)──2E──>(6,-3)
                                |
                                3N
                                v
                              (6,0)  facing N

Distance from start (= 6) km; facing North. Option (d).

UPSC CSAT 2026 · Question 37

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

1. 29
2. 37
3. 43
4. 47

Explanation

The rule: from the third term onward, each term equals the sum of the two immediately preceding terms (Fibonacci-style). Given (a_1 = 4) and (a_2 = 7).

Step 1: (a_3 = a_1 + a_2 = 4 + 7 = 11).

Step 2: (a_4 = a_2 + a_3 = 7 + 11 = 18).

Step 3: (a_5 = a_3 + a_4 = 11 + 18 = 29).

Step 4: (a_6 = a_4 + a_5 = 18 + 29 = 47).

 n :  1   2   3   4   5   6
 a :  4   7  11  18  29  47
             ↑   ↑   ↑   ↑
            4+7 7+11 11+18 18+29

The sixth term is (47). Option (d).

UPSC CSAT 2026 · Question 38

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?

1. 24 : 23
2. 23 : 24
3. 18 : 17
4. 27 : 18

Explanation

Both companies have the same number of female workers — call it (F). Express each company’s totals using the given ratios.

Step 1: Company A has Male : Female (= 13 : 10). Let (F = 10k). Then males in A (= 13k) and total in A (= 13k + 10k = 23k).

Step 2: Company B has Male : Female (= 7 : 5). Rewrite by multiplying both parts by 2 so that the female part also equals (10): (7 : 5 equiv 14 : 10). With (F = 10k), males in B (= 14k) and total in B (= 14k + 10k = 24k).

Step 3: Ratio of totals.

          Males   Females   Total
   A  :   13k      10k       23k
   B  :   14k      10k       24k

\(\text{Total in A} : \text{Total in B} = 23k : 24k = 23 : 24\).

Hence the ratio is (23 : 24). Option (b).

UPSC CSAT 2026 · Question 39

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?

1. I. The difference of the numbers is an even number.
2. II. One of the numbers is a perfect square.

Select the answer using the code given below.

1. I only
2. II only
3. Both I and II
4. Neither I nor II

Explanation

Let the two distinct positive integers be (a) and (b).

Key identity: \(\text{HCF}(a, b) \times \text{LCM}(a, b) = a \cdot b\) for any two positive integers.

Given: this product equals the cube of one of the numbers. Say it equals \(a^{3}\).

Step 1: From \(a \cdot b = a^{3}\), divide both sides by (a) (nonzero): \(b = a^{2}\).

Step 2: Test Statement II: ‘One of the numbers is a perfect square.’ Since \(b = a^{2}\), (b) is the perfect square of (a). Statement II is TRUE.

Step 3: Test Statement I: ‘The difference of the numbers is an even number.’ Difference \(= b – a = a^{2} – a = a(a – 1)\). The product of two consecutive integers (a) and (a-1) is always even (exactly one of them is even). So the difference is always even. Statement I is TRUE.

Quick sanity check: (a = 3, b = 9): HCF (= 3), LCM (= 9), product \(= 27 = 3^{3}\) ✓; difference (= 6) (even); \(9 = 3^{2}\) (perfect square). Works.

Both statements are correct. Option (c).

UPSC CSAT 2026 · Question 40

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?

1. 1
2. 3
3. 5
4. 7

Explanation

Apply the divisibility rule for (11): a number is divisible by (11) iff the alternating sum of its digits is a multiple of (11) (including (0)).

Step 1: Write the digits of (4x5y790) from left to right with their positions.

  position : 1  2  3  4  5  6  7
  digit    : 4  x  5  y  7  9  0
  sign     : +  -  +  -  +  -  +

Step 2: Alternating sum (odd positions minus even positions):

(S = (4 + 5 + 7 + 0) – (x + y + 9) = 16 – 9 – (x + y) = 7 – (x + y)).

Step 3: For divisibility by (11), require (S equiv 0 pmod{11}), i.e. (7 – (x + y) equiv 0 pmod{11}), so ((x + y) equiv 7 pmod{11}).

Step 4: With (x, y) each a single digit ((0) to (9)), the possible sums are (0) to (18). Solutions to ((x + y) equiv 7 pmod{11}) are (x + y = 7) or (x + y = 18).

Step 5: Compute the remainder when (x + y) is divided by (11). If (x + y = 7), remainder (= 7). If (x + y = 18), remainder (= 18 – 11 = 7).

In both cases the remainder is (7). Option (d).

Passage — for Questions 41–42

Read the following passage and answer the items that follow. Your answers should be based solely on the passage.

Was it the sun-dappled ambience, the strawberries and cream, the frustration of Flavio Cobolli's unforced errors against Serbian Novak Djokovic on Centre Court or simply the crushing weight of being a 64-year-old man in the third act of a very public life? Whatever the reason, Hugh Grant, the actor, deserves empathy. There he was, in the Royal Box at Wimbledon, flanked by Britain's well-dressed and well-rested spectators, watching the men's singles quarterfinals, when the actor did something quietly radical: head at a tilt, eyes closed, utterly unbothered, he took a nap. So praise be to Grant for serving up an unexpected ace. In that small, delicious moment, he didn't merely catch forty winks, he made an elegant case for surrender. Not to laziness, but to limits. To the body's quiet wisdom over society's relentless performance metrics. Wimbledon had its tennis. The perpetually sleep-deprived discovered a leading man, not of action, but of rest.

UPSC CSAT 2026 · Question 41

Which of the following statements is/are correct?

1. 1. Radical action can also be attributed to mild surrender where one acts against societal expectations.
2. 2. Submitting to one’s limitations, given the effect of age and other factors, ought not to be conflated with laziness.
3. 3. ‘Leading man’ usually refers to one who plays the lead role in a movie; in this instance the implication is that Hugh Grant is performing the role of not an action hero, but that of a resting one!

Select the answer using the code given below.

1. 1 and 2 only
2. 2 and 3 only
3. 1, 2 and 3
4. 3 only

Explanation

The passage frames Grant’s nap as ‘something quietly radical’ and ‘an elegant case for surrender. Not to laziness, but to limits.’ Statement 1 captures this: a mild surrender (falling asleep) performed against the social expectation of the Royal Box becomes radical action. So 1 is correct.

Statement 2 mirrors the explicit line ‘Not to laziness, but to limits’ and links it to age via ‘a 64-year-old man in the third act of a very public life.’ Yielding to one’s bodily limits as one ages is therefore wisdom, not sloth, so 2 is correct.

Statement 3 reads the closing line: ‘discovered a leading man, not of action, but of rest.’ The passage plays on Grant’s film-actor identity (leading man) and inverts it: he leads in rest, not action. So 3 is correct.

All three statements correctly interpret the passage, so the answer is (c).

UPSC CSAT 2026 · Question 42

Which of the following statements is/are correct?

1. 1. Hugh Grant was watching, from the Royal Box, the men’s semifinal match on Centre Court between Flavio Cobolli and Novak Djokovic.
2. 2. The phrase ‘unexpected ace’ in the context uses a term from the game of tennis to highlight Hugh Grant’s somewhat uncharacteristic act of ‘catching forty winks’; an act that is viewed with opprobrium.
3. 3. Grant subjects the demands of society to the wisdom of his body.

Select the answer using the code given below.

1. 1 and 2
2. 3 only
3. 2 and 3
4. 2 only

Explanation

Statement 1 says Grant was watching the men’s semifinal. The passage explicitly says ‘watching the men’s singles quarterfinals.’ Quarterfinal not semifinal, so 1 is wrong.

Statement 2 reads ‘unexpected ace’ as ‘viewed with opprobrium.’ The passage actually says ‘praise be to Grant for serving up an unexpected ace’ it is admiring praise, not disapproval. So 2 is wrong.

Statement 3 paraphrases ‘the body’s quiet wisdom over society’s relentless performance metrics.’ Grant places bodily wisdom above social demand, exactly what statement 3 says, so 3 is correct.

Only 3 is correct, hence (b).

Passage — for Questions 43–44

Read the following passage and answer the items that follow. Your answers should be based solely on the passage.

The process by which countries close their labour-productivity gap with the technology leader is based on convergence theory. The convergence model divides economic eras into three phases: the breakaway, the catch-up, and the fine-tuning phase. It also divides economic entities into two categories: the technology leaders and the technology followers. The process begins with the development of a new technology, such as scavenging—three million years ago (MYA), hunting—one MYA, farming—12 thousand years ago, and industrial technology—a little more than 200 years ago. During the breakaway phase, the per capita income of the technology leaders (e.g., Western Europe and North America in the industrial era) rises, but is unchanged for the technology followers. In the catch-up phase, the followers adopt the new technology and close their per capita income gap with the technology leaders. In the fine-tuning phase, where participants try to extract the remaining benefits from an increasingly exhausted technology, leaders and followers have similar per capita incomes.

UPSC CSAT 2026 · Question 43

Which of the following conclusions are correct?

1. 1. In the breakaway phase, economic progress is slow for the technology followers.
2. 2. In the catch-up phase, leaders stagnate and followers, therefore, close the gap between them and the leaders.
3. 3. In the fine-tuning phase, technology is exhausted, as it were, and both leaders and followers attempt to extract leftover benefits, leading to more or less similar per capita income levels.
4. 4. Industrial technology followed scavenging, which preceded hunting, which itself was followed by farming.

Select the answer using the code given below.

1. 3 and 4 only
2. 1, 2, 3 and 4
3. 1 and 2 only
4. 2, 3 and 4 only

Explanation

Statement 1 says followers’ progress is ‘slow’ in the breakaway phase. The passage says per capita income of followers ‘is unchanged’ during breakaway, not slow. ‘Slow’ implies some progress, so 1 is inaccurate.

Statement 2 says leaders ‘stagnate’ in catch-up. The passage only states that followers close the gap by adopting the new technology; it does not say leaders stagnate. 2 is wrong.

Statement 3 paraphrases the fine-tuning phase: ‘participants try to extract the remaining benefits from an increasingly exhausted technology’ and ‘leaders and followers have similar per capita incomes.’ Matches exactly, so 3 is correct.

Statement 4 orders the technologies scavenging (3 MYA) → hunting (1 MYA) → farming (12 KYA) → industrial (200 yrs). Statement 4 says ‘industrial followed scavenging, which preceded hunting, which itself was followed by farming.’ That ordering is correct. So 4 is correct.

Only 3 and 4 are correct, hence (a).

UPSC CSAT 2026 · Question 44

Which of the following statements is/are correct?

1. 1. The convergence model divides nations into three phases of economic progress.
2. 2. At the heart of the convergence theory is the closing of the gap between labour and productivity.
3. 3. Technology leaders typically have arrived earlier at different economic eras.
4. 4. The time period covered by the convergence theory presented herein encompasses, as mentioned, 4012200 years.

Select the answer using the code given below.

1. 2 and 4
2. 2 and 3
3. 3 only
4. 1, 3 and 4

Explanation

Statement 1: The model divides ‘economic eras into three phases’ (breakaway, catch-up, fine-tuning), not nations. Economic entities fall into two categories, leaders and followers. So 1 is wrong.

Statement 2: The passage opens by saying that the process by which countries ‘close their labour-productivity gap with the technology leader is based on convergence theory’. Closing that gap is the core idea of the theory. The statement’s wording, ‘the gap between labour and productivity’, is loose, but it points to the same labour-productivity gap. So 2 is correct.

Statement 3: Each era ‘begins with the development of a new technology’, and the technology leaders gain first in the breakaway phase while followers catch up later. So leaders typically arrive earlier at each economic era. 3 is correct.

Statement 4: 4012200 is what you get by adding all four figures (3,000,000 + 1,000,000 + 12,000 + 200). The passage gives separate starting points, not a span to add up, and the period it covers reaches back about three million years. 4 is wrong.

Statements 2 and 3 are correct, hence (b).

UPSC CSAT 2026 · Question 45

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?

1. 1 : 3
2. 3 : 1
3. 2 : 3
4. 3 : 2

Explanation

Let (P:Q = x:y) by weight. Q is 60% copper, 40% zinc; P is 20% copper, 80% zinc.

Copper in mix = (0.2x + 0.6y).
Zinc in mix = (0.8x + 0.4y).

For R to contain equal copper and zinc:
(0.2x + 0.6y = 0.8x + 0.4y)
(0.6y – 0.4y = 0.8x – 0.2x)
(0.2y = 0.6x)
(y = 3x).

So (P:Q = x : 3x = 1:3). Answer (a).

UPSC CSAT 2026 · Question 46

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?

1. 6
2. 9
3. 12
4. 18

Explanation

Let A = 10a + b and B = 10b + a. Both are 2-digit numbers, so a is at least 1 and b is at least 1 (if b = 0, B would be the single digit a). A > B means a > b. Then A – B = 9(a – b).

A – B must be a multiple of 27, so 9(a – b) is a multiple of 27, which means a – b is a multiple of 3. Since b is at least 1 and a is at most 9, a – b is at most 8. So a – b can only be 3 or 6.

a-b = 3: (a,b) in {(4,1),(5,2),(6,3),(7,4),(8,5),(9,6)}  -> 6 pairs
a-b = 6: (a,b) in {(7,1),(8,2),(9,3)}                    -> 3 pairs
a-b = 9: only (9,0), but B = 09 is not 2-digit          -> 0 pairs
Total A's = 6 + 3 = 9

Each pair gives a different A: 41, 52, 63, 74, 85, 96, 71, 82, 93. That is 9 numbers. Answer (b).

The trap is 30, 60 and 90. They pass the 27 test, but reversing them gives 03, 06 and 09, which are not 2-digit numbers.

UPSC CSAT 2026 · Question 47

If ZERO is encoded as ADSN, then how do you encode STOP?

1. SPOT
2. TSPO
3. TSOP
4. POST

Explanation

Compare ZERO and ADSN letter by letter using alphabet positions:

Pos 1: Z(26) -> A(1)   shift +1  (mod 26)
Pos 2: E(5)  -> D(4)   shift -1
Pos 3: R(18) -> S(19)  shift +1
Pos 4: O(15) -> N(14)  shift -1

So odd positions shift (+1) and even positions shift (-1). Apply to STOP:

Pos 1: S(19) +1 -> T(20)
Pos 2: T(20) -1 -> S(19)
Pos 3: O(15) +1 -> P(16)
Pos 4: P(16) -1 -> O(15)
Result: T S P O

Answer (b) TSPO.

UPSC CSAT 2026 · Question 48

There are three types of rectangular tiles: 3′ × 3′, 3′ × 7′ and 3′ × 11′. An area of rectangular shape of dimensions 3’x100′ 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

1. 20
2. 12
3. 10
4. 7

Explanation

Each tile occupies a \(3′ \times L\) strip with (L in {3,7,11}). Tiles are laid end-to-end along the 100′ length, so we need non-negative integers (a,b,c) with (3a+7b+11c=100), where (a,b,c) count tiles of length 3, 7, 11.

Maximum tiles ((x)): use the shortest tile as much as possible. Try (b=1,c=0): \(3a+7=100 \Rightarrow 3a=93 \Rightarrow a=31\), total (= 31+1 = 32) tiles. ((b=0,c=0) gives (3a=100), not integer, so 32 is the max.)

Minimum tiles ((y)): use the longest tile as much as possible. Try (c=8): \(11\cdot 8 = 88\), remainder \(12 = 3 \cdot 4\), so (a=4,b=0,c=8) gives (4+0+8 = 12) tiles. Try (c=9): (99), remainder (1), not feasible. So 12 is the minimum.

Max (32 tiles):  3+3+...+3 (x31) | 7   = 100
Min (12 tiles):  11+11+...+11 (x8) | 3+3+3+3 = 100

(x – y = 32 – 12 = 20). Answer (a).

UPSC CSAT 2026 · Question 49

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?

1. 200
2. 190
3. 150
4. 100

Explanation

Let the original speed be (v) km/h. ‘Accident’ halves the current speed; ‘mechanical defect’ reduces the current speed to one-fourth.

Original trip: 200 km at (v), then accident, 400 km at (v/2), then defect, 200 km at ((v/2)/4 = v/8).
Time \(T_1 = \dfrac{200}{v} + \dfrac{400}{v/2} + \dfrac{200}{v/8} = \dfrac{200}{v} + \dfrac{800}{v} + \dfrac{1600}{v} = \dfrac{2600}{v}\).

Alternate trip: 200 km at (v), then defect, 400 km at (v/4), then accident, 200 km at ((v/4)/2 = v/8).
Time \(T_2 = \dfrac{200}{v} + \dfrac{400}{v/4} + \dfrac{200}{v/8} = \dfrac{200}{v} + \dfrac{1600}{v} + \dfrac{1600}{v} = \dfrac{3400}{v}\).

Difference \(T_2 – T_1 = \dfrac{800}{v} = 4\) hours, so (v = 200) km/h. Answer (a).

UPSC CSAT 2026 · Question 50

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 cut-off score if exactly three candidates are to be selected:–

1. 5.1
2. 5.2
3. 5.3
4. 5.4

Explanation

Weighted score \(= 0.2 \cdot C_1 + 0.3 \cdot C_2 + 0.5 \cdot C_3\). Compute for each candidate:

Cand | C1 | C2 | C3 | 0.2*C1 + 0.3*C2 + 0.5*C3 = Total
  1  |  5 |  4 |  6 | 1.0 + 1.2 + 3.0 = 5.2
  2  |  4 |  6 |  5 | 0.8 + 1.8 + 2.5 = 5.1
  3  |  3 |  2 |  8 | 0.6 + 0.6 + 4.0 = 5.2
  4  |  9 |  4 |  3 | 1.8 + 1.2 + 1.5 = 4.5
  5  |  8 |  8 |  2 | 1.6 + 2.4 + 1.0 = 5.0

Ranking the scores in descending order: 5.2, 5.2, 5.1, 5.0, 4.5. To select exactly three candidates, the cut-off must include the top three (5.2, 5.2, 5.1) and exclude the fourth (5.0). The cut-off score is therefore 5.1. Answer (a).

UPSC CSAT 2026 · Question 51

Consider the following three statements, namely S1, S2 and S3:

1. S1: Protecting the environment is an existential exigency for humans, given the impact of environmental degradation on climate change.
2. S2: Scientific consensus has not been achieved with regard to the extent of the contribution of human intervention to climate change.
3. S3: Environmental activism includes climate alarmism and other extremist points of view that often become the focus of climate change deniers.
4. Which of the following relationships based on the statements given above is/are correct?
5. 1. S3 is a counterpoint to S1
6. 2. S3 is unconnected to S1 and S2
7. 3. S2 could be the reason for S3

Select the answer using the code given below.

1. 2 and 3 only
2. 1, 2 and 3
3. 1 and 2 only
4. 3 only

Explanation

S1 frames environmental protection as an existential exigency tied to climate change. S2 says there is no scientific consensus on the extent of human contribution. S3 says environmental activism includes alarmism that climate-change deniers focus on.

(1) ‘S3 is a counterpoint to S1’ is wrong: S3 critiques activism’s extremism, but does not deny the existential exigency that S1 asserts. They sit on different planes.

(2) ‘S3 is unconnected to S1 and S2’ is wrong: all three sit inside the climate-change/environment debate; S3 explicitly references ‘climate change deniers,’ which connects it tightly to S1’s climate-change framing and to S2’s no-consensus claim.

(3) ‘S2 could be the reason for S3’ is reasonable: if there is no scientific consensus (S2), that gives climate-change deniers room to seize on activist extremism (S3) as cover. So S2 plausibly motivates the dynamic described in S3.

Only relationship 3 holds, hence (d).

UPSC CSAT 2026 · Question 52

Match List-I with List-II and select the answer using the code given below the Lists:

Code:

1. A-1, B-3, C-4, D-2
2. A-1, B-4, C-3, D-2
3. A-2, B-4, C-3, D-1
4. A-2, B-3, C-4, D-1

Explanation

Decode each relationship by formality (formal vs informal) and decision style (firm directive vs open-ended discussion).

A. Captain <-> team   : team-mates, locker-room tone -> Informal,
                        but on-field instructions are firm  -> A-1 (Informal & firm)
B. Judge <-> lawyers  : courtroom etiquette -> Formal,
                        rulings are firm                    -> B-4 (Formal & firm)
C. VC <-> Dy Registrar: institutional hierarchy -> Formal,
                        admin discussions are open-ended    -> C-3 (Formal & open-ended)
D. Peers <-> coworkers: same-level colleagues -> Informal,
                        chats are open-ended                -> D-2 (Informal & open-ended)

Mapping A-1, B-4, C-3, D-2 corresponds to option (b).

UPSC CSAT 2026 · Question 53

Match List-I with List-II and select the answer using the code given below the Lists:

Code:

1. A-1, B-4, C-2, D-3
2. A-1, B-2, C-4, D-3
3. A-3, B-4, C-2, D-1
4. A-3, B-2, C-4, D-1

Explanation

Match each communication tool with its core purpose.

A. Memorandum -> internal written directive to staff       -> A-3 (intimate a directive)
B. Flyer      -> printed/digital handout, broad audience    -> B-4 (disseminate non-targeted info)
C. Bcc        -> hidden copy; recipients don't see each other-> C-2 (inform confidentially)
D. Minutes    -> formal record of meeting resolutions       -> D-1 (record decisions)

A-3, B-4, C-2, D-1 matches option (c).

UPSC CSAT 2026 · Question 54

Which among the following actions would constitute the most appropriate directive(s) in resolving interpersonal conflict in an office with culturally diverse personnel?

1. 1. Direct personnel to practise activities that are the cultural markers of diverse groups
2. 2. Allow conflicts to resolve naturally over time to set an appropriate precedent of leadership
3. 3. Encourage personnel to seek each other’s perspectives

Select the answer using the code given below.

1. 1 and 3
2. 2 only
3. 3 only
4. 2 and 3

Explanation

Evaluate each directive for resolving interpersonal conflict in a culturally diverse office.

(1) Directing personnel to perform other cultures’ rituals is tokenistic, can feel coercive, and risks trivialising cultural identity. It does not address the interpersonal conflict at its root, so it is not appropriate.

(2) Letting conflicts ‘resolve naturally’ is an abdication of managerial responsibility. Unaddressed conflicts tend to fester, harden into factions, and damage team trust. Not appropriate.

(3) Encouraging personnel to seek each other’s perspectives builds mutual understanding across cultural lines, which reduces in-group/out-group bias and rebuilds working relationships. This is a sound, appropriate directive.

Only 3 is appropriate, hence (c).

UPSC CSAT 2026 · Question 55

Match List-I with List-II and select the answer using the code given below the Lists:

Code:

1. A-3, B-1, C-4, D-2
2. A-3, B-4, C-1, D-2
3. A-2, B-1, C-4, D-3
4. A-2, B-4, C-1, D-3

Explanation

Match each barrier type with the example that fits it.

A. Semantic       -> about the meaning of words  -> A-2 (Misunderstanding the meaning of a word)
B. Cognitive      -> limits of mental processing -> B-4 (Information overload)
C. Organisational -> structure and process gaps  -> C-1 (Lack of feedback)
D. Affective      -> emotions and feelings        -> D-3 (Fear of social stigma)

A-2, B-4, C-1, D-3 matches option (d).

Watch the first two codes: options (a) and (b) both start with A-3, pairing Semantic with fear of social stigma. Stigma is an emotional block, so it belongs with Affective, not Semantic.

UPSC CSAT 2026 · Question 56

You are required to design a ‘questionnaire’ to be filled on-location by visitors, based on the following objective while writing a report:

1. “To determine the feasibility of setting up a family-oriented vacation resort in the vicinity of a lake destination in the mountains”
2. Which of the following heads would you include in the questionnaire to make it most appropriate for your purpose?
3. 1. Size of family
4. 2. Budget
5. 3. Number of earners in the family
6. 4. Food allergies and dietary restrictions

Select the answer using the code given below.

1. 1 and 3
2. 1, 2 and 4
3. 1 and 2 only
4. 2 and 3

Explanation

The objective is feasibility of a family-oriented vacation resort. The questionnaire goes to on-site visitors and must inform decisions about whether to build the resort and how to size it.

Head                            | Feasibility relevance
1. Size of family               | Directly sizes rooms, dining, activity capacity. YES
2. Budget                       | Sets price point, room categories, F&B tier.     YES
3. Number of earners in family  | Already proxied by 'budget'; adds little.        NO
4. Food allergies / dietary     | Operational menu detail for a running resort,
                                  not a feasibility question.                       NO

Only 1 and 2 belong in the feasibility questionnaire, so the answer is (c).

UPSC CSAT 2026 · Question 57

With reference to ‘circular letters’, which of the following statements is/are correct?

1. 1. Circular letters are usually addressed to a group of people.
2. 2. Non-standard and customized content is typical of circular letters.
3. 3. Circular letters are used to intimate appraisals and increments of employees within organisations.
4. 4. Circular letters are less cost-effective than personalised and specific recipient-directed letters.

Select the answer using the code given below.

1. 1 and 2
2. 2 and 4
3. 1, 3 and 4
4. 1 only

Explanation

Evaluate each claim about circular letters.

(1) ‘Usually addressed to a group of people’ is the very definition of a circular: one message circulated to many recipients. Correct.

(2) ‘Non-standard and customized content’ contradicts the nature of a circular, which uses standardised, identical content for all recipients. Wrong.

(3) Appraisals and increments are confidential and individualised; they are communicated through personal letters/emails, not circulars. Wrong.

(4) Circular letters are more cost-effective than personalised letters (one draft, mass distribution), not less. Wrong.

Only 1 is correct, hence (d).

UPSC CSAT 2026 · Question 58

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?

1. ₹12,000
2. ₹10,000
3. ₹12,500
4. ₹10,750

Explanation

Profit shares are proportional to (capital share) \(\times\) (time share). Take total capital (= 1) and total time (= 1).

A invests \(\dfrac{1}{3}\) for \(\dfrac{1}{3}\) time: weight \(= \dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9}\).
B invests \(\dfrac{1}{4}\) for \(\dfrac{1}{4}\) time: weight \(= \dfrac{1}{4} \cdot \dfrac{1}{4} = \dfrac{1}{16}\).
C invests the remaining capital \(= 1 – \dfrac{1}{3} – \dfrac{1}{4} = \dfrac{12 – 4 – 3}{12} = \dfrac{5}{12}\), for the whole duration ((1)): weight \(= \dfrac{5}{12}\).

Bring to common denominator 144: \(A : B : C = \dfrac{16}{144} : \dfrac{9}{144} : \dfrac{60}{144} = 16 : 9 : 60\). Sum (= 85).

C’s share \(= \dfrac{60}{85} \times 17000 = \dfrac{60 \times 17000}{85} = 60 \times 200 = 12000\). Answer (a) ₹12,000.

UPSC CSAT 2026 · Question 59

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)?

1. 25
2. 27
3. 29
4. 31

Explanation

Container starts with 10 L of pure A. The standard replacement-mixture formula says: if each step removes 1 L of mixture and replaces with 1 L of B, then after each step the amount of A is multiplied by \(\dfrac{10-1}{10} = \dfrac{9}{10}\).

Note that step 1 of this problem removes 1 L of pure A first, then adds 1 L of B; this is algebraically the same first step ((A) goes from 10 to 9, factor 9/10). Steps 2 and 3 use the standard ‘remove mixture, add B’ procedure, each multiplying (A) by 9/10.

After 3 steps: \(A = 10 \cdot \left(\dfrac{9}{10}\right)^3 = 10 \cdot \dfrac{729}{1000} = 7.29\) L.

Then (B = 10 – 7.29 = 2.71) L. Percentage of B in the 10-L container \(= \dfrac{2.71}{10} \times 100 = 27.1\%\), approximately 27. Answer (b).

UPSC CSAT 2026 · Question 60

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?

1. 80
2. 85
3. 90
4. The question cannot be answered due to insufficient data

Explanation

Daily earning (E = 5d + 2k), where (d) is number of deliveries and (k) is kilometres travelled, with the constraint \(k \le 100\). We want (E > 500) guaranteed.

The question gives us (d) but does not tell us (k) for the day. Kilometres travelled depend on the route, customer locations and traffic on a given day; (d) alone does not fix (k).

d  | min E (k=0) | max E (k=100) | always > 500?
80 |   400       |   600         | NO (could be 400)
85 |   425       |   625         | NO
90 |   450       |   650         | NO

For any of 80, 85, 90 deliveries, if the day happened to involve very few kilometres, earnings could fall below ₹500. So none of (a), (b), (c) ‘definitely’ meets the target without knowing (k).

Hence the question cannot be answered with the data provided, so the answer is (d) — insufficient data.

Directions for the next 5 items

Each item in this section contains a question followed by two statements. Answer each item using the following instructions and mark your response on the Answer Sheet accordingly.

1. Select this option if the question can be answered using one of these statements alone, but cannot be answered using other statement
2. Select this option if the question can be answered using either statement alone
3. Select this option if the question can be answered using both the statements together, but cannot be answered using either statement alone
4. Select this option if the question cannot be answered even using any of the statements

UPSC CSAT 2026 · Question 61

Question: X receives three coins of different denominations: 1, 2, 5, 10 and 20. If the total amount received by X is m, does X receive a coin of denomination 5?

1. Statement I: m is not a prime number.
2. Statement II: The sum of the digits of m is greater than 5.

1. The question can be answered using one of the statements alone, but cannot be answered using the other statement.
2. The question can be answered using either statement alone.
3. The question can be answered using both the statements together, but cannot be answered using either statement alone.
4. The question cannot be answered even using both the statements together.

Explanation

Step 1 — Enumerate all sums of 3 distinct coins from {1, 2, 5, 10, 20}. The (binom{5}{3}=10) triples give sums: (1+2+5=8), (1+2+10=13), (1+2+20=23), (1+5+10=16), (1+5+20=26), (1+10+20=31), (2+5+10=17), (2+5+20=27), (2+10+20=32), (5+10+20=35).

Step 2 — Test Statement I ((m) not prime). Non-prime sums in the list are 8, 16, 26, 27, 32, 35. The triple ({2,10,20}) gives (m=32) (no 5-coin), while ({1,2,5}) gives (m=8) (with 5-coin). Both are non-prime, so Statement I alone does not decide whether a 5-coin is present — INSUFFICIENT.

Step 3 — Test Statement II (digit-sum of (m) greater than 5). Compute digit-sums: 8→8 ✓, 13→4, 23→5, 16→7 ✓, 26→8 ✓, 31→4, 17→8 ✓, 27→9 ✓, 32→5, 35→8 ✓. The qualifying sums are 8, 16, 17, 26, 27, 35 — corresponding to triples ({1,2,5}, {1,5,10}, {2,5,10}, {1,5,20}, {2,5,20}, {5,10,20}). Every one contains the 5-coin.

Step 4 — Conclusion: Statement II alone answers YES definitively; Statement I alone cannot. Hence the answer is (a).

UPSC CSAT 2026 · Question 62

Question: For two distinct real numbers x and y, which of them is bigger?

1. Statement I: (x^2 < y < 1)
2. Statement II: \(y < \sqrt{x} < 1\)

1. The question can be answered using one of the statements alone, but cannot be answered using the other statement.
2. The question can be answered using either statement alone.
3. The question can be answered using both the statements together, but cannot be answered using either statement alone.
4. The question cannot be answered even using both the statements together.

Explanation

Step 1 — Statement I says (x^2 < y < 1). Take (x=0.5), so (x^2=0.25); choose (y=0.3) ((yx)). Both satisfy the inequality, so we cannot decide which is bigger — INSUFFICIENT.

Step 2 — Statement II says \(y < \sqrt{x} < 1\). Take (x=0.25), so \(\sqrt{x}=0.5\); choose (y=0.1) ((yx)). Again ambiguous — INSUFFICIENT.

Step 3 — Combine. From I: (x^2 < y). From II: \(y < \sqrt{x}\). So \(x^2 < y < \sqrt{x}\), implying (0 < x < 1). The interval \((x^2,\sqrt{x})\) contains (x) itself (since for (0

Step 4 — Concrete check: (x=0.25). Range for (y): ((0.0625, 0.5)). Pick (y=0.1) → (yx). Even combined, the data are INSUFFICIENT. Hence (d).

UPSC CSAT 2026 · Question 63

Question: If x and y are integers, then is x even?

1. Statement I: x²y² is even.
2. Statement II: 1 + x² + y² is odd.

1. The question can be answered using one of the statements alone, but cannot be answered using the other statement.
2. The question can be answered using either statement alone.
3. The question can be answered using both the statements together, but cannot be answered using either statement alone.
4. The question cannot be answered even using both the statements together.

Explanation

Step 1 — Restate Statement I: (x^2 y^2 = (xy)^2) is even ⟹ (xy) is even ⟹ at least one of (x,y) is even. But (x) could be odd (with (y) even) or even. INSUFFICIENT.

Step 2 — Statement II: (1+x^2+y^2) is odd ⟹ (x^2+y^2) is even ⟹ (x^2) and (y^2) have the same parity ⟹ (x) and (y) have the same parity. They could both be odd or both even. INSUFFICIENT alone.

Step 3 — Combine. From II: same parity. From I: at least one even. The only consistent case is both even. Therefore (x) is even — DEFINITE YES.

Step 4 — Verify with examples: (x=2,y=2) gives ((xy)^2=16) even ✓ and (1+4+4=9) odd ✓. Try (x=1,y=1) (both odd): (x^2y^2=1) odd, fails I. So combined statements force both even. Answer (c).

UPSC CSAT 2026 · Question 64

Question: X is a collection of certain odd numbers whereas Y is a collection of certain even numbers. T consists of the numbers all of which are either from X or from Y. Is every number of T from Y?

1. Statement I: The sum of any two numbers belonging to T is even.
2. Statement II: If both p and q are picked from T, then (p – 1)q is even.

1. The question can be answered using one of the statements alone, but cannot be answered using the other statement.
2. The question can be answered using either statement alone.
3. The question can be answered using both the statements together, but cannot be answered using either statement alone.
4. The question cannot be answered even using both the statements together.

Explanation

Step 1 — (T) consists entirely of odd numbers (all from (X)) OR entirely of even numbers (all from (Y)). Question: is every element of (T) from (Y) (i.e., is (T) entirely even)?

Step 2 — Statement I: sum of any two members of (T) is even. If (T) is all-odd: odd+odd = even ✓. If (T) is all-even: even+even = even ✓. Both cases satisfy Statement I — INSUFFICIENT.

Step 3 — Statement II: ((p-1)q) is even for any (p,qin T). Case all-odd: (p) odd ⟹ (p-1) even ⟹ product even ✓. Case all-even: (q) even ⟹ product even ✓. Both cases satisfy — INSUFFICIENT.

Step 4 — Combine I and II: Both statements are simultaneously consistent with EITHER scenario, so even together they fail to decide. Answer (d).

UPSC CSAT 2026 · Question 65

Question: If x, y and z are integers, each greater than 1, then is x a prime number?

1. Statement I: xy² = 116
2. Statement II: xz = 261

1. The question can be answered using one of the statements alone, but cannot be answered using the other statement.
2. The question can be answered using either statement alone.
3. The question can be answered using both the statements together, but cannot be answered using either statement alone.
4. The question cannot be answered even using both the statements together.

Explanation

Step 1 — Statement I: (xy^2 = 116) with (x, y > 1). Factorise \(116 = 2^2 \times 29\).

Step 2 — Since (y > 1), the value (y^2) must be a perfect-square divisor of 116 that is greater than 1. The divisors of 116 are ({1, 2, 4, 29, 58, 116}); among these, only (4) is a perfect square greater than 1. So \(y^2 = 4 \Rightarrow y = 2\), and (x = 116/4 = 29).

Step 3 — Check primality of (x = 29). Trial-divide by primes up to \(\sqrt{29} < 6\): 29 is not divisible by 2, 3, or 5. So 29 is prime — definite YES. SUFFICIENT.

Step 4 — Statement II: \(xz = 261 = 3^2 \times 29\), with (x, z > 1). Factor pairs (both factors (> 1)): ((3, 87), (9, 29), (29, 9), (87, 3)). So (x in {3, 9, 29, 87}).

Step 5 — Primality check: (3) ✓ prime, (9 = 3^2) ✗ not prime, (29) ✓ prime, \(87 = 3 \times 29\) ✗ not prime. Mixed answers, so we cannot decide. INSUFFICIENT.

Step 6 — Statement I alone is sufficient; Statement II alone is not. By the Data Sufficiency convention for Q61–Q65 (option (a) = Statement I alone sufficient, Statement II alone not), the answer is (a).

Directions for the next 2 items

Read the following passage and answer the items that follow. Your answers should be based solely on the passage.

‘Kalagram’, the cultural village set up at the Maha Kumbh Mela, unfolded as a mosaic of India’s diverse regions, each represented by seven meticulously crafted ‘Sanskriti Angans’. Stepping through the grand portal was like entering another world. These thematic zones, inspired by iconic temples like the Dakshineshwar Kali Temple and the Brahma Mandir, were treasure troves of regional artistry. Bengal’s Pattachitra paintings, Assam’s bamboo crafts, Tamil Nadu’s Thanjavur paintings, and Madhya Pradesh’s tribal sculpture—all were showcased in these living galleries where 230 master artisans breathed life into them using age-old techniques, their hands shaping India’s ancient history into creations to behold.

UPSC CSAT 2026 · Question 66

Which of the following conclusions are valid?

1. 1. Seven Sanskriti Angans, representing different regions of India, had been showcased in Kalagram.
2. 2. Regional artistry was recognised via the inspiration drawn from iconic temples.
3. 3. India’s ancient history had been crafted by the contemporary craftsmanship of 230 artisans into creations to behold.
4. 4. Art forms from all regions of India had been showcased in these living galleries.

Select the answer using the code given below.

1. 1 and 2 only
2. 2, 3 and 4
3. 1, 2 and 4
4. 1 and 3

Explanation

Step 1 — Conclusion 1 says seven Sanskriti Angans represented different regions of India. The passage explicitly states ‘seven meticulously crafted Sanskriti Angans’ as a ‘mosaic of India’s diverse regions’. VALID.

Step 2 — Conclusion 2: regional artistry was recognised via inspiration from iconic temples. The passage describes ‘thematic zones, inspired by iconic temples … treasure troves of regional artistry’. VALID.

Step 3 — Conclusion 3 calls the craftsmanship ‘contemporary’. The passage explicitly uses ‘age-old techniques’ — the OPPOSITE of contemporary. INVALID.

Step 4 — Conclusion 4 generalises to ‘all regions of India’. The passage names only Bengal, Assam, Tamil Nadu, Madhya Pradesh — a small subset. Overreach. INVALID.

Step 5 — Only 1 and 2 are valid. Answer (a).

UPSC CSAT 2026 · Question 67

Which one of the following statements is not correct?

1. Paintings from four States of India have been mentioned.
2. Stepping into Kalagram is likened to stepping into another world.
3. The Angans have been described as living galleries.
4. Kalagram is divided into thematic zones, inspired by well-known temples of India.

Explanation

Step 1 — Identify which item is NOT supported. Option (a) claims ‘Paintings from four States’ are mentioned.

Step 2 — The passage lists: Bengal’s Pattachitra paintings, Assam’s bamboo crafts, Tamil Nadu’s Thanjavur paintings, Madhya Pradesh’s tribal sculpture. Of these, only Bengal and Tamil Nadu are PAINTINGS; Assam (bamboo) and MP (sculpture) are not paintings. So ‘paintings from four States’ is INCORRECT.

Step 3 — Option (b): ‘Stepping through the grand portal was like entering another world’ — directly supported. CORRECT statement about the passage.

Step 4 — Option (c): ‘living galleries’ is verbatim from the passage. CORRECT.

Step 5 — Option (d): ‘thematic zones, inspired by iconic temples’ is verbatim. CORRECT.

Step 6 — The not-correct statement is (a). Answer (a).

UPSC CSAT 2026 · Question 68

The weight of X, in kg, is denoted by [X]. The weights of A, B, C, D, P, Q, R and S are measured.

1. Given:
2. A + B + C + D = 17
3. A + C = 6
4. P + Q + S + D = 15
5. P + Q + R + B = 17
6. P = R and Q = S
7. Which one of the following statements is correct?

1. B and D together weigh less than the total weight of P and Q.
2. P and Q together weigh more than the total weight of A and C.
3. P weighs more than Q.
4. Q weighs more than P.

Explanation

Step 1 — Given: (A+B+C+D=17) and (A+C=6), so (B+D = 17-6 = 11).

Step 2 — Substitute (R=P) and (S=Q) into the P-equations. (P+Q+S+D=15) becomes (P+2Q+D=15). (P+Q+R+B=17) becomes (2P+Q+B=17).

Step 3 — Add the two transformed equations: ((P+2Q+D)+(2P+Q+B) = 15+17), giving (3P+3Q+(B+D)=32). Substitute (B+D=11): (3(P+Q) = 32-11 = 21), so (P+Q=7).

Step 4 — Compare totals: (A+C=6), (P+Q=7), (B+D=11).

Step 5 — Check options. (a) (B+D=11 > P+Q=7), so (a) is FALSE. (b) (P+Q=7 > A+C=6) — TRUE. (c) and (d) compare (P) vs (Q) individually, which the system does not determine.

Step 6 — Answer (b).

UPSC CSAT 2026 · Question 69

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.

1. 6
2. 8
3. 10
4. 12

Explanation

Step 1 — QUEUE has letters Q, U, E, U, E. Constraint: ‘Q is always followed by U’ ⟹ glue Q and one U into a single block [QU].

Step 2 — After gluing, the units to arrange are: [QU], E, U, E — that is, 4 units containing two identical E’s.

Step 3 — Number of distinct arrangements of 4 items where two are identical: \(\frac{4!}{2!} = \frac{24}{2} = 12\).

Step 4 — Sanity check: the leftover U (not glued) is still distinct as a separate unit; the two E’s are identical so we divide by 2!. Total = 12.

Step 5 — Answer (d).

UPSC CSAT 2026 · Question 70

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?

1. 11
2. 12
3. 13
4. 14

Explanation

Step 1 — X jumps 4′, Y jumps 6′, Z jumps 5′. All start at mark 199 at 8 AM. They share a mark whenever the additional distance from 199 is a common multiple of 4, 6, 5.

Step 2 — Compute \(\mathrm{lcm}(4,6,5)\). \(\mathrm{lcm}(4,6)=12\); \(\mathrm{lcm}(12,5)=60\). So common landing marks occur every 60 feet from 199.

Step 3 — Common marks: \(199, 259, 319, 379, 439, 499, 559, 619, 679, 739, 799, 859, 919, 979, 1039,\ldots\)

Mark sequence (every +60 from 199):
199, 259, 319, 379, 439, 499, 559, 619, 679, 739, 799, 859, 919, 979 | 1039
  1    2    3    4    5    6    7    8    9   10   11   12   13   14    (excluded: >1000)
Range asked: between 195' and 1000' — all 14 marks above lie in (195, 1000).

Step 4 — Count marks (m) with (195 < m < 1000) from the arithmetic progression: (m = 199 + 60k) for \(k=0,1,\ldots\) with (m<1000). Largest valid: \(199+60\cdot 13 = 199+780 = 979 < 1000\) ✓; next (1039 > 1000).

Step 5 — Number of marks (= 13-0+1 = 14). Answer (d).

Passage — for Questions 71–72

Read the following passage and answer the items that follow. Your answers should be based solely on the passage.

Sport is not just about winning medals or getting jobs through a sports quota. Today's generation is struggling with issues like depression and anxiety. Parents often say that their children are inactive and rarely leave the house. Sport can help tackle these problems. From sport you can learn time management, fitness, teamwork, coordination, and so much more. We need to develop a culture in which sport is seen as a way of life—a path to building a healthier, happier society.

UPSC CSAT 2026 · Question 71

Which of the following conclusions are valid?

1. 1. Sport is more than just games; it is a way of life.
2. 2. Sport can help mitigate the problems of seclusion among the young.
3. 3. Earning laurels in sport can help one secure a job on the basis of an assigned quota.
4. 4. Standard corporate sector skills cannot be learnt through sport.

Select the answer using the code given below.

1. 1 and 3 only
2. 2 and 4 only
3. 1, 2 and 4
4. 1, 2 and 3

Explanation

Step 1 — Conclusion 1: ‘Sport is … a way of life’. The passage literally says ‘sport is seen as a way of life’. VALID.

Step 2 — Conclusion 2: sport mitigates seclusion among the young. The passage notes children ‘rarely leave the house’ (seclusion) and that sport can ‘tackle these problems’. VALID.

Step 3 — Conclusion 3: earning laurels in sport secures a job via quota. The opening line acknowledges ‘getting jobs through a sports quota’. VALID.

Step 4 — Conclusion 4: corporate skills CANNOT be learnt through sport. The passage lists exactly such skills — time management, teamwork, coordination — as learnable. INVALID.

Step 5 — Valid: 1, 2, 3. Answer (d).

UPSC CSAT 2026 · Question 72

Which of the following statements is/are correct?

1. 1. Parents are not encouraging enough when it comes to children playing sport.
2. 2. Participation in sporting activities can help develop life skills.
3. 3. Sport as a way of life can help evolve the very nature of society itself.

Select the answer using the code given below.

1. 1 and 3
2. 2 and 3
3. 2 only
4. 3 only

Explanation

Step 1 — Statement 1: parents are not encouraging enough. The passage only reports that parents complain about inactivity; it makes no claim about whether they encourage or discourage sport. UNSUPPORTED.

Step 2 — Statement 2: sport develops life skills. The passage explicitly lists time management, fitness, teamwork, coordination — clearly life skills. CORRECT.

Step 3 — Statement 3: sport-as-a-way-of-life can evolve society. The passage’s closing line — ‘a path to building a healthier, happier society’ — supports this. CORRECT.

Step 4 — Correct items: 2 and 3 only. Answer (b).

UPSC CSAT 2026 · Question 73

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?

1. 6
2. 7
3. 8
4. 9

Explanation

Step 1 — Let (f) = forward jumps (each +5′), (b) = backward jumps (each −2′).

Equations:
  f + b      = 31    (total jumps)
  5f - 2b    = 15    (net forward displacement)

Step 2 — From the first equation: (b = 31 – f). Substitute into the second: (5f – 2(31-f) = 15).

Step 3 — Expand: (5f – 62 + 2f = 15) ⟹ (7f = 77) ⟹ (f = 11).

Step 4 — Then (b = 31 – 11 = 20).

Step 5 — Verify: (5(11) – 2(20) = 55 – 40 = 15) ✓ and (11+20=31) ✓.

Step 6 — Required difference (= |b – f| = |20 – 11| = 9). Answer (d).

UPSC CSAT 2026 · Question 74

Eight persons P, Q, R, S, T, U, V and W sit around a round table in eight different seats placed with equal distance between any two consecutive seats. Both P and R are adjacent to Q. Both T and R are adjacent to S. Both U and W are adjacent to V. S and W are on opposite chairs. If while going in the clockwise direction around the table from P, one meets R before T, then how many persons shall Q cross while moving in the clockwise direction around the table before meeting W?

1. 5
2. 4
3. 2
4. 1

Explanation

Step 1 — Decode adjacencies. ‘Both P and R are adjacent to Q’ ⟹ Q is between P and R. ‘Both T and R are adjacent to S’ ⟹ S is between T and R. So R is sandwiched between Q and S, yielding the consecutive arc P—Q—R—S—T (5 people in a row).

Step 2 — ‘Both U and W are adjacent to V’ ⟹ V is between U and W, giving the block U—V—W. These three fill the remaining three seats.

Step 3 — Place on an 8-seat round table using positions 1..8 clockwise. The clockwise hint ‘from P, meet R before T’ fixes the chain CLOCKWISE as P→Q→R→S→T. Put P at seat 1, Q at 2, R at 3, S at 4, T at 5.

Step 4 — S and W are diametrically opposite (4 seats apart on 8). S is at seat 4, so W is at seat 8. The U-V-W block must occupy seats 6, 7, 8 with V in the middle: U at 6, V at 7, W at 8.

Clockwise round table (8 seats):

              1 = P
         8=W       2 = Q
       7=V           3 = R
         6=U       4 = S
              5 = T

S (seat 4) and W (seat 8) are diametrically opposite ✓
Clockwise order from P: P → Q → R → S → T → U → V → W → P

Step 5 — Verify all adjacencies: Q(2) neighbors P(1) and R(3) ✓; S(4) neighbors R(3) and T(5) ✓; V(7) neighbors U(6) and W(8) ✓; T(5) and U(6) are adjacent (fine — no rule against it); W(8) and P(1) are adjacent (fine).

Step 6: The question asks how many persons Q crosses, not P. Q sits at seat 2. Moving clockwise, Q crosses R (3), S (4), T (5), U (6) and V (7), and then meets W (8). That is 5 persons. Counting from P would give 6, which is the trap here.

Step 7 — Answer: (a) 5.

UPSC CSAT 2026 · Question 75

The top of a table is rectangular and its dimensions are 6′ × 10′. Two rectangular portions of the table top are painted in blue colour; both these portions have dimensions 2.5′ × 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?

1. 2
2. 4
3. 6
4. 8

Explanation

Step 1 — Table top: 6′ × 10′. Two blue rectangles, each 2.5′ × 8′, must each have exactly two sides common with two edges of the table — i.e., each blue rectangle sits in a CORNER.

Step 2 — Orientation check. A blue side of length 8 cannot lie along a 6-foot edge ((8>6)). So the 8-side must align with a 10-foot edge, and the 2.5-side with a 6-foot edge. This fixes orientation at every corner.

Table (6' × 10'), top view. Corners: TL, TR, BL, BR.

   +----------- 10' -----------+
   |TL                       TR|
   | 2.5'                  2.5'|
   |  x  along             x   |
   |  8' edge             8'   |
   |                           |  6'
   |                           |
   |BL                       BR|
   +---------------------------+

Each blue rect occupies a corner with 8 along the long edge.
Valid pairs of corners (the two rectangles must not overlap):

Step 3 — Two corners on the SAME long edge (e.g., TL+TR or BL+BR) would force two 8-foot strips along the same 10-foot edge, totalling 16′ but only 10′ is available — overlap. INVALID.

Step 4 — Remaining corner pairs are: {TL, BL}, {TR, BR}, {TL, BR}, {TR, BL}. Each is a distinct pattern from above.

Step 5 — Count: 4 distinct patterns. Answer (b).

Passage — for Questions 76–77

Read the following passage and answer the items that follow. Your answers should be based solely on the passage.

How is deflation done? Most countries use a method called 'double deflation', where input and output prices are deflated separately. Consider a manufacturer importing oil for use in production. If oil prices fall, output prices do not and quantities remain the same, real value added should not change. But if the same deflator is used for inputs and outputs, as in India, it would look as if the manufacturer had become more productive.

UPSC CSAT 2026 · Question 76

Which of the following statements is/are correct?

1. 1. Real value should not change in the instance of static output cost and unchanged quantities against falling oil prices.
2. 2. Deflators are to be used separately for inputs and outputs, and this is a practice universally adopted by all economies.

Select the answer using the code given below.

1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2

Explanation

Step 1: Statement 1 changes the passage’s example in two places. The passage says that if oil prices fall, ‘output prices do not’ and quantities stay the same, then ‘real value added should not change’.

Step 2: First change: statement 1 speaks of ‘static output cost’, not static output prices. Oil is an input, so when oil prices fall, the cost of producing the output falls too. Output cost cannot stay static in this scenario; only the output price does.

Step 3: Second change: statement 1 says ‘real value’, while the passage is about real value added, which is output minus inputs. These are different measures. So statement 1 is INCORRECT.

Step 4: Statement 2 claims double deflation is adopted by all economies. The passage says ‘Most countries’ use it, and names India as one that uses a single deflator. INCORRECT.

Step 5: Neither statement is correct. Answer (d).

UPSC CSAT 2026 · Question 77

Which of the following assumptions is/are valid?

1. 1. Deflation strategies can be used to make manufacturers appear to be doing better than they actually are.
2. 2. When input and output prices are both deflated against a single input price, it is referred to as ‘double deflation’.

Select the answer using the code given below.

1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2

Explanation

Step 1 — Assumption 1: deflation strategies can make manufacturers appear to do better than they actually do. The passage says using a single deflator for inputs and outputs ‘would look as if the manufacturer had become more productive’ — a misleading appearance. VALID.

Step 2 — Assumption 2: ‘double deflation’ means deflating input and output prices against a SINGLE input price. The passage defines double deflation as deflating input and output prices SEPARATELY. The assumption misstates the definition. INVALID.

Step 3 — Only Assumption 1 is valid. Answer (a).

UPSC CSAT 2026 · Question 78

A pattern formed by two characters a and b is repeated more than once in the following string:

1. × b × a × a × × a × a × bab
2. What is ×× in the 7th and 8th positions from the left in the above string?

1. aa
2. ab
3. ba
4. bb

Explanation

Step 1 — The string has 15 characters with some unknowns. The known characters provided are: positions 2=b, 4=a, 6=a, 9=a, 11=a, 13=b, 14=a, 15=b (the trailing ‘bab’).

Step 2 — A pattern using only {a,b} is repeated more than once and the total length is 15, so the pattern length (L) divides 15 with (L<15). Divisors: 1, 3, 5. Length 1 (all a or all b) contradicts the mix of a and b in the string. Length 3 conflicts with positions (e.g., position 13=b would imply position 1=b, but the constraints don't all fit; checking exhaustively rules it out). So (L=5), repeated 3 times.

Step 3 — The last 5 characters (positions 11–15) read ‘a-?-bab’ i.e. ‘a_bab’, and this equals the 5-letter pattern.

Position:   1  2  3  4  5 | 6  7  8  9 10 | 11 12 13 14 15
Given:      _  b  _  a  _ | a  ?  ?  a  _ |  a  _  b  a  b
Pattern P:  P1 P2 P3 P4 P5  P1 P2 P3 P4 P5  P1 P2 P3 P4 P5

Step 4 — From positions 11–15: (P_1=a, P_2=?, P_3=b, P_4=a, P_5=b). From position 2: (P_2=b). So the pattern is (P = a,b,b,a,b) = ‘abbab’.

Step 5 — Reconstruct full string: ‘abbab abbab abbab’. Verify against givens: pos 4=a ✓, pos 6=a ✓ ((P_1)), pos 9=a ✓ ((P_4)), pos 11=a ✓ ((P_1)), pos 13=b ✓ ((P_3)), pos 14=a ✓, pos 15=b ✓. All consistent.

Step 6 — Positions 7 and 8 correspond to (P_2) and (P_3) = b and b. So the ‘××’ is ‘bb’. Answer (d).

UPSC CSAT 2026 · Question 79

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

1. 101
2. 111
3. 121
4. 131

Explanation

Step 1 — Rewrite: \(10^m \times 1000 \times n = 75^{25} \times 25^{32} \times 32^{75}\). Note (1000 = 10^3), so the left side is \(10^{m+3} \cdot n\).

Step 2 — Factor each base on the right: \(75 = 3 \cdot 5^2\), so \(75^{25} = 3^{25} \cdot 5^{50}\). (25 = 5^2), so \(25^{32} = 5^{64}\). (32 = 2^5), so \(32^{75} = 2^{375}\).

Step 3 — Multiply: RHS \(= 2^{375} \cdot 3^{25} \cdot 5^{50+64} = 2^{375} \cdot 3^{25} \cdot 5^{114}\).

Step 4 — LHS \(= 10^{m+3} \cdot n = 2^{m+3} \cdot 5^{m+3} \cdot n\). Since (n) is not divisible by 10, (n) contains no pair of (2, 5) factors. In particular, all the 5’s on the RHS must come from \(10^{m+3}\) (because (n) cannot supply 5’s without also supplying 2’s, which would let (10) be factored out — actually, the cleaner reasoning: if (n) contained any factor of 5, then because RHS has \(2^{375}\) ample 2’s left over, (n) could absorb a 2 alongside the 5 making (n) divisible by 10, contradicting the condition; the standard convention is to make (m+3) maximal so (n) has no factor of 5).

Step 5 — Set (m+3 = 114) ⟹ (m = 111).

Step 6 — Then \(n = 2^{375-114} \cdot 3^{25} = 2^{261} \cdot 3^{25}\), which has no factor of 5 and hence is not divisible by 10. ✓

Step 7 — Answer (b) 111.

UPSC CSAT 2026 · Question 80

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?

1. 13.51
2. 13.65
3. 13.85
4. 14.05

Explanation

Step 1 — Same-direction case (given). Train T at 100 km/h, person P at 4 km/h, both moving the same way. Relative speed (= 100 – 4 = 96) km/h.

Same direction (given):
   P --4→         T --100→
   Relative speed of T w.r.t. P = 100 - 4 = 96 km/h
   Time to cross  = 15 s  ⟹  Length L = 96 × (15/3600) km

Step 2 — Convert 15 s into hours: (15/3600 = 1/240) hour. Train length \(L = 96 \times \frac{1}{240} = 0.4\) km (= 400) m.

Step 3 — Opposite-direction case.

Opposite direction (asked):
   ←4-- P         T --100→
   Relative speed = 100 + 4 = 104 km/h
   Time = L / relative speed = 0.4 / 104  (hours)

Step 4 — Time in hours: \(t = \frac{0.4}{104}\) hr. Convert to seconds: \(t = \frac{0.4}{104} \times 3600 = \frac{1440}{104}\) s.

Step 5 — Divide: \(1440 / 104 = 13.846\ldots\) seconds.

Step 6 — Approximately 13.85 s. Answer (c).