UPSC Prelims 2024 CSAT (Paper II · Set A)

The passage lets us test each statement.

Statement 1: claims distribution methods are *solely* responsible. The passage says food is lost "throughout the supply chain, from initial agricultural production to final household consumption" — many stages, not distribution alone. Wrong (too absolute).

Statement 2: wastage causes "land degradation by about 45%" via deforestation and unsustainable practices; degraded land lowers productivity. Supported.

Statement 3: "energy spent over wasted food results in about $3.5$ billion tonnes of carbon dioxide" — cutting wastage cuts this footprint. Supported.

Statement 4: nothing says post-harvest technologies are unavailable. Not supported.

Only 2 and 3 are valid.

Option B

An assumption is something taken for granted that underlies the passage.

Assumption 1: The passage says food is lost across the whole supply chain and that "addressing the loss and wastage of food in all forms is critical." This presupposes the distribution/handling mechanism must be improved. Valid.

Assumption 2: The passage never speaks of a "social and moral responsibility of all citizens"; it discusses the FAO, land, energy and gases only. Introducing citizens' moral duty is outside the passage. Invalid.

Option A

Statement 1: "Without the fiscal backup, monetary policy eventually loses traction" and higher rates become inflationary — i.e. central banks cannot curb inflation without budgetary backing. Supported.

Statement 2: The whole passage argues rate hikes work only if governments provide fiscal austerity; so monetary policy's effect depends on fiscal policy. Supported.

Both are logical inferences.

Option C

Assumption 1: claims fiscal policies are *solely* responsible for higher prices. The passage treats inflation as a given and discusses how monetary and fiscal policy interact — it never says fiscal policy alone causes inflation. Invalid (too absolute).

Assumption 2: claims higher prices do not affect long-term government bonds. The passage says rising rates raise debt-service costs and force more borrowing — it neither states nor assumes bonds are unaffected. Invalid.

Neither assumption holds.

Option D

Interpretation: Under the intended convention, the pieces are not rearranged or stacked between cuts, and the target pieces are the $4\times4\times4$ smaller cubes. Without this convention, the wording is ambiguous.

$64=4^3$, so each edge must be divided into four equal parts.

• Along one direction, four equal parts require three parallel cuts.
• The same is required in each of the three mutually perpendicular directions.
• Hence the minimum is $3+3+3=9$ cuts.

With only eight cuts, the best allocation is $3,3,2$ cuts across the three directions. This gives $4\times4\times3=48$ pieces, not $64$ ✗

Nine cuts give $4\times4\times4=64$ identical cubes ✓

Option B

We fill four operators in $5*4*3*2*1$ using $+,-,\times$, each at most twice, and want the smallest value $\ge 0$.

Try to hit $0$ (remember $\times$ acts first):

• $5 - 4 - 3 + 2 \times 1$
• $2 \times 1 = 2$
• $5 - 4 - 3 + 2 = 0$ ✓

Operators used: $-$ twice, $+$ once, $\times$ once — all within the "at most two" limit.

Value $0$ is non-negative and cannot be beaten.

Option D

Men $\times$ days $=$ constant work.

• Let men $= M$ for $6k$ days.
• New men $M'$ for $5k$ days: $M \times 6k = M' \times 5k$
• $M' = \frac{6}{5}M = 1.2M$

Increase $= 0.2M$, i.e. $20\%$. Note $k$ cancels out.

Option C

Rates per hour are $X=\frac{1}{6}$ and $Y=Z=\frac{1}{8}$. Since nobody may work in two consecutive hours, place the faster worker $X$ in alternate hours.

Use the six-hour schedule $X,Y,X,Z,X,Y$:

• Work by $X$ is $3\times\frac{1}{6}=\frac{1}{2}$.
• Work by $Y$ and $Z$ is $3\times\frac{1}{8}=\frac{3}{8}$.
• Total work is $\frac{1}{2}+\frac{3}{8}$.
• Thus, work completed is $\frac{7}{8}$.

In six hourly slots, $X$ can work at most three times, so $\frac{7}{8}$ is the maximum possible work. The remaining $\frac{1}{8}$ should be done by $X$, who is eligible after the sixth hour and is the fastest.

• Required time is $\frac{1/8}{1/6}$.
• This equals $\frac{3}{4}$ hour, or $45$ minutes.

Total time is $6$ hours $45$ minutes.

Option C

The general term is $n^{2n}$. The number of trailing zeros is the smaller of the total powers of $2$ and $5$.

Only multiples of $5$ contribute factors of $5$:

• From $5^{10}$: $10$ factors of $5$.
• From $10^{20}$: $20$ factors of $5$.
• From $15^{30}$: $30$ factors of $5$.
• From $20^{40}$: $40$ factors of $5$.
• Since $25^{50}=(5^2)^{50}$, it contributes $100$.

Thus $v_5=10+20+30+40+100$, so $v_5=200$.

Every even base contributes at least one factor of $2$ for each copy of that base. Therefore, even before counting extra factors of $2$,

• $v_2\ge 4+8+\cdots+48$
• $v_2\ge312>200$

Hence the number of trailing zeros is $200$.

Option D

Savings form $1,3,5,\dots$ (first term $1$, common difference $2$).

• Total after $n$ days $=$ sum of first $n$ odd numbers $=n^2$.

So the total is always a perfect square. We need it to also be a perfect cube:

• $n=7\Rightarrow 49$, not a cube ✗
• $n=8\Rightarrow 64=4^3$ ✓ (also $8^2$)
• $n=9\Rightarrow 81$, not a cube ✗

Note that the total on January 1 is also $1$, which is both a perfect square and a perfect cube. Therefore, the literal wording admits more than one date. Among the listed options, January 8 is the only valid choice: the total $64$ is both a perfect square and a perfect cube.

Option B

The passage credits great Western planners to a curriculum that absorbed "computing, statistics, optimization" while having "its roots in the social sciences, geography and architecture" — i.e. a diverse, interdisciplinary mix.

Option A captures exactly this: planning curricula should be diverse and interdisciplinary. Supported.

• B (bureaucracy lacks training), C (local affair, insufficient funds), D (density/poverty make planning hard) all bring in facts about India that the passage never states. Unsupported.

Option A

Assumption 1: The passage contrasts the West's system-trained planners with "Planning in India, and its education differ from the West," implying India needs professionals equipped for modern urban practice. This underlies the passage. Valid.

Assumption 2: claims Indian universities have *no capacity or potential* for systems training. The passage only says Indian planning education *differs* — "differs" is not the same as "incapable." This is too strong. Invalid.

Option A

The passage directly establishes three points: money can buy a louder voice; digital rivalry can damage the quality of politics; and social media promotes engagement even when the content is inflammatory. These points show that social media is not an ideal or morally neutral public forum.

Option B is the best available answer because its critical description is consistent with those incentives. However, the passage does not expressly say that companies built social media "to make profits"; that clause is an inference from paid amplification and engagement-driven promotion, so the option is broader than the passage.

• A mentions instant access to information, which is absent from the passage.
• C says social media was created to strengthen democracy, contrary to the passage's criticism of its political effects.
• D calls social media inevitable for an informed life, which the passage does not suggest.

Option B

Assumption 1: "Not every voice on the internet commands the same kind of audience" and money buys a louder voice — this presupposes the internet is not equally inclusive. Valid.

Assumption 2: "the quality of politics suffers" directly shows the internet can adversely affect politics. Valid.

Both assumptions hold.

Option C

Check each factor of $222^{333}+333^{222}$:

By 2: $222^{333}$ is even, $333^{222}$ is odd; even $+$ odd $=$ odd. Not divisible.

By 3: $222=3\times74$ and $333=3\times111$, so each term is a multiple of $3$; their sum is too. Divisible.

By 37: $222=6\times37$ and $333=9\times37$, so each term is a multiple of $37$; their sum is too. Divisible.

So it is divisible by $3$ and $37$ but not $2$.

Option B

Let honey cost ₹ 1 per litre and take $1$ litre of honey; water is free.

• Add $w$ litres of water $\Rightarrow$ mixture $=(1+w)$ litres.
• Sold at honey's cost price ₹ 1/litre $\Rightarrow$ revenue $=(1+w)$.
• Cost = ₹ 1 (honey only).
• Profit $=w$, so profit% $=\frac{w}{1}\times100$.

For $20\%$: $w=0.2$ litre $=20\%$ of the honey.

Option A

First split the power:

• $30^{30}=(3\times10)^{30}$
• $=3^{30}\times10^{30}$

Thus the number is $3^{30}$ followed by $30$ zeros. The digit immediately before those zeros is the unit digit of $3^{30}$.

The unit digits of powers of $3$ cycle through $3,9,7,1$, with period $4$:

• $30 \bmod 4=2$
• The required digit is that of $3^2$, namely $9$.

Option D

Same remainder $1$ means the divisor divides $421-1=420$ and $427-1=426$, and must be greater than the remainder ($>1$).

• Common divisors are the divisors of $\gcd(420,426)$.
• $426-420=6\Rightarrow \gcd=6$.
• Divisors of $6$: $1,2,3,6$.
• Discard $1$ (divisor must exceed remainder $1$).

Usable divisors: $2,3,6$ — that is $3$ numbers. (Check: $421=2\cdot210+1$, $421=3\cdot140+1$, $421=6\cdot70+1$, all remainder $1$.)

Option C

The bottle size must divide both $399$ and $532$ (each fully bottled), so it divides $\gcd(399,532)$.

• $399=3\times7\times19$
• $532=2^2\times7\times19$
• $\gcd=7\times19=133$

Common bottle sizes $=$ divisors of $133$: $1,7,19,133$ — that is $4$ sizes.

Option B

Primes below $10$: $\{2,3,5,7\}$. Sums of three distinct ones:

• $2+3+5=10$ (unit $0$)
• $2+3+7=12$ (unit $2$)
• $2+5+7=14$ (unit $4$)
• $3+5+7=15$ (unit $5$)

Statement 1 (unit $0$): $10$ works ✓

Statement 2 (unit $9$): units available are $0,2,4,5$ — no $9$ ✗

Statement 3 (unit $5$): $15$ works ✓

Only 1 and 3 are correct.

Option C

Assumption 1: claims parents have a *greater role than governments*. The passage mentions parents' expectations but never compares parents' and governments' roles. Invalid.

Assumption 2: claims a *uniform national* curriculum would fix things. The passage notes abilities fall below curricular expectations but says nothing about a single uniform-for-the-whole-country curriculum. Invalid.

Neither assumption is valid.

Option D

The passage's core tension: students leave class 8 with high aspirations (college, not agriculture) but with competencies "far lower than they should be." The gap is between expectations and abilities.

Option C — public policy should align young people's competencies/achievements with their expectations — captures this. Best.

• A (eradicate poverty), B (monetary incentives), D (return to agriculture) are solutions/claims the passage does not advance. Wrong.

Option C

The author notes that today science is expected to have a "social responsibility," whereas Newton and Galileo acknowledged only the responsibility "to tell the truth." The author presents both facets — pursuit of truth and social responsibility.

Option D — science must pursue truth but be responsible for social welfare — blends both, matching the author's view. Best.

• A (value scientists' commitment), B (product of civilized society, spread awareness), C (industrial revolution caused by science) are not the author's stated position. Wrong.

Option D

Split the string into blocks of five ending in $D$. Each block is $ABCD$ with exactly one letter doubled, and the doubled letter marches $A\to B\to C\to D$:

• Block 1: $A\,\_\,BCD \to AABCD$ (double $A$) $\Rightarrow$ blank $=A$
• Block 2: $\_\,BBCD \to ABBCD$ (double $B$) $\Rightarrow$ blank $=A$
• Block 3: $ABC\,\_\,D \to ABCCD$ (double $C$) $\Rightarrow$ blank $=C$
• Block 4: $ABC\,\_\,D \to ABCDD$ (double $D$) $\Rightarrow$ blank $=D$

So the blanks are $A, A, C, D$.

Option C

Let the profit shares of $P$ and $Q$ be $p$ and $q$.

• $p+q=2000$
• $p=q+400$
• Hence $q=800$ and $p=1200$.
• Therefore, $p:q=3:2$.

Profit is divided in the ratio of capital multiplied by time. Let $Q$'s capital be $x$; then $P$'s capital is $x+14000$.

• $8(x+14000):10x=3:2$
• $16(x+14000)=30x$
• $16x+224000=30x$
• $14x=224000$
• Hence $x=16000$.

Therefore, $P$ invested $16000+14000$, or ₹ 30,000.

Check:

• $P$'s weighted capital is $30000\times8=240000$.
• $Q$'s weighted capital is $16000\times10=160000$.
• Their ratio is $3:2$ ✓

Option A

Take $R=100$.

$Q$ is $20\%$ below $R$, so $Q=80$.

$P$ is $20\%$ below $Q$, so $P=64$.

Calculation:

• Difference: $100-64=36$
• Percentage increase: $\frac{36}{64}\times100$
• Result: $56.25\%$ ✓

Option B

Let the number be $N$.

• Intended result: $4N$
• Mistaken result: $\frac{N}{4}$

Relative to the intended result:

• Actual fraction: $\frac{N/4}{4N}=\frac{1}{16}$
• Percentage drop: $(1-\frac{1}{16})\times100$
• Result: $93.75\%$ ✓

Option D

Failed English $=100-80=20\%$; failed Hindi $=100-70=30\%$; failed both $=15\%$.

• Failed only English $=20-15=5\%$
• Failed only Hindi $=30-15=15\%$

Failed in exactly one subject $=5+15=20\%$.

Option B

Let the present ages of the father and son be $F$ and $S$ years, respectively.

First condition: $n$ years ago, the father’s age equalled the son’s present age.

• $F-n=S$
• Hence $n=F-S$

Second condition: The father’s present age is four times the son’s age $n$ years ago.

• $F=4(S-n)$

Substitute $n=F-S$:

• $F=4(2S-F)$
• $5F=8S$

Together with $F+S=130$, this gives $F=80$ and $S=50$.

Therefore, the age difference is $80-50=30$ years.

Check: $n=30$, the son was $20$, and $4\times20=80$ ✓

Option A

• (1) $1000$ litre $=1\,\text{m}^3$ since $1$ litre $=10^{-3}\,\text{m}^3$ ✓
• (2) $1$ metric ton $=1000$ kg ✓
• (3) $1$ hectare $=10000\,\text{m}^2$ ✓

All three are correct.

Option D

The passage only asserts that science and mankind cannot be divorced — the epigram merely links science with its background.

• Statement 1 (mankind could not exist without science): the passage never claims science is necessary for survival; it discusses the relation, not dependence for existence. Unsupported.
• Statement 2 (science will determine mankind's destiny): the closing lines ask whether conflict is 'the inevitable result of the progress of science or does the fault lie elsewhere?' — it questions science's role, it does not assert science decides destiny. Unsupported.

Neither is emphatically conveyed.

Option D

• Assumption 1: The passage does NOT assume the horrors are the inevitable result of science; it explicitly asks 'Is this the inevitable result of the progress of science or does the fault lie elsewhere?' — leaving it open. Invalid.
• Assumption 2: The 'aspect of truth easily overlooked' is the relation between science and mankind — that science is not an abstract third party but is bound to man, i.e. 'science is what man has made it.' This matches the passage. Valid.

Only 2 is correct.

Option B

The passage says the merchant learns 'what is parochial and what is universal in his choice' — parochial (local) versus universal values.

• (A) imagination/understanding — not mentioned.
• (B) communities/nationalities — not the distinction drawn.
• (C) local vs universal values — directly stated. ✓
• (D) friends/foes — not in the passage.

Option C

• (1) 'learning fundamental common values is the greatest gain of travel to those who wish to live at ease among their fellows' — travel leads to understanding of humans. Valid.
• (2) The same line supports that travel helps those who wish to learn fundamental common values. Valid.
• (3) The passage speaks of learning to live 'at ease among their fellows', not of resolving differences among other people. Not supported.

So 1 and 2 only.

Option A

Let $X=10a+b$, so $Y=10b+a$.

$X+Y=11(a+b)$ — always a multiple of $11$.

Greatest two-digit multiple of $11$ is $99$, so $a+b=9$.

Both $X,Y$ must be two-digit, so $a\ge1$ and $b\ge1$:

• $a=1,b=8$; $a=2,b=7$; ... up to $a=8,b=1$

That is $a=1$ to $8$ → $8$ values of $X$.

Option D

The total weight of $6$ boys equals that of $4$ women.

• Weight of $4$ women: $4\times63=252$ kg
• Average weight of boys: $\frac{252}{6}=42$ kg ✓

Option B

Hands coincide $11$ times every $12$ hours (once about every $65.45$ min).

Starting after $10{:}00$, coincidences fall at about:

• $10{:}54{:}33$ ✓
• $12{:}00{:}00$ ✓
• $1{:}05{:}27$ ✓

The next is $\approx2{:}10{:}54$, which is after $2{:}00$ p.m.

So they coincide $3$ times.

Option A

A calendar repeats when the weekday shift is $0$ modulo $7$ and both years are of the same type. The year $2025$ is non-leap.

From $2025$ through $2030$:

• Five non-leap years contribute $5$ odd days.
• Leap year $2028$ contributes $2$ odd days.
• Total: $5+2=7\equiv0\pmod{7}$ ✓

Thus $2031$ begins on the same weekday as $2025$. Since $2031$ is also non-leap, the calendars are identical.

Option C

$p,q$ are odd and $r,s$ are even.

• (1) Since $q$ is odd and $s$ is even, $qs$ is even; hence the full product is even. ✓
• (2) The product contains the even factor $s$, so it is even. ✓
• (3) Both $q+r$ and $p+s$ are odd. Therefore $(q+r)^2$ and the full product are odd. ✓

All three statements are correct.

Option D

At $4{:}25$:

• Minute-hand angle: $25\times6=150°$
• Hour mark for $4$: $4\times30=120°$
• Extra hour-hand movement: $25\times0.5=12.5°$
• Hour-hand angle: $120+12.5=132.5°$
• Smaller angle: $150-132.5=17.5°$ ✓

Option C

• Statement 1: The passage says conventional classrooms emphasize fixed duration over learning effectiveness. Therefore duration, rather than attainment of competency, is the controlling goal. ✓
• Statement 2: Every learner receives the same lectures, in the same way, for the same duration. This directly describes a one-size-fits-all system that removes differentiated instruction. Different learner outcomes do not disprove this point. ✓

Both statements follow.

Option C

• Assumption 1: The passage argues the current model fails ('10 percent employability', 'repeating the same ineffectual script...will not produce different results') — it implies India MUST change, the opposite of 'does not need to change.' Invalid.
• Assumption 2: The passage says formal education yields only '10 percent employability' — i.e. it does NOT produce sufficient skilled workers. Invalid.

Neither assumption is valid.

Option D

The supplied excerpt starts mid-sentence. On the intended reading, the omitted subject refers to dogmatic parents or adults in whom the young seldom confide.

• Assumption 1: The young avoid confiding in such parents and instead follow their own way secretly. This supports the inference that adolescents are uncomfortable with dominating and assertive parents. Valid. ✓
• Assumption 2: Lack of confidence and secret conduct do not establish lack of respect for parents. Unsupported. ✗

Only 1 is valid.

Option A

The supplied excerpt is truncated. Under its intended reading, ‘them’ refers to dogmatic parents or adults; the young seldom confide in them and instead follow their own way secretly. This makes (A) the closest available statement of the central idea.

• (B) The passage does not say parental involvement is unnecessary. ✗
• (C) Nuclear families are not mentioned. ✗
• (D) The young are not described as stubborn, disobedient or careless. ✗

The missing opening context should be restored for the inference to be fully self-contained.

Option A

Count the digit $5$ in page numbers $1$–$260$.

Units place (ending in 5): $5,15,\dots,255$ → $\frac{255-5}{10}+1=26$.

Tens place (5 in tens): $50$–$59$, $150$–$159$, $250$–$259$ → $10+10+10=30$.

Hundreds place: none ($\le260$).

Total $=26+30=56$.

Option B

Group the sequence into blocks:

• Block $k$: $1,k,k-1,\dots,2$ (length $k$)

Block lengths $1+2+\dots+7=28$, so the first $28$ terms $=$ blocks $1$–$7$.

Sum of block $k=\frac{k(k+1)}{2}$ (a triangular number):

• $1,3,6,10,15,21,28$

Total $=1+3+6+10+15+21+28=84$.

Option B

Since $R$ costs $20\%$ more than $Q$, take $Q=10k$ and $R=12k$.

Since $P$ costs $25\%$ more than $R$:

• $P=15k$
• Total ratio $P:Q:R=15:10:12$
• $37k=3330\Rightarrow k=90$
• $P=15\times90=1350$ ✓

Option D

Write $AB=10A+B$ and $CD=10C+D$. Since $AB+CD=1CE$:

• $10A+B+10C+D$
• $=100+10C+E$

Cancel $10C$:

• $10A+B+D=100+E$
• $10A=100+E-B-D$

Because $B+D\le18$ and $0\le E\le9$:

• $82\le10A\le109$

Since $A$ is a digit, $10A\le90$. Thus the only possible multiple of $10$ in this range is $90$, so $A=9$.

Feasibility check: $92+38=130$; the digits $9,2,3,8,0$ are distinct. ✓

Option A

Need $x>2y$ and $2y>3z$ with $x,y,z\in\{1,\dots,7\}$.

From $x>2y$ and $x\le7$: $y\le3$; from $2y>3z\ge3$: $y\ge2$.

• $y=2$ ($2y=4$): $x>4\Rightarrow x\in\{5,6,7\}$; $4>3z\Rightarrow z=1$ → $(5,2,1),(6,2,1),(7,2,1)$
• $y=3$ ($2y=6$): $x>6\Rightarrow x=7$; $6>3z\Rightarrow z=1$ → $(7,3,1)$

Total $=4$ triplets.

Option D

Let the prices of an orange, a mango and an apple be $o$, $m$ and $a$.

Given: $4o+6m+8a=2(o+2m+5a)$

$4o+6m+8a=2o+4m+10a$

$2o+2m-2a=0$

$o+m=a$

Statement 1: The difference between $4o+6m+8a$ and $3o+5m+9a$ is $o+m-a=0$. Therefore, the two totals are equal. ✓

Statement 2: $o+m=a$ means that one orange and one mango together cost as much as one apple. ✓

Both statements are correct. — Option C

The passage states that 'carbon dioxide is converted into stone.' Immobilizing $CO_2$ in a solid form directly supports its use as a method of carbon sequestration.

• Option A — the passage gives no evidence that the method is cheap, commercially practical or intended for building material.
• Option C — calcium and magnesium are dissolved during limestone formation; the passage does not propose extracting them as mineral products.
• Option D — the acidic mixture is part of the process, but producing a rock-dissolving acid is not the suggested application.

— Option B

Assumption 1: The passage says iceberg meltwater supports 'unexpectedly high levels of phytoplankton growth.' Because phytoplankton drives primary productivity and marine food chains, giant icebergs bear on both. Valid ✓

Assumption 2: The passage links giant icebergs with storage of 'up to 20 percent of carbon' and with phytoplankton growth. Carbon storage has climate effects, while changes at the base of marine food chains can affect fisheries. Valid ✓

Both assumptions follow. — Option C

The passage says the finding could illuminate 'compounds associated with plant response to pathogens like fungi or bacteria.' A possible plant-protection application is therefore the closest match among the choices, so C is the best available answer.

Question-quality caveat: The passage does not establish that any resulting pesticide would be organic or ecologically sustainable. Thus, C is plausible but not fully entailed by the passage.

• Option A — the caterpillars feed on corn, not weeds.
• Option B — a plant response to pathogens does not by itself imply clinically useful antimicrobial compounds.
• Option D — genetic modification and predation on other pests are not mentioned.

— Option C

$32=2^5$, so $32^5=2^{25}$.

$32^5+2^{27}$

$=2^{25}(1+2^2)$

$=2^{25}\times 5$

$=2^{24}\times 10$

Hence the number is divisible by $10$. Its prime factors are only $2$ and $5$, so it is not divisible by $3$, $7$ or $11$. — Option C

We need the smallest $k$ for which more than one ordered pair satisfies $p

• $k=3$: only $(1,2)$ — unique
• $k=4$: only $(1,3)$; $(2,2)$ violates $p
• $k=5$: $(1,4)$ and $(2,3)$ — not unique ✓

Thus the smallest such value is $k=5$. — Option C

Relative to the initial direction:

• the first and third legs give a net movement of $200$ m backward
• the second and fourth legs give a total movement of $200$ m to the left

For the net movement to be North-East, backward must point North and left must point East. Both occur when the initial direction is South. — Option C

Start facing West. Turn left: the left of West is South, so after the 2nd leg he faces South.

Bearing of South $=180°$. A $225°$ clockwise turn adds:

$180°+225°=405°\equiv 45°$

Bearing $45°$ is North-East. — Option D

Being the largest producer does not imply being the largest exporter (most milk may be consumed at home) — Conclusion I does not follow.

Being the largest producer says nothing about imports; a country can produce the most and still import — Conclusion II does not follow.

Neither follows. — Option D

Statement I: $m+n>mn$.

$\Rightarrow (m-1)(n-1)<1$

Because $m,n$ are positive integers, $(m-1)(n-1)$ is a non-negative integer; hence it is $0$. Thus $m=1$ or $n=1$. With $m>n$, we get $n=1$ and $m\ge2$. Both $(m,n)=(2,1)$ and $(3,1)$ work. Insufficient ✗

Statement II: $mn=24$ allows, for example, $(m,n)=(1,24)$ and $(2,12)$. Insufficient ✗

Combined: Statement I gives $n=1$. Then $mn=24$ gives $m=24$. The unique pair is $(24,1)$. Sufficient ✓

— Option C

Statement I: The size is known, but the transfer rate is not. The download time cannot be determined. Insufficient ✗

Statement II: The transfer rate is known, but the software size is not. The download time cannot be determined. Insufficient ✗

Combined: After converting the units consistently, time is obtained by dividing size by transfer rate. Both required quantities are known. Sufficient ✓

— Option C

The passage's core: a 'normatively responsible media... has the responsibility to circulate the content of reality without distortion' in a robust democracy. That is exactly option A.

• B — the passage never says fake news is inherent.
• C — it speaks of responsibility, not absence of restrictions.
• D — controlling irresponsible media is not discussed.

— Option A

The passage says manufacturers may gather 'more of our personal data—with or without our agreement' and observe device owners' buying needs and preferences. This directly supports the privacy-risk part of D.

• Option A — manufacturing capital expenditure is not discussed.
• Option B — growth of trade or national economic development is not discussed.
• Option C — comfort and quality-of-life benefits are not mentioned.

Question-quality caveat: The passage does not establish D's separate claim that the devices themselves face a security risk. No option is fully supported; D is the best available answer only because its privacy-risk claim is directly grounded in the passage.

— Option D

Crux: 'green growth' rethinks growth around environmental sustainability, replacing resource/energy-intensive processes with 'resource frugal' ones to avoid an 'economic dead end' and a world of elite affluence amid poverty — i.e. sound environmental standards underpin steady, sufficient, inclusive growth. Option C.

• A — the passage treats sustainability as enabling, not hostile, to growth.
• B — it warns against environment-degrading growth, not endorses it.
• D — it argues degradation/inequality can be avoided, not that they are inevitable.

— Option C

Statement I: If $x/y$ is an odd integer, both $(x,y)=(3,1)$ and $(9,3)$ satisfy it. Insufficient ✗

Statement II: $xy=12$ permits multiple ordered pairs, for example:

• $(x,y)=(1,12)$
• $(x,y)=(2,6)$

Insufficient ✗

Combined: Let $x=ky$, where $k$ is odd. Then $ky^2=12$. The only possible odd divisors $k$ are checked separately:

• $k=1$: $y^2=12$, impossible for natural $y$ ✗
• $k=3$: $y^2=4$, so $y=2$ and $x=6$ ✓

The unique ordered pair is $(6,2)$. Sufficient ✓

— Option C

Let the total amount be $T$.

Statement I: $X=\frac{4}{5}(Y+Z)$, so $X=\frac{4}{9}T$. For $T=9$, both distributions satisfy the statement:

• $(X,Y,Z)=(4,1,4)$: $Y$ is least
• $(X,Y,Z)=(4,3,2)$: $Z$ is least

Insufficient ✗

Statement II: $Y=\frac{2}{7}(X+Z)$, so $Y=\frac{2}{9}T$. For $T=9$:

• $(X,Y,Z)=(1,2,6)$: $X$ is least
• $(X,Y,Z)=(3,2,4)$: $Y$ is least

Insufficient ✗

Combined: $X=\frac{4}{9}T$ and $Y=\frac{2}{9}T$, hence $Z=\frac{3}{9}T$. Therefore $Y$ receives the least amount. Sufficient ✓

— Option C

Statement I: The average is already $60$. These two classes both have lowest marks $50$ and highest marks $70$:

• $3$ students: $50,60,70$
• $4$ students: $50,60,60,70$

The number of students differs. Insufficient ✗

Statement II: If removing one highest and one lowest mark leaves the average at $60$, those two removed marks sum to $120$. This condition does not determine the class size; the same two examples satisfy it. Insufficient ✗

Combined: Both examples satisfy both statements but have different class sizes. Insufficient ✗

— Option D

A prime sum of three distinct primes cannot include $2$: the other two primes would be odd, making the sum even and greater than $2$. Hence all three numbers are odd primes.

Statement I: For a sum below $23$, all possible triples are:

• $3+5+7=15$ ✗
• $3+5+11=19$ ✓
• $3+5+13=21$ ✗
• $3+7+11=21$ ✗

Only $\{3,5,11\}$ works. Sufficient ✓

Statement II: If one number is $5$, multiple triples work:

• $\{3,5,11\}$ gives $19$
• $\{5,7,11\}$ gives $23$
• $\{5,11,13\}$ gives $29$

Insufficient ✗

Statement I alone is sufficient, but Statement II alone is not. — Option A

Statement I: $2x+y$ can be an integer in both cases:

• $x=0,y=0$: $x+y=0$ is an integer ✓
• $x=0.5,y=0$: $2x+y=1$, but $x+y=0.5$ ✗

The answer varies. Insufficient ✗

Statement II: $x+2y$ can also be an integer in both cases:

• $x=0,y=0$: $x+y=0$ is an integer ✓
• $x=0,y=0.5$: $x+2y=1$, but $x+y=0.5$ ✗

The answer varies. Insufficient ✗

Combined: Adding the two integer expressions shows that $3(x+y)$ is an integer, but $x+y$ need not be an integer:

• $x=y=1$: both given expressions are $3$ and $x+y=2$ ✓
• $x=y=\frac{1}{3}$: both given expressions are $1$, but $x+y=\frac{2}{3}$ ✗

The answer still varies. Insufficient ✗

— Option D

From $p+q+r=50$ and $q=16$, we get $p+r=34$. Since $q$ is least, $p,r\ge16$.

Statement I: $p\le r$. Both price pairs satisfy it:

• $(p,r)=(17,17)$
• $(p,r)=(16.5,17.5)$

Thus $p$ is not unique. Insufficient ✗

Statement II: $r\le p$. Both price pairs satisfy it:

• $(p,r)=(17,17)$
• $(p,r)=(17.5,16.5)$

Again $p$ is not unique. Insufficient ✗

Combined: $p\le r$ and $r\le p$, so $p=r$. Since $p+r=34$, $p=17$. Sufficient ✓

— Option C

Statement I: $P+Q=R+S$ equates two sums but does not determine whether $P>Q$.

• $P=10,Q=5,R=8,S=7\Rightarrow$ $P>Q$
• $P=5,Q=10,R=4,S=11\Rightarrow$ $P

Both keep $P+Q=R+S=15$, so Insufficient. ✗

Statement II: $P+S>Q+R$ also involves $R,S$, so it does not determine whether $P>Q$.

• $P=10,S=10,Q=5,R=3\Rightarrow$ yes, $P>Q$
• $P=5,S=10,Q=8,R=3\Rightarrow$ $15>11$ but $P

So Insufficient. ✗

Combined: From I, $R=P+Q-S$. Put into II: $P+S>Q+(P+Q-S)$ $\Rightarrow 2S>2Q\Rightarrow S>Q$. This proves $S>Q$, but does not determine whether $P>Q$.

• $P=10,Q=5,R=8,S=7$: I, II hold, $P>Q$
• $P=5,Q=10,R=4,S=11$: I, II hold, $P

Still Insufficient. ✗

— Option D

The passage states that poetry remains a "partisan of democratic reality" against the "tyranny of truth" and proposes "defiance" against the arrogance of power. This directly supports poetry as resistance.

• Option A: The passage says truth admits no imperfections and poetry accepts plurality; it does not say that only poetry recognizes imperfections in human beings. ✗
• Option B: It neither says truth is revealed through science or religion nor calls poetry anathema to truth. ✗
• Option C: It does not describe poetry as romantic, imaginary or merely emotional. ✗
• Option D: Its central idea—poetry siding with democratic reality and defying tyranny and arrogant power—supports "dynamic resistance." Violence and bigotry are not separately named, but this is still the closest statement of the message. ✓

— Option D

Assumption 1 (Nature’s luxury): Invalid. ✗ The passage says the flower "flaunted its petals... to attract an insect" so pollen "is not wasted" — the passage assigns the flower a useful role rather than describing it as luxury.

Assumption 2 (no usefulness except beauty): Invalid. ✗ This is the opposite of the passage. Beauty is called "merely a by-product and a human invention," while the passage explicitly stresses the useful role of attracting an insect to carry pollen so that it is not wasted.

Both assumptions are unsupported.

— Option D

Let $P=10a+b$ and $Q=10b+a$. Since both ages exceed 10, $a,b\in\{1,\dots,9\}$. Their age difference is $9|a-b|$.

Statement I: $P>Q$ gives only $a>b$.

• $21-12=9$
• $73-37=36$

The difference is not unique. Insufficient. ✗

Statement II:

• Sum: $P+Q=11(a+b)$
• Difference: $|P-Q|=9|a-b|$

Substitution gives $2(a+b)=3|a-b|$.

• If $a>b$, then $a=5b$; the only digit solution is $(a,b)=(5,1)$.
• If $b>a$, then $b=5a$; the only digit solution is $(a,b)=(1,5)$.

Both cases give ages 51 and 15, so the difference is $36$. Sufficient. ✓

Only Statement II answers the question.

— Option A

Main statement: Pradeep is a Director or Producer ($D\lor Pr$).

Pair $SP$: $S=$ "not a Producer" $\Rightarrow$ by $D\lor Pr$, he must be a Director $=P$. So $S\Rightarrow P$, and consistent with the main. ✓

Pair $RQ$: $R=$ "not a Director" $\Rightarrow$ by $D\lor Pr$, he must be a Producer $=Q$. So $R\Rightarrow Q$, consistent with the main. ✓

Both ordered pairs satisfy "first implies second" and agree with the main statement.

— Option C

Swap operators simultaneously: $+\to-$, $-\to\times$, $\times\to\div$, $\div\to+$.

Expression $10+30-100\times50\div25$ becomes:
$10-30\times100\div50+25$.

Now apply normal BODMAS:

• $30\times100=3000$
• $3000\div50=60$
• $10-60+25=-25$

— Option D

Symbol meanings:

• $P$: $>$
• $Q$: $<$
• $R$: $\le$
• $S$: $\ge$
• $T$: $=$

Statement 1: From $3x=4z$, we get $x=\frac{4z}{3}$. Substituting in $2x\ge3y$ gives $8z\ge9y$. Hence $9y\le8z$, so the claim $9y>8z$ is incorrect. ✗

Statement 2: The premises are $x<2y$ and $y\le z$.

• Take $y=10$ and $z=10$.
• Take $x=15$; then $x<2y$ is true.
• But $x>z$, so $x\le z$ is false.

Statement 2 is incorrect. ✗

Neither statement is correct.

— Option D

Each letter takes its alphabet position; multiply the four values.

• $ABCD=1\times2\times3\times4=24$ ✓
• $EFGH=5\times6\times7\times8=1680$ ✓

So $IJKL=9\times10\times11\times12$:

• $9\times10=90$
• $90\times11=990$
• $990\times12=11880$

— Option A

Rule: reverse the word, then prefix the letter A.

• $POT\to TOP\to ATOP$ ✓
• $TRAP\to PART\to APART$ ✓

For $ARENA$, the reverse is $ANERA$; prefixing $A$ gives $AANERA$.

• Option A: $AARENA$ is not reversed. ✗
• Option B: $AANREA$ has the wrong internal order. ✗
• Option C: $AANEAR$ has the wrong internal order. ✗
• Option D: $AANERA$ follows the rule. ✓

— Option D

Look at the differences of consecutive terms:

• $14-3=11$
• $39-14=25$
• $84-39=45$

Second differences: $25-11=14$, $45-25=20$ — rising by $6$ each time ($14,20,26,32,\dots$).

So the next first difference $=45+26=71\Rightarrow *=84+71=155$.

Check: next second diff $32\Rightarrow$ diff $71+32=103\Rightarrow 155+103=258$ ✓.

— Option B

Values: $\{4,5,10,12,15\}$.

From $Q-S=2S$, we get $Q=3S$. There are two candidates:

• $S=4$ and $Q=12$
• $S=5$ and $Q=15$

Now use $T=R+S+3$.

• If $S=4$, then $T=R+7$. No assignment from the remaining values $\{5,10,15\}$ satisfies this. ✗
• If $S=5$, then $T=R+8$. Taking $R=4$ gives $T=12$, both available. ✓

Thus $Q=15$, $S=5$, $R=4$ and $T=12$; the remaining value is $P=10$.

$P+R-T=10+4-12=2$.

— Option B