IPMAT Indore 2021 · Question Paper & Solutions

Factorise: $1890=2\cdot 3^3\cdot 5\cdot 7$, $130=2\cdot 5\cdot 13$, $170=2\cdot 5\cdot 17$, so
$$N=2^3\cdot 3^3\cdot 5^3\cdot 7\cdot 13\cdot 17.$$

Total divisors: $4\cdot 4\cdot 4\cdot 2\cdot 2\cdot 2=512$.

Divisors divisible by $45=3^2\cdot 5$: the power of 3 must be 2 or 3 (2 choices) and the power of 5 must be 1, 2 or 3 (3 choices), the rest free: $4\cdot 2\cdot 3\cdot 2\cdot 2\cdot 2=192$.

Required count $=512-192=320$.

The $k$-th term is $(4k-3)(4k-1)=16k^2-16k+3$.

$$\sum _{k=1}^{10}(16k^2-16k+3)=16\cdot 385-16\cdot 55+3\cdot 10=6160-880+30=5310.$$

Add 1 to both sides:
$$1+x_{n+1}=(1+x_n)+2\sqrt{1+x_n}+1={\left(\sqrt{1+x_n}+1\right)}^2.$$

So $\sqrt{1+x_{n+1}}=\sqrt{1+x_n}+1$: the sequence $\sqrt{1+x_n}$ is an AP with first term $\sqrt{1+0}=1$ and difference 1, i.e. $\sqrt{1+x_n}=n$.

Hence $x_n=n^2-1$ and $x_{31}=961-1=960$.

A rectangle needs 2 of the 5 parallel lines and 2 of the $n$ perpendicular lines:
$$\left(\genfrac{}{}{0ex}{}{5}{2}\right)\left(\genfrac{}{}{0ex}{}{n}{2}\right)=360\Rightarrow 10\cdot \frac{n(n-1)}{2}=360\Rightarrow n(n-1)=72.$$

So $n=9$.

Rates (tank per minute): T1 $=\frac{1}{30}$, together $=\frac{1}{20}$.

T2 $=\frac{1}{20}-\frac{1}{30}=\frac{3-2}{60}=\frac{1}{60}$.

So T2 alone takes 60 minutes.

Total maximum marks across everyone $=30\times 5\times 200=30000$.

The average rises from $80\%$ to $80.02\%$, i.e. by $0.02\%$ of 30000:
$$\frac{0.02}{100}\times 30000=6\text{ marks in total}.$$

Marks cannot fall, so the combined increase of the two students is 6. The largest one student can gain is when the other gains 0, which is 6 marks.

The weights all go on one pan, so each weight is either used or not. With $k$ weights you can make at most $2^k-1$ different non-zero totals.

• $k=6$: at most $63<100$ totals, so 6 weights are not enough.
• $k=7$: the weights $1,2,4,8,16,32,64$ give every integer from 1 to $127$ (binary representation), which covers 1 to 100.

Minimum $=7$.

Put $y=mx$. The slopes of the first pair satisfy $3m^2+am+2=0$, and those of the second pair satisfy $-3m^2+bm+2=0$, i.e. $3m^2-bm-2=0$.

Let the common slope be $m_1$, the other slope of the first pair $m_2$ and of the second pair $m_3$. From the products of roots:
$$m_1{m}_2=\frac{2}{3},m_1{m}_3=-\frac{2}{3}.$$

Multiplying and using $m_2{m}_3=-1$ (perpendicular): $m_1^2(-1)=-\frac{4}{9}$, so $m_1=\pm \frac{2}{3}$.

Take $m_1=\frac{2}{3}$: then $m_2=1$, $m_3=-1$.

• Sum of roots, first pair: $m_1+m_2=-\frac{a}{3}\Rightarrow \frac{5}{3}=-\frac{a}{3}\Rightarrow a=-5$.
• Sum of roots, second pair: $m_1+m_3=\frac{b}{3}\Rightarrow -\frac{1}{3}=\frac{b}{3}\Rightarrow b=-1$.

(The choice $m_1=-\frac{2}{3}$ just flips both signs.) So $a^2+b^2=25+1=26$.

Split by where the arguments turn non-negative.

• $x<-1$: both terms are 0, so the left side is $0\ne 8$.
• $-1\le x<3$: $f(x-3)=0$, so $2(x+1)=8\Rightarrow x=3$, which is outside this range.
• $x\ge 3$: $(x-3)+2(x+1)=8\Rightarrow 3x-1=8\Rightarrow x=3$, which is valid.

The only solution is $x=3$, so the smallest integer value is 3.

$n(P\cup M)\le 100\%$, so
$$n(P\cap M)=n(P)+n(M)-n(P\cup M)\ge 60+68-100=28.$$

This minimum is reached when every student passed at least one subject. At least 28% passed both.

Putting $x=0$ gives $f(0)=2f(0)$, so $f(0)=0$.

For any $a\ne 0$ and any $b$, set $x=a$, $y=b/a$: then $f(a+b)=f(a)+f(b)$. For $a=0$ this is also true because $f(0)=0$. So $f$ is additive.

An additive function satisfies $f(nx)=nf(x)$ for integers $n$, and therefore $f(qx)=qf(x)$ for every rational $q$. Writing $2021=\frac{2021}{2020}\cdot 2020$:
$$f(2021)=\frac{2021}{2020}f(2020)=\frac{2021}{2020}.$$

(This is what $f(x)=\frac{x}{2020}$ gives, but the argument does not need to assume $f$ is linear.)

Remove one logarithm at a time, using ${\log }_{k}X=0\Rightarrow X=1$ and ${\log }_{k}X=1\Rightarrow X=k$.

• ${\log }_3({\log }_{4}a)=1\Rightarrow {\log }_{4}a=3\Rightarrow a=64$.
• ${\log }_4({\log }_{2}b)=1\Rightarrow {\log }_{2}b=4\Rightarrow b=16$.
• ${\log }_2({\log }_{3}c)=1\Rightarrow {\log }_{3}c=2\Rightarrow c=9$.

$a+b+c=64+16+9=89$.

$S_9-S_5=a_6+a_7+a_8+a_9=0$. These four terms are symmetric about the middle of $a_7$ and $a_8$, so $a_7+a_8=0$, i.e. $2a+13d=0$ and $a=-\frac{13d}{2}$.

$$a_3=a+2d=-\frac{9d}{2},a_5=a+4d=-\frac{5d}{2}.$$

So $a_3:a_5=9:5$.

Because $AB=BA$, both $A$ and $B$ commute with $A+B$ and so with $(A+B{)}^{-1}$. Therefore
$$B(A+B{)}^{-1}B-A(A+B{)}^{-1}A=(B^2-A^2)(A+B{)}^{-1}=(B-A)(B+A)(A+B{)}^{-1}=B-A.$$

(Here $B^2-A^2=(B-A)(B+A)$ holds because $A$ and $B$ commute.)

The whole expression is then
$$2A-B+(B-A)=A,$$
which is option A.

The official key gives option D ($I$). That cannot be right as the question is printed: if $A$ and $B$ are replaced by $tA$ and $tB$, every term of the expression is multiplied by $t$, so the result must scale with $A$ and $B$. A constant such as $I$ does not. A quick numerical check with random commuting matrices also returns $A$.

Angles in AP give $A+C=2B$, and $A+B+C=180^{\circ }$, so $B=60^{\circ }$ and $A+C=120^{\circ }$.

Since $0<A<120^{\circ }$, we have $60^{\circ }<2A+60^{\circ }<300^{\circ }$. The only angle in this range with sine $\frac{1}{2}$ is $150^{\circ }$, so $2A+60^{\circ }=150^{\circ }$, giving $A=45^{\circ }$ and $C=75^{\circ }$.

$$\sin (B+2C)=\sin (60^{\circ }+150^{\circ })=\sin 210^{\circ }=-\frac{1}{2}.$$

Unit digits repeat with period 4 for bases ending in 3 and 7.

• $3^{85}$: $85\equiv 1(\mathrm{mod}4)$, so the unit digit is 3.
• $525^{37}$ ends in 5.
• $7^{96}$: $96\equiv 0(\mathrm{mod}4)$, so the unit digit is 1.

The value is positive ($987^{96}$ is by far the largest term), and its unit digit is $(3-5+1)\mathrm{mod}10=-1\mathrm{mod}10=9$.

Let $c=\cos x\in [-1,1]$. Then
$$p=3(1-c^2)+12c-3=-3c^2+12c=-3(c-2{)}^2+12.$$

The vertex is at $c=2$, outside $[-1,1]$, so $p$ increases over the whole interval:

• $c=-1$: $p=-3-12=-15$
• $c=1$: $p=-3+12=9$

Hence $p\in [-15,9]$, which is option C.

The official key gives D, $[-15,12]$. The value 12 is the vertex value at $c=2$, but $\cos x$ can never equal 2, so $p=12$ has no solution. For example, $p=10$ would need $-3c^2+12c=10$, whose roots are $c\approx 1.18$ and $c\approx 2.82$, both outside $[-1,1]$.

Triangles AOB and AOD share the height from A to BD, so $\frac{[AOB]}{[AOD]}=\frac{OB}{OD}$. Similarly $\frac{[BOC]}{[COD]}=\frac{OB}{OD}$. Hence
$$[AOB]\cdot [COD]=[AOD]\cdot [BOC]=36.$$

By AM–GM, $[AOD]+[BOC]\ge 2\sqrt{36}=12$, with equality when both equal 6 (a trapezium with $AB\parallel CD$).

Minimum area $=4+9+12=25$.

Let the leading digit be $a$, so the number $N<(a+1)\cdot 1000$.

• If the other three digits are all 0, then $N=1000a$ and the digit sum is $a$, so the ratio is exactly 1000.
• Otherwise the digit sum is at least $a+1$, so the ratio is below $\frac{(a+1)\cdot 1000}{a+1}=1000$.

The maximum is 1000 (for example 1000, 2000, …, 9000). Checking all four-digit numbers confirms this.

$g$ passes through $(a,b)$, so $b=-2a$.

$f(-2)=0$ gives $4a-2b+c=0\Rightarrow 4a+4a+c=0\Rightarrow c=-8a$.

So $f(x)=a(x^2-2x-8)=a\left[(x-1{)}^2-9\right]$. Since $a>0$, the minimum of $f$ is $-9a$, at $x=1$.

Minimum of $f(x)+9a+1=-9a+9a+1=1$, which is option B, whatever the value of $a$. A numerical check with several values of $a$ also gives 1.

The official key gives D (3). We could not reach 3 from the question as printed. It may come from a misprint in the constant or in $g(x)$.

Take the population as 100.

• Exactly one language: 50; at least two: 40. So speakers of none $=100-90=10$.
• None is twice 'all three', so all three $=5$ and exactly two $=40-5=35$.
• 'Exactly one language, that language being English or Tamil' $=25$, so only Hindi $=50-25=25$.

Let $x$ be the number who speak Hindi and exactly one of English/Tamil (that is, H&E only or H&T only). Then
$$\text{Hindi}=\text{only H}+x+\text{all three}\Rightarrow 52=25+x+5\Rightarrow x=22.$$

Answer: 22%. (The other reading, 'speaks exactly one of English and Tamil regardless of Hindi', gives 11%, which is not an option. So the reading above is the intended one.)

Let the speeds be $u$ (first train) and $v$ (second). At 8 PM both have covered the same distance: the first has run 8 h and the second 6 h, so $8u=6v\Rightarrow v=\frac{4u}{3}$.

$u+v=140\Rightarrow \frac{7u}{3}=140\Rightarrow u=60$, $v=80$ km/h.

New case: the second train starts at 5 PM, when the first is $5\times 60=300$ km ahead. The relative speed is $20$ km/h, so catching up takes $\frac{300}{20}=15$ h.

5 PM + 15 h = 8 AM next day.

The odd places (1st, 3rd, 5th) must hold distinct odd digits from $\{1,3,5,7,9\}$: $5\times 4\times 3=60$ ways. The leading digit is odd, so it is never 0.

The even places (2nd, 4th) take any 2 of the remaining 7 digits in order: $7\times 6=42$ ways.

Total $=60\times 42=2520$. A brute-force count confirms this.

If no three points were collinear there would be $\left(\genfrac{}{}{0ex}{}{10}{2}\right)=45$ lines.

The 5 collinear points give $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ pairs that all lie on one line, so replace those 10 lines by 1:
$$45-10+1=36.$$

Intersection: from $x=4-2y$, substitute into $2x+3y=6$ to get $8-4y+3y=6$, so $y=2$ and $x=0$. The point is $(0,2)$.

$3x-y=2$ has slope 3, so the perpendicular slope is $-\frac{1}{3}$.

Line: $y-2=-\frac{1}{3}x$. Setting $y=0$ gives $x=6$.

Completing the table. Each team plays 5 matches (win 2, draw 1).

• A: $L=0$, $W+D=5$, $2W+D=8$ → $W=3,D=2$.
• B: $L=2$, $W+D=3$, $2W+D=6$ → $W=3,D=0$.
• C: $L=2$, $W+D=3$, $2W+D=5$ → $W=2,D=1$.
• D: $L=1$, $W+D=4$, $2W+D=5$ → $W=1,D=3$.
• E: $L=1$, $W+D=4$, points $=W+4<5$ → $W=0,D=4$, 4 points.
• Total points $=15\times 2=30$, so F has $30-28=2$ points.

Fixing the results.

• A never lost and B never drew, so A beat B.
• B never drew and E never won, so B beat E. That is E's only loss, so E drew with A, C, D and F.
• C's single draw is with E. C beat D (given), lost to B (given), and lost to A (A never lost). So C beat F.
• D's only loss is to C. D cannot draw with B, so D beat B. D's other results (vs A and vs F) are draws.
• A: drew with D and E, beat B, C and F.
• B: lost to A and D, beat C, E and F.
• F: lost to A, B and C; drew with D and E (2 points ✓).

Drawn matches: A–D, A–E, C–E, D–E, D–F, E–F, a total of 6. (A brute-force search over all $3^{15}$ outcomes finds this as the only consistent table.)

From the completed table in Q26, the draws are A 2, B 0, C 1, D 3, E 4, F 2.

Team E has the most draws (4). It had no wins, since any win would give it at least 5 points.

The 15 matches distribute $15\times 2=30$ points. A, B, C, D and E have $8+6+5+5+4=28$, so F has 2 points, from its draws with D and E (see Q26).

A has 3 wins and 2 draws. From Q26, A's draws are with D (D's only loss is to C, and D beat B) and with E (E's only loss is to B).

So A beat B, C and F but did not defeat D. E is also undefeated by A but is not among the options.

The table fixes the losses for A, B, C, D and E and the points for A, B, C and D; E and F must finish below 5 points; B beat C and C beat D. Checking every one of the $3^{15}$ ways the fifteen matches could go, exactly one tournament satisfies all of it (W-L-D, points):

• A 3-0-2, 8 points
• B 3-2-0, 6 points
• C 2-2-1, 5 points
• D 1-1-3, 5 points
• E 0-1-4, 4 points
• F 0-3-2, 2 points

B has no draws, so B's match against E was decisive, and E has no wins — so B beat E. That is E's single loss, which means every other team, A included, drew with E.

Answer: only Team B (option C).

The official key gives B, "Only Team A". That would need E to lose to A as well as to B (B cannot draw), giving E two losses where the table shows one.

The same unique tournament answers the rest of the set: 6 drawn matches (Q26), E with the most draws (Q27), F on 2 points (Q28) and D as the team A did not beat (Q29) — all four agree with the official key, which is a good check that the table has been read correctly.

The opening paragraph says logic and rhetoric once trained young men to spot spurious arguments. It then says Italy wants its youth prepared 'as their ancient counterparts were' to deal with social media. The aim, learning to recognise false arguments and facts, is the same then and now, so B is correct.

• A contradicts that parallel.
• C misreads 'privileged': the students were already privileged, and education was not meant to win them privilege.
• D (the basis of education has expanded) is not argued in the passage.

The third paragraph says legacy media is ordered by a need for balance and at least tries to tell both sides of a story. On social media the user is the only editor and algorithms reinforce existing biases. So B is correct.

• A (reach) is never compared.
• C reverses the passage.
• D is wrong because the passage links fake stories to online media, not to traditional media.

The passage says the effect of social media on opinion goes 'far beyond fake news'. The key mechanism is selection: algorithms show people more of what they 'like', so biases are reinforced rather than challenged. That is the choice of which items are presented, option A.

• B describes traditional media, not social media.
• C sounds plausible, but the passage does not say social media steers readers to motivated conclusions. It says it filters what they see.
• D covers only fake news, which the passage explicitly treats as just one part of the problem.

'Vested interests' are parties with a personal stake who act to benefit themselves. Here it means groups that hijack public debate for their own ends. C, self-serving intentions, captures this.

• A (hidden purposes): such interests need not be hidden.
• B (popular interests) is the opposite of a narrow stake.
• D (profitable motives) is too narrow, because the stake can be political, as with the propaganda example.

'Spurious' means false, fake or not what it claims to be, so its antonym is genuine.

• 'Dubious' is close to a synonym.
• 'Inferior' is about quality, not authenticity.
• 'Incisive' means sharp or penetrating and is unrelated.

The Roman Forum was the public space for debate and discussion. The passage talks about youth needing to 'engage meaningfully' there and about protecting public debate. So the comparison is to a space for discussing ideas.

• A and D describe misuses of that space (propaganda, spurious arguments), not what makes it a forum.
• C (political ambitions) is not mentioned.

'Domineer' means to control someone in an overbearing, bossy way, which fits an impatient and intolerant mother. Option A is the intended choice.

• B uses 'dominate', but 'the child's lives' is faulty: a singular possessive with plural 'lives' and no agreement with 'her'.
• C, 'denominate', means to name or express in a unit (e.g. currency) and does not fit.
• D, 'dominion', is a noun and cannot be used as a verb.

(Note: in modern usage 'domineer' usually takes 'over', as in 'domineer over her children', but A is still clearly the best of the four options.)

The verb meaning 'to influence' is affect, so B is correct.

• A: 'effect' as a verb means to bring about, as in 'effect a change', which is wrong here.
• C: 'affective' relates to emotions and moods.
• D: 'affectionate' means loving. Neither fits.

To keep someone at arm's length is an idiom for avoiding close involvement with them, which explains the gain in productive time.

• A ('a foot away') is literal and not an idiom.
• C ('at bay's length') confuses 'keep at bay' with 'at arm's length'.
• D ('out of harm's way') means keeping someone safe, the wrong sense.

To make ends meet is to earn just enough to cover expenses. Cutting costs is exactly how one does that.

• B means finishing off remaining details.
• C and D do not follow logically from cutting expenses.

To cut corners is to do something cheaply or carelessly, often cutting quality to save money. That matches poor materials and a collapse.

• A is not an idiom.
• C ('bit the bullet') means facing something unpleasant bravely.
• D ('made a fortune') gives no cause for the collapse.

On a roll means enjoying a continuous run of success, which fits an unbroken winning streak.

• 'Rolling in money' means very rich.
• 'Rolling in luck' is not a standard idiom.
• 'On a roller coaster' suggests ups and downs, the opposite of steady wins.

'Described as' takes a noun phrase directly: 'described as a mere junior member'. 'Being' is unnecessary, and 'merely being' is wordy.

• A repeats the original error.
• B puts 'merely' in front of the adjective 'junior', which is awkward, and keeps 'being'.
• C changes the meaning: 'described for' is unidiomatic.

D is concise and idiomatic.

The subject is a city, so the pronoun must be the singular possessive its. The present perfect 'has depended' fits a situation that continues into recent years.

• A uses 'it's' (short for 'it is') and the past perfect 'had depended', which clashes with 'recent years'.
• C ('having depended') leaves the main clause without a finite verb.
• D keeps the plural 'their'.

B is correct.

Two errors need fixing:

• 'People' is countable, so use fewer, not 'less'.
• A comparison needs than, not 'then'.

Only C fixes both. A fixes only 'fewer', and B fixes neither. D uses 'than have ever before', where 'have' does not match the verb 'die' (it would need 'did'), so it is ungrammatical.

The paragraph gives a benefit from reintroducing a herbivore (bandicoots) and a risk from reintroducing predators. A conclusion that covers both is that success depends on what the introduced animals eat, which is C.

• A draws a policy recommendation (revive more herbivores) that goes beyond the passage.
• B addresses only the carnivore half and exaggerates it ('so many').
• D contradicts the warning about predators.

The sentences before the blank list worries: postponement, wary locals, anxious athletes, cut-back training and restrictions. The blank should sum these up by naming their cause, and the next sentence ('Yet Japan… safe choice') pushes back against that concern. B does both.

• A is positive in tone, which weakens the contrast that 'Yet' sets up.
• C is also positive, so 'Yet' would make no sense after it.
• D adds a new, unsupported issue (recession, funds) that the following sentences ignore.

• Blank 1: a mistaken belief is a misconception.
• Blank 2: if listening were just hearing, it would come automatically, i.e. instinctive.
• Blank 3: making little effort to develop a skill is to neglect it.

Other options:

• A: 'forget a function' is odd, and 'blunder' is an act, not a belief.
• B: 'impulsive' means acting rashly, the wrong sense.
• C: 'natural' works, but 'reject' is too strong. We do not actively reject listening.

• Governments relax pandemic regulations (rules and restrictions).
• People working in event management are professionals.
• 'Reaching and potentially surpassing' earlier levels moves logically from matching to exceeding.

Other options:

• B: 'confounding' levels makes no sense.
• C: 'reducing' contradicts a recovery, and one does not relax 'resolutions'.
• D: 'relax their pandemic systems' and 'event management masters' are unidiomatic.

The two adjectives should describe the kind of person who would make a top manager, and the last blank needs a superlative. Assiduous (hard-working) and analytical explain why he was the best manager.

• A pairs 'ludicrous' (absurd) with 'assiduous', an incoherent description.
• C and D put non-superlative adjectives ('compassionate', 'friendly') in the last blank, which does not fit 'the ___ manager the company ever had'.

Performance appraisals are delivered orally and in writing, and achievements should be recognised the same way. 'Deliver it orally as well as in ___' needs a mode of communication, and only 'writing' is one.

• A: 'in awards' is not a mode of delivery.
• B: 'indictment' (a formal accusation) is the wrong context, and 'realise achievements' is unidiomatic.
• D: 'in increments' is not a mode of delivery, and 'advertise' is inappropriate.

• The oldest surviving geological feature, i.e. still in existence despite its age.
• Over hundreds of millions of years mountains are eroded to low hills.
• The hills have been witness to events in history (the idiom is 'witness to').

Other options:

• A: 'sentinel to' is unidiomatic.
• B: 'destroyed to low hills' is wrong, since destroyed things do not remain as hills, and 'testament to events' is off.
• D: 'broken down' is less precise, and 'observers to' is unidiomatic.

Climatic refers to climate or weather; climactic refers to a climax.

In D, the change from summer to winter is a change in weather, so it should be 'climatic changes'. 'Climactic' is misused.

• A ('climatic changes') is correct.
• B ('climatic conditions') is correct.
• C ('climactic scene' of a movie) is correct.

Luxurious means rich, costly and comfortable. Luxuriant means growing thickly and abundantly, as with foliage or hair.

Trimmings of gold, silk and velvet are costly, so B should say 'luxurious'.

• C ('luxuriant, golden tresses') correctly describes thick hair.
• A and D correctly use 'luxurious'.

Intense describes a strong degree of a feeling or force, as in intense pressure, loathing or debate. Intensive describes concentrated, thorough effort or activity, as in intensive study or intensive care.

Pressure is a force of high degree, so A should be 'intense pressure'.

• B ('intense' loathing) is correct.
• C ('intensive study') is correct.
• D ('intense debate') is correct.

Appraise means to assess the value or quality of something. Apprise means to inform, and is used as 'apprise someone of something'.

In D the soldiers inform the officer 'of the enemy's location', so it should be 'apprise'.

• A (appraise a car) is correct.
• B (apprise the board of progress) is correct.
• C (appraise their own performance) is correct.

• Lighting means setting alight: 'lighting of the lamp' (A) is correct.
• Lightning is the electrical flash in a storm: 'struck by lightning' (B) is correct.
• Lightening means making lighter in colour or weight: 'lightening her heavy load' (D) is correct.

In C, products that make skin lighter are 'skin lightening' products, so 'lightning' is misused.

• 4 opens with the general thesis: to succeed globally you must learn about conditions in overseas markets.
• 1 gives the past: companies only learned legal and procedural differences.
• 2 turns with 'however': modern trade demands more.
• 3 explains that 'more': today companies must also understand the people. 'Today' contrasts with 'In the past' in 1.

'However' in 2 must follow the past practice in 1, and 3 spells out 2, so the order is 4-1-2-3.

• A (2314) starts with 'however', which has nothing to contrast with.
• B and D begin with 'Today… also', which assumes an earlier point.

• 2 gives the general observation: social media makes us expect every post to be irreproachable.
• 3 introduces a specific article asking whether celebrities should always have a take.
• 1 says 'It mentions a video'. 'It' needs 3's 'article' as its antecedent.
• 4 is the quoted question, which reads as the comment on that video.

The pairs 3→1 ('It' = the article) and 1→4 (video → quoted reaction) are fixed, and 2 frames them, so the order is 2-3-1-4.

• B (2413) puts the quote before its context.
• C (3124) leaves 2 dangling at the end.
• D (1324) starts with 'It', which has no referent.

• 3 is the general context: leopards stray into villages because their habitat is shrinking.
• 1 is a specific recent incident: a cub found in a trench.
• 4 follows: a vet examined the cub ('it').
• 2 ends it: 'The cub was then taken back' and reunited with her mother.

The story runs discovery → examination → release (1-4-2), with the general statement 3 introducing it, so the order is 3-1-4-2.

• A (3421) has the vet examining before the cub is found.
• B and C start with the incident, and C puts the general statement in the middle of the story.