IPMAT Indore 2022 · Question Paper & Solutions

The sum of coefficients of a polynomial $p(x)$ is $p(1)$.

$$p(1)=(5-9{)}^4=(-4{)}^4=256$$

$A$ swaps the 2nd and 3rd coordinates, so $A^2=I$. Hence odd powers equal $A$ and even powers equal $I$:

$$A^9+A^6+A^3+A=A+I+A+A=3A+I=\left[\begin{array}{ccc}4& 0& 0\\ 0& 1& 3\\ 0& 3& 1\end{array}\right]$$

$$\det =4(1\cdot 1-3\cdot 3)=4(-8)=-32$$

Absolute value $=32$.

The boundary $2|x|+3|y|=6$ is a rhombus with vertices $(\pm 3,0)$ and $(0,\pm 2)$.

Diagonals are $6$ and $4$, so

$$\text{Area}=\frac{1}{2}\cdot 6\cdot 4=12\text{ sq. units}$$

Square and add the two equations:

$$(\sin \alpha +\sin \beta {)}^2+(\cos \alpha +\cos \beta {)}^2=\frac{2}{3}+\frac{1}{3}=1$$

The left side is $2+2\cos (\alpha -\beta )$, so $\cos (\alpha -\beta )=-\frac{1}{2}$.

$${\cos }^2\frac{\alpha -\beta }{2}=\frac{1+\cos (\alpha -\beta )}{2}=\frac{1}{4}$$

$${\left(20\cos \frac{\alpha -\beta }{2}\right)}^2=400\cdot \frac{1}{4}=100$$

Eight people at a round table.

Sharmas together: treat them as one block, giving 7 units in a circle: $(7-1)!\times 2=1440$.

Of these, Ahujas also together: 6 units in a circle, each couple can swap: $(6-1)!\times 2\times 2=480$.

$$1440-480=960$$

If a set has $k$ elements, its power set has $2^k$.

• $|A|=0$
• $|P(A)|=2^0=1$
• $|P(P(A))|=2^1=2$
• $|P(P(P(A)))|=2^2=4$
• $|P(P(P(P(A))))|=2^4=16$

Let $T$ = sum of the largest 25 and $B$ = sum of the other 25. The average condition gives $\frac{T}{25}-\frac{B}{25}=50$, so $T=B+1250$.

Total $=T+B=2T-1250$, so maximise $T$.

The largest 25 distinct integers not exceeding 100 are $76,77,\dots ,100$:

$$T_{\max }=25\times \frac{76+100}{2}=2200\Rightarrow B=950$$

$B=950$ is achievable with 25 distinct integers below 76 (their sum can range from $1+\cdots +25=325$ up to $51+\cdots +75=1575$).

$$\text{Total}=2200+950=3150$$

Let $a$ apples and $m$ mangoes. Cost: $20a$ and $25m$.

Scenario 1 (apples +10%, mangoes −20%, no net profit/loss):
$$0.1(20a)-0.2(25m)=0\Rightarrow 2a=5m\Rightarrow m=0.4a$$

Scenario 2 (apples −20%, mangoes +10%, loss of 150):
$$-0.2(20a)+0.1(25m)=-150\Rightarrow -4a+2.5m=-150$$

Substitute $m=0.4a$: $-4a+a=-150$, so $a=50$.

Let $t={\log }_{x}y$. Then ${\log }_{x^2}y=\frac{t}{2}$ and ${\log }_{y^2}x=\frac{1}{2t}$.

$$\frac{t}{2}+\frac{1}{2t}=1\Rightarrow t^2-2t+1=0\Rightarrow t=1\Rightarrow y=x$$

Then $x=x^2-30\Rightarrow x^2-x-30=0\Rightarrow x=6$ or $x=-5$.

$x=-5$ gives $y=-5$, and the logarithm of a negative number is undefined, so $x=y=6$.

$$x^2+y^2=36+36=72$$

As printed (four terms) there is no solution: equal first and second differences would need $2^{x+3}+16=-2^{2x-1}-16-2^{x+3}$, i.e. $2^{2x-1}+2^{x+4}+32=0$, which is impossible because the left side is positive. (Check $x=3$: $-16,64,-48,48$ is not an AP.)

Intended reading (three terms): $-16,2^{x+3}-2^{2x-1}-16,2^{2x-1}+16$. A stray comma turned one term into two.

For an AP, twice the middle term equals the sum of the outer terms:

$$2\left(2^{x+3}-2^{2x-1}-16\right)=-16+2^{2x-1}+16=2^{2x-1}$$

Let $u=2^x$. Then $16u-u^2-32=\frac{u^2}{2}$, so

$$3u^2-32u+64=0\Rightarrow (u-8)(3u-8)=0\Rightarrow u=8\text{ or }u=\frac{8}{3}$$

$u=8$ gives $x=3$, and the terms are $-16,16,48$ (common difference 32). This matches the key.

$u=\frac{8}{3}$ gives $x={\log }_2\frac{8}{3}\approx 1.415$, which also satisfies the equation.

A triangle needs 3 non-collinear points, so take 2 points from one line and 1 from the other.

$$\left(\genfrac{}{}{0ex}{}{7}{2}\right)\cdot 5+\left(\genfrac{}{}{0ex}{}{5}{2}\right)\cdot 7=21\cdot 5+10\cdot 7=105+70=175$$

Check: $\left(\genfrac{}{}{0ex}{}{12}{3}\right)-\left(\genfrac{}{}{0ex}{}{7}{3}\right)-\left(\genfrac{}{}{0ex}{}{5}{3}\right)=220-35-10=175$.

The official key says 75. No standard reading of the question gives 75, so the key looks like a typo for 175.

Among $1,\dots ,n$ there are $\lfloor \sqrt{n}\rfloor$ perfect squares, so $n$ is the $\left(n-\lfloor \sqrt{n}\rfloor \right)$-th non-square (if $n$ is not a square).

Try $n\approx 2022+45=2067$: $\lfloor \sqrt{2067}\rfloor =45$ (since $45^2=2025\le 2067<2116=46^2$).

$$2067-45=2022$$

$2067$ is not a perfect square, so it is the 2022nd term.

Let the distance be $d$ km.

$$\frac{d}{12}-\frac{d}{20}=1\Rightarrow \frac{5d-3d}{60}=1\Rightarrow d=30$$

Let the AP have first term $a$ and difference $d\ne 0$. Terms: $a+2d$, $a+13d$, $a+68d$.

GP condition:
$$(a+13d{)}^2=(a+2d)(a+68d)\Rightarrow 169d^2+26ad=136d^2+70ad\Rightarrow 33d^2=44ad\Rightarrow d=\frac{4}{3}a$$

Take $a=3$, $d=4$: the terms are $11,55,275$ with ratio $5$. The next GP term is $1375$.

$$3+(n-1)\cdot 4=1375\Rightarrow n-1=343\Rightarrow n=344$$

(Any $a\ne 0$ scales everything the same way, so $n$ does not change.)

A sum of weighted distances $\sum w_i|x-p_i|$ is minimised at a weighted median of the points.

Points with weights: $0(1),1(2),2(1),4(1),6(1),10(2)$. Total weight $8$. Cumulative weight reaches $4$ at $x=2$ and stays there until $x=4$, so $f$ is constant and minimal on $[2,4]$.

$$f(2)=2+2(1)+0+2+4+2(8)=26$$

(Check: $f(3)=3+4+1+1+3+14=26$.)

$S_{15}=200$ and $S_{30}=200+350=550$.

Each of terms 16 to 30 is $15d$ more than the matching term among terms 1 to 15. So

$$350-200=15\times 15d=225d\Rightarrow d=\frac{150}{225}=\frac{2}{3}$$

For $0<\theta <\frac{\pi }{4}$: $0<\sin \theta <\cos \theta <1$. So ${\log }_2\sin \theta <{\log }_2\cos \theta <0$.

Write $s=\sin \theta$ and $k=\cos \theta$. All four numbers are (positive factor) × (negative log), so a bigger positive factor makes the number more negative.

Positive factors: since $s<1$ and $s<k$, we get $s^k<s^s<k^s$.

• $a=s^s{\log }_{2}k$ and $c=s^k{\log }_{2}k$. Same log, and $s^k<s^s$, so $c>a$.
• $d=s^s{\log }_{2}s$ and $b=k^s{\log }_{2}s$. Same log, and $k^s>s^s$, so $b<d$.
• $a=s^s{\log }_{2}k$ and $d=s^s{\log }_{2}s$. Same factor, and $|{\log }_{2}s|>|{\log }_{2}k|$, so $d<a$.

Order: $b<d<a<c$. With four values, the median is the mean of the middle two:

$$\text{median}=\frac{a+d}{2}$$

(Numeric check at $\theta =0.5$: $b\approx -0.996$, $d\approx -0.746$, $a\approx -0.132$, $c\approx -0.099$.)

By the midpoint theorem, the segment joining the midpoints $D$ (of $AB$) and $E$ (of $AC$) is parallel to $BC$ and half its length.

$$BC=2\times DE=2\times 6=12$$

Consistency check: the right angle is at $B$, so $AB=\sqrt{13^2-12^2}=5$.

$$160n+172m=164(m+n)\Rightarrow 8m=4n\Rightarrow m:n=1:2$$

(Alligation: distances from 164 are $4$ and $8$, so $n:m=8:4$.)

Divisible by 2 and 3 means divisible by 6. Multiples of 6 from 1002 to 9996:

$$\frac{9996-1002}{6}+1=1500$$

Of these, those also divisible by 5 are multiples of 30, from 1020 to 9990:

$$\frac{9990-1020}{30}+1=300$$

$$1500-300=1200$$

MANAGEMENT has 10 letters. Vowels: A, A, E, E (4). Consonants: M, M, N, N, G, T (6).

Arrange the consonants: $\frac{6!}{2!2!}=180$.

They create 7 gaps. Pick 4 gaps for the vowels: $\left(\genfrac{}{}{0ex}{}{7}{4}\right)=35$. Arrange the vowels: $\frac{4!}{2!2!}=6$.

$$180\times 35\times 6=37800$$

$81^{f(x)}=3-81^x$. Since $81^x$ takes every value in $(0,\infty )$ and $81^{f(x)}>0$, the right side must lie in $(0,3)$.

$$f(x)={\log }_{81}(3-81^x)<{\log }_{81}3=\frac{1}{4}$$

As $81^x\to 3$, $f(x)\to -\infty$. So the range is $(-\infty ,0.25)$.

Intersect the first two lines: $x=y+1$, so $2(y+1)+3y=12$, giving $y=2$ and $x=3$.

The third line must pass through $(3,2)$:

$$2(3)-3(2)+k=0\Rightarrow k=0$$

Note $\sqrt{2}\approx 1.414$ and $\sqrt{3}\approx 1.732$, and $\frac{\sqrt{2}+\sqrt{3}}{2}\approx 1.573$ is just above $\frac{\pi }{2}$.

$$\sin \sqrt{2}-\sin \sqrt{3}=2\cos \frac{\sqrt{2}+\sqrt{3}}{2}\sin \frac{\sqrt{2}-\sqrt{3}}{2}$$
Both factors are negative, so this is positive (about $0.0007$).

$$\cos \sqrt{2}-\cos \sqrt{3}=-2\sin \frac{\sqrt{2}+\sqrt{3}}{2}\sin \frac{\sqrt{2}-\sqrt{3}}{2}>0(\approx 0.3165)$$

Both denominators are positive, so the curve is an ellipse. The larger denominator is under $y^2$, so the major axis, and hence the foci, lie on the y-axis.

$|A\times B|=3\times 2=6$. Every subset is a relation, so there are $2^6=64$ relations.

A function assigns exactly one image to each of the 3 elements: $2^3=8$ functions.

$$P=\frac{8}{64}=\frac{1}{8}$$

Ashok beats Bipin by 150 m, which is also 15 s. So Bipin covers 150 m in 15 s, a speed of 10 m/s.

Bipin's time for 400 m $=40$ s, so Ashok's time $=40-15=25$ s.

Chandan's time $=25+25=50$ s, a speed of $400/50=8$ m/s.

When Bipin finishes at 40 s, Chandan has run $8\times 40=320$ m. Bipin wins by $400-320=80$ m.

Case $|x|\le 4$: $|x+4|+|x-4|=8$, so $x^2=27$ and $|x|=3\sqrt{3}\approx 5.2>4$. Rejected.

Case $x>4$: $x^2+2x-35=0\Rightarrow (x+7)(x-5)=0$, so $x=5$.

Case $x<-4$: $x^2-2x-35=0\Rightarrow (x-7)(x+5)=0$, so $x=-5$.

Roots are $\pm 5$, and the sum of squares is $25+25=50$.

Let $f(0)=a$.

Put $x=0$: $f(f(y))=y$ for all $y$. In particular $f(a)=f(f(0))=0$.

Put $y=a$: $f(x^2+f(a))=xf(x)+a$, i.e. $f(x^2)=xf(x)+a$.

Put $x=1$ in this: $f(1)=f(1)+a$, so $a=0$.

$$[f(0){]}^2+f(0)=0$$

Left inequality: ${\log }_{27}8=\frac{3{\log }_{3}2}{3}={\log }_{3}2$, so ${\log }_{3}x\ge {\log }_{3}2$, i.e. $x\ge 2$.

Right inequality as printed: $\frac{1}{9{\log }_{2}3}=\frac{{\log }_{3}2}{9}$, which gives $x<2^{1/9}\approx 1.08$. Combined with $x\ge 2$, the set is empty, and no option matches.

Likely intended right side: $9^{{\log }_{3}2}=3^{2{\log }_{3}2}=4$. Then

$${\log }_{3}2\le {\log }_{3}x<4\Rightarrow 2\le x<3^4=81$$

This gives $[2,81)$, which matches option C (printed without brackets as "2, 81").

As printed: $f(x)=\frac{|x|}{|x|}=1$ for every $x\ne 0$, so the set of values is $\{1\}$ (option D).

Intended function (standard question): $f(x)=\frac{x}{|x|}$, the sign function:

• $x>0$: $f(x)=1$
• $x<0$: $f(x)=-1$

Its set of values is $\{1,-1\}$, which is the key's option B. The numerator's absolute-value bars are almost certainly a typesetting error.

The punch is $60+40+100=200$ ml, of which orange juice is $60$ ml ($30\%$).

After 50 ml is drunk, $150$ ml remains, containing $0.3\times 150=45$ ml orange juice.

Add $v$ ml of orange juice so it makes up half the mixture:

$$\frac{45+v}{150+v}=\frac{1}{2}\Rightarrow 90+2v=150+v\Rightarrow v=60\text{ ml}$$

Check: $105$ ml of $210$ ml is $50\%$. With 600 ml added, orange would be $645/750=86\%$. The key's 600 ml looks like a typo for 60 ml, which is not among the options.

Expand along the first row:

$$\det A=a(bc-a^2)-b(b^2-ac)+c(ab-c^2)=3abc-a^3-b^3-c^3$$

Use the identity

$$a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)\left[(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\right]$$

With $a,b,c>0$ and distinct, the right side is strictly positive. So $\det A<0$: negative.

$(1,4)\in A\times B$ gives $1\in A$ and $4\in B$. $(4,1)\in A\times B$ gives $4\in A$ and $1\in B$.

So $\{1,4\}\subseteq A$ and $\{1,4\}\subseteq B$. Since $|A|\cdot |B|=4$ and each has at least 2 elements, $A=B=\{1,4\}$.

Hence $A\times B=B\times A$.

• $\varphi \in A\times B$ is false: the empty set is a subset, not an element, of $A\times B$.
• $A\times B\ne B\times A$ is false because $A=B$.

$abcde$ divisible by 6 means the last digit $e$ is even and $a+b+c+d+e$ is a multiple of 3.

• eee: ends in even $e$, digit sum $3e$ is a multiple of 3. Divisible.
• cdbae: same digits, so same digit sum, and ends in $e$. Divisible.
• bbadcacede: digit sum $2(a+b+c+d+e)$ is a multiple of 3, and ends in $e$. Divisible.
• edcba: same digit sum, but it ends in $a$, which may be odd (e.g. $12342\to 24321$, which is odd). Not necessarily divisible.

$$\text{Cost}=3600\times {\left(\frac{4}{10}\right)}^2=3600\times 0.16=576$$

$$A^2=\left[\begin{array}{cc}1& 0\\ \frac{1}{2}& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ \frac{1}{2}& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ \frac{1}{2}& 0\end{array}\right]=A$$

$A$ is idempotent, so $A^n=A$ for all $n\ge 1$, and $A^{2022}=\left[\begin{array}{cc}1& 0\\ \frac{1}{2}& 0\end{array}\right]$.

This matches none of the matrices in options A, B, D. So the answer is None of these.

Total factors: $(7+1)(8+1)(3+1)=288$.

A perfect-square factor needs even exponents:

• power of 3: $0,2,4,6$ (4 choices)
• power of 2: $0,2,4,6,8$ (5 choices)
• power of 17: $0,2$ (2 choices)

$$\frac{4\cdot 5\cdot 2}{288}=\frac{40}{288}=\frac{5}{36}$$

Triangle inequality: $x+21>40$, so $x>19$. Because $x$ is the shortest side, $x\le 21$.

Of the options, only $20$ lies in $(19,21]$. The others (16, 18, 19) fail $x+21>40$.

$$\frac{(a+b{)}^2-(a-b{)}^2}{4}=\frac{4ab}{4}=ab$$

For any two natural numbers, $ab=\text{LCM}(a,b)\times \text{HCF}(a,b)$.

Distance from the foot of the tower:

• at $45^{\circ }$: $200\cot 45^{\circ }=200$ m
• at $30^{\circ }$: $200\cot 30^{\circ }=200\sqrt{3}\approx 346.41$ m

Distance moved $\approx 146.41$ m in 2 minutes:

$$\text{speed}=\frac{146.41}{120}\text{ m/s}\times 3.6\approx 4.39\text{ km/h}$$

Setting up the roster. 15 shifts in total (2 per day Mon–Sat, 3 on Sunday), and $5\times 3=15$, so every slot is filled. Each person's three days are two consecutive days plus one separate day. The week is treated as cyclic (Sunday is followed by Monday), which option 45C assumes.

• Alex and Dilip share Tue, Wed. Neither Bhabha nor Cathy works with Alex, so Tue and Wed are fully staffed by Alex and Dilip.
• Cathy avoids Mon, Sat, Tue, Wed, so her days are from Thu, Fri, Sun. She takes all three: Thu, Fri, Sun (Thu–Fri consecutive).
• Alex's third day cannot be Mon or Thu (that would make three consecutive days), and it cannot be Fri or Sun because Cathy works then. So Alex's third day is Sat, and Dilip's third day differs from Sat.
• Dilip's third day cannot be Mon or Thu (three in a row) or Sat (it must differ from Alex's), so it is Fri or Sun. Bhabha and Ethan fill the rest. Bhabha cannot take Sat (Alex), so Ethan must.
• Dilip on Fri: Fri is full, so Sun's two open slots and Mon's two slots all go to Bhabha and Ethan. Ethan then has Sat, Sun, Mon, which is three consecutive days. Impossible.
• Dilip on Sun: Mon goes to Bhabha and Ethan. Ethan has Mon, Sat plus one more. Sun would make three in a row, and Thu would give no consecutive pair, so it is Fri (Fri–Sat). Bhabha gets the remaining Thu and Sun (Sun–Mon consecutive).

Unique roster:

• Alex: Tue, Wed, Sat
• Bhabha: Sun, Mon, Thu
• Cathy: Thu, Fri, Sun
• Dilip: Tue, Wed, Sun
• Ethan: Fri, Sat, Mon

Daily staff: Mon B,E; Tue A,D; Wed A,D; Thu B,C; Fri C,E; Sat A,E; Sun B,C,D. (A full computer search of all schedules confirms this is the only solution under a cyclic week.)

Q41. Dilip (Tue, Wed, Sun) and Ethan (Mon, Fri, Sat) share no day. The other pairs do overlap: Bhabha–Dilip on Sun, Bhabha–Ethan on Mon, Cathy–Dilip on Sun.

From the unique roster derived in Q41, Sunday's three employees are Bhabha, Cathy and Dilip.

Alex never works on Sunday (his days are Tue, Wed, Sat), and Ethan works Mon, Fri, Sat.

Alex works Tue, Wed and Saturday.

Monday and Thursday would give him three consecutive days. Friday and Sunday are Cathy's days, and Cathy cannot work with Alex.

Bhabha works Sun, Mon, Thu. Cathy works Thu, Fri, Sun.

They are together on Thursday and Sunday, so 2 days.

Ethan works Mon, Fri, Sat. His consecutive pair is Friday and Saturday, and Monday is the separate day.

(Note: the unique solution needs Sunday and Monday to count as consecutive, as for Bhabha's Sun–Mon pair. Option C in this question shows the paper treats the week that way.)

The first paragraph says the difference matters to fruit growers and greengrocers, who need their produce in top condition when it reaches the market. That is a trade concern, so the context is commercial.

• Public health and environment are never mentioned.
• "Gastronomical" (taste or cuisine) is not the reason the passage gives.

Case for B (key). The biologists' explanation starts from the idea that fruits exist to solve a problem *all* plants face: spreading their seeds. Climacterism is presented as one way fruit-bearing plants tackle that shared problem, by targeting the right seed dispersers. On this reading, B states the premise the whole argument rests on.

Why the others are weaker.

• C says propagation "cannot be studied accurately", which overstates the passage. It only notes one limitation of this particular study.
• D ("better protected… propagate easily") has no support in the passage.
• A picks up the last paragraph's call to study wild fruits, not just ones humans eat. That is the writer's own verdict on the research, so it is a strong rival. The passage's framing (a problem faced by all plants) is what supports B.

The first paragraph says how the differences came about "remains unclear". The Tokyo study only *offers a suggestion* (a hypothesis) backed by a correlation, and the writer notes a flaw in its sample and says more work is needed. So the origins are still nebulous (unclear) despite the research.

• A ("clear now") contradicts the tentative tone.
• C ("trivial") is not suggested.
• D is vague and does not describe the origins themselves.

The last paragraph says the main limitation is that the researchers looked at fruits eaten by people. Centuries of selective breeding may have blurred the natural signal.

• A (wild fruits) is what the writer recommends as the *next* step, not what was studied.
• B and C are not supported by the passage.

• Arboreal means living in trees. The passage pairs arboreal animals with aerial ones that can reach fruit still on the plant.
• Camouflage means disguise or blending in. Climacteric fruits are "better camouflaged", so they are hard to spot until they fall.

D reverses the first meaning (ground-dwelling). A and C match neither word.

• Finding (ii): $15$ of $19$ ground-dweller fruits are climacteric and $21$ of $26$ arboreal/aerial fruits are non-climacteric, and $15+21=36$. This is the core result of the 80-variety study, which the writer calls a "suggestively strong correlation". It is the main evidence.
• Finding (i): vivid colours of non-climacteric fruits is introduced as "other evidence" that *fortifies* the hypothesis. It is supplementary.

So (i) is supplementary and (ii) is main, which is option A. C swaps the roles, and B and D treat both findings alike.

The passage turns on a lucky break. Jenner could not find volunteers in London, and vaccination was saved only when Cline happened to try it on a sick child. The author says the child's rundown state could have caused a serious reaction, that "the fate of vaccination hung in the balance", and that good luck favoured it. So chance and risk shaped the outcome.

• A (urban centres) is not argued. London was actually where he failed.
• B: Jenner's records and pamphlet appear, but they did not win acceptance on their own.
• D: opposition by Ingenhouz and Watt is one episode, not the focus.

He stayed in London almost three months without finding anyone willing to be vaccinated, then went home disappointed. So he could not repeat his demonstration for lack of volunteers.

• B and C: the passage never says his friends rejected the findings or reacted badly. They actually kept his vaccine material.
• D: recognition and rewards came later. Patents are never mentioned.

The last paragraph contrasts the few "illogical" opponents with the many honours Jenner received. These show how much his contemporaries valued the discovery, i.e. most recognised its effectiveness.

• A copies a phrase from the passage, but having few opponents is not the *reason* for the honours.
• B: Watt's claim is mentioned, but nothing says it was disproved or that this led to recognition.
• C ("tokenism") misreads "token of recognition".

Ingenhouz's faction said vaccination was dangerous and unjustifiable because it passed on a disease (cowpox) without protecting against any other. In other words, it made people ill and gave no immunity. That is option B.

• A and C read "unjustifiable" as a matter of fairness or law, which the passage does not support.
• D exaggerates. He claimed it transmitted a disease, not that it spread one through a large population.

Jenner recorded the 23 cases to show that cowpox protects against smallpox, a different and deadly disease.

• A and D mention details (dairy workers, cows and horses) that describe where some cases came from, not why they were recorded.
• C (testing on his son) is a detail, not the purpose.

In medicine a "local reaction" is a reaction at the inoculation site (swelling, inflammation). The passage worries that in a weakened child the vaccine "might very well have provoked a rather serious local reaction", an adverse bodily reaction to the inoculation.

A, C and D wrongly read "local" as local doctors or local people.

Pride in a rich, aristocratic background suggests she thinks herself superior. "Look down on" means to regard someone as inferior.

• "Looks up to" means to admire, the opposite.
• "Looks away from" and "looks towards" are literal and do not fit the cause given.

The typical change after winning a big prize, as the sentence implies, is acting superior: condescendingly.

• "Overly" is an intensifier that needs an adjective ("overly proud").
• "Ambitiously" and "controversially" do not follow naturally from winning a prize.

"Hiding in plain sight" is the idiom for being concealed while openly visible. That is exactly the case of spies disguised as ordinary secretaries.

• A garbles the idiom.
• B and C describe literal concealment, which contradicts the "innocuous guise" idea.

The standard idiom for focusing on the positive is "look on the bright side".

• "Think about/upon the sunny side" are not idioms.
• "Look forward to the bright side" mixes two expressions.

A culture where "everyone is welcome", with support for the LGBTQ community as one example, shows inclusivity and diversity.

• "Discriminatory practices" is the opposite.
• "Mindfulness and objectivity" and "discretionary practices" do not fit the idea of welcoming everyone.

The idiom "a taste of one's own medicine" means being treated the way you treat others.

The other options ("a spoon", "an unpleasant medicine dose", "a dose of bitter medicine") are distorted versions of the idiom.

"A piece of cake" is the idiom for something very easy.

"Slice of cake", "left-handed game" and "left-handed sport" are not idioms with this meaning.

"Picked up my interest" is wrong. It is probably a confusion with "piqued my interest". The standard formal collocation among the options is "aroused my interest".

• "Brightened" and "adopted" do not collocate with interest.
• "Caught my interest" is also an ordinary idiom, so option D is a close rival. The key prefers the more formal collocation "aroused".

A negative adverbial at the front of a clause ("never once", "not once") causes subject–auxiliary inversion: "not once *did he* complain".

• B keeps normal word order, which is the original error.
• A ("give the complaint") and C ("any complaining happened") are unidiomatic.

In conditional clauses introduced by "provided (that)", English uses the simple present for a present or future condition, not "will". So: "provided she has a good attendance record".

• B repeats the error.
• C is a clumsy passive.
• D (past "had") does not match the present rule "are authorized".

This is a third conditional (an unreal past): "If + had + past participle" needs "would have + past participle" in the main clause, so "I would have driven you there".

• B and D use present or future forms, which clash with "yesterday".
• A ("should have") changes the meaning to obligation.

"Revert" already means "go back", so "reverted back" is redundant. The possessive is its ("it's" means "it is").

• A uses "it's".
• C and D keep "back", and D also uses "it's".

The paragraph gives examples ("for example", Southeast Asia vs Europe) and then "Then there are gender variations". So the opening must state a general claim that shopping behaviour varies with background factors, and "culture" covers the regional example.

• A (discounts) and C (people enjoy shopping) do not introduce regional differences.
• D fits only the gender sentence and would make "for example" about Southeast Asia pointless.

The blank needs a consequence of studying most of the time that can be called "this situation", whose causes are then discussed. Little time for hobbies or socialising follows directly and is the "situation".

• A jumps to a solution before causes are discussed.
• C (exhaustion) and D (applying knowledge) do not follow from the first sentence and do not fit "this situation… exams".

The previous sentence says getting to Goa from Mumbai "is not a major problem". The natural next step is how to get there: buses, trains, car. That is option B.

• A, C and D describe what to do in Goa. They would break the link between "reaching there" and the travel details, and C also overlaps with the following sentence about beaches.

A text in two scripts, one of which can already be read, works like a Rosetta Stone. It lets scholars decipher the unknown script, and the paragraph opened by saying the Indus script has not been deciphered.

B (trade), C (which is older) and D (relationships) are side issues the paragraph never raises.

• Silent films of the 1920s were visually strong because they had no sound.
• Surviving copies being poorly preserved and cut explains why people misjudge them.
• The belief that they were primitive is framed as mistaken ("visually strong"), so misconception fits.

A has "barely preserved" and "observation" (too neutral). C's "accusation" and D's "dispassionately preserved" do not fit.

"Thought-provoking questions", "different thinkers" and "life in general" all point to philosophy. It makes us think about ourselves. Pondering ideas gives a better understanding of the world.

• B: linguistics is not about "life in general", and "gain a better picture" is weaker.
• C: "gain a better acquaintance of the world" is unidiomatic, and "animals" does not fit.
• D: orthography (spelling) makes no sense here.

• "Anthropocentric worldview" is the natural collocation.
• "Humane treatment" is paired with "necessary suffering", the standard phrase in animal-ethics debates. It matches the later "as long as animals are not made to suffer".
• The view rests explicitly on the idea that animals have no right to life.

A's "necessary comfort" and B's "necessary well-being" are not parallel to "humane treatment" in an economic framing. D's "anthropocentric universe" and "necessary gain" do not fit.

Bulk commercial messages that people try to stop are unsolicited (spam). Regulators try to stop them, and spammers find ways around the rules.

• A: firewalls do not "use technology", they are technology.
• B: advertisers dodging governments is weaker, and "inescapable" clashes with efforts to stop them.
• C: "informative" gives no reason to stop the messages.

• Investigators allege wrongdoing before it is proven, and "alleged that" is grammatical ("accused that" is not).
• Insider information is the technical term for confidential deal information.
• Profiting from it is done illicitly.

A ("treacherously deal") and D ("dramatically deal") give odd adverbs. B's "discovered… enormously" is weaker, and "internal information" is less precise.

• "The solar system and the rest of the universe"
• "objects and phenomena better observed from space"
• "the application of that understanding to serve human purposes"
• all are elements of space exploration.

A (ideologies, enormity) and D (configuration, vastness) do not make sense. C has "principles of that understanding to serve…" and "space identification", both awkward.

Flout means to openly disregard a rule. Flaunt means to show off.

D should be "flaunted her designer wear". In A, B and C, rules, a dress code and tradition are correctly "flouted".

As an adjective, cross means annoyed or irritable, and it describes people, not jokes. "Vulgar, cross, and crude jokes" misuses it (the writer probably meant "coarse").

B (cross the river, verb), C ("a cross to bear", idiom) and D ("cross and cranky", irritable person) are all correct.

The idiom for a stain on someone's reputation or career is a blot ("a blot on his career"). "A spot on his career" is unidiomatic.

A (a spot of ink), B (a scenic spot) and D ("spot on", exactly right) are standard uses.

Evidence corroborates (confirms) a thesis. Collaborate means to work together. D should read "to corroborate her thesis".

A (worked with film-makers), B (collaborated with the enemy) and C (refused to work together) use "collaborate" correctly.

The storage compartment is a drawer, so "the draw of his cupboard" is wrong.

A (draw water), B (a lottery draw) and D (a match heading for a draw) are correct uses of "draw".

• A: "state the grounds" (verb, declare) is correct.
• B: "the state of his health" (condition) is correct.
• C: Uttar Pradesh as a "state" (political unit) is correct usage of the word. (The claim "most densely populated" is factually shaky, but the question tests usage of "state", not facts.)
• D: "a demeanour of great state" is unidiomatic. The sense "state = pomp, grandeur" survives only in set phrases like "lie in state" or "travel in state", and does not describe someone's manner. "Dignity" or "poise" was needed.

• 4 opens with the topic: the monsoon is the lifeblood of the economy.
• 3 "This is because…" gives the reason: half the population depends on farm income.
• 5 "Unfortunately, however…": 40% of sown area lacks irrigation, so dependence on rain is high.
• 1 Therefore a sub-par monsoon cuts yields and raises import dependence.
• 2 Conversely, a robust monsoon curbs food inflation.

Order: 43512. B, C and D start with 1, but sentence 3's "This is because" needs 4's claim just before it, which only A provides.

• 4 introduces Japanese ceramics, called Yakimono.
• 1 "these creations" refers back to ceramics in 4. They grew popular with the 16th-century tea ceremony.
• 3 "them" again means ceramics: modern shops display thousands of them.
• 2 "However", bargain-hunters shop in Kitchen Town instead of touristy shops, which contrasts with 3.

Order: 4132. Options starting with 1 or 3 leave "these creations" or "them" without a referent.

• 1 states the event: Bitcoin approved as official currency.
• 4 "this move" refers to 1: economists criticise it as a risk to financial stability.
• 3 "Others claim…" adds more critics (money laundering, electricity use).
• 2 "Experts foresee that this will damage the environment as well as the economy" sums up both kinds of concern: environment (electricity, from 3) and economy (from 4).

Order: 1432. A, B and C put 2, 4 or "Others" before the event is stated.

Key links in the sentences.

• 5 defines saccades (jiggles in response to changes in view). This is a natural opener.
• 4 our eyes scan the surroundings and send data to the brain. This expands 5.
• 2 youthful eyes jiggle regularly to take in new stimuli.
• 3 "As the person attached to *those eyes* ages…" continues 2: the youthful eyes grow old and muscles slow.
• 1 "This means the brain receives less input" is the result of 3's slowing and damage.

The most coherent order is 5 4 2 3 1, which is not offered.

The key's 54321 puts 2 after 3 and 1 after 2. Then "This means the brain receives less input" would follow "youthful eyes jiggle regularly to take in new stimuli", which is the opposite of what 1 says. Options A (23145) and D (24315) keep the strong 2→3 or 3→1 links but end awkwardly. No option is fully coherent. B is the official answer.

• 3 introduces the news and names the subject: the corporation will rejuvenate Sion Talao (Lake).
• 1 "the lake": location and heritage status.
• 5 "Earlier the talao was…": its past as a natural water tank.
• 4 "Over the years, the lake has lost its prominence" moves from past to decline.
• 2 "Once home to marine creatures, now surrounded by high-rises" illustrates the decline.

Order: 31542. B and C open with 2, which says "the lake" before it has been named. D (31425) puts the decline before the "earlier" history, which breaks the time sequence.