UPSC Prelims 2026 · CSAT PaperSet D Answer Key

Answer: B (II only)

Statement I — Not always true.

Consider $f(t) = t + \tfrac{1}{t}$. Its derivative is $f'(t) = 1 - \tfrac{1}{t^2}$, which is negative on $(0,1)$ and positive on $(1,\infty)$. So $f$ decreases on $(\tfrac{1}{3}, 1)$ and increases on $(1, 2)$ — not monotonic on the whole interval.

Counterexample: $x = \tfrac{1}{2}, y = \tfrac{3}{4}$. Then $x + \tfrac{1}{x} = 2.5$ but $y + \tfrac{1}{y} = 0.75 + 1.333\ldots \approx 2.083$. So $x + \tfrac{1}{x} > y + \tfrac{1}{y}$, contradicting I.

Statement II — Always true.

Rewrite:

$\frac{\sqrt{1+t^2}}{t} = \sqrt{\frac{1+t^2}{t^2}} = \sqrt{\frac{1}{t^2} + 1}$

For $t > 0$, as $t$ increases, $\tfrac{1}{t^2}$ decreases, so $\sqrt{\tfrac{1}{t^2}+1}$ decreases. Thus $g(t) = \tfrac{\sqrt{1+t^2}}{t}$ is strictly decreasing on $(0, \infty)$.

Since $x < y$ (both positive), $g(x) > g(y)$, i.e.

$\frac{\sqrt{1+y^2}}{y} < \frac{\sqrt{1+x^2}}{x}. \checkmark$

To find the minimum number of measurements required to get exactly $298$ litres, we can use a combination of drawing water from the tank (adding to our container) and pouring water back into the tank (subtracting from our container).

Approach 1: Only Adding Water If we only draw water, we would use the greedy approach (using the largest possible cylinders first):

• $2$ times $100$ litres = $200$ litres
• $3$ times $25$ litres = $75$ litres
• $3$ times $6$ litres = $18$ litres
• $5$ times $1$ litre = $5$ litres Total volume = $200 + 75 + 18 + 5 = 298$ litres. Total measurements = $2 + 3 + 3 + 5 = 13$ times.

Approach 2: Adding and Subtracting Water (Optimal) To minimize the number of measurements, we can overshoot the target volume and then remove the excess water by pouring it back into the tank.

• We can draw $100$ litres $3$ times: $3 \times 100 = 300$ litres. (This takes $3$ measurements)
• We have an excess of $2$ litres ($300 - 298 = 2$).
• We can remove this excess by using the $1$-litre cylinder to pour water back into the tank $2$ times: $2 \times 1 = 2$ litres. (This takes $2$ measurements)

Total volume = $300 - 2 = 298$ litres. Total measurements = $3$ (drawing) $+ 2$ (pouring back) = $5$ times.

Can we do it in $4$ measurements? To get close to $298$ with $4$ measurements, we must use the $100$-litre cylinder $3$ times ($300$ litres). We would then have only $1$ measurement left to subtract the excess. The closest volumes we could achieve are:

• $3 \times 100 - 1 \times 1 = 299$ litres
• $3 \times 100 - 1 \times 6 = 294$ litres Since neither gives exactly $298$ litres, $4$ measurements are not enough.

Thus, the minimum number of times one needs to measure is $5$.

Answer: 8

Let $a, b, c, d$ be the number of 1, 2, 5, 10 kg weights. We need $a + 2b + 5c + 10d = 20, \quad a \in \{8, 9, 10, 11\},\ b,c,d \ge 0.$

Case $a = 8$: $2b + 5c + 10d = 12$

• $d=0$: $(b,c) = (6,0), (1,2)$ → 2 ways
• $d=1$: $(b,c) = (1,0)$ → 1 way

Subtotal: 3

Case $a = 9$: $2b + 5c + 10d = 11$ (odd, so $c$ must be odd)

• $d=0$: $(b,c) = (3,1)$ → 1 way

Subtotal: 1

Case $a = 10$: $2b + 5c + 10d = 10$

• $d=0$: $(b,c) = (5,0), (0,2)$ → 2 ways
• $d=1$: $(b,c) = (0,0)$ → 1 way

Subtotal: 3

Case $a = 11$: $2b + 5c + 10d = 9$ ($c$ odd)

• $d=0$: $(b,c) = (2,1)$ → 1 way

Subtotal: 1

Total: $3 + 1 + 3 + 1 = 8$

Answer: A — 16

The condition says that after the third cut, the pieces are identical. It does not necessarily mean that there are $8$ pieces. For example, three parallel equally spaced cuts can produce $4$ identical slabs, and other arrangements can produce $6$ or $8$ identical pieces.

A fourth cut can only increase the number of pieces by the number of existing pieces it actually cuts. Hence:

$x_4 = x_3 + \left(\text{number of existing pieces cut by the 4th plane}\right)$

Now check the options:

• $x_4 = 5$ is possible: after three parallel cuts, there are $4$ identical slabs; the fourth cut may cut only one slab, giving $4+1=5$.
• $x_4 = 8$ is possible: after three parallel cuts, there are $4$ identical slabs; the fourth cut may cut all $4$ slabs, giving $4+4=8$.
• $x_4 = 12$ is possible: after three cuts producing $6$ identical pieces, the fourth cut may cut all $6$ pieces, giving $6+6=12$.
• $x_4 = 16$ would require $x_3=8$ and the fourth cut to cut all $8$ pieces. But after three cuts producing $8$ identical pieces, the cube is divided into a $2\times 2\times 2$ arrangement, and a single plane cannot pass through the interiors of all $8$ pieces.

Therefore, $16$ is not a possible value of $x_4$.

Answer: B — $3$

Let the number of students in the class be $n$.

The first correction changes a student's score from $0$ to $100$, so the total score increases by $100$. Since the average increases by $4$, we have:

$\frac{100}{n}=4$

Therefore:

$n=25$

Now, another student's score was recorded as $81$ instead of $56$. Correcting this decreases the total score by:

$81-56=25$

Since there are $25$ students, the average decreases by:

$\frac{25}{25}=1$

After the first correction, the average becomes $x+4$. After the second correction, the final corrected average is:

$y=x+4-1=x+3$

Hence:

$y-x=3$

Therefore, the correct answer is $3$.

Answer: B — 2 only

Statement 1 — Invalid. The passage explicitly states that HMPV and SARS-CoV-2 "belong to two very different virus families with fundamentally different characteristics and epidemiology." Calling them "similar viruses" directly contradicts the passage. While the passage does support that SARS-CoV-2 was novel and no one had specific immunity to it (which fits the second half of statement 1), the premise that the two viruses are "similar" undermines the conclusion as stated.

Statement 2 — Valid. The passage establishes that the two viruses have fundamentally different characteristics, different epidemiology (with strong seasonality in HMPV but not SARS-CoV-2), different severity of symptoms — particularly over the long term — and "affected population segments [that] do not fully overlap." This collectively supports the inference that they have different impacts on human populations and therefore should not be dealt with in a similar manner.

Answer: D — None of the above

Statement 1 — Incorrect. The passage describes existing knowledge about the two viruses (exposure history, epidemiology, symptom severity). It does not propose or develop methodologies for analysis. So "evolving methodologies for objective analysis" is not the intent.

Statement 2 — Plausible but not paired correctly. The passage does contrast the two viruses' epidemiology (HMPV has strong seasonality, SARS-CoV-2 was novel, families differ, exposure history differs), so this somewhat reflects the author's intent. However, no option offers "2 only".

Statement 3 — Incorrect. The passage discusses neither remedies nor a comparative treatment framework. It is purely descriptive about how the two viruses differ.

Since 1 and 3 are clearly not the writer's intent, the available options A (1 and 2), B (2 and 3) and C (1 and 3) all fail. The only consistent choice is D — None of the above.

Answer: B — 1, 2 and 3

Statement 1 — Correct. The passage states people have been exposed to HMPV "for decades" and the virus "is well-studied," whereas SARS-CoV-2 was "a truly novel virus" in 2019. So HMPV has a longer history of documentation and study than SARS-CoV-2.

Statement 2 — Correct. The passage notes that the two viruses cause "different severity of symptoms" and that "affected population segments do not fully overlap" — i.e., the differences between the viruses result in markedly different outcomes among those affected.

Statement 3 — Correct. The passage explicitly says symptoms differ "particularly over the long term," and that HMPV has "no long-term post-viral symptoms" (contrasted with SARS-CoV-2's known long-term sequelae). This makes long-term impact a key differentiator between the two.

Statement 4 — Incorrect. The passage notes "strong seasonality seen in HMPV, unlike SARS-CoV-2." So seasonality is a difference, not a common factor between the two viruses.

To find the largest and smallest possible values for the expression $(x \times y) + z$, we must choose distinct values for $x$, $y$, and $z$ from the given set $\{2, 3, 4, 5\}$.

Step 1: Finding the largest possible value ($M$) To maximize the value of $(x \times y) + z$, we should maximize the product $(x \times y)$ because multiplication of numbers greater than $1$ yields a larger result than addition.

• The two largest numbers available are $4$ and $5$. Therefore, we assign $x = 4$ and $y = 5$ (or vice versa), giving a product of $4 \times 5 = 20$.
• The remaining numbers are $2$ and $3$. To maximize the overall sum, we assign the largest remaining number to $z$, which is $3$.
• Calculating the maximum value: $M = (4 \times 5) + 3 = 20 + 3 = 23$

Step 2: Finding the smallest possible value ($N$) To minimize the value of $(x \times y) + z$, we should minimize the product $(x \times y)$.

• The two smallest numbers available are $2$ and $3$. Therefore, we assign $x = 2$ and $y = 3$ (or vice versa), giving a product of $2 \times 3 = 6$.
• The remaining numbers are $4$ and $5$. To minimize the overall sum, we assign the smallest remaining number to $z$, which is $4$.
• Calculating the minimum value: $N = (2 \times 3) + 4 = 6 + 4 = 10$

(Self-check: If we tried $x=2, y=4, z=3$, the result would be $(2 \times 4) + 3 = 11$, which is larger than $10$. Thus, $10$ is indeed the absolute minimum.)

Step 3: Finding the difference ($M - N$) Now, we subtract the smallest value from the largest value: $M - N = 23 - 10 = 13$

Therefore, the difference between the largest and smallest possible values is $13$.

Answer: $\frac{1}{5}$

Set up the proportionalities.

• $y \propto x^2 \Rightarrow y = k x^2$
• $x \propto \frac{1}{z} \Rightarrow x = \frac{c}{z}$, i.e. $xz = c$

Find the constants using $z = \tfrac{7}{25},\ x = 5,\ y = 50$.

• From $y = kx^2$: $50 = k \cdot 25 \Rightarrow k = 2$
• From $xz = c$: $c = 5 \cdot \tfrac{7}{25} = \tfrac{7}{5}$

So $y = 2x^2$ and $xz = \tfrac{7}{5}$.

Use $y = 98$ to find $x$.

$2x^2 = 98 \Rightarrow x^2 = 49 \Rightarrow x = 7$

Find $z$.

$z = \frac{c}{x} = \frac{7/5}{7} = \frac{1}{5}$

Correct option: $\frac{1}{5}$

Based on the passage, let us evaluate each statement:

1. Statement 1 is invalid: The passage mentions that in the US and Europe, generative AI tools have started offering precision farming at scale. It does not state or imply that the overall agricultural productivity in the West has "marched ahead" historically because of AI tools. The causal link is an overreach and cannot be assumed from the text.

2. Statement 2 is valid: The passage explicitly suggests that for "at-scale integration and accessibility" of AI tools in India, it is helpful to develop them in Indian languages and partner with AgTechs to create affordable solutions. This logically assumes that affordability and local language availability are key drivers to help these solutions reach a wider base of smallholder farmers.

3. Statement 3 is valid: The passage directly states that "India is starting to deploy AI for critical use cases such as weather forecasting, pest detection and control..." and follows it up by noting that "penetration is limited to a small subset of tech-savvy farmers." This confirms that while penetration is low, critical applications of AI are indeed already in use in India.

Since Statements 2 and 3 are logically supported by the passage, the correct answer is D.

Based on the logical analysis of the statements in the context of artificial intelligence and agriculture:

1. Statement 1 is incorrect: In the context of agricultural technology, it is typically the innovations and advancements by AgTech companies that drive the transformation of agriculture and empower farmers (making them tech-savvy), rather than the farmers driving the AgTech companies. The statement reverses the general cause-and-effect relationship highlighted in such contexts.
2. Statement 2 is correct: The development of government advisory services to train and assist farmers in using AI tools is a highly practical and positive step toward agricultural modernization.
3. Statement 3 is incorrect: This is an extreme statement. Even in developed regions like the US and Europe, AI tools and precision agriculture augment and assist traditional agricultural practices to improve efficiency and yield; they have not completely replaced traditional farming practices.
4. Statement 4 is correct: The integration of large datasets (Big Data) for real-time agronomic analysis (such as weather forecasting, soil health monitoring, and crop predictive analytics) is already a reality in modern precision agriculture.

Since the question asks for the statements that are not correct, the correct combination is 1 and 3.

Answer: D — Neither I nor II

Given: $P=4$, $R=1$, $Q\ne 5$, $S\ne 2$, $T\ne 3$. The remaining ranks $\{2,3,5\}$ are distributed among $Q$, $S$, $T$.

Statement I: If $S=3$, then $T=2$.

If $S=3$, then $Q$ and $T$ take $\{2,5\}$. Since $Q\ne 5$, we must have $Q=2$, which forces $T=5$. So $T=5$, not $2$. Statement I is incorrect.

Statement II: If $Q=3$, then $T=5$.

If $Q=3$, then $S$ and $T$ take $\{2,5\}$. Since $S\ne 2$, we must have $S=5$, which forces $T=2$. So $T=2$, not $5$. **Statement II is incorrect.

Let's break down the given information step by step:

1. Understanding the Alignment: Two identical straight rods are painted in five distinct colours. Because they are identical, the sequence of colours on both rods is exactly the same. When they are kept parallel side by side and their portions "match", it means they align perfectly and have the same colour. If they were shifted, none of the distinct colours would match. If they were placed in opposite directions, at most only the middle portion would match in colour. However, the problem gives us three matching pairs, which means the rods must be aligned perfectly in the same direction (Position 1 to 1, Position 2 to 2, etc.).

2. Analyzing the Conditions: Let the 5 positions on the rods be 1, 2, 3, 4, and 5. At each position, there is one $P$ label (from Rod 1) and one $Q$ label (from Rod 2). The given matching conditions tell us which labels share the same position:

• $P1$ and $Q3$ are at the same position.
• $P4$ and ($Q1$ or $Q2$) are at the same position.
• $Q4$ and ($P3$ or $P5$) are at the same position.
• $Q3$ and $Q5$ are adjacent (their position numbers differ by 1).

3. Evaluating the Statements (Checking for Possibility): The question asks which of the statements is/are possible. We just need to find if there is at least one valid arrangement that satisfies all conditions for each statement.

Let's try to construct an arrangement where $Q3$ is in the middle (Position 3) and $P2$ is at an extreme (Position 1 or 5):

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

Let's verify if this arrangement violates any rules:

• Does $P1$ match $Q3$? Yes, both are at Position 3.
• Does $P4$ match $Q1$ or $Q2$? Yes, $P4$ matches $Q1$ at Position 5.
• Does $Q4$ match $P3$ or $P5$? Yes, $Q4$ matches $P3$ at Position 2.
• Are $Q3$ and $Q5$ adjacent? Yes, $Q3$ is at Position 3 and $Q5$ is at Position 4.
• Are all labels $P1-P5$ and $Q1-Q5$ used exactly once? Yes.

4. Conclusion: In our perfectly valid arrangement:

• $Q3$ is at the middle portion (Position 3). Thus, Statement I is possible.
• $P2$ is at an extreme portion (Position 1). Thus, Statement II is possible.

Since both scenarios can occur without violating any given conditions, both I and II are possible.

Answer: I and III only

Deduce the seating.

• $A$ is in $X$ (given).
• "Only $E$ travels with $G$" → $E$ and $G$ share a car alone (just the two of them).
• $C$ is in $Z$.
• $B \notin Y$ and $B \neq A$'s car, so $B$ is in $Z$.
• $E, G$ cannot be in $Z$ (else $C$ would be with them, contradicting "only $E$ with $G$"). So $E, G$ are in $Y$.
• $X$ must hold $A$ + two others. The only people left are $D$ and $F$, so $X = \{A, D, F\}$.

Final assignment:

• $X = \{A, D, F\}$
• $Y = \{E, G\}$
• $Z = \{B, C\}$

Check the statements.

• I. No one travels alone. Smallest car has 2 people. ✓ True
• II. Only $D$ travels with $F$. $F$ is in $X$ with both $A$ and $D$, so $F$ has two companions. ✗ False
• III. Only $C$ travels with $B$. $Z = \{B, C\}$, so $B$'s sole companion is $C$. ✓ True

Let us represent the statements being correct as $X$, $Y$, $Z$, and $W$, and the statements being incorrect as $\neg X$, $\neg Y$, $\neg Z$, and $\neg W$.

According to the question, the given relations can be written as logical implications:

1. If $X$ is incorrect, then $Z$ is incorrect: $\neg X \implies \neg Z$
2. If $Y$ is incorrect, then $W$ is correct: $\neg Y \implies W$
3. If $W$ is correct, then $X$ is incorrect: $W \implies \neg X$

By combining these three relations, we can form a continuous chain of implications: $\neg Y \implies W \implies \neg X \implies \neg Z$

In logic, the contrapositive of a statement $A \implies B$ is $\neg B \implies \neg A$, and both always hold the exact same truth value. Taking the contrapositive of our entire chain of implications (reversing the order and negating each term), we get: $Z \implies X \implies \neg W \implies Y$

Now, let us evaluate the given conclusions based on this contrapositive chain:

Statement I: If $X$ is correct, then so is $Y$. From our contrapositive chain, we can clearly see that $X \implies Y$. This means that if $X$ is correct, $Y$ must definitely be correct. Therefore, Statement I is correct.

Statement II: If $Z$ is correct, then it is not necessary that $Y$ is correct. From our contrapositive chain, we can see that $Z \implies Y$. This means that if $Z$ is correct, it is absolutely necessary that $Y$ is correct. Therefore, Statement II is incorrect.

Since only Statement I is correct, the correct option is A.

Let's break the problem into two phases: before and after $Y$ changes direction.

Phase 1: Running in the same direction

• We are given that when $X$ completes $7$ rounds, $Y$ completes exactly $5$ rounds.
• Let this duration be $T$.
• The speeds of $X$ and $Y$ are $V_X = 7$ rounds per $T$ and $V_Y = 5$ rounds per $T$.
• Since they are running in the same direction, their relative speed is $V_X - V_Y = 7 - 5 = 2$ rounds per $T$.
• The number of times they meet is equal to the relative distance covered in rounds. In time $T$, they cover a relative distance of $2$ rounds, meaning they meet exactly $2$ times.
• These meetings occur at $t = 0.5T$ and $t = 1.0T$. The meeting at $t = 1.0T$ happens exactly when $X$ completes $7$ rounds and $Y$ completes $5$ rounds (both are at the starting point).

Phase 2: Running in opposite directions

• After $t = 1.0T$, $Y$ changes direction and doubles his speed.
• $Y$'s new speed is $V_Y' = 2 \times 5 = 10$ rounds per $T$.
• $X$ continues in the same direction with the same speed, $V_X = 7$ rounds per $T$.
• Because they are now running in opposite directions, their new relative speed is $V_X + V_Y' = 7 + 10 = 17$ rounds per $T$.
• $X$ stops when he completes exactly $21$ rounds. Since he already completed $7$ rounds in Phase 1, he must run $21 - 7 = 14$ more rounds in Phase 2.
• The time taken for Phase 2 is the distance $X$ runs divided by his speed: $\frac{14 \text{ rounds}}{7 \text{ rounds per } T} = 2T$.
• In this duration of $2T$, the number of times they meet is their relative speed multiplied by the time: $17 \times 2 = 34$ times.
• These $34$ meetings happen strictly after $t = 1.0T$. The very last meeting (the 34th in this phase) occurs exactly at the end of the $2T$ duration (at $t = 3.0T$), which is when $X$ completes his 21st round and they both finally stop.

Total Number of Meetings:

• The question asks for the number of meetings after they had started and before they finally stopped.
• We exclude the start ($t = 0$) and the final stop ($t = 3.0T$).
• Meetings in Phase 1: $2$ (at $t = 0.5T$ and $t = 1.0T$).
• Meetings in Phase 2: $34 - 1 = 33$ (excluding the final stop at $t = 3.0T$).
• Total valid meetings = $2 + 33 = 35$.

Therefore, $X$ and $Y$ met $35$ times.

Let the number of correctly answered questions be $x$ and the number of wrongly answered questions be $y$. Since the student attempted all questions, the total number of questions in the paper is $x + y$.

According to the first condition, $5$ marks are awarded for a correct answer and $2$ marks are deducted for a wrong answer, resulting in a score of $69$: $5x - 2y = 69 \quad \text{--- (Equation 1)}$

According to the second condition, if $4$ marks were awarded for a correct answer and $1$ mark deducted for a wrong answer, the score would be $84$: $4x - y = 84 \quad \text{--- (Equation 2)}$

From Equation 2, we can express $y$ in terms of $x$: $y = 4x - 84$

Substitute this value of $y$ into Equation 1: $5x - 2(4x - 84) = 69$ $5x - 8x + 168 = 69$ $-3x = 69 - 168$ $-3x = -99$ $x = 33$

Now, substitute the value of $x$ back into the expression for $y$: $y = 4(33) - 84$ $y = 132 - 84$ $y = 48$

The student answered $33$ questions correctly and $48$ questions wrongly. The total number of questions in the paper is: $x + y = 33 + 48 = 81$

Therefore, there were $81$ questions in the question paper.

Let us evaluate the given conclusions based on the passage:

Statement 1 is invalid: The passage explicitly states that the JJ Act allows for the possibility of trying adolescents above 16 as adults if they are accused of committing a heinous offence, not a serious offence. Furthermore, the passage notes that in the opinion of activists, a serious offence does not merit the transfer of a case to the adult criminal justice system. Therefore, the claim that only a serious offence justifies this transfer is completely contradictory to the passage.

Statement 2 is valid: The passage mentions, "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." This clearly indicates that for an offence to be classified as a "serious offence," the determining factor is the maximum sentence it carries (seven years or more), given that there is no prescribed minimum sentence.

Since only the second conclusion logically follows from the text, the correct option is B.

Answer: C — Both 1 and 2

Statement 1 is correct: The passage defines a heinous offence as "one with a minimum punishment of seven years," and explicitly cites culpable homicide and death by negligence as offences that "are not heinous offences because they do not have a prescribed minimum punishment." So if an offence has no minimum prescribed punishment, it cannot be classified as heinous under the JJ Act, 2015.

Statement 2 is correct: The passage states, "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." This is exactly what Statement 2 paraphrases.

Since both statements are directly supported by the passage, Option C is correct.