IPMAT Indore 2025 · Question Paper & Solutions

Let Arpita alone take $x$ days and Nikita alone $y$ days.

• Together: $\frac{1}{x}+\frac{1}{y}=\frac{1}{12}$
• Split job: $0.4x+0.6y=24\Rightarrow 2x+3y=120\Rightarrow x=\frac{120-3y}{2}$

Substitute: $\frac{2}{120-3y}+\frac{1}{y}=\frac{1}{12}$, so $12(120-y)=y(120-3y)$, i.e. $3y^2-132y+1440=0\Rightarrow y^2-44y+480=0\Rightarrow (y-20)(y-24)=0$.

• $y=20\Rightarrow x=30$: check $\frac{1}{30}+\frac{1}{20}=\frac{1}{12}$ and $12+12=24$ days.
• $y=24\Rightarrow x=24$: check $\frac{1}{24}+\frac{1}{24}=\frac{1}{12}$ and $9.6+14.4=24$ days.

The official answer is 20 (Nikita faster than Arpita).

Total members $=75$, so each skill set has $25$ members.

Explorers. They are distinct and decrease A > B > C > D > E, each at least 2. Upper bounds:

• C has 6 biologists and more biologists than explorers, so $E_C\le 5$; hence $E_D\le 4$, $E_E\le 3$.
• B: biologists > explorers and geologists $\ge 2$, so $2E_B+1\le 13\Rightarrow E_B\le 6$.
• A: 6 geologists and at least 2 biologists, so $E_A\le 7$.

Maximum possible total $=7+6+5+4+3=25$, which must be reached exactly. So explorers are A 7, B 6, C 5, D 4, E 3.

Complete the grid (Biologists, Geologists, Explorers):

• A: $15-6-7=2$ biologists → (2, 6, 7)
• B: biologists $>6$ and geologists $\ge 2$ with $B+G=9$ → (7, 2, 6)
• C: (6, 4, 5); D: (6, 5, 4)
• E: biologists $=25-(2+7+6+6)=4$, geologists $=15-4-3=8$ → (4, 8, 3)

Check geologists: $6+2+4+5+8=25$. ✓

Biologists in team E $=$ 4.

By the triangle inequality, $|a-b|+|b-c|\ge |a-c|$, with equality exactly when $b$ lies between $a$ and $c$.

Since the numbers are distinct, $b$ must be strictly between them, so one of $a,c$ is larger than $b$. The largest allowed number is $99$, so $b\le 98$. Example: $a=1,b=98,c=99$ gives $97+1=98$. ✓

Maximum $b=$ 98.

A full round robin of 8 teams has $\left(\genfrac{}{}{0ex}{}{8}{2}\right)=28$ matches.

The suspended team would have played 7 matches but played only 3, so $7-3=4$ matches are cancelled. All other matches go ahead.

Matches played $=28-4=$ 24.

Write each term as $\ln a-k\ln b$, where $k=1,\frac{3}{2},2,\frac{5}{2},\dots$, i.e. $k_i=1+\frac{i-1}{2}$.

Sum of 21 terms $=21\ln a-\left({\sum }_{i=1}^{21}{k}_i\right)\ln b$.

$$\sum _{i=1}^{21}{k}_i=21+\frac{0+1+\cdots +20}{2}=21+\frac{210}{2}=126$$

So the sum is $\ln \frac{a^{21}}{b^{126}}$: $m=21$, $n=126$, and $m+n=$ 147.

Let $x$ = absent from Math, so $2x$ = absent from English. Let $F_E,F_M$ = appeared but failed; $P_E,P_M$ = passed.

Given $P_M=2F_E$ and $P_E=2F_M$:

• English: $P_E+F_E=120-2x\Rightarrow 2F_M+F_E=120-2x$
• Math: $P_M+F_M=120-x\Rightarrow 2F_E+F_M=120-x$

Add: $3(F_E+F_M)=240-3x\Rightarrow F_E+F_M=80-x$.

Subtract (Math − English): $F_E-F_M=x$.

Adding these two results: $2F_E=80\Rightarrow F_E=$ 40 (independent of $x$).

$$A^2={\left[\begin{array}{cc}2& n\\ 4& 1\end{array}\right]}^2=\left[\begin{array}{cc}4+4n& 3n\\ 12& 4n+1\end{array}\right]$$

$$A^3=A^{2}A=\left[\begin{array}{cc}8+20n& 4n^2+7n\\ 28+16n& 16n+1\end{array}\right]$$

Top-left must equal $27\times 4=108$: $8+20n=108\Rightarrow n=5$.

Then $A^3=\left[\begin{array}{cc}108& 135\\ 108& 81\end{array}\right]=27\left[\begin{array}{cc}4& 5\\ 4& 3\end{array}\right]$, so $q=5,p=4,r=3$.

$p+q+r=$ 12.

From the grid derived in Q2 (Biologists, Geologists, Explorers):

• A (2, 6, 7), B (7, 2, 6), C (6, 4, 5), D (6, 5, 4), E (4, 8, 3)

The explorer totals are forced: the upper bounds A ≤ 7, B ≤ 6, C ≤ 5, D ≤ 4, E ≤ 3 add to exactly 25, so every team is at its bound, and the rest follows.

Geologists > biologists: A ($6>2$) and E ($8>4$). B, C, D have more biologists.

Answer: 2 teams.

AP condition: $2{\log }_3(2x^2+1)={\log }_3(x^2-1)+{\log }_3(6x^2+3)$, so

$$(2x^2+1{)}^2=(x^2-1)\cdot 3(2x^2+1)$$

Since $2x^2+1>0$, divide: $2x^2+1=3x^2-3\Rightarrow x^2=4$.

Terms: ${\log }_{3}3,{\log }_{3}9,{\log }_{3}27=1,2,3$ (common difference 1).

Next three terms: $4+5+6=$ 15.

Place $B=(0,0)$, $A=(0,18)$, $C=(18,0)$, $D=(18,18)$. A circle tangent to $AB$ ($x=0$) and $BC$ ($y=0$) with radius 13 has centre $(13,13)$, so $M=(0,13)$.

On $CD$ ($x=18$): $5^2+(y-13{)}^2=169\Rightarrow y-13=\pm 12$; inside the square $y=1$, so $P=(18,1)$.

Triangle $PMD$: $P$ and $D$ lie on $x=18$, so base $PD=18-1=17$; the height is the distance from $M$ to $x=18$, which is 18.

Area $=\frac{1}{2}\times 17\times 18=$ 153 sq. cm.

Father's age $=3\times 18=54$, so the age gap is $54-18=36$ years, and it never changes.

If Monica is $m$ when she is half his age: $2m=m+36\Rightarrow m=$ 36 (father 72).

From the grid in Q2, biologists are A 2, B 7, C 6, D 6, E 4.

Sorted: $2,4,6,6,7$. The middle value is 6.

Work mod 7: $200\equiv 4$ and $11n\equiv 4n$, so $4n\equiv 4\Rightarrow n\equiv 1(\mathrm{mod}7)$.

• $n=1$: $m=27$, $m+n=28$
• $n=8$: $m=16$, $m+n=24$
• $n=15$: $m=5$, $m+n=20$
• $n=22$: $11n>200$, not allowed

Each step raises $n$ by 7 and lowers $m$ by 11, so $m+n$ falls by 4. The largest valid $n$ gives the minimum: 20.

A square factor $3^a{5}^b{7}^c$ needs every exponent even:

• $a\in \{0,2,4\}$: 3 choices
• $b\in \{0,2,4,6,8\}$: 5 choices
• $c\in \{0,2\}$: 2 choices

Total $=3\times 5\times 2=$ 30.

Remainder theorem:

• $f(1)=a+b+5=3\Rightarrow a+b=-2$
• $f(-1)=a-b+5=2\Rightarrow a-b=-3$

So $a=-\frac{5}{2}$, $b=\frac{1}{2}$, and $2b-4a=1+10=$ 11.

Let $S=\sum \frac{1}{n^2}=\frac{{\pi }^2}{6}$. The even terms are $\sum \frac{1}{(2n{)}^2}=\frac{1}{4}S$.

Odd terms $=S-\frac{1}{4}S=\frac{3}{4}\cdot \frac{{\pi }^2}{6}=\frac{{\pi }^2}{8}$.

$y-a=b\ln x=\ln x^b\Rightarrow e^{y-a}=x^b\Rightarrow e^y=e^a{x}^b$.

Since $e^a$ is a constant, $e^y\propto x^b$.

• (B) fails: $y-a=\ln x^b$, a logarithm of $x^b$, not proportional to it.
• (A), (D) do not follow from the relation.

$x^8+x^7+\cdots +1=\frac{x^9-1}{x-1}$, so the roots are the ninth roots of unity other than 1. Each satisfies $a_i^9=1$.

$2025=9\times 225$, so $a_i^{2025}=(a_i^9{)}^{225}=1$.

Sum $=8\times 1=8$.

Let the numbers in order be $p\le m\le q$, with $p+q=15$.

$m-\frac{p+m+q}{3}=2\Rightarrow 3m-(15+m)=6\Rightarrow 2m=21\Rightarrow m=10.5$.

Peel off one layer at a time:

• $5{\log }_3(1+{\log }_3(1+2{\log }_{2}x))=25^{1/2}=5$
• ${\log }_3(1+{\log }_3(1+2{\log }_{2}x))=1\Rightarrow 1+{\log }_3(1+2{\log }_{2}x)=3$
• ${\log }_3(1+2{\log }_{2}x)=2\Rightarrow 1+2{\log }_{2}x=9$
• ${\log }_{2}x=4\Rightarrow x=16$

Hundreds digit is 1, 2 or 3; the last two digits must add to 9, 8 or 7.

• $h=1$ (sum 9): 10 numbers; last two digits divisible by 4: 36, 72 → 2
• $h=2$ (sum 8): 9 numbers; 08, 44, 80 → 3
• $h=3$ (sum 7): 8 numbers; 16, 52 → 2

Total $=27$ numbers, favourable $=7$. Probability $=\frac{7}{27}$.

By the sine rule, a side equals $2R$ times the sine of the opposite angle. Side $AC=x$ is opposite $\angle B=\theta$:

$x=2y\sin \theta \Rightarrow \frac{x}{y}=2\sin \theta$.

Let the roots be $r$ and $pr$.

• Sum: $r(1+p)=\frac{2k}{8}=\frac{k}{4}$
• Product: $p{r}^2=\frac{k}{8}$

Divide the square of the sum by the product: $\frac{(1+p{)}^2}{p}=\frac{k^2/16}{k/8}=\frac{k}{2}$.

$k=\frac{2(1+p{)}^2}{p}=2{\left(\sqrt{p}+\frac{1}{\sqrt{p}}\right)}^2$.

Slope of $AB=\frac{1-3}{5-1}=-\frac{1}{2}$.

$\tan 45^{\circ }=\left|\frac{m-(-\frac{1}{2})}{1+m(-\frac{1}{2})}\right|=1\Rightarrow m+\frac{1}{2}=\pm \left(1-\frac{m}{2}\right)$.

• $+$: $\frac{3m}{2}=\frac{1}{2}\Rightarrow m=\frac{1}{3}$
• $-$: $m+\frac{1}{2}=-1+\frac{m}{2}\Rightarrow m=-3$

Determinant $=P(0)P(2)-P(0)P(1)=P(0)[P(2)-P(1)]=0$. Since $P(0)=2\ne 0$, $P(2)=P(1)$.

Let $P(x)=a{x}^2+bx+2$. $P(2)-P(1)=3a+b=0\Rightarrow b=-3a$.

• $P(1)=P(2)=2-2a$, $P(3)=9a-9a+2=2$
• Sum: $4-4a+2=14\Rightarrow a=-2,b=6$

$P(4)=16(-2)+4(6)+2=-6$.

Tangent to the y-axis at $(0,4)$ means the centre is $(h,4)$ and the radius is $|h|$.

Through $(-2,0)$: $(h+2{)}^2+4^2=h^2\Rightarrow 4h+20=0\Rightarrow h=-5$.

Radius $=5$.

The path is parallel to each side at 1 m (straight parts totalling the perimeter $4+6+9=19$), joined at each vertex by a circular arc of radius 1.

The arcs' angles are the exterior angles of the triangle, which add to $360^{\circ }$, so together they form one full circle of length $2\pi (1)=2\pi$.

Total $=19+2\pi$ m.

Let $B+C=70000$. First-year interest: $3000+0.08B+0.12C=10600$.

$0.08(70000-C)+0.12C=7600\Rightarrow 5600+0.04C=7600\Rightarrow C=50000,B=20000$.

Second year, interest is on the year-1 amounts:

• A: $33000\times 10\%=3300$
• B: $21600\times 8\%=1728$
• C: $56000\times 12\%=6720$

Total $=3300+1728+6720=11748$.

The leading coefficient $a^2>0$ (since $a\ne 0$), so the graph is an upward parabola.

It has a minimum at $x=-\frac{b}{a^2}$ and grows without bound, so there is no maximum.

Vertices:

• $y=0$ and $12x=5y$: $(0,0)$
• $y=0$ and $3x+4y=7$: $\left(\frac{7}{3},0\right)$
• $y=\frac{12x}{5}$ in $3x+4y=7$: $3x+\frac{48x}{5}=7\Rightarrow x=\frac{5}{9},y=\frac{4}{3}$

Base on the x-axis $=\frac{7}{3}$, height $=\frac{4}{3}$. Area $=\frac{1}{2}\cdot \frac{7}{3}\cdot \frac{4}{3}=\frac{14}{9}$.

Joining the centre to the vertices gives 8 congruent triangles, each with two sides 1 and included angle $\frac{360^{\circ }}{8}=45^{\circ }$.

Area $=8\times \frac{1}{2}\cdot 1\cdot 1\cdot \sin 45^{\circ }=4\cdot \frac{\sqrt{2}}{2}=2\sqrt{2}$.

Take the pool length as 30 units: Ankit's speed is 3 units/min, Bipul's 2 units/min.

First meeting: they close 30 units at 5 units/min, so they meet at $t=6$ min.

Later meetings: after each meeting they must together cover $2\times 30=60$ units before the next head-on meeting, which takes $60/5=12$ min. Meetings: $t=6,18,30,42,54,66,78$.

Check $t=30$: Ankit has covered 90 units (at the far end), Bipul 60 units (back at his start, the same far end), so they are together. The next would be at $t=90>80$.

They never overtake each other at any other time, so they meet 7 times.

$S_7=S_5\Rightarrow T_6+T_7=0\Rightarrow a{r}^5(1+r)=0\Rightarrow r=-1$ (since $a,r\ne 0$).

With $r=-1$ the terms are $a,-a,a,\dots$, so $S_9=a=24$.

$T_4=a{r}^3=-24$.

Base: $x+\frac{1}{x}>0$ requires $x>0$, and then $x+\frac{1}{x}\ge 2>1$, so the base is always greater than 1.

With base $>1$, the inequality means ${\log }_2\frac{x-1}{x+2}>1$, i.e. $\frac{x-1}{x+2}>2$.

For $x>0$, $x+2>0$, so $x-1>2x+4\Rightarrow x<-5$, which contradicts $x>0$.

(Also, for $x>0$ we always have $\frac{x-1}{x+2}<1$, so the inner log is negative and the outer log is not even defined.)

No $x$ works: null set.

$a_k=100+5(k-1)$, $b_k=100-5(k-1)$, so $a_k{b}_k=10000-25(k-1{)}^2$.

$$\sum _{k=1}^{20}{a}_k{b}_k=20\times 10000-25\sum _{j=0}^{19}{j}^2=200000-25\cdot \frac{19\cdot 20\cdot 39}{6}$$

${\sum }_{j=0}^{19}{j}^2=2470$, so the sum $=200000-61750=138250$.

A wins on his 1st, 2nd, 3rd … shot, each round both missing first:

$$P=0.4+(0.6)(0.4)(0.4)+(0.6\cdot 0.4{)}^2(0.4)+\cdots =\frac{0.4}{1-0.6\times 0.4}=\frac{0.4}{0.76}=\frac{10}{19}$$

(Miss probabilities: A $0.6$, B $0.4$.)

Let $y=3^{10}$. Then $3^{30}+8=y^3+2^3=(y+2)(y^2-2y+4)$, which is divisible by $y+2=3^{10}+2$.

The others fail, since $y\equiv -2(\mathrm{mod}y+2)$:

• $y^2+4\equiv 8$, $y^3+2\equiv -6$, $y^2+8\equiv 12$ — all non-zero remainders, as $y+2$ is far larger.

Let the three disjoint parts be $t-d,t,t+d$. Their total is $n(A\cup B)=3t=18\Rightarrow t=n(A\cap B)=6$.

$n(A)+n(B)=n(A\cup B)+n(A\cap B)=18+6=24$.

Divisible by 5 means the last digit is 5 (no 0 available). The other digits come from $\{1,3,7,8,9\}$.

• 4-digit (must exceed 5000): first digit from $\{7,8,9\}$: 3 ways; middle two from the remaining 4: $4\times 3=12$ → $36$
• 5-digit: any 4 of the 5 digits in order: $5\times 4\times 3\times 2=120$
• 6-digit: all 5 digits arranged: $5!=120$

Total $=36+120+120=276$.

• $11\equiv 2(\mathrm{mod}9)$, and $2^6\equiv 1(\mathrm{mod}9)$. $1011=6\times 168+3$, so $11^{1011}\equiv 2^3=8$.
• $1011$ has digit sum 3, so $1011\equiv 3(\mathrm{mod}9)$, and $3^{11}$ is divisible by 9, so $1011^{11}\equiv 0$.

Remainder $=8+0=8$.

Apparel month-on-month declines (sales in ₹ lakh):

• March: $279\to 236$, $\frac{43}{279}\approx 15.4\%$
• June: $249\to 230$, $\frac{19}{249}\approx 7.6\%$
• October: $288\to 222$, $\frac{66}{288}\approx 22.9\%$
• December: $228\to 221$, $\frac{7}{228}\approx 3.1\%$

(May also falls, by about 3.5%.) Largest decline: October.

Look for three consecutive months with rising return percentages:

• Apparel: Jan 13% → Feb 16% → Mar 20% ✓
• Footwear: Oct 8% → Nov 9% → Dec 10% ✓
• Electronics: 2, 3, 2, 1, 4, 3, 4, 2, 5, 2, 4, 1 — every rise is followed immediately by a fall ✗

Answer: both Apparel and Footwear.

Net sales = sales × (1 − return %).

• May: $69\times 0.94=64.86$
• June: $111\times 0.95=105.45$

Increase $=\frac{105.45-64.86}{64.86}=\frac{40.59}{64.86}\approx 62.58\%$.

Apparel ÷ (Apparel + Footwear + Electronics):

• January: $\frac{262}{655}=40.00\%$
• April: $\frac{258}{641}\approx 40.25\%$
• August: $\frac{252}{648}\approx 38.89\%$
• December: $\frac{221}{575}\approx 38.43\%$

Highest: April (just ahead of January).

Value returned = Footwear sales × return %:

• September: $118\times 9\%=10.62$
• July: $119\times 9\%=10.71$
• June: $111\times 5\%=5.55$
• March: $121\times 7\%=8.47$

Highest: July (same rate as September, slightly higher sales).

The third paragraph says unfiltered content can push users away from social media towards legacy forms that have better moderation. That is option D.

• A: the writer suggests the opposite — unmoderated content can drive users away.
• B: the passage calls informed choice social media's *weakness* (Achilles' heel), not users' inability to fact-check its strength.
• C: "never be discarded" contradicts the idea that users may move to legacy media.

Paragraph two calls social media the connective tissue for mass consumption of entertainment, i.e. the main route through which people reach it.

• A: lawmakers "are unlikely to give social media a free run" and protections already exist, so it is not unaffected.
• C: the writer does not credit its success to publishing anything; unfiltered content is shown as a risk.
• D: the writer says mediums *must* adapt (through moderation), not that social media cannot.

The passage frames Meta's content changes as a response to the political tide turning in Washington, with Washington leaning on foreign governments and Republicans pulling US platforms rightwards. What appears on social media follows whoever holds power.

• A: the writer argues for moderation, not deregulation.
• B: the conclusion states moderation, not free speech, is the adaptation mechanism.
• C: lawmakers worldwide are shown putting protections against misinformation in place, not backing free speech.

The passage says social media now has enough control over all other forms of media, and alternative platforms are reworking how they engage with it. So it controls media that depend on it.

• A: technology (synthetic content) is shown helping social media, not being ignored.
• B: community review is "not a significant improvement" over fact-checking.
• C: it is the connective tissue *linking* other media, not independent of them.

Technologies drive social media's advantage further through synthetic content targeted precisely at its audience. Wider reach plus precise targeting means more influence and persuasion.

• B, C: the passage does not say algorithms cannot control content, or that social media is "accepting" trends.
• D: it is *other* platforms that rework their engagement with social media, not social media adjusting towards competitors.

The first paragraph calls this "the nature of the social media beast, designed to amplify bias". An inherent design that amplifies bias matches C.

• A: the digital divide is never mentioned.
• B: "unremitting expansion of freedom of expression" is not presented as its inherent flaw; the writer discounts free speech as the answer.
• D: amplifying bias is the opposite of reinforcing objectivity.

The passage ends by saying consumption feels like a miracle because commodities are divorced from the social processes that produce them. Reconnecting consumption with production removes the magical expectation that causes disappointment.

• A: banning commercials is an extreme step the passage never proposes.
• C: believing in miraculous change is exactly the problem.
• D: colonialism and oppression are not discussed.

Baudrillard argues consumers have replaced 'real' happiness with 'signs' of happiness. That is option A.

• B: commercials present goods as miracles, but unhappiness is attributed to heightened expectations, not placed "at the heart" on ads.
• C: consumption is *experienced* as magical; he does not say production and consumption are magical.
• D: coping with colonial inequality is not his point; the myth is an analogy.

The cargo myth "holds an important analogy" for how consumers set out sham signs and wait for happiness to arrive — a critique of modern consumption.

• A: he shows consumption fails to deliver happiness, not that it is a blessing.
• C: the myth is not used to describe suffering.
• D: "more serious objects" is never suggested.

The passage says consumers should blame their heightened expectations for their lack of happiness, but blame the commodity instead.

• B: commercials over-hype goods; lack of hype is the reverse.
• C: the "Law of Diminishing Marginal Commodities" is invented and not mentioned.
• D: they think a newer version would be better, but the root cause given is expectation, not functionality.

Replacing real happiness with signs results in "the endless deferment" of total happiness — it is perpetually delayed.

• B: consumers already wait (for the next version) and happiness still never comes.
• C, D: production-over-consumption priorities and ethical consumption are not discussed as routes to total happiness.

Modern consumers view consumption in the same magical way the Melanesians viewed aeroplanes, and expect happiness from sham signs. That is magical, irrational thinking.

• A: it is not driven by practical utility but by signs of happiness.
• B: tempting because the passage opens by saying commodities are not *inherently* negative, but that is about objects; the process of consumption is critiqued.
• C: equality is not discussed.

"Put up with" means tolerate — we tolerate an unpleasant person for his sister's sake.

• A: "put him down" means insult or belittle.
• B: "put him aside" means set aside, which does not fit a person here.
• D: "put along with" is not an idiom.

"Usher in" means to begin or introduce a new period — policies ushering in growth.

• B: "set in" is intransitive and usually used for something unwelcome (winter, decay).
• C: "turn up" means appear or increase.
• D: "set forth" means state or begin a journey.

To appraise is to assess the value of something, which is what you do with jewellery of unknown worth.

• D: "apprised" means informed — a confusable.
• B: "appreciated" means gained in value or valued emotionally, not assessed.
• A: "get an approval" does not tell her its worth.

The list describes deer moving over the road surface; "bounding across the road" (leaping across) fits naturally.

• A, B: deer cannot jump or forage *under/beneath* the road.
• C: "staggering with the road" is not meaningful English.

Itinerant (adjective) means travelling from place to place — exactly a travel vlogger's lifestyle.

• B: "itinerary" is a noun (a travel plan) and cannot describe a lifestyle.
• C: "iterative" means repeating steps.
• D: "manage his itinerary in his life style" is wordy and changes the meaning.

A silver bullet is a single, complete solution to a hard problem. Renewables alone are not enough, since conservation is also needed.

• A: "dead ringer" means an exact look-alike.
• B: "silver lining" is a hopeful side of a bad situation.
• C: "red herring" is a misleading distraction — renewables are called a key solution, not a distraction.

Vandalism makes matters worse; exacerbate means worsen.

• B: "extended their troubles" is unidiomatic.
• C: "exaggerated their hardships" means overstated them, which they did not do.
• D: "extenuated" means made less serious — the opposite.

Two errors: the idiom is regarded as, and "one of the …" takes a plural noun (discoveries). Only B fixes both.

• A: "regarded for being" is wrong.
• C: "regarded like" is wrong and keeps the singular "discovery".
• D: "regarded being" drops the needed "as".

Negligible means so small it can be ignored — right for slight damage (hence "thank goodness").

• D: "negligent" and the original "neglectful" mean careless; they describe people, not damage.
• B: "neglecting" is a verb form.
• C: "neglectable" is non-standard.

Items after "aims to" must be parallel: to increase, (to) reduce, (to) enhance. Only D keeps all three as bare verbs.

• A: mixes increase / reducing / enhanced.
• B: gerunds after "aims to" are ungrammatical, and "reducing of costs" is awkward.
• C: mixes reduce / enhancing and adds needless "the".

The result clause "would have handled" is a past unreal (third) conditional, which needs a past perfect condition: "If he had known", or inverted, "Had the President known".

• A: "Had the President knowledge" is archaic and unidiomatic.
• C: "could know beforehand" does not form the third conditional.
• D: "knew … can" mismatches tenses.

The causative have + person + bare infinitive: "had my brother feed my dog".

• D: "to feed" wrongly adds "to" (same error as the original).
• B: "giving" changes it to a continuous sense.
• C: "who fed" makes it a relative clause and loses the causative meaning.

The gap follows the claim that BMI causes both under- and over-diagnosis and precedes details (fat distribution, true health). D explains *why* ("This is because …") and its two parts are expanded by the next sentences.

• A: says BMI can pinpoint the cause, contradicting the paragraph.
• B: social media misinformation is off-topic.
• C: praises BMI's accuracy, contradicting the argument.

The paragraph describes turning manure methane into hydrogen that powers a zoo. C sums this up as sustainable energy from innovative technology.

• A: shifts to advice for India, which the paragraph gives no basis for.
• B: a flippant reading (one animal powering others) that misses the point about energy.
• D: "always used technology to help animals" is an unsupported generalisation and misreads the aim.

The gap must connect "not merely a mode of communication" to the conclusion "a way of thinking dies". C says language carries a community's mindset and culture, which justifies the "So".

• A: revival movements do not lead to "a way of thinking dies".
• B: whether endangered languages look healthy is a different point.
• D: natural change over time does not explain why loss kills a way of thinking.

After "twenty-three States are heatwave-prone", the paragraph gives health effects (heatstroke cases, deaths) and economic effects (reservoirs, power demand). D introduces exactly these two strands.

• A: contradicts the study's picture of widespread risk and is not what follows.
• B: the monsoon is never taken up afterwards.
• C: compensation is not mentioned and breaks the flow to heatstroke data.

Processing feelings and reducing stress are healing effects, so art is therapeutic, and it would enhance (improve) self-awareness.

• A: "decrease" self-awareness is negative, clashing with the benefits listed.
• B: "disturb" is negative, and "acceptable" is too weak.
• C: "avoidable" and "mitigate" contradict the positive tone.

"An extended period" is the natural collocation; "physical endurance" pairs with "mental resilience", both positive to match "remarkable".

• A: "dysfunctional endurance" contradicts "remarkable".
• B: "explicit period" makes no sense; "mental integrity" is odd.
• C: "expanded period" and "stern endurance" are unidiomatic.

Curcumin is an ingredient (compound) of turmeric known for reducing inflammation, and side effects are induced (brought on).

• A: curcumin is not an enzyme; you do not "infuse" a headache.
• B: it is not an alkali; "promote headache" is odd.
• C: "alchemy" is not a substance.

"Premised on" (based on) fits the preposition; capacities emerge from the larger-scale structural organisation of genetic material; and consciousness belongs in a list with thought and moral agency.

• A: "disorganisation" and "deconstruction" cannot give rise to capacities.
• B: "interdependent on" is wrong usage; "differentiation" is not a capacity like thought.
• C: "calibration" is not a mental capacity; "structural distribution" is weak.

Stress compromises (weakens) the immune system, impairs well-being, and good habits counter the negative effects. All three fit the logic.

• A: stress does not "elevate" well-being, and practices do not "impede" negative effects in this sense.
• C: "improve" well-being contradicts the negative sense.
• D: "preserve" well-being also contradicts it.

One maintains an online persona; it generates unrealistic standards; and it triggers negative self-comparisons. Each word collocates correctly.

• A: one does not "endorse" one's own persona (you endorse products or views).
• C: "advocate a persona" is wrong usage.
• D: "profess a persona" is wrong, "inspire" carries a positive sense, and "release self-comparisons" is unidiomatic.

Bradley measures dynamism with the "shuffle rate": how much the bottom moves to the top. That means frequent changes in market position and leadership.

• A: slow, gradual growth is the opposite.
• C, D: never-changing practices and stability describe the traditional "muggle" industries.

Rahilly introduces arenas as "the industries that will matter most", and Bradley calls them new "competitive sectors". They are industry categories.

• A: arenas are sectors, not individual companies.
• B: regulation is not discussed.
• C: they are not physical locations.

Bradley says the muggles play by a more traditional set of economic rules.

• A, B, D: rapid change, high market-cap growth and R&D intensity are traits of the "wizard" industries.

The shuffle rate measures how much the bottom moves to the top — changes in rank within an industry. Profit is discussed separately as economic profit.

• A (volatility of leadership), B (churn within the arena) and D (relative change within an industry) all describe that reshuffling, so they *are* measured.

Bradley says economic profit is in the wizard industries and they are two times more R&D intensive.

• A: they capture 45% of market-cap growth, so growth is faster, not slower.
• C: traditional rules belong to muggles.
• D: R&D is higher, not lower.

The whole exchange contrasts wizard industries (about 10% of industries taking 45% of market-cap growth, higher shuffle rate, most economic profit) with traditional muggles. B captures this.

• A, C: both reverse the finding.
• D: the Harry Potter terms are a nickname for economic differences, not about magic.

• 5 opens: it names the Indus Valley Civilisation and introduces it as an early urban society.
• 2 follows: "this ancient culture" refers back to it and introduces planned cities and engineering.
• 1: "this architectural prowess" needs 2, and gives Harappa and Mohenjo-Daro as examples.
• 3: "Economically" moves from architecture to trade and agriculture.
• 4 closes: "Despite its impressive achievements" sums up and turns to the civilisation's decline.

Order: 52134.

• 2 opens: it sets the scene (robots cooking at the Toyota lab; robot chefs are nothing new).
• 4: "But these robots are more proficient" contrasts with "nothing special".
• 5: "The difference is" explains what makes them special — taught only a basic set of skills.
• 1: "improved upon those skills" refers to the basic skills in 5.
• 3 closes: "Despite their extraordinary culinary capabilities" sums up and says they are not headed for catering.

Order: 24513.

• 2 opens: the specific fact that India was the sixth worst-affected country (1993–2022).
• 1: widens the view — seven of the ten worst-affected countries "(including India)" are low- and middle-income.
• 4: "This reinforces" draws the developing world's argument from that pattern.
• 3 closes: "meanwhile" gives the rich nations' counter-position.

Order: 2143.

• 2 opens: it defines creativity.
• 4: states the common misconception that creativity needs big ideas or adventures.
• 3: "In fact" corrects it — inspiration comes from small daily details (examples given).
• 1 closes: "these moments" refers to 3's examples and concludes that creativity is finding the special in the ordinary.

Order: 2431.

• 3 opens: the general point that plants use roots to seek water.
• 1: the problem — in drought, water is only in deeper subsoil.
• 4: "these challenging conditions" refers to the drought in 1 and introduces abscisic acid.
• 2 closes: "the acid" refers back to 4, and the new study explains how it steers roots deeper.

Order: 3142.