UPSC Prelims 2025 CSAT (Paper II · Set A)

The passage frames a three-way conflict: conservation vs. livelihoods vs. commercial exploitation, and notes that deforestation harms both livelihoods and the environment ("floods, siltation problems and microclimatic instability").

• Option A: Passage stresses biodiversity matters; it never says concern can be dropped. ✗

• Option B: Passage does not invoke any "fundamental rights" framework. ✗

• Option C: Captures that livelihood pressure and ecosystem degradation reinforce each other, producing the very conflicts the passage describes. ✓

• Option D: Passage never calls for a complete stop; it only flags trade-offs. ✗

Option C

• Assumption I: Passage explicitly says commercial exploitation "has the potential to boost national income" — i.e., countries DO rely on ecosystems for income. Contradicts I. ✗

• Assumption II: Passage says nothing about inter-country resource-sharing. No basis. ✗

Neither assumption is supported.

Option D

Passage's punchline: "Moving early and gradually... A late, more chaotic transition is costlier." Timing is the central thread.

• Option A: Economies of scale are a mechanism, not the central message. ✗

• Option B: Too general; passage is specifically about green/renewable tech, not all modern tech. ✗

• Option C: Passage never compares large vs. small economies. ✗

• Option D: Captures the early-vs-late timing argument directly. ✓

Option D

• Assumption I: Passage states early path-dependent investment lets economies grow "without much disruption" while a late transition is costlier — implying timely action helps growth and avoids fiscal burden. Reasonable for a developing economy like India. ✓

• Assumption II: Passage uses solar as the "strongest evidence" of the falling-cost path-dependent pattern. If others replicate that pattern, an easier transition follows directly. ✓

Option C

Let the three factors be $N/a$, $N/b$, $N/c$. Then $1/a+1/b+1/c=1$, with $a,b,c$ distinct divisors of $N$.

The positive integer solutions of this reciprocal equation are:

• $(3,3,3)$

• $(2,4,4)$

• $(2,3,6)$

Only $(2,3,6)$ gives three distinct factors. Hence $N$ must be divisible by $6$.

Multiples of $6$ below $50$ are:

• $6,12,18,24$

• $30,36,42,48$

Each works by using $N/2$, $N/3$, and $N/6$.

Total $=8$.

Option C

$p-q=q-r \Rightarrow p+r=2q$ (AP with middle term $q$). Sum $=p+q+r=3q$. Primes $<20$: $\{2,3,5,7,11,13,17,19\}$.

For each prime $q$, check if some valid $(p,r)$ exists (including $p=q=r$):

• $q=2 \Rightarrow$ sum $=6$

• $q=3 \Rightarrow$ sum $=9$

• $q=5$: $(3,5,7)$ ✓ sum $=15$

• $q=7$: $(3,7,11)$ ✓ sum $=21$

• $q=11$: $(3,11,19),(5,11,17)$ ✓ sum $=33$

• $q=13$: $(7,13,19)$ ✓ sum $=39$

• $q=17 \Rightarrow$ sum $=51$

• $q=19 \Rightarrow$ sum $=57$

Distinct sums: $\{6,9,15,21,33,39,51,57\}=8$ values, which is more than 6.

Option D

Assume $p \le q \le r$.

• $p=1$ is impossible, because the first term is already $1$.

• Since $3/p \ge 1$, we get $p \le 3$.

So $p=2$ or $p=3$.

Case $p=3$:

• Need $1/q+1/r=2/3$.

• With $q,r \ge 3$, only $q=r=3$.

• Sum $=9$.

Case $p=2$:

• Need $1/q+1/r=1/2$.

• $q=3 \Rightarrow r=6$, sum $=11$.

• $q=4 \Rightarrow r=4$, sum $=10$.

• For $q \ge 5$, the left side is too small.

Distinct sums are $9,10,11$.

Option C

Successive backward gaps grow by $1$ each step: $0,1,2,3,4,5,6,7$ months back.

• Jan $\to$ Jan: $-0$ ✓

• Jan $\to$ Dec: $-1$ ✓

• Dec $\to$ Oct: $-2$ ✓

• Oct $\to$ X: $-3 \Rightarrow$ X $=$ July

• July $\to$ Mar: $-4$ ✓

• Mar $\to$ Oct: $-5$ ✓ (Feb, Jan, Dec, Nov, Oct)

• Oct $\to$ Y: $-6 \Rightarrow$ Y $=$ April

• Apr $\to$ Sep: $-7$ ✓

Option B

Y tied in $20\times0.9=18$ overs, so Y's actual rate is $N/18$.

If the rate were $50\%$ higher:

• new rate $=1.5\times N/18$

• new rate $=N/12$

Runs in $12$ overs at this rate are $N$. This is true for every $N$.

The second condition is an identity; it gives no fixed value of $N$.

Option D

One increase-decrease cycle has multiplier:

• $(100+k)(100-k)/10^4$

• $(10^4-k^2)/10^4$

There are two such cycles, so:

• divide $q$ by $p$

• square the above multiplier

After multiplying by $10^8$, the equality is:

• left side: $p(10^4-k^2)^2$

• right side: $q\times10^8$

This is exactly option A.

Option A

Passage lists state-specific challenges (coastal, drier districts, Himalayan, peninsular) and concludes "these various aspects need to be considered for developing clear and sustainable goals."

• Option A: "essentially" the State's responsibility — passage doesn't say only States; too narrow. ✗

• Option B: Passage offers a constructive way forward, not pessimism. ✗

• Option C: Passage doesn't single out the Union Government. ✗

• Option D: Calls for state/region-level strategies — exactly what passage urges. ✓

Option D

• Assumption I: Passage cites wind power (green energy) as an opportunity "for reducing emissions" — directly linking green energy to mitigation. Valid. ✓

• Assumption II: Passage lists challenges in coastal, drier and Himalayan regions equally — never says coastal/mountain effects are "much more severe." Invalid. ✗

Option A

Passage's key sentence: "insecurity, more than poverty, is the most acutely felt economic problem" and "Government policies are actually intended towards mitigating these insecurities."

• Option A: Conflates social inequality with the economic problem; passage keeps them separate. ✗

• Option B: Passage says insecurity is felt MORE than poverty, not that poverty is the cause. ✗

• Option C: "unmanageable" contradicts passage's mention of policies that address it. ✗

• Option D: Mirrors passage almost verbatim. ✓

Option D

• Assumption I: Passage states "even those just over the poverty line are subject to multiple economic insecurities." Directly supports I. ✓

• Assumption II: Passage keeps social inequality and economic insecurity as separate problems and never claims eradicating poverty brings social equality. Invalid. ✗

Option A

If the cube is cut into $a\times b\times c$ equal cuboids, the number of cuts is $a+b+c-3$.

For $abc=60$, the closest factor split gives the fewest cuts:

• $3\times4\times5=60$

• cuts $=3+4+5-3=9$

Other unordered splits give larger sums, e.g. $2\times5\times6$ gives $10$ cuts.

Statement I: $9$ cuts, so it is correct. ✓

Unpainted pieces are strictly internal:

• count $=(3-2)(4-2)(5-2)$

• count $=1\times2\times3=6$

Statement II: $6$ unpainted pieces, so it is correct. ✓

Option C

Pattern: $a*b=\sqrt{a^2+b^2}$, using Pythagorean triples.

• $7^2+24^2=625=25^2$ ✓

• $12^2+16^2=400=20^2$ ✓

• $16^2+63^2=4225=65^2$ ✓

Therefore $16*63=65$.

Option C

Old monthly fuel: $2200/16 = 137.5$ L.

Old cost: $137.5 \times 90 = ₹12{,}375$.

New price: $90 \times 1.10 = ₹99$/L.

New fuel at same cost: $12{,}375 / 99 = 125$ L.

New distance: $125 \times 16 = 2000$ km.

Reduction: $2200 - 2000 = 200$ km.

Option B

In every case, divisor minus remainder is $2$:

• $3-1=2$

• $5-3=2$

• $6-4=2$

• $9-7=2$

So $N+2$ is divisible by $3,5,6,9$, hence by $90$.

For the smallest 4-digit $N$:

• $N+2\ge1002$

• $90\times11=990$ is too small

• $90\times12=1080$ works

Thus $N=1080-2=1078$.

Option C

Statement I: There is no fixed order between $B$ and $E$.

Counter-example:

• $A=1,B=2,C=1$

• $D=10,E=5,F=4$

• $G=H=3$

This satisfies the whole chain, but $B

Statement II: From $S=T\le U=V$, we get only $S\le V$. Equality is possible.

Counter-example:

• $P=6,Q=R=5$

• $S=T=4,U=V=4$

• $W=3$

This satisfies the whole chain, but $S=V$, not $S

Option D

Product of odd numbers $1,3,5,\ldots,999$.

• Contains $5$ as a factor (also $15, 25, \ldots$) — divisible by $5$.

• All factors are odd — no factor of $2$.

So the product is odd and divisible by $5$ $\Rightarrow$ unit digit $=5$.

Quick check: $1 \times 3 \times 5 = 15$ (ends in 5); $15 \times 7 = 105$ (ends in 5); any odd $\times 5 \to$ ends in 5.

Option C

The passage says agriculture never had a high-growth phase, reached only $3.5\%$ CAGR in the 1980s, and then stagnated. In contrast, non-agricultural growth consistently increased from the 1980s to the 2000s.

• Option A: Closest corollary: economy-wide gains reached agriculture more slowly than other sectors. ✓

• Option B: Green revolution is not mentioned. ✗

• Option C: No comparison with other countries is made. ✗

• Option D: Rural-to-urban migration is not mentioned. ✗

Caveat: the passage does not explicitly use the phrase "economic reforms"; A is still the best available option.

Option A

Passage only describes the divergence — it never proposes or hints at any specific remedy.

• Assumption I: Commercial crops / corporate farming — not mentioned. ✗

• Assumption II: Free insurance / heavy subsidies — not mentioned. ✗

Neither can be validly drawn from the passage.

Option D

Passage argues the "single, linear" tradition narrative is contested; promoting handloom only as traditional makes large sections "choose to ignore it."

• Option A: Free handloom from the narrow tradition-only framing — matches the passage's diagnosis. ✓

• Option B: Passage challenges the tradition basis of State support, doesn't endorse it. ✗

• Option C: Modernization into organized industry — not the passage's argument. ✗

• Option D: Machine-made conversion contradicts the very idea of handloom. ✗

Option A

• Assumption I: Passage questions the *basis* of State support (single tradition narrative), not State involvement per se. Doesn't say State should withdraw entirely. ✗

• Assumption II: Passage attributes the ignoring to the framing/narrative, not to the products being unappealing. ✗

Neither valid.

Option D

Anything divisible by $9$ is divisible by $3$, so dropping $9$ doesn't change the set. Count those NOT divisible by $2, 3, 5, 7$ in $1$–$100$.

Inclusion–exclusion ($|.|$ = count in $1$–$100$):

• Singles: $50+33+20+14 = 117$

• Pairs ($6,10,14,15,21,35$): $16+10+7+6+4+2 = 45$

• Triples ($30,42,70,105$): $3+2+1+0 = 6$

• Quadruple ($210$): $0$

Divisible by at least one: $117 - 45 + 6 - 0 = 78$.

Not divisible: $100 - 78 = 22$.

Option C

Maximum value of $x+y$ is $8+7=15$.

Minimum value of $x-y$ is $4-7=-3$.

So the required ratio is $15/(-3)=-5$.

The listed numerical options are $6$, $15/2$, and $-15/2$. None equals $-5$.

Option D

Need $p^2+k$ to be prime and less than $30$.

Case $p=2$:

• $p^2=4$, so $k<26$.

• $k=3,7,13,19$ give $7,11,17,23$. ✓

• $k=5,11,17,23$ give $9,15,21,27$, all composite. ✗

Case $p=3$:

• $p^2=9$, so $k<21$.

• $k=2$ gives $11$. ✓

• Other possible prime $k$ values are odd, so $9+k$ is even and greater than $2$. ✗

Case $p=5$:

• $k=2$ gives $27$. ✗

• $k=3$ gives $28$. ✗

For $p\ge7$, $p^2>30$.

Distinct $k$ values are $2,3,7,13,19$. Total $=5$.

Option B

$1001 = 7 \times 11 \times 13$. Each element is a divisor of $1001$, so it corresponds to a subset of $\{7,11,13\}$ — $8$ divisors total.

For a 3-element set:

• $\mathrm{LCM}=1001$: union of subsets $= \{7,11,13\}$

• $\mathrm{HCF}=1$: intersection of subsets $= \emptyset$

Total 3-subsets of 8 divisors: $\binom{8}{3} = 56$.

Bad subsets (for each prime, all 3 lack it OR all 3 contain it):

• All 3 lack prime $i$: $\binom{4}{3} = 4$ ways; $3$ primes $\Rightarrow 12$.

• All 3 contain prime $i$: $4$ ways; $3$ primes $\Rightarrow 12$.

• Pairwise overlaps: choose 3 from 2 codes $= 0$.

Bad $= 24$; Good $= 56 - 24 = 32 > 8$.

Option D

Write the numbers as:

• $PQR=100P+30+R$

• $PS=10P+S$

• $PPT=110P+T$

Using $PQR-PS=PPT$ gives:

• $R-S=20P+T-30$

Case $P=1$: Then $R-S=T-10$, with $T=2,4,5$.

• $T=2$: only $(R,S)=(1,9)$, but $R=P$. ✗

• $T=4$: valid pair $(2,8)$. ✓

• $T=5$: valid pairs $(2,7)$ and $(4,9)$. ✓

Case $P\ge2$: Since $T\ge1$, $R-S\ge11$, impossible for digits.

Valid $(R,S)$ pairs are $(2,8)$, $(2,7)$, and $(4,9)$. Total $=3$.

Option B

Try the period ABBCCC (length 6) repeated three times $\to$ total 18 chars.

Lay out vs. given (blanks shown as `_`):

• Block 1 (1–6): A B B C C C $\to$ given A B _ C C _ $\to$ blanks at $3, 6$ = B, C

• Block 2 (7–12): A B B C C C $\to$ given A _ B C C C $\to$ blank at $8$ = B

• Block 3 (13–18): A B B C C C $\to$ given _ B B C _ C $\to$ blanks at $13, 17$ = A, C

Fills in order: B, C, B, A, C.

Option C

Passage's closing line: "Teaching has to be re-invented" — capping a call for new modes (critical thinking, learning by doing, etc.).

• Option A: Resources/funds not mentioned. ✗

• Option B: Passage warns against being a "servant of govt policy" but doesn't demand institutions be outside Government control. ✗

• Option C: "Done away with classroom" too extreme; passage urges reinvention, not abolition. ✗

• Option D: Matches "teaching has to be re-invented" directly. ✓

Option D

• Assumption I: Passage says "new demands of society and the future of work require..." — higher education must align with new developments. Valid. ✓

• Assumption II: Funds/allocation never mentioned. Invalid. ✗

Option A

The passage's central thrust is the inefficiency of plant-to-animal food conversion.

• Passage: only $16\%$ of calories fed to chickens are recovered; $5$–$7\%$ for large animals — direct evidence of low efficiency.

• Option A ✗ — the passage doesn't call for any public policy.

• Option B ✓ — directly mirrors the calorie-recovery data: animal-based food is far less efficient than plant-based.

• Option C ✗ — no recommendation to replace animal protein in diets.

• Option D ✗ — passage notes inequality exists, but never calls it *inevitable*.

— Option B

Assumption I: Food industries must align with changing societal needs.

• The passage describes inefficiency; it makes no prescription for what industries *should* do. Invalid.

Assumption II: Wealthier societies lose calories via indirect (animal-mediated) use of grain.

• Passage: wealthier societies feed grain to animals; only $5$–$16\%$ of those calories return as food. This is the indirect-utilization loss. Valid.

— Option B

Since $35=5\times7$, each factor of $35$ needs one $5$ and one $7$.

Factorize the product:

• $7=7^1$

• $343=7^3$

• $385=5\times7\times11$

• $1000=2^3\times5^3$

• $2401=7^4$

• $77777=7\times11111$

Powers of 7: $1+3+1+0+4+1=10$

Powers of 5: $0+0+1+3+0+0=4$

The smaller count is $4$, so the product is divisible by $35^4$ but not by $35^5$.

— Option B

Look at successive ratios (each term ÷ previous):

• $18/12 = 1.5$

• $36/18 = 2$

• $90/36 = 2.5$

Ratios form an AP with common difference $0.5$. Extending backwards:

• $12/X = 1.0 \Rightarrow X = 12$

• $X/24 = 0.5 \Rightarrow X = 12$ ✓

Both checks give the same value, confirming $X = 12$.

— Option B

Opposite directions on a circular track give relative speed $5+3=8$ rounds/hr.

So they meet every $1/8$ hr, i.e. every $7.5$ minutes after 5:00 a.m.

Meeting times after 5:00 are:

• $7.5$, $15$, $22.5$, $30$, $37.5$, $45$

• $52.5$, $60$, $67.5$, $75$, $82.5$

• $90$, $97.5$, $105$, $112.5$, $120$

The question says between 5:20 a.m. and 7:00 a.m.; under the strict endpoint reading, 7:00 a.m. is not counted. Thus the valid meetings are from $22.5$ to $112.5$ minutes.

Count $=13$.

— Option B

Test each option using BODMAS:

• A (SPRQ): $60\div15+3\times20-4$ = $4+60-4=60$ ✗

• B (QRPS): $60-15\times3+20\div4$ = $60-45+5=20$ ✓

• C (QRSP): $60-15\times3\div20+4$ = $60-2.25+4=61.75$ ✗

• D (SPQR): $60\div15+3-20\times4$ = $4+3-80=-73$ ✗

Only QRPS gives $20$.

— Option B

Let tram speed be $T$ km/hr and tram length be $L$ m.

For overtaking in the same direction:

• For X, relative speed is $T-3$.

• For Y, relative speed is $T-4$.

Since the same tram length is covered, cancel the common conversion factor $5/18$:

• $8(T-3)=9(T-4)$

• $8T-24=9T-36$

• Hence $T=12$ km/hr.

Now use X's case:

• $L=9\times5/18\times8$

• $L=20$ m

— Option C

The number has the standard repunit-square pattern:

• $1^2=1$

• $11^2=121$

• $111^2=12321$

• $1111^2=1234321$

Continuing the same pattern, the square of the 9-digit repunit is the given number.

So $N=111111111$, which has 9 digits.

Cross-check: $N^2$ has $17$ digits, so $N$ must have $9$ digits.

— Option B

The passage's central claim: behavioural change is only half of the route to net-zero, and even that half requires incentives (supply-side investments) from companies and governments.

• Option A ✗ — passage doesn't say household emissions alone are sufficient; it actually says they get at most half.

• Option B ✓ — directly captures: low-carbon behaviour needs incentivising ("would require companies and governments to provide more incentives to change").

• Option C ✗ — passage mentions cheaper *low-carbon options*, not generic goods/services.

• Option D ✗ — subsidies to manufacturers are nowhere mentioned.

— Option B

Assumption I: Supply-side investments in companies can result in low-carbon behaviour in people.

• The passage says incentives through supply-side investments can make low-carbon options cheaper and more widely available. That can support low-carbon behaviour. Valid.

Assumption II: People are not capable of adapting low-carbon behaviour without Government and Companies.

• The passage says incentives are required for achieving the demand-side reduction at scale. It does not say people are inherently incapable of low-carbon behaviour without government and company involvement. Invalid.

Thus only I is valid.

— Option A

Passage: "Reusing materials makes sense" + battery recycling example $\Rightarrow$ a circular approach to materials is sensible.

• Option A ✗ — "not possible without" is too absolute; passage doesn't claim impossibility.

• Option B ✗ — "every sector ... immediately" overstates; passage discusses one example (batteries), not a universal mandate.

• Option C ✓ — "can be beneficial for sustainable growth" matches the measured, pragmatic tone ("makes sense").

• Option D ✗ — passage doesn't make survival of industries hinge on circular use.

— Option C

Assumption I: Auto factories are examples of circular economy.

• Passage says batteries *should* go back to a factory — a prescription, not a description of current practice. Auto factories are not held up as existing examples. Invalid.

Assumption II: Growth is compatible with circular mineral use.

• Passage: consumption has quadrupled (growth) and "reusing materials makes sense" — i.e., growth and circular reuse can co-exist. The whole argument presupposes this compatibility. Valid.

— Option B

Compute rates as tank-fraction per minute.

• Set X: $70\%$ in $14$ min, so set rate $=1/20$. Per pipe $=1/400$.

• Set Y: $3/8$ in $6$ min, so set rate $=1/16$. Per pipe $=1/160$.

• Set Z: half in $20$ min, so emptying rate $=1/40$.

Now only 10 pipes of X and 5 pipes of Y are open, and all Z pipes are open:

• X fills at $10/400=1/40$.

• Y fills at $5/160=1/32$.

• Z empties at $1/40$.

X and Z cancel each other. Net filling rate $=1/32$ tank/min.

Time to fill half the tank $=(1/2)/(1/32)=16$ minutes.

— Option D

Always $1^n=1$.

Now check $2^n$ modulo $4$:

• For $n=1$, $1^n+2^n=3$, so the remainder is $3$.

• For $n\ge2$, $2^n\equiv0\pmod4$, so the remainder is $1$.

Thus the possible remainders are $3$ and $1$.

Number of distinct remainders $=2$.

— Option C

$P=QQQ=111Q$, where $Q$ is one of $1$ to $9$.

Factorize:

• $111=3\times37$

• $481=13\times37$

So every valid $P$ has the common factor $37$ with $481$.

There is no extra factor $13$, because $13$ does not divide $3Q$ for any digit $Q$ from $1$ to $9$.

Hence HCF $=37$.

— Option C

Count digits in chunks of equal length:

• $1$ to $9$: $9$ digits (cumulative $9$).

• $10$ to $99$: $90 \times 2 = 180$ digits (cumulative $189$).

• $100$ onwards: each number contributes $3$ digits.

Remaining: $489 - 189 = 300$ digits into the 3-digit block.

$300 / 3 = 100$ — exactly the 100th three-digit number, which is $199$.

The $489$th digit is the last (third) digit of 199, i.e., $9$.

— Option D

The problem gives no rule about which suspect tells the truth or lies (no "each tells one truth and one lie", no "innocent always tells truth"). Check each case for internal consistency:

• P is thief: P's two statements are both false ("I didn't steal", "Q stole it"). Q's and R's statements remain true. Consistent ✓

• Q is thief: P's both statements true. Q lies in "I didn't steal"; says "R didn't steal" (true). Consistent ✓

• R is thief: P's "Q stole it" is false; Q's "R didn't steal" is false; R lies in both. Consistent ✓

All three scenarios are internally consistent, so the thief cannot be uniquely determined.

— Option D

Points pattern:

• $P=3$ means one win and one draw.

• $R=1$ means one draw and one loss.

• $Q=2$ can be either two draws or one win and one loss.

First test the case where Q has one win and one loss. Then P must beat Q and draw with R, while Q must beat R.

Let P-Q be $p-q$, P-R be $a-a$, and Q-R be $u-v$.

• P total: $p+a=1$

• R total: $a+v=1$

• Q must beat R, so $u>v$

If $a=0$, then $p=1$ and $v=1$, but Q's total allows $u\le1$, so $u>v$ is impossible. If $a=1$, then $p=0$, so P cannot beat Q. Thus this points pattern is impossible.

Therefore Q must have two draws: P-Q is a draw, Q-R is a draw, and P beats R.

Let P-Q be $a-a$, Q-R be $b-b$, and P-R be $p-r$.

• P total: $a+p=1$

• Q total: $a+b=1$

• R total: $b+r=1$

• P beats R, so $p>r$

The only valid solution is:

• P-Q: $0-0$

• P-R: $1-0$

• Q-R: $1-1$

So Statement I is true and Statement II is also true.

— Option C

Passage chain: high-value crops have been *genetically honed* $\Rightarrow$ useful variations bred out $\Rightarrow$ "a minor disease or even an unknown one could suddenly rampage through a genetically honed crop" — i.e., dependence on such crops creates risk.

• Option A ✗ — too generic; the passage's core point is *genetic narrowing*, not just climate impact on productivity.

• Option B ✓ — directly mirrors the warning: reliance on genetically honed crops $\to$ vulnerability $\to$ possible food insecurity.

• Option C ✗ — passage isn't about diversifying away from agriculture.

• Option D ✗ — passage doesn't recommend replacing with wild varieties.

— Option B

Assumption I: Global climate change can result in the migration of several plant diseases to new areas.

• The passage says climate and atmospheric changes may make some plant infections more common in all climatic zones. This supports spread into new areas. Valid.

Assumption II: Scientific understanding of the wild relatives of our present crops would enable us to strengthen food security.

• The passage says selective breeding has removed genetic variations that are not immediately useful, making crops vulnerable to changes and diseases. Wild relatives are a likely source of such genetic variation; understanding them can help strengthen food security. Valid.

Thus both I and II are valid.

— Option C

Passage: opponents keep the statesman on "the path of reason and good sense"; national unity rests on "sufficiently even balance of political power".

• Option A ✗ — exact opposite of the passage's thesis.

• Option B ✗ — the passage explicitly warns against opposition being "revolutionary and irreconcilable".

• Option C ✗ — passage is about *good governance*, not the ruler's *political survival*.

• Option D ✓ — captures both halves: opposition is indispensable, and the balance of political power yields good governance.

— Option D

Assumption I: Strong opposition needed only if the Head is indifferent.

• Passage says opposition is *always* needed — "he ought also to pray never to be left without opponents". No conditional on the Head's attitude. Invalid.

Assumption II: More aggressive opposition $\Rightarrow$ better governance.

• Passage warns against opposition being "revolutionary and irreconcilable"; it wants *balance*, not aggression. Invalid.

— Option D

Given: P, Q, R are siblings (P is brother of Q and R). S is R's mother. T is P's father. Hence S and T are parents of all three siblings. P is male (a brother). Q's and R's genders are not fixed (one can be a brother to a sister or to a brother).

Check each:

• I. S and T are a couple (parents of the same children) ✓

• II. Q is T's *son* — Q's gender unknown ✗ (could be daughter)

• III. T is Q's father — T is Q's parent regardless of Q's gender ✓

• IV. S is P's mother ✓

• V. R is T's *daughter* — R's gender unknown ✗

• VI. P is S's son — P is male, S is P's mother ✓

Definitely true: I, III, IV, VI = 4 statements.

— Option C

Use letter positions $A=1, B=2, \ldots, Z=26$.

• $N=14$, $O=15$, $T=20$, $E=5$, $S=19$.

The code is the product of positions:

• NO: $14 \times 15 = 210$ ✓

• NOT: $210 \times 20 = 4200$ ✓

• NOTE: $4200 \times 5 = 21000$ ✓

• NOTES: $21000 \times 19 = 399000$

— Option A

Build the letter→digit map from the two codings.

FRANCE = 654321
=> F=6, R=5, A=4, N=3, C=2, E=1.

GERMANY = 9158437
=> G=9, E=1, R=5, M=8, A=4, N=3, Y=7.

Both codings agree on shared letters (E=1, R=5, A=4, N=3). Now read YEMEN letter-by-letter:

• Y=7, E=1, M=8, E=1, N=3

• YEMEN = 71813

— Option D

Translate constraints:

• $T \in \{1, 2, 3\}$ (so $P = 3T$ is a single digit, $\le 9$, and $T \ne 0$).

• $R \in \{0, 1, 2\}$ (so $S = R + 7 \le 9$).

• $Q = R + 3$, $S = R + 7$.

Enumerate $(T, P)$ with each $R$:

$T = 1, P = 3$:

• $R=0$: $Q=3=P$ ✗

• $R=1$: $R=T$ ✗

• $R=2$: digits $\{3,5,2,9,1\}$ all distinct ✓

$T = 2, P = 6$:

• $R=0$: $\{6,3,0,7,2\}$ ✓

• $R=1$: $\{6,4,1,8,2\}$ ✓

• $R=2$: $R=T$ ✗

$T = 3, P = 9$:

• $R=0$: $Q=3=T$ ✗

• $R=1$: $\{9,4,1,8,3\}$ ✓

• $R=2$: $S=9=P$ ✗

Valid count: $1 + 2 + 1 = \mathbf{4}$.

— Option B

Compute per-day rates:

• X: $1/3$ work in $6$ days, so rate $=1/18$.

• Y: $1/3$ work in $8$ days, so rate $=1/24$.

• Z: $3/4$ work in $12$ days, so rate $=1/16$.

Combined rate:

• $1/18=8/144$

• $1/24=6/144$

• $1/16=9/144$

• Sum $=23/144$ per day

In $n$ days, work done $=23n/144$.

Y alone works for $8\frac{2}{3}$ days, i.e. $26/3$ days.

Work by Y then:

• $(26/3)\times(1/24)=26/72$

• $26/72=52/144$

Total work is $1$:

• $23n/144+52/144=1$

• $23n=92$

• $n=4$

— Option B

Take successive differences of the sequence:

• $3 - 1 = 2$

• $6 - 3 = 3$

• $11 - 6 = 5$

• $18 - 11 = 7$

• $X - 18 = ?$

• $42 - X = ?$

The differences $2, 3, 5, 7$ are the first four primes, so the next two should be $11$ and $13$:

• $X = 18 + 11 = 29$

• Check: $42 - 29 = 13$ ✓ (next prime)

— Option C

Passage opens: "no democracy that has grown economically without corporate capitalism" — pairing economic growth with democracy. Then it lists benefits: modernization, jobs, competitiveness, welfare for the marginalized.

• Option A ✓ — captures both anchors of the passage: economic growth and democracy.

• Option B ✗ — passage doesn't speak of "political objectives" of a modern state.

• Option C ✗ — passage talks about democracies growing, not about state survival generally.

• Option D ✗ — passage says democracy needs corporate capitalism, not the other way around — no *mutual* dependence is asserted.

— Option A

Assumption I: Corporate capitalism promotes the growth of labour force and provides more employment opportunities.

• The passage says corporate capitalism "generates jobs" and that unemployment cannot be fixed without further impetus to private business. This supports more employment opportunities and a larger engaged labour force. Valid.

Assumption II: Poor and marginalized sections benefit due to trickle-down effect.

• The passage says a vibrant corporate capitalist base leads to additional State revenues, which can be used for welfare of the marginalized and for a more level playing field. This supports the assumed benefit reaching the poor and marginalized. Valid.

Thus both I and II are valid.

— Option C

The passage says voluntary associations stand between the relatively powerless individual and the potentially powerful State as a buffer.

• Option A ✗ — the passage does not mainly say that the State is unable to enforce its will against groups.

• Option B ✗ — this describes welfare and personality development inside an association, not the buffer role.

• Option C ✓ — associations reduce the isolation of individuals and let them participate in a common endeavour; that is how they become a buffer between the individual and the State.

• Option D ✗ — the passage is not about multiple loyalties or allegiances.

— Option C

Statement I: Take $N = 2$. Increase by 50%: $2 \times 1.5 = 3$.

• $d(2) = 2$ (factors 1, 2)

• $d(3) = 2$ (factors 1, 3)

Factor count unchanged. ✓

Statement II: Take $N = 2$. Increase by 150%: $2 \times 2.5 = 5$.

• $d(2) = 2$

• $d(5) = 2$ (factors 1, 5)

Factor count unchanged. ✓

Both statements have explicit witnesses. — Option C

Place coordinates: $A=(0,0)$, $B=(0,6)$, $C=(10,0)$, $D=(10,-6)$, $E=(5,-6)$, $F=(-9,6)$, $G=(-9,-6)$.

Allowed roads: AB, AC, CD, DE, BF, EG, FG.

Route 1: D → C → A → B → F

• $CD=6$, $AC=10$, $AB=6$, $BF=9$

• Total $=6+10+6+9=31$ km

Route 2: D → E → G → F

• $DE=5$

• $EG=\sqrt{14^2+0}=14$ km

• $FG=12$

• Total $=5+14+12=31$ km

Both available routes give 31 km. — Option C

Recognize the cubes:

• $64=4^3 \to 343=7^3$, so base $4 \to 7$

• $216=6^3 \to 729=9^3$, so base $6 \to 9$

Pattern: cube of $n$ maps to cube of $n+3$.

For $512=8^3$: code $=(8+3)^3=11^3=1331$.

— Option B

Modulo 6, $9 \equiv 3$.

For every $n \ge 1$:

• $9^n \equiv 3$

Terms from $9^3$ to $9^{100}$:

• Count $=98$

• Sum mod 6 $=98 \times 3$

• $98 \times 3=294$

• $294$ is divisible by $6$

Remainder $=0$. — Option A

List 1-digit numbers with exactly 4 distinct factors:

• $6$: factors $\{1,2,3,6\}$ → 4 ✓

• $8$: factors $\{1,2,4,8\}$ → 4 ✓

All others (1,2,3,4,5,7,9) have 1, 2, or 3 factors. The smallest is $6$.

The question already uniquely fixes the answer as 6 — no statement is required.

Statement I: Both 6 and 8 contain 2, doesn't add info that's needed.

Statement II: Confirms 6, but smallest was already 6 anyway.

The answer is reachable without either statement. — Option D

$PP=11P$, $PQ=10P+Q$, and $RRSS=11(100R+S)$.

Dividing by 11 gives:

• $P(10P+Q)=100R+S$

Check $P \in \{1,2,3\}$:

• $P=1$: left side $\le 14$, impossible

• $P=2$: left side $\le 48$, impossible

• $P=3$: check $Q=1,2,4$

For $P=3$:

• $Q=1$: value $93$, impossible

• $Q=2$: value $96$, impossible

• $Q=4$: value $102$, so $R=1,S=2$ ✓

Unique solution: $33 \times 34=1122$. Hence $Q=4$ from the question alone. No statement is required. — Option D

Statement I: P has sisters R, S — no mention of Q. Insufficient.

Statement II: R's father is brother of Q — but R's link to P is unknown. Insufficient.

Both together: R is P's sister, so R's father = P's father. Then Q is the sibling of P's father.

But the gender of Q is not given. 'Brother of Q' tells us R's father is male and a sibling of Q; Q itself can be male or female.

• Q male $\Rightarrow$ paternal uncle of P

• Q female $\Rightarrow$ paternal aunt of P

The exact relation cannot be pinned down even with both. — Option D

Passage explicitly says: 'price rise differs for different items of consumption' and the single number is just a weighted aggregate; further, 'CPI is different for each of these classes'. The logical takeaway is that the headline single number cannot correctly capture the actual price rise of individual items/commodities.

• (a) Passage discusses WPI and CPI neutrally; never recommends scrapping CPI. ✗

• (b) Matches the line about price rise differing across items and the aggregate hiding it. ✓

• (c) Passage never compares goods vs services representation. ✗

• (d) Passage doesn't claim the inflation number is useless. ✗

— Option B

Passage stresses two parallel ideas: trust is essential — 'essence of globalizing behaviour', 'super-commodity in the social market' — and it inherently involves risk — 'being vulnerable to others', 'willingness to expose oneself to betrayal'.

• (a) Passage says entrepreneurship implies trust in others, not absence of trust. ✗

• (b) Passage defines trust as vulnerability of the truster; never claims trustworthy people are the most vulnerable. ✗

• (c) Too absolute; passage doesn't say every economic activity requires exposure to betrayal. ✗

• (d) Captures importance plus risk — the two strands of the passage. ✓

— Option D

Let P's score at the 10-min-left mark be $x$, so Q's score then $= x+3$.

Statement I: P adds 4 goals, so $P_{\text{final}}=x+4$. Q's last-10-min goals unknown $\Rightarrow$ $Q_{\text{final}}$ unknown. Insufficient.

Statement II: $Q_{\text{final}}=4$ but $P_{\text{final}}$ unknown. Insufficient.

Both together: $P_{\text{final}}=x+4$, $Q_{\text{final}}=4$. Also Q at 10-min mark $=x+3 \le 4$, so $x \le 1$.

• $x=0$: $P=4$, $Q=4$ $\Rightarrow$ draw, not win

• $x=1$: $P=5$, $Q=4$ $\Rightarrow$ P wins

Two different outcomes remain possible. Insufficient even combined. — Option D

Expand the expression:

• $(p+q)^2=p^2+2pq+q^2$

• Subtracting $4pq$ gives $(p-q)^2$

So the expression is positive exactly when $p \ne q$.

Without any statement, $p=q=2$ gives $0$, so the question cannot be answered.

Statement I: $pSufficient.

Statement II: $p>q$, so $p \ne q$ and the value is positive. Sufficient.

Either one alone settles the question. — Option B

The ratio condition gives:

• $X/Y=Y/Z$

• so $Y^2=XZ$

With total runs $37$, two valid triples are enough to test determinacy.

• $(X,Y,Z)=(9,12,16)$: sum $=37$ and $12^2=9 \times 16$ ✓. Here $I

• $(X,Y,Z)=(16,12,9)$: sum $=37$ and $12^2=16 \times 9$ ✓. Here $III

Both satisfy every condition but give different orderings. The ordering is not determined. — Option D

$p+q=10$, so $pq=q(10-q)=10q-q^2$, a downward parabola maximized at $q=5$, giving $pq=25$.

Quantity I: $p, q$ positive integers. $q=5$ is allowed, $p=5$. So $I=25$.

Quantity II: $p \ge -6, q \ge -4$, so $-4 \le q \le 16$. The unconstrained max $q=5$ lies inside this range, so again $pq=25$. So $II=25$.

Thus $I=II=25$. — Option C

Quantity I: 11 consecutive integers, each $\ge -5$. To minimize the average, start as low as possible: $\{-5,-4,\ldots,5\}$.

• Sum $=0$, average $=0/11=0$. So $I=0$.

Quantity II: 11 consecutive non-negative integers. Smallest start $=0$, giving $\{0,1,\ldots,10\}$.

• Any consecutive run of 11 non-negative integers starting at 0 contains 0, so product $=0$. (Starting from 1 onwards the product is positive and larger.) So $II=0$.

Thus $I=II=0$. — Option C

Average condition: $p+q+r = 3k$.

Given: $p-k = k-q \Rightarrow p+q = 2k$.

Substitute: $2k + r = 3k \Rightarrow r = k$.

Quick check: $p=k+d$, $q=k-d$, then $r=k$, average $=\frac{(k+d)+(k-d)+k}{3}=k$ ✓.

— Option A

$0

Statement I: $x^2>x^3$?

• $x^2-x^3=x^2(1-x)$

• Here $x^2>0$ and $1-x>0$

• Therefore $x^2>x^3$ ✓

Statement II: $x>\sqrt{x}$?

For $0

• $x^2

• hence $x<\sqrt{x}$

So Statement II is false. ✗

Only I is correct. — Option A

Let the two numbers be $n$ and $n+10$. Count multiples of 5 strictly between them:

• $n=5$ (multiple of 5): between 5 and 15 only $\{10\}$ → 1 number

• $n=1$: between 1 and 11 are $\{5,10\}$ → 2 numbers

• $n=3$: between 3 and 13 are $\{5,10\}$ → 2 numbers

The count is either 1 or 2 depending on whether the endpoints are multiples of 5.

• (a) 'Only one' fails for $n=1$ (gives 2). ✗

• (b) 'Only two' fails for $n=5$ (gives 1). ✗

• (d) At least one always exists. ✗

• (c) 'Can be more than one' — true ($n=1$ gives 2). ✓

— Option C