UPSC Prelims 2026 CSAT (Paper II · Set A)

Three distinct coins from $\{1,2,5,10,20\}$ give 10 possible sums.

• Without a 5: $m\in\{13,23,31,32\}$

• With a 5: $m\in\{8,16,17,26,27,35\}$

Statement I: $m$ not prime $\Rightarrow m\in\{8,16,26,27,32,35\}$.

• $m=8$ has a 5 ✓

• $m=32$ has no 5 ✗

• Insufficient

Statement II: digit-sum $>5$.

• $m=13\to 4$ ✗, $m=23\to 5$ ✗, $m=31\to 4$ ✗, $m=32\to 5$ ✗

• Remaining: $\{8,16,17,26,27,35\}$ — every one contains a 5

• Sufficient

Only Statement II answers the question.

— Option A

Statement I: $x^2

• $x=0.5, y=0.3$: $0.25<0.3<1$ ✓, $y

• $x=0.1, y=0.5$: $0.01<0.5<1$ ✓, $y>x$

• Insufficient

Statement II: $y<\sqrt{x}<1$ (so $0\le x<1$).

• $x=0.25, y=0.3$: $0.3<0.5<1$ ✓, $y>x$

• $x=0.25, y=0.1$: $0.1<0.5<1$ ✓, $y

• Insufficient

Both together: $x^2

• $x=0.25, y=0.3$: both hold; $y>x$

• $x=0.25, y=0.1$: both hold; $y

• Insufficient

— Option D

Statement I: $x^2y^2=(xy)^2$ even $\Rightarrow xy$ even $\Rightarrow$ at least one of $x,y$ is even.

• $x=2, y=1$: product $=4$ ✓, $x$ even

• $x=1, y=2$: product $=4$ ✓, $x$ odd

• Insufficient

Statement II: $1+x^2+y^2$ odd $\Rightarrow x^2+y^2$ even $\Rightarrow x,y$ have same parity.

• $x=1, y=1$: $1+1+1=3$ ✓, $x$ odd

• $x=2, y=2$: $1+4+4=9$ ✓, $x$ even

• Insufficient

Combined: same parity AND at least one even $\Rightarrow$ both even $\Rightarrow x$ even. Sufficient

— Option C

The base information only says each element of $T$ comes from $X$ or $Y$. So $T$ may be all odd, all even, or mixed before the statements are applied. The question asks whether every element of $T$ is from $Y$.

Statement I: Sum of any two elements of $T$ is even.

• This rules out a mixed odd-even set.

• $T=\{3,5\}$: all sums are even; answer is No. ✓

• $T=\{2,4\}$: all sums are even; answer is Yes. ✓

• Insufficient

Statement II: $(p-1)q$ is even for any $p,q\in T$.

• If all elements are odd, $p-1$ is even. ✓

• If all elements are even, $q$ is even. ✓

• A mixed set fails by taking even $p$ and odd $q$.

• Both all-odd and all-even cases remain possible. Insufficient

Combined: The statements still cannot distinguish all-odd from all-even. Insufficient

— Option D

Statement I: $xy^2=116=2^2\cdot 29$, with $y\ge 2$ so $y^2\ge 4$.

• $y=2 \Rightarrow y^2=4 \Rightarrow x=29$ ✓ (prime)

• No other $y\ge 2$ gives $y^2 \mid 116$

• $x=29$ uniquely $\Rightarrow$ prime. Sufficient

Statement II: $xz=261=3^2\cdot 29$, both $>1$.

• $x=3, z=87$: $x$ prime ✓

• $x=9, z=29$: $x=9$ not prime ✗

• $x=29, z=9$: $x$ prime ✓

• $x=87, z=3$: not prime ✗

• Insufficient

— Option A

1. Passage says Kalagram had 'seven meticulously crafted Sanskriti Angans' representing India's diverse regions — supported. ✓

2. The passage says the zones were inspired by temples and were treasure troves of regional artistry, but it does not say artistry was 'recognised via' that inspiration. Unsupported/overstated. ✗

3. Passage says 230 master artisans, using age-old techniques, shaped India's ancient history into creations to behold. The artisans are present-day; 'contemporary craftsmanship' does not contradict the passage. Supported. ✓

4. Passage names selected regions and says 'diverse regions', not all regions. ✗

Valid: 1 and 3.

— Option D

Option A: Passage mentions paintings from only TWO states — Bengal (Pattachitra) and Tamil Nadu (Thanjavur). Assam contributes bamboo crafts and MP tribal sculpture (not paintings). 'Paintings from four States' is incorrect. ✓ (this is the wrong statement)

Option B: 'stepping through the grand portal was like entering another world' — correct.

Option C: 'showcased in these living galleries' refers to the Angans/zones — correct.

Option D: 'These thematic zones, inspired by iconic temples' — correct.

— Option A

From $A+B+C+D=17$ and $A+C=6$: $B+D=11$.

Using $P=R$ and $Q=S$:

• $P+2Q+D=15$

• $2P+Q+B=17$

Add the two equations:

• Left side: $3P+3Q+B+D$

• Right side: $32$

Since $B+D=11$:

• $3(P+Q)+11=32$

• $3(P+Q)=21$

• $P+Q=7$

Check options:

• A: $B+D=11 > P+Q=7$, so $B+D$ is NOT less. ✗

• B: $P+Q=7 > A+C=6$ ✓

• C, D: $P$ and $Q$ individually are not determined.

— Option B

QUEUE has letters Q, U, E, U, E.

Since Q must be immediately followed by U, glue them as a single block $[QU]$.

Remaining entities: $[QU], E, U, E$ — total 4 units, with $E$ repeated twice.

Arrangements: $\dfrac{4!}{2!} = \dfrac{24}{2} = 12$.

— Option D

A common landing mark $p$ satisfies $p-199$ divisible by $\text{lcm}(4,6,5)=60$.

So $p = 199 + 60k$ for $k=0,1,2,\ldots$

Marks between 195 and 1000:

• $k=0$: 199 ✓

• $k=13$: $199+780=979$ ✓

• $k=14$: $199+840=1039 > 1000$ ✗

Valid $k$: $0,1,2,\ldots,13$ — total 14 marks.

— Option D

1. Passage says sport should be seen as 'a way of life' and is not merely about medals or quota jobs. Supported. ✓

2. Passage says children are inactive and rarely leave the house, and that sport can help tackle these problems. Supported. ✓

3. Passage expressly refers to 'getting jobs through a sports quota'. This supports the limited point that sport-related achievement can help in quota-based employment. ✓

4. Passage says sport teaches time management, fitness, teamwork and coordination. So the claim that such skills cannot be learnt through sport is not supported. ✗

Valid: 1, 2, 3.

— Option D

1. Passage only reports parents' observation that children are inactive — no claim about parents' encouragement. ✗

2. 'From sport you can learn time management, fitness, teamwork, coordination' — life skills supported. ✓

3. 'a path to building a healthier, happier society' — sport as a way of life evolves society. ✓

Correct: 2 and 3.

— Option B

Let $f$ = forward jumps, $b$ = backward jumps.

• $f + b = 31$

• $5f - 2b = 15$

Substitute $b = 31 - f$:

• $5f - 2(31-f)=15$

• $7f - 62 = 15$

• $7f = 77$

• $f = 11$

So $b = 20$.

Difference: $|f - b| = |11 - 20| = 9$.

— Option D

Build the arrangement clockwise.

• $Q$ has $P$ and $R$ as neighbours.

• $S$ has $T$ and $R$ as neighbours.

• So $R$ is between $Q$ and $S$.

• Clockwise from $P$, $R$ comes before $T$.

• Hence the order is $P, Q, R, S, T$.

Place them at seats 1–5: $P=1, Q=2, R=3, S=4, T=5$.

$W$ is opposite $S$, so $W$ is at seat 8. Then $V$ must be between $U$ and $W$, so seats 6, 7, 8 are $U, V, W$.

From $Q$ clockwise before meeting $W$, the persons crossed are:

• $R$ at seat 3

• $S$ at seat 4

• $T$ at seat 5

• $U$ at seat 6

• $V$ at seat 7

So $Q$ crosses 5 persons.

— Option A

Each $2.5\times 8$ rectangle, having two sides on table edges meeting at a corner, must sit with $2.5$ along the $6$-side and $8$ along the $10$-side (since $8>6$ won't fit).

Four corner positions available; choose 2 of 4: $\binom{4}{2}=6$ pairs.

Check overlaps (table coords $x\in[0,10], y\in[0,6]$):

• Bottom-left occupies $x\in[0,8], y\in[0,2.5]$

• Bottom-right: $x\in[2,10], y\in[0,2.5]$ — overlaps BL on $x\in[2,8]$ ✗

• Top-left: $x\in[0,8], y\in[3.5,6]$

• Top-right: $x\in[2,10], y\in[3.5,6]$ — overlaps TL ✗

Invalid pairs: (BL,BR) and (TL,TR). Other 4 pairs are valid (y-ranges disjoint).

Valid patterns: 4.

— Option B

1. Direct quote: 'If oil prices fall, output prices do not and quantities remain the same, real value added should not change' — exactly statement 1. ✓

2. Passage says 'Most countries use... double deflation' (not 'all'), and India is explicitly cited as NOT doing so. 'Universally adopted' is contradicted. ✗

— Option A

1. Passage: India's same-deflator method makes 'it... look as if the manufacturer had become more productive' — i.e., the strategy makes performance appear better. ✓

2. 'Double deflation' is defined in the passage as input and output prices deflated separately, NOT both against a single input price. Statement 2 inverts the definition. ✗

— Option A

The string has 15 characters. A repeating block must divide 15: candidates are lengths 3 or 5.

Length-3 block: positions 1,4,7,10,13 would all match — known values $\{\times, a, \times, \times, b\}$ force $a=b$. ✗

Length-5 block: check positions known.

• Pos 4, 9, 14 = $a, a, a$ ✓

• Pos 2, 13 give pos 2 = $b$; pos 13 of block = $b$ ✓

• Reading the known letters back: block must be $a, b, b, a, b$ (i.e., $abbab$).

Apply pattern $abbab$:

• Positions 1–5: $a, b, b, a, b$

• Positions 6–10: $a, b, b, a, b$

• Positions 11–15: $a, b, b, a, b$ ($\to$ ends ...$bab$ ✓)

Positions 7, 8 = $b, b$.

— Option D

Factorise the RHS.

• $75=3\cdot 5^2$

• $75^{25}=3^{25}\cdot 5^{50}$

• $25^{32}=5^{64}$

• $32^{75}=2^{375}$

So RHS has:

• power of $2$: $375$

• power of $5$: $114$

LHS is $10^{m+3}\cdot n$. Since $n$ is not divisible by 10, all possible paired factors of 2 and 5 must be in $10^{m+3}$.

Therefore:

• $m+3=114$

• $m=111$

Check: $n=2^{261}\cdot 3^{25}$, which is not divisible by 10. ✓

— Option B

Same direction: relative speed $=100-4=96$ km/h.

Convert speed:

• $96$ km/h $=96000/3600$ m/s

• $=26.67$ m/s approximately

Train length:

• $L=26.67\times 15$

• $L=400$ m

Opposite direction: relative speed $=100+4=104$ km/h.

Convert speed:

• $104$ km/h $=104000/3600$ m/s

• $=28.89$ m/s approximately

Time:

• $t=400/28.89$

• $t\approx 13.85$ s

So the closest option is 13.85.

— Option C

Statement I: Let $f(t)=t+1/t$.

$f$ decreases on $(1/3,1)$ and increases on $(1,2)$, so it is not monotonic on the whole interval.

Counter-example:

• $x=0.5, y=1.5$

• $f(x)=2.5$

• $f(y)\approx 2.17$

• So $f(x)>f(y)$ ✗

I is NOT always correct.

Statement II: Let $g(t)=\sqrt{1+t^2}/t$.

Since $t>0$:

• $g(t)^2=1+1/t^2$

• as $t$ increases, $1/t^2$ decreases

• hence $g(t)$ strictly decreases

Therefore:

• $x

• $g(x)>g(y)$

• so statement II holds ✓

— Option B

The intended option works only if a measured amount may also be poured back into the tank. Under that interpretation:

Construction:

• Take $100$ L three times: $300$ L.

• Return $1$ L twice: subtract $2$ L.

• Net amount: $298$ L.

• Total measurements: $5$.

Why four is impossible: four signed measurements cannot make $298$.

• To reach $298$, at least three $100$ L measurements are needed.

• Then the fourth measurement would have to be $-2$ L, which is unavailable.

So the intended answer is Option B.

Question-quality note: If only one-way drawing from the tank is allowed, the minimum is $12$ measurements, which is not listed.

Let $a,b,c,d$ = number of 10, 5, 2, 1 kg weights. $10a+5b+2c+d=20$, $d\in\{8,9,10,11\}$.

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

• $(1,0,1)$: $10+0+2=12$ ✓

• $(0,0,6)$: $12$ ✓

• $(0,2,1)$: $10+2=12$ ✓ → 3 ways

$d=9$: $10a+5b+2c=11$

• $(0,1,3)$: $5+6=11$ ✓ → 1 way

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

• $(1,0,0)$ ✓

• $(0,2,0)$ ✓

• $(0,0,5)$ ✓ → 3 ways

$d=11$: $10a+5b+2c=9$

• $(0,1,2)$: $5+4=9$ ✓ → 1 way

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

— Option B

After 3 cuts producing identical pieces, $x_3\in\{4, 6, 8\}$:

• 3 parallel cuts $\Rightarrow$ 4 identical slabs.

• 2 parallel + 1 perpendicular at midpoint $\Rightarrow$ 6 identical boxes.

• 3 mutually perpendicular through midpoints $\Rightarrow$ 8 identical small cubes.

Check options:

• $x_4=5$: from 4 slabs, 4th cut parallel within one slab adds 1 piece $\to 5$ ✓

• $x_4=8$: from 4 slabs, 4th cut perpendicular slices all $\to 8$ ✓

• $x_4=12$: from 8 small cubes, 4th cut parallel to an existing plane at a new position cuts 4 cubes $\to 8+4=12$ ✓

• $x_4=16$: would need a single plane to slice ALL 8 small cubes. In a $2\times 2\times 2$ arrangement, a flat plane cannot pass through the interior of all 8 octant cubes simultaneously (the diagonally opposite octants force contradictory inequalities on the plane). Not achievable. ✗

— Option A

Let $n$ = number of students; original average = $x$.

First correction: $0 \to 100$ increases total by 100. Average rises by 4: $\dfrac{100}{n}=4 \Rightarrow n=25$.

Second correction: $81 \to 56$ decreases total by 25. Change in average: $-\dfrac{25}{25}=-1$.

Net change from original $x$ to final $y$: $+4 - 1 = +3$.

$\Rightarrow y - x = 3$.

— Option B

1. Passage states the viruses belong to 'two very different virus families with fundamentally different characteristics and epidemiology'. Calling them 'similar' contradicts this. ✗

2. Passage repeatedly emphasises fundamental differences — characteristics, epidemiology, severity (esp. long term), affected segments. This supports differential handling. ✓

— Option B

1. Passage discusses neither methodologies nor any prescription for objective analysis. ✗

2. Passage primarily contrasts the two viruses — but emphasises differences, with no claim of similarities. Calling intent 'similarities and differences' overstates. ✗

3. Passage mentions no remedies for HMPV. ✗

None of the listed combinations (A, B, C) accurately captures the writer's intent.

— Option D

1. HMPV exposed 'for decades' and 'well-studied' vs SARS-CoV-2 first detected in 2019 $\Rightarrow$ HMPV has longer documentation. ✓

2. Different families, different characteristics; 'affected population segments do not fully overlap' $\Rightarrow$ different outcomes. ✓

3. 'different severity of symptoms, particularly over the long term' and 'no long-term post-viral symptoms' for HMPV $\Rightarrow$ long-term impacts dissimilar. ✓

4. 'strong seasonality seen in HMPV, unlike SARS-CoV-2' $\Rightarrow$ seasonality is NOT a common factor. ✗

Valid: 1, 2, 3.

— Option B

Values $x, y, z$ are distinct, drawn from $\{2, 3, 4, 5\}$.

Maximum $M$: maximise $x\times y$ first.

• Largest product of two distinct values: $4\times 5=20$.

• $z$ from remaining $\{2,3\}$; max $z=3$.

• $M = 20 + 3 = 23$.

Minimum $N$: minimise $x\times y$ first.

• Smallest product of two distinct values: $2\times 3=6$.

• $z$ from remaining $\{4,5\}$; min $z=4$.

• $N = 6 + 4 = 10$.

$M - N = 23 - 10 = 13$.

— Option C

Let $y=kx^2$ and $x=c/z$.

Find $c$:

• $5=c/(7/25)$

• $5=25c/7$

• $c=7/5$

So $x=7/(5z)$.

Find $k$:

• $50=25k$

• $k=2$

So $y=2x^2$.

If $y=98$:

• $2x^2=98$

• $x^2=49$

• $x=7$

From $x=7/(5z)$:

• $7=7/(5z)$

• $5z=1$

• $z=1/5$

— Option B

Statement 1: The passage only says AI tools have *started* offering precision farming in the US/EU. It makes no claim that Western productivity has marched ahead *because of* AI economies of scale. ✗

Statement 2: Directly supported — the passage prescribes "Indian languages-based AI tools for smallholder farmers" and "affordable AI solutions" via AgTech partnerships. ✓

Statement 3: Directly supported — "India is starting to deploy AI for critical use cases such as weather forecasting, pest detection... However, penetration is limited". ✓

Only 2 and 3 are valid.

— Option D

Statement 1: Passage notes tech-savvy farmers are early adopters of AI, but says nothing about them *driving* AgTech companies. NOT correct. ✗

Statement 2: Passage explicitly recommends "disseminate AI-based advisory services through government programmes". Correct. ✓

Statement 3: Passage says generative AI has only *started* offering precision farming at scale — no claim of *replacing* traditional practices. NOT correct. ✗

Statement 4: Passage confirms "integrating large datasets to provide real-time agronomic insights". Correct. ✓

NOT correct: 1 and 3.

— Option A

$P=4$ and $R=1$ are fixed; ranks $\{2,3,5\}$ go to $Q,S,T$ with $Q\ne 5$, $S\ne 2$, $T\ne 3$.

Statement I: Assume $S=3$. Then $Q,T\in\{2,5\}$. Since $Q\ne 5$, $Q=2$ and $T=5$. So $T=5$, not $2$. Incorrect. ✗

Statement II: Assume $Q=3$. Then $S,T\in\{2,5\}$. Since $S\ne 2$, $S=5$ and $T=2$. So $T=2$, not $5$. Incorrect. ✗

Neither statement holds.

— Option D

Try a layout with $Q3$ at position 3 (middle), so $P1=3$.

• Place $Q5=2$ (adjacent to $Q3$). ✓

• Let $Q4=1$, $Q1=4$, $Q2=5$.

• $P4$ must match $Q1$ or $Q2$: choose $P4=4$. ✓

• $Q4$ must match $P3$ or $P5$: place $P3=1$. ✓

• Remaining slots 2 and 5 for $P2,P5$: set $P5=2$, $P2=5$.

Check: $P1{=}Q3{=}3$, $P4{=}Q1{=}4$, $Q4{=}P3{=}1$, $Q3,Q5$ at 3,2 adjacent. ✓

In this layout, I ($Q3$ at middle) ✓ and II ($P2$ at position 5, extreme) ✓ both hold.

— Option C

Deduce the seating:

• $B\ne Y$ and $B\ne$ car of $A$ ($X$), so $B\in Z$.

• $E$ is the *only* companion of $G$, so $\{E,G\}$ alone occupy one car. Not $X$ (would join $A$), not $Z$ (already has $C,B$). So $\{E,G\}=Y$.

• $X=\{A,D,F\}$ (the remaining), $Z=\{C,B\}$.

Check the claims:

• I. $X$ has 3, $Y$ has 2, $Z$ has 2 — no one alone. ✓

• II. "Only $D$ travels with $F$" — but $A$ is also with $F$ in $X$. ✗

• III. $Z$ has just $B,C$ — so $C$ is the only one with $B$. ✓

I and III only.

— Option B

Rules and contrapositives:

• $\neg X \Rightarrow \neg Z$, so $Z \Rightarrow X$.

• $\neg Y \Rightarrow W$, so $\neg W \Rightarrow Y$.

• $W \Rightarrow \neg X$, so $X \Rightarrow \neg W$.

Statement I: If $X$ is correct, then $\neg W$ is true. From $\neg W \Rightarrow Y$, $Y$ is correct. ✓

Statement II: If $Z$ is correct, then:

• $Z \Rightarrow X$

• $X \Rightarrow \neg W$

• $\neg W \Rightarrow Y$

So $Y$ is forced to be correct. The claim "not necessary" is false. ✗

— Option A

Take $X$'s speed as $7$ laps/unit, $Y$'s as $5$ laps/unit. Total run time = $3$ units (X does 21 laps).

Phase 1 ($t\in[0,1]$, same direction):

• Relative speed $= 7-5 = 2$ laps/unit.

• Meetings = lap-difference at $t=1$, i.e. $2$. (Meetings at $t=0.5$ and $t=1$.)

Phase 2 ($t\in[1,3]$, $Y$ reverses, speed $=10$):

• Relative speed (opposite) $= 7+10 = 17$ laps/unit.

• Over $2$ units they cover $34$ relative laps $\Rightarrow 34$ meeting events.

• The 34th meeting is at $t=1+\frac{34}{17}=3$ — the *stopping instant*, excluded by "before they finally stopped".

Total counted: $2 + 33 = 35$.

— Option A

Let $c$ = correct and $w$ = wrong.

• Scheme 1: $5c - 2w = 69$

• Scheme 2: $4c - w = 84$

• From Scheme 2: $w = 4c - 84$

Substitute in Scheme 1:

• $5c - 2(4c-84) = 69$

• $5c - 8c + 168 = 69$

• $-3c = -99$

• $c = 33$

Then $w = 48$.

Total $= 33 + 48 = 81$.

— Option B

Conclusion 1: Passage says transfer to adult system is permitted for *heinous* offences (min. 7 yrs). It explicitly notes a *serious* offence "does not merit the transfer of a case to the adult criminal justice system". So 1 inverts the rule. ✗

Conclusion 2: Directly supported — "an offence that has no minimum sentence, but has a maximum sentence of seven years or more as a serious offence". The classifier is the *maximum* sentence. ✓

— Option B

Statement 1: Passage: "culpable homicide... are not heinous offences because they do not have a prescribed minimum punishment." No minimum $\Rightarrow$ not heinous under 2015 Act. ✓

Statement 2: Passage: 2021 amendment categorises an offence "that has no minimum sentence, but has a maximum sentence of seven years or more as a serious offence". Exact restatement. ✓

Both correct.

— Option C

Let $d$ be the distance in metres. Take the explosion time as $0$.

• Sound reaches the camp at $d/300$ seconds.

• Drone 1 then takes $d/30$ seconds.

• Drone 2 also takes $d/30$ seconds after picture 1 is received.

The two pictures are received 3 minutes apart:

• $d/30 = 180$

• $d = 5400$ m

Time from explosion to picture 1:

• Sound time $= 5400/300 = 18$ s

• Drone time $= 5400/30 = 180$ s

• Total $= 198$ s $= 3$ min $18$ s

Explosion time = 5:02:00 PM - 3 min 18 s = 4:58:42 PM.

— Option C

Unit digit of $6^{129}$: Any positive power of 6 ends in $6$. So unit digit $=6$.

Unit digit of $7^{307}$: Cycle of length 4 — $7, 9, 3, 1$.

• $307 \bmod 4 = 3$

• So unit digit of $7^{307}$ is the 3rd in cycle $= 3$.

Product unit digit: $6 \times 3 = 18 \Rightarrow$ unit digit $= 8$.

— Option C

Take salary $=100$. Saving $=10$, expenditure $=90$.

After change:

• New salary $= 112$.

• New expenditure $= 99$.

• New saving $= 13$.

Change in saving:

• Increase $= 13 - 10 = 3$

• Percentage increase $= 3/10 \times 100\%$

• Saving increases by $30\%$.

— Option B

• $P$ has one son and one daughter.

• $Q$ and $R$ are $P$'s children, so they are those two children.

• $S$ is the mother of $T$, so $S$ is female.

• $S$ is $R$'s spouse, so $R$ is male.

• Therefore, $R$ is $P$'s son and $Q$ is $P$'s daughter.

• $S$'s husband is $R$; $Q$ is $R$'s sister.

Hence $Q$ is the sister of $S$'s husband.

— Option D

Three-digit range: $100 \le 2^n \le 999$.

• $2^6 = 64$ — too small ✗

• $2^7 = 128$ ✓

• $2^8 = 256$ ✓

• $2^9 = 512$ ✓

• $2^{10} = 1024$ — four digits ✗

Exactly 3 three-digit powers of 2.

— Option C

Inference 1: Opening line — "The key source of the battle for clean skies and clear lungs is the fuel we burn". ✓

Inference 2: Passage links effectiveness of bans to availability of *alternatives* ("people need alternatives for a ban to be effective"), not to public will or conviction. ✗

Inference 3: Supported by "The Supreme Court agreed that fiscal measures were needed to keep clean fuel cheaper than dirty fuel" — judicial approval for fiscal pricing intervention. ✓

Only 1 and 3.

— Option C

Statement 1: Passage says plants "would clean up *or* shut down" — not a summary shutdown. ✗

Statement 2: Directly stated — "When diesel buses were stopped, CNG supply had to be assured". ✓

Statement 3: Passage credits the Delhi Government with the coal ban (later extended to NCR), not the Supreme Court. ✗

Only 2.

— Option C

The inclusions are:

• Red $\subset$ Blue

• Blue $\subset$ Green

• Green $\subset$ Yellow

By transitivity:

• P. Blue $\subset$ Yellow. ✓

• Q. Red $\subset$ Green. ✓

• R. Red $\subset$ Yellow. ✓

All three follow.

— Option D

Range: $10$ to $99$.

5 at tens place:

• $50$ to $59$

• This gives $10$ occurrences.

5 at units place:

• $15, 25, 35, 45, 55$

• $65, 75, 85, 95$

• This gives $9$ occurrences.

$55$ is counted once in each place, correctly, because it has two 5's.

Total: $10 + 9 = 19$.

— Option B

Convert each leg to minutes ($1$ hour $= 60$ min).

$X$'s time:

• $1$ km at $5$ km/h $= 12$ min

• $2$ km at $10$ km/h $= 12$ min

• $3$ km at $4$ km/h $= 45$ min

• Total $= 69$ min

$Y$'s time:

• $1$ km at $4$ km/h $= 15$ min

• $2$ km at $10$ km/h $= 12$ min

• $3$ km at $5$ km/h $= 36$ min

• Total $= 63$ min

Difference: $69 - 63 = 6$ min. $Y$ finishes 6 min earlier.

— Option D

Strategy (worst-case 2 weighings):

Weighing 1: Split 7 cubes into groups of $3, 3, 1$. Put the two groups of 3 on the pans.

• Balanced $\Rightarrow$ the set-aside cube is the light cube. ✓

• Unbalanced $\Rightarrow$ the light cube is in the lighter group of 3.

Weighing 2: Needed only in the unbalanced case. Weigh 2 of the 3 suspects, one on each pan.

• Balanced $\Rightarrow$ the third suspect is the light cube.

• Unbalanced $\Rightarrow$ the lighter pan has the light cube.

Thus 2 attempts are sufficient in the worst case, and 1 attempt cannot distinguish among 7 cubes with certainty.

— Option A

The passage closes with the salutary view that "Human agents could be freed up to spend more time on creative and rewarding tasks" — i.e., human work is *redirected* to more meaningful tasks.

• A. Opposite of passage; the passage says effects "may be salutary". ✗

• B. Passage doesn't compare *extent* of redundancy. ✗

• C. Captures the central salutary point — redirection to fulfilling tasks. ✓

• D. Passage notes previous waves *also* stoked job-loss concerns. ✗

— Option C

1. "Previous waves of customer-service technology... stoked concerns of job losses" — supports the claim of past fears. ✓

2. "only for them to fail to materialise" + "Human agents could be freed up to spend more time on creative and rewarding tasks" — alternative engagement, not job loss. ✓

3. Passage describes effects as potentially *salutary*; it does NOT call outcomes inevitably stressful. ✗

Only 1 and 2.

— Option D

1. Passage contrasts Rai Sahib with "selfish Zamindars" — implying not all landlords have goodness. ✗

2. Passage says "Cattle from the *nearby villages* came to the common ground" — usage is broader than one village; no claim of intent. ✗

3. Passage: "the Rai Sahib still held to the old values, feeling that any landlord who didn't look after his tenants was less than human" — tradition correlates with tenant care. ✓

4. Passage: cool freshness *now* (morning), "scorching 'loo' wind the rest of the day" — later wind is hotter, not cooler. ✗

Only 3.

— Option C

1. Passage discusses *one* plot (grazing land) — it doesn't claim *all* of Rai Sahib's landholdings are unfarmed. The sweeping claim isn't supported. NOT correct. ✗

2. Passage clearly contrasts Rai Sahib with "selfish Zamindars" — he is not greedy. NOT correct. ✗

3. Passage says "a number of farmers... had offered a good price" — so it *can* be ascertained he could have earned money. The denial is wrong. NOT correct. ✗

4. Passage shows he refused money to protect tenants' grazing — clearly correct. ✓

NOT correct: 1, 2 and 3.

— Option D

Track coordinates with east $=+x$, north $=+y$. Start at $(0,0)$, facing East.

• Walk 4 km East: $(4, 0)$, facing East.

• Turn right $\Rightarrow$ South, walk 3 km: $(4, -3)$, facing South.

• Turn left $\Rightarrow$ East, walk 2 km: $(6, -3)$, facing East.

• Turn left $\Rightarrow$ North, walk 3 km: $(6, 0)$, facing North.

Distance from origin: $\sqrt{6^2+0^2} = 6$ km. Final facing: North.

— Option D

Generate term by term:

• $a_1 = 4$

• $a_2 = 7$

• $a_3 = 4 + 7 = 11$

• $a_4 = 7 + 11 = 18$

• $a_5 = 11 + 18 = 29$

• $a_6 = 18 + 29 = 47$

— Option D

Equalise female counts. Use LCM$(10,5)=10$.

• $A$: $M:F = 13:10$ — take $M_A=13, F_A=10$, total $=23$.

• $B$: $M:F = 7:5$ — scale by 2: $M_B=14, F_B=10$, total $=24$.

With $F_A = F_B = 10$:

Ratio of totals $= 23 : 24$.

— Option B

For any two numbers:

• HCF $\times$ LCM $= a \times b$

Given that this product is the cube of one of the numbers. If it is $a^3$, then:

• $ab = a^3$

• $b = a^2$

If it is $b^3$, the same argument gives $a = b^2$. So the two numbers are of the form $n$ and $n^2$.

Statement I: Difference $= n^2 - n = n(n-1)$. Product of consecutive integers is always even. ✓

Statement II: One number is $n^2$, a perfect square. ✓

Both correct.

— Option C

Digits from left: $4, x, 5, y, 7, 9, 0$.

Divisibility rule for 11: alternating sum must be divisible by $11$.

• Odd-position digits: $4 + 5 + 7 + 0 = 16$

• Even-position digits: $x + y + 9$

• Alternating sum $= 16 - (x+y+9)$

• So alternating sum $= 7 - (x+y)$

For divisibility by $11$:

• $7 - (x+y) \equiv 0$

• $x+y \equiv 7 \pmod{11}$

Since $0 \le x+y \le 18$, possible sums are $7$ or $18$.

In both cases, the remainder on division by $11$ is 7.

— Option D

Statement 1: The passage calls the nap *"quietly radical"* — a surrender to limits in defiance of *"society's relentless performance metrics"*. So a mild surrender against societal expectations is indeed framed as radical. ✓

Statement 2: *"Not to laziness, but to limits. To the body's quiet wisdom..."* — the passage explicitly separates submitting to limits (age etc.) from laziness. ✓

Statement 3: *"discovered a leading man, not of action, but of rest"* — the cinematic phrase is repurposed to cast Grant as a hero of rest rather than action. ✓

All three are supported by the passage. — Option C

Statement 1: The passage says Grant was watching the men's singles quarterfinals, not the semifinals. ✗

Statement 2: The phrase *"unexpected ace"* does use tennis language for Grant's nap. But the added claim that the act is viewed with *opprobrium* is not supported by the passage. The passage says he *"deserves empathy"* and even says *"praise be to Grant"*. ✗

Statement 3: *"To the body's quiet wisdom over society's relentless performance metrics"* directly places bodily wisdom above social demands. ✓

Only 3 is supported. Option B

Statement 1: Passage says followers' per capita income is *"unchanged"* in the breakaway phase — not merely slow. ✗

Statement 2: Passage says followers *"adopt the new technology and close their per capita income gap"*. It does not say leaders stagnate; the gap closes because followers rise. ✗

Statement 3: *"participants try to extract the remaining benefits from an increasingly exhausted technology, leaders and followers have similar per capita incomes"* — directly supported. ✓

Statement 4: Chronology — scavenging (3 MYA) $\to$ hunting (1 MYA) $\to$ farming (12 KYA) $\to$ industrial (200 yrs). Industrial followed scavenging ✓, scavenging preceded hunting ✓, hunting followed by farming ✓.

Only 3 and 4. — Option A

Statement 1: The model divides economic eras into three phases. It divides economic entities into two categories: leaders and followers. So this misstates the passage. ✗

Statement 2: The passage speaks of a labour-productivity gap with the technology leader, i.e. a productivity gap between countries. It is not a gap between labour and productivity. ✗

Statement 3: Technology leaders are the entities associated with the new technology in each era, while followers adopt it later in the catch-up phase. So leaders arrive earlier at those eras. ✓

Statement 4: The passage ranges from roughly 3 million years ago to the present. The stated figure wrongly adds separate date markers:

• 3,000,000 years

• 1,000,000 years

• 12,000 years

• about 200 years

That sum is not the time period covered. ✗

Only 3 is correct. Option C

Required copper in $R$ is $50\%$.

Let $P:Q = x:y$. Copper balance:

• Copper from $P$: $0.2x$

• Copper from $Q$: $0.6y$

• Copper in mixture: $0.5(x+y)$

So:

• $0.2x + 0.6y = 0.5(x+y)$

• $0.1y = 0.3x$

• $y = 3x$

Therefore $P:Q = 1:3$.

Alligation check: distances from $50\%$ are:

• for $P$: $30$

• for $Q$: $10$

So $P:Q = 10:30 = 1:3$. ✓

Option A

Let $A = 10a+b$ and $B = 10b+a$.

Since $B$ is also two-digit, $b \ne 0$. Also $A>B$, so $a>b$.

• $A-B = 9(a-b)$

• For a multiple of $27$, $a-b$ must be a multiple of $3$

• Thus $a-b$ can be $3$, $6$, or $9$

Case $a-b=3$: $b=1$ to $6$ gives 6 values. ✓

Case $a-b=6$: $b=1$ to $3$ gives 3 values. ✓

Case $a-b=9$: only $(a,b)=(9,0)$, but then $B=09$, not a two-digit number. ✗

Total $= 6+3 = 9$.

Option B

Pattern in ZERO → ADSN:

• $Z \to A$: $+1$ (wrap-around)

• $E \to D$: $-1$

• $R \to S$: $+1$

• $O \to N$: $-1$

So positions alternate $+1, -1, +1, -1$.

Apply to STOP:

• $S \to T$ ($+1$)

• $T \to S$ ($-1$)

• $O \to P$ ($+1$)

• $P \to O$ ($-1$)

Result: TSPO.

— Option B

All tiles have width $3'$, so only the $100'$ length must be tiled by pieces of length $3$, $7$, and $11$.

Let the number of tiles be $n$.

Maximum: Start with all $3'$ pieces. Since $100$ is not divisible by $3$, $n=33$ is impossible. For $n=32$:

• take 31 tiles of $3'$

• take 1 tile of $7'$

Length $= 93+7 = 100$, so $x=32$. ✓

Minimum: Use as many $11'$ pieces as possible.

• With 9 tiles, maximum length is $99$, so impossible.

• With 10 tiles, replacing $11'$ by $7'$ or $3'$ reduces length by $4$ or $8$; a reduction of $10$ is impossible.

• With 11 tiles, the total remains odd, so it cannot be $100$.

• With 12 tiles, 8 tiles of $11'$ and 4 tiles of $3'$ give $88+12 = 100$.

Thus $y=12$.

Therefore $x-y = 32-12 = 20$.

Option A

Let original speed $= v$ km/h.

Case 1 (accident first, then defect):

• 200 km at $v$ → $200/v$

• 400 km at $v/2$ → $800/v$

• 200 km at $v/8$ (defect on already halved speed) → $1600/v$

• Total $= 2600/v$

Case 2 (defect first, then accident):

• 200 km at $v$ → $200/v$

• 400 km at $v/4$ → $1600/v$

• 200 km at $v/8$ (accident halves $v/4$) → $1600/v$

• Total $= 3400/v$

Difference: $3400/v - 2600/v = 800/v = 4$

$\Rightarrow v = 200$ km/h.

— Option A

Weighted score formula: $0.2c_1+0.3c_2+0.5c_3$.

Scores:

• C1: $1.0+1.2+3.0 = 5.2$

• C2: $0.8+1.8+2.5 = 5.1$

• C3: $0.6+0.6+4.0 = 5.2$

• C4: $1.8+1.2+1.5 = 4.5$

• C5: $1.6+2.4+1.0 = 5.0$

Sorted scores: $5.2, 5.2, 5.1, 5.0, 4.5$.

The third-highest score is $5.1$. A cut-off of $5.1$ selects exactly three candidates.

Option A

Relation 1 (S3 counterpoint to S1): S1 calls protection an existential need; S3 only highlights extremism *within* activism — it does not deny the need to protect the environment. Not a counterpoint. ✗

Relation 2 (S3 unconnected): S3 explicitly references *climate change deniers*, who feed off S2's *"no consensus on extent of human contribution"*. Clearly connected. ✗

Relation 3 (S2 → S3): Lack of scientific consensus (S2) gives deniers a platform; this absence of settled science could plausibly fuel both alarmism and the deniers' use of it — so S2 could be a reason behind the dynamic in S3. ✓

(Also, 2 and 3 are mutually contradictory — if S2 causes S3, S3 cannot be unconnected.)

Only 3 holds. — Option D

• A. Captain & team members: dressing-room camaraderie (informal) but instructions are decisive — Informal and firm (1).

• B. Judge & lawyers in court: courtroom protocol with binding rulings — Formal and firm (4).

• C. VC & Deputy Registrar: hierarchical office relationship but consultative on administration — Formal and open-ended (3).

• D. Peers & coworkers: equals, free exchange — Informal and open-ended (2).

Matching: A-1, B-4, C-3, D-2. — Option B

• A. Memorandum (ज्ञापन): formal internal instruction — intimate a directive (3).

• B. Flyer: mass-distributed leaflet to general audience — disseminate non-targeted information (4).

• C. Bcc (blind carbon copy): recipients hidden from each other — inform confidentially (2).

• D. Minutes: written record of a meeting — record decisions (1).

Matching: A-3, B-4, C-2, D-1. — Option C

• Action 1: *Directing* people to practise others' cultural rituals is coercive and may amplify discomfort/stereotyping. ✗

• Action 2: Letting conflicts "resolve themselves" is leadership abdication — issues fester. ✗

• Action 3: Encouraging perspective-taking builds empathy and is the standard remedy for cross-cultural friction. ✓

Only 3 is appropriate. — Option C

• A. Semantic (अर्थगत): about meaning of words — Misunderstanding a word (2).

• B. Cognitive (संज्ञानात्मक): processing capacity limits — Information overload (4).

• C. Organisational (संगठनात्मक): structural channels failing — Lack of feedback (1).

• D. Affective (भावात्मक): emotional/feeling-based — Fear of social stigma (3).

Matching: A-2, B-4, C-1, D-3. — Option D

Goal: assess *feasibility* of a family resort — i.e., is there demand at a viable price-point and the right scale.

• 1. Size of family: decides room mix, capacities, family-suite design — directly feasibility-relevant. ✓

• 2. Budget: willingness-to-pay drives pricing and viability. ✓

• 3. Number of earners: intrusive proxy, already captured by budget — not appropriate on a visitor questionnaire. ✗

• 4. Food allergies/dietary restrictions: operational menu detail; relevant only after resort exists, not for feasibility study. ✗

Only 1 and 2. — Option C

• 1. A circular letter, by definition, goes to a group/wide audience. ✓

• 2. Circulars are standardised with identical content, not customised. ✗

• 3. Appraisal/increment letters are confidential and individual — sent as personalised letters, not circulars. ✗

• 4. Circulars are more cost-effective (one drafting, mass distribution) than individualised letters. ✗

Only 1. — Option D

Take total capital $=1$ and total time $=1$.

Weights:

• $A$: capital $1/3$, time $1/3$, so weight $1/9$

• $B$: capital $1/4$, time $1/4$, so weight $1/16$

• $C$: remaining capital $5/12$, time $1$, so weight $5/12$

Using denominator $144$:

• $A = 16/144$

• $B = 9/144$

• $C = 60/144$

Ratio $A:B:C = 16:9:60$. Total parts $=85$.

$C$'s profit:

• one part $= 17000/85 = 200$

• $60$ parts $= 12000$

Option A

After $n$ replacements of $1$ L from a $10$ L container, fraction of $A$ left is $(1-\tfrac{1}{10})^n$.

• After 3 operations: $A$ fraction $= (0.9)^3 = 0.729$

• $A$ volume $= 7.29$ L; $B$ volume $= 2.71$ L

• $B\% = 27.1\% \approx 27$

— Option B

Earning $= 5d + 2k$, with constraint $k \le 100$. Target: $5d + 2k > 500$.

The distance per delivery is not specified. For the same number of deliveries, total distance can be small or large, as long as it does not exceed $100$ km.

Check A/B/C with a valid low-distance case, say $k=1$ km:

• $d=80, k=1 \Rightarrow 402 < 500$ ✗

• $d=85, k=1 \Rightarrow 427 < 500$ ✗

• $d=90, k=1 \Rightarrow 452 < 500$ ✗

Thus none of A/B/C definitely meets the target without knowing $k$. The data is insufficient.

Option D