IPMAT Indore 2024 · Question Paper & Solutions

Use $a^{{\log }_{b}c}=c^{{\log }_{b}a}$: $4^{{\log }_{2}x}=x^{{\log }_{2}4}=x^2$ and $9^{{\log }_{3}y}=y^2$.

The equation becomes $x^2-4x+y^2-16y+68=0$, i.e.

$$(x-2{)}^2+(y-8{)}^2=0.$$

A sum of squares is zero only if each is zero, so $x=2$, $y=8$ (both positive, so the logs are defined). Hence $y-x=6$.

Let the counts be $3k,6k,7k$, total $16k$.

Oranges $3k$ must be a multiple of $\text{lcm}(5,6)=30$, so $k$ must be a multiple of $10$. The smallest is $k=10$ (30 oranges, 60 apples, 70 bananas).

Minimum total $=16\times 10=160$.

$a^b=1$ (real) happens in three ways.

• Base $=1$: $x^2-15x+54=0\Rightarrow x=6,9$.
• Exponent $=0$, base $\ne 0$: $x^2-5x+6=0\Rightarrow x=2,3$. Bases are $29$ and $19$, both non-zero.
• Base $=-1$, exponent even: $x^2-15x+56=0\Rightarrow x=7,8$. Exponents are $49-35+6=20$ and $64-40+6=30$, both even.

All six values are distinct, so there are $6$ solutions.

Table: A 32 (24), B 28 (30), C 43 (39), D 39 (45), E 35 (49), F 29 (54), G 23 (59), H 16 (63) — employees (median age).

B has 28 employees, so its median is the average of its 14th and 15th ages, which equals 30. Hence the 14th age is at most 30, so the youngest person in B is at most 30.

Every A employee is strictly younger than every B employee, so A's oldest is at most $30-1=29$.

This is achievable: let all of B's first 15 employees be 30, and A's ages be anything with median 24 and maximum 29 (e.g. first 16 at 24, the rest 24–29). Answer: $29$.

Let $x$ be borrowed at 3%. Interest over 4 years:

$$x\cdot 0.03\cdot 4+(4000-x)\cdot 0.04\cdot 4=520$$

$$0.12x+640-0.16x=520\Rightarrow 0.04x=120\Rightarrow x=3000.$$

Check: $360+160=520$.

Let sides $a\le b\le c$ with $a+b+c=15$. Triangle inequality $a+b>c$ gives $15-c>c$, so $c\le 7$; also $c\ge 5$.

• $c=7$, $a+b=8$, $b\le 7$: $(1,7),(2,6),(3,5),(4,4)$ → 4
• $c=6$, $a+b=9$, $b\le 6$: $(3,6),(4,5)$ → 2
• $c=5$, $a+b=10$, $b\le 5$: $(5,5)$ → 1

Total $=7$.

$A=\left[\begin{array}{ccc}{x}_1& x_2& 7\\ y_1& y_2& y_3\\ z_1& 8& 3\end{array}\right]$ is a magic square with common sum $S$.

Adding the middle row, middle column and both diagonals counts every cell once and the centre four times: $4S=S_{\text{total}}+3y_2=3S+3y_2$, so $y_2=S/3$.

Row 3: $z_1=S-11$. Anti-diagonal: $7+y_2+z_1=S\Rightarrow 7+\frac{S}{3}+S-11=S\Rightarrow S=12$, $y_2=4$, $z_1=1$.

Then $y_3=12-10=2$, $x_1=12-4-3=5$, $x_2=12-4-8=0$, $y_1=12-4-2=6$. Check row 1: $5+0+7=12$; column 1: $5+6+1=12$.

$$\det A=5(4\cdot 3-2\cdot 8)-0+7(6\cdot 8-4\cdot 1)=-20+308=288.$$

$1800=2^3\cdot 3^2\cdot 5^2$.

A factor divisible by 6 has the form $2^a{3}^b{5}^c$ with $a\in \{1,2,3\}$, $b\in \{1,2\}$, $c\in \{0,1,2\}$.

Count $=3\times 2\times 3=18$.

Total employees $=32+28+43+39+35+29+23+16=245$, so the overall median is the $123$rd age.

Because ages increase strictly from company to company, sorting everyone just lists A, then B, and so on. Cumulative counts: A 32, B 60, C 103, D 142.

The 123rd person is the $123-103=20$th in D. D has 39 employees, so its 20th is exactly D's median, $45$.

Put $B$ at the origin, legs along the axes; the hypotenuse is $x+y=18$.

A rectangle with vertex $B$ has its opposite vertex $(x,y)$ on the hypotenuse, so its area is $xy$ with $x+y=18$.

By AM–GM, $xy\le {\left(\frac{18}{2}\right)}^2=81$, with equality at $x=y=9$. Maximum area $=81$.

Shift: $u=x-5$, $v=y-5$; count integers with $|u|+|v|\le 6$.

For each $u$ from $-6$ to $6$, $v$ ranges over $2(6-|u|)+1$ values:

$$1+3+5+7+9+11+13+11+9+7+5+3+1=85.$$

(General formula $2n^2+2n+1$ with $n=6$.)

F has 29 employees, so its median (54) is the 15th age.

Lower bound for F's ages. E has 35 employees with median 49 = its 18th age, so E's oldest is at least 49. Every F employee is older than every E employee, so each F age is at least 50.

Minimise. Employees 1–14 of F: as low as possible, $50$ each. Employee 15: $54$. Employees 16–29: at least 54, so $54$ each.

$$14\times 50+15\times 54=700+810=1510.$$

This is consistent (E's ages can top out at 49, G's start at 55 with median 59).

Let $a,b,c$ be the numbers liking exactly one, two, three drinks.

$$a+b+c=150,a+2b+3c=52+48+62=162.$$

Subtracting: $b+2c=12$. The number liking more than one is $b+c=12-c$, maximised at $c=0$, giving $12$.

Feasible: e.g. 12 students like exactly tea and coffee, all others exactly one drink.

$$\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right)=1-\frac{x^2}{10000}=0.9676$$

So $\frac{x^2}{10000}=0.0324$, $x^2=324$, $x=18$.

$f(x)=2x$ for $x\ge 0$ and $f(x)=0$ for $x<0$.

$g(f(a))$ needs $f(a)\ne 0$, so $a>0$ and $f(a)=2a=t$. Then

$$t+\frac{1}{t}=\frac{13}{6}\Rightarrow 6t^2-13t+6=0\Rightarrow t=\frac{3}{2}\text{ or }\frac{2}{3}.$$

So $a=\frac{3}{4}$ or $a=\frac{1}{3}$. Since $g(a)=1/a>0$, $f(g(a))=2/a=\frac{8}{3}$ or $6$.

Maximum $=6$.

Let first term $a$, ratio $r$, $L=\log a$, $R=\log r$.

$$L+(p-1)R=\log q,L+(q-1)R=\log p.$$

Subtracting: $(p-q)R=\log q-\log p$, so $R=\frac{\log q-\log p}{p-q}$.

$$L=\log q-(p-1)R=\frac{(p-q)\log q-(p-1)\log q+(p-1)\log p}{p-q}=\frac{(1-q)\log q-(1-p)\log p}{p-q}.$$

This is option B.

Both extreme distances lie along the line through the point and the centre.

• Point outside: longest − shortest $=2r$, so $2r=9-4=5$, $r=2.5$ cm.
• Point inside: longest + shortest $=2r$, so $2r=9+4=13$, $r=6.5$ cm.

Both are possible, so the radius is 2.5 cm or 6.5 cm.

Both terms are non-negative and sum to 0, so $x=-1$ and $y=-2$.

Then $a(-1)-3a(-2)=-a+6a=5a=1$, so $a=\frac{1}{5}$.

Put $t=|x|\ge 0$: $t^2-10t-56=0\Rightarrow (t-14)(t+4)=0$.

Only $t=14$ is valid, so $x=\pm 14$: 2 real solutions.

Write everything with exponent 100:

• $2^{300}=8^{100}$
• $3^{200}=9^{100}$
• $4^{100}$
• $2^{100}+3^{100}<2\cdot 3^{100}<9^{100}$

Since $9^{100}>8^{100}>4^{100}$, the greatest is $3^{200}$.

(The printed stem list is garbled, but the four options are as above.)

$\frac{a}{1-r}=80$ and $a(1+r)=35$. Substituting $a=80(1-r)$: $80(1-r^2)=35$, so $r^2=\frac{9}{16}$, $r=\pm \frac{3}{4}$.

• $r=\frac{3}{4}$ ($a=20$): partial sums rise towards 80 and never get near 100 at any particular $n$, so this GP cannot be the intended one.
• $r=-\frac{3}{4}$ ($a=140$): $S_n=80\left(1-{\left(-\frac{3}{4}\right)}^n\right)$ oscillates around 80.

$$S_3=113.75,S_4\approx 54.7,S_5=80\left(1+\frac{243}{1024}\right)\approx 98.98,S_6\approx 65.8,S_7\approx \mathrm{90.7.}$$

Later odd sums get closer to 80, so the sum closest to 100 is $S_5$: $n=5$. (The printed stem drops the letter $n$.)

The condition means every girl gets at least as many balloons as the boy with the most. Let $M$ be the largest boy share. Writing each girl's share as $M+e_i$ with $e_i\ge 0$, the girls' part can be done in $\left(\genfrac{}{}{0ex}{}{20-S_b-5M+4}{4}\right)$ ways, where $S_b$ is the boys' total.

• $M=1$: boys $(1,1,1)$, $S_b=3$, remainder $12$: $\left(\genfrac{}{}{0ex}{}{16}{4}\right)=1820$.
• $M=2$: boys in $\{1,2{\}}^3$ with at least one 2. $S_b=4$ (3 ways), $5$ (3 ways), $6$ (1 way), remainders $6,5,4$: $3\cdot 210+3\cdot 126+70=1078$.
• $M=3$: needs $S_b\le 5$, so boys are a permutation of $(3,1,1)$ (3 ways), remainder $0$: $3$.
• $M\ge 4$: $5M+S_b>20$, impossible.

$$n=1820+1078+3=2901$$

(confirmed by computer enumeration). This lies in none of the printed ranges (6000–10000), which are also misprinted as "$9000\ge n<10000$" etc. The official key lists E, i.e. none of options A–D.

${\log }_{7}4=2{\log }_{7}2$, ${\log }_{7}5-{\log }_{7}2={\log }_7\frac{5}{2}$, ${\log }_{7}25=2{\log }_{7}5$, ${\log }_{7}8-{\log }_{7}4={\log }_{7}2$.

$$a=\frac{2{\log }_{7}2\cdot {\log }_7\frac{5}{2}}{2{\log }_{7}5\cdot {\log }_{7}2}=\frac{{\log }_7\frac{5}{2}}{{\log }_{7}5}={\log }_5\frac{5}{2}.$$

Hence $5^a=\frac{5}{2}$.

We need an integer $g$ with $0.48N<g<0.5N$.

• $N$ even: $g\le \frac{N}{2}-1>0.48N$ needs $0.02N>1$, so $N>50$.
• $N$ odd: $g\le \frac{N-1}{2}>0.48N$ needs $0.02N>0.5$, so $N>25$. The smallest such odd $N$ is $27$.

Check $N=27$, $g=13$: $13/27\approx 48.1\%$. (For $N=25$ we'd need $12<g<12.5$, impossible.) Answer: $27$.

The paper does not define $E$; the options only make sense if $E$ is the foot of the perpendicular from $B$ to $AC$, so $\theta$ is the angle between the median and $AC$. Take $E$ between $A$ and $D$.

In right triangle $BED$: $BE=k\sin \theta$, $ED=k\cos \theta$.

In right triangle $BEA$: $AE=\sqrt{c^2-k^2{\sin }^2\theta }$.

So $AD=AE+ED$ and $AC=2AD$. Then

$$[ABC]=\frac{1}{2}\cdot AC\cdot BE=AD\cdot k\sin \theta =k\sin \theta \sqrt{c^2-k^2{\sin }^2\theta }+k^2\sin \theta \cos \theta .$$

Since $k^2\sin \theta \cos \theta =\frac{k^2\sin 2\theta }{2}$, this is option B.

Let $\angle B=\angle C=\beta$ and $\angle CDE=x$.

• $\angle ADC$ is exterior to $△ABD$: $\angle ADC=\beta +30^{\circ }$.
• $AD=AE$, so $\angle ADE=\angle AED$.
• $\angle AED$ is exterior to $△EDC$: $\angle AED=x+\beta$.

Also $\angle ADC=\angle ADE+\angle EDC$:

$$\beta +30^{\circ }=(x+\beta )+x\Rightarrow x=15^{\circ }.$$

${\log }_{2}x=2{\log }_{4}x=2a$ and ${\log }_{5}x=2{\log }_{25}x=2b$.

$${\log }_{x}10={\log }_{x}2+{\log }_{x}5=\frac{1}{2a}+\frac{1}{2b}=\frac{a+b}{2ab}.$$

The condition means no two girls sit next to each other.

Total circular arrangements of 8 people: $7!=5040$.

Favourable: seat the 5 boys in a circle ($4!=24$ ways), creating 5 gaps; put the 3 girls in 3 different gaps: $\left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot 3!=60$ ways. Total $24\times 60=1440$.

$$P=\frac{1440}{5040}=\frac{2}{7}.$$

Let apples $=a$, bananas $=a+10$, oranges $=o$.

$$0.7a+0.6(a+10)+0.5o=0.55(2a+10+o)\Rightarrow 0.2a+0.5=0.05o\Rightarrow o=4a+10.$$

Constraints:

• $2a+10$ divisible by 11 $\Rightarrow 2a\equiv 1(\mathrm{mod}11)\Rightarrow a\equiv 6(\mathrm{mod}11)$.
• Fruits sold must be whole: $0.7a$ integer $\Rightarrow a$ a multiple of 10 (then $0.6(a+10)$ and $0.5o$ are also whole).

The smallest $a$ with $a\equiv 6(\mathrm{mod}11)$ and $10\mid a$ is $a=50$, so $o=4(50)+10=210$.

Check: 50 apples, 60 bananas (total 110), 210 oranges; sold $35+36+105=176=55\%$ of 320.

Upstream speed $=96/8=12$ km/h; downstream speed $=96/6=16$ km/h.

Each pair (1 h down, 1 h up) moves the boat $16-12=4$ km downstream net.

After 5 pairs: $5\times 4=20$ km downstream of A.

Substitute $y_1=x_1-1$, $y_2=x_2-2$: $y_1+y_2+x_3+x_4=47$, all $\ge 0$.

By stars and bars: $\left(\genfrac{}{}{0ex}{}{47+3}{3}\right)=\left(\genfrac{}{}{0ex}{}{50}{3}\right)=\frac{50\cdot 49\cdot 48}{6}=19600$.

If $2^n$ has $p$ digits and $5^n$ has $q$ digits, then $10^{p-1}<2^n<10^p$ and $10^{q-1}<5^n<10^q$ (the inequalities are strict since neither is a power of 10).

Multiplying: $10^{p+q-2}<10^n<10^{p+q}$, so $n=p+q-1$, i.e. $p+q=n+1$.

With $n=2024$: total digits $=2025$.

Centre $(7,-8)$, radius $r=\sqrt{49+64-88}=5$.

Distance² from $P(0,3.5)$ to the centre: $7^2+11.5^2=181.25$. Tangent length $L=\sqrt{181.25-25}=12.5$.

Half the angle between the tangents satisfies $\tan \frac{\theta }{2}=\frac{r}{L}=\frac{2}{5}$.

$$\tan \theta =\frac{2\cdot \frac{2}{5}}{1-\frac{4}{25}}=\frac{4/5}{21/25}=\frac{20}{21}.$$

Let $t=x^2-6$: $(t+1{)}^4+(t-1{)}^4=16$.

Expanding, $2t^4+12t^2+2=16\Rightarrow t^4+6t^2-7=0\Rightarrow (t^2+7)(t^2-1)=0$, so $t=\pm 1$.

Thus $x^2=7$ or $x^2=5$; the real roots are $\pm \sqrt{7},\pm \sqrt{5}$.

Max − min $=\sqrt{7}-(-\sqrt{7})=2\sqrt{7}$.

With height $h$, the distances from the base are $h\cot 30^{\circ }=h\sqrt{3}$ and $h\cot 45^{\circ }=h$.

$$h\sqrt{3}-h=20\Rightarrow h=\frac{20}{\sqrt{3}-1}=\frac{20(\sqrt{3}+1)}{2}=10(\sqrt{3}+1).$$

${}^n{C}_r$ needs integers $n\ge r\ge 0$.

• $17-x$ an integer forces $x$ to be an integer.
• $3x+1\ge 0\Rightarrow x\ge -\frac{1}{3}$, so $x\ge 0$.
• $17-x\ge 3x+1\Rightarrow x\le 4$.

So $x\in \{0,1,2,3,4\}$: 5 values.

Let $O$ be the midpoint of $AB$ (centre, radius $k/2$). The circle meets $AC$ and $BC$ where the angle subtended by $AB$ is $90^{\circ }$, i.e. at the feet of the altitudes, which are the midpoints $E$ of $AC$ and $F$ of $BC$.

The common region splits into:

• $△AOE$: $OA=OE=AE=k/2$, equilateral, area $\frac{\sqrt{3}}{16}{k}^2$.
• $△BOF$: likewise $\frac{\sqrt{3}}{16}{k}^2$.
• Sector $EOF$: $\angle EOF=180^{\circ }-60^{\circ }-60^{\circ }=60^{\circ }$, area $\frac{1}{6}\pi \frac{k^2}{4}=\frac{\pi k^2}{24}$. (The arc $EF$ lies inside the triangle, since $C$ is outside the circle.)

$$\text{Area}=\frac{\sqrt{3}}{8}{k}^2+\frac{\pi k^2}{24}=(3\sqrt{3}+\pi )\frac{k^2}{24}.$$

Let $x$ go into A. After one year:

• A: $x(1.05{)}^2=1.1025x$
• B: $1.12(10000-x)$

$$11200-1.12x-1.1025x=2310\Rightarrow 2.2225x=8890\Rightarrow x=4000.$$

Interest: A $=0.1025\times 4000=410$; B $=0.12\times 6000=720$. Total $=1130$.

The four groups (only 4-wheeler, only 2-wheeler, both, neither) add up to 500.

Both $=500-250-100-100=50$.

$ac>0$ means $a,c$ have the same sign, so $\frac{c}{a}=4$.

$\frac{b}{a}=\frac{b}{c}\cdot \frac{c}{a}=-\frac{3}{4}\cdot 4=-3$ (consistent with $|a/b|=1/3$).

$$\frac{b+c}{a}=-3+4=1.$$

Solving the caselet (used for Q41–45).

1. C wins S2: $c_2>a_2$ and $c_2>b_2$, so $20-c_2=a_2+b_2\le 2(c_2-1)$, giving $c_2\ge 8$. Likewise $c_3\ge 8$. Since C's total is 16, $c_2=c_3=8$ and C gets 0 in S1, S4, S5.
2. So S4 and S5 (not won by A or C) are won by B. In S4, $b_4>a_4=20-b_4$, so $b_4\ge 11$; then $b_5\ge 12$.
3. In S2 and S3, $a+b=12$ with both at most 7, so $b_2,b_3\in \{5,6,7\}$ with $b_2<b_3$, giving $b_2+b_3\ge 11$.
4. B's total: $b_1+(b_2+b_3)+(b_4+b_5)\ge 1+11+23=35$. The total is exactly 35, so every bound is tight: $B=(1,5,6,11,12)$.
5. Then $A=(19,7,6,9,8)$ (total 49) and $C=(0,8,8,0,0)$.

Votes (A, B, C) by constituency:

• S1: 19, 1, 0
• S2: 7, 5, 8
• S3: 6, 6, 8
• S4: 9, 11, 0
• S5: 8, 12, 0

Q41. B is below both A and C only in S2 ($5<7$, $5<8$). In S1, C has 0; in S3, B ties A at 6.

From the unique solution derived in Q41, $B=(1,5,6,11,12)$ across S1–S5. So B got 5 votes in S2.

From Q41, S5 has B = 12 and C = 0, so A $=20-12-0=8$.

From Q41, A's votes across S1–S5 are $19,7,6,9,8$. The lowest is 6, in S3.

Alliance votes (A + C) vs B, using the table in Q41:

• S1: $19$ vs $1$ → alliance
• S2: $15$ vs $5$ → alliance
• S3: $14$ vs $6$ → alliance
• S4: $9$ vs $11$ → B
• S5: $8$ vs $12$ → B

The alliance wins 3 seats.

The opening is ironic: India leads the world, "but this time for the wrong reason", namely having the most polluted capital. The irony sets up a critique of how India has handled air quality, and the rest of the passage calls for government action.

• A (negative) is vaguer. The writer isn't just gloomy; he is pointing out a failure.
• C (confusion) doesn't fit; the point is made clearly.
• D (judgmental) suggests passing moral verdicts on people. The writer criticises a situation and policy backed by data.

Paragraph 3 says many judicial pronouncements have upheld the right to clean air as a corollary to the right to life. Judicial pronouncements come from the courts.

The media (A), government (B) and Parliament (D) are not described as endorsing this right. The governments are the ones being pushed to act.

The last paragraph lists what must change and says every level of government must join the fight. The closing line stresses how important and urgent that fight is.

• A (ambivalent): the writer is clearly for action.
• B (melodramatic): the claim follows from evidence about health and cities, so it's emphatic but not overdone.
• D (bleak future): the line calls for action and doesn't predict failure.

GRAP "kicks in" when air quality turns bad and grows stricter as it worsens. The plan is activated, i.e. it comes into force.

"Causes violence" (B) takes "kick" literally. "Gains importance" (C) and "gets rejected" (D) don't fit a plan that is triggered by a threshold.

Check each option against the last two paragraphs:

• A is made: the fight is not only for the Centre or the courts; states and municipal governments must take part.
• B is made: governments must turn to science to understand, measure and monitor the problem.
• C is made: behaviour change needs a combination of incentives and penalties.
• D is not made. The passage describes clean air as a right linked to the right to life and as a common public good. It never frames it as a consumer right.

Rule out the clearly false options:

• A is false. New Delhi is the most polluted *capital*, but Begusarai is the most polluted *city*, and Delhi UT is only third among regions.
• B is false. The passage says the problem is not unique to metros; smaller cities and towns are choking too.
• C is false. 42 of the world's 50 most polluted cities are Indian, not all 50.

That leaves D, which is the official answer. The part the passage does support is that Begusarai is the world's most polluted city and is partly industrialised, with an oil refinery. But the passage never says industrialisation is the *cause*, and it calls Begusarai primarily agrarian. So D is only the best of the four, chosen by elimination.

Paragraph 2: anxiety becomes a problem when the danger is imagined or exaggerated, out of proportion to the real danger, and persistent and disabling. D captures this.

• A: scientists are studying both genes and environment; no primacy is claimed.
• B: "always imagined" is too strong; exaggerated real dangers count too.
• C: reverses the passage, which says *out of* proportion.

The passage covers anxiety's evolutionary survival role, when it becomes a disorder, research into causes and the brain, psychotherapy and medication, the benefits of exercise, and a closing note that normal anxiety is adaptive. Only D is broad enough for all of this.

A (therapy vs medication), B (exercise) and C (genes vs environment) each describe just one paragraph.

The exercise paragraph says physical tiredness brings faster, deeper sleep, and that for anxious people who struggle with insomnia a good night's rest can change their perspective. That is C.

• A: no link between endorphins and CBT is stated.
• B: the passage cites "plenty of evidence" that exercise helps; it doesn't call it a temporary distraction.
• D: the amygdala is mentioned only in the research paragraph, not in connection with exercise.

The first paragraph says anxiety is built into us, inherited from our evolutionary past, because it generally has a survival function. D restates this.

A (limited relevance now), B (an impediment to survival) and C (not integral to human nature) all contradict the passage.

The treatment section goes through exposure therapy (the first shown to work), group CBT, medication (controls but doesn't cure) and exercise (evidence that it helps). That is a discussion of options and how well they work.

• B: types of anxiety disorders are not compared.
• C: apart from calling exposure therapy "the first", there is no historical timeline.
• D: no opposing viewpoints are presented.

The passage says exposure therapy encourages people, in a graded way, to enter feared situations and stay in them. "Graded" means gradually, so A is correct.

• B contradicts the evidence that exercise helps with anxiety and sleep.
• C is wrong: medication won't cure the disorder; it only keeps it under control during psychotherapy.
• D is unsupported: no treatment is called the most preferred.

Fall through means a plan fails or doesn't happen, which is what heavy rain does to a picnic.

• Drop out means to withdraw (used for people).
• Drop off means to decline or fall asleep.
• Fall behind means to lag behind schedule or others.

Comes down to means "depends ultimately on": success comes down to fostering teamwork.

• Comes about means happens and takes no object.
• Come along with is ungrammatical here (and the verb doesn't agree with "success").
• Comes with means is accompanied by, which reverses the intended cause and effect.

Put up with means tolerate. He tolerates the interruptions because she's his younger sister.

• Puts along with is not an English idiom.
• Puts her off means discourages or repels her, which doesn't fit "but … because she is his sister".
• Puts her on means teases or deceives her.

Walk the talk means act on what you say, which is what earns credibility on issues you campaign about.

• Rule the roost means to be in charge.
• Go around the bend means to go mad.
• Read the lines is not the idiom ("read between the lines" means infer a hidden meaning).

Under the weather is the idiom for feeling unwell. The other prepositions (beneath, beside, outside) do not form idioms.

Bite someone's head off means to reply angrily and sharply, which matches a bad mood.

"Cut my voice" and "hit my ear off" are not idioms. "Ate my head" is a literal translation of an Indian-language expression and is not standard English.

Bring about means cause to happen, as in "brought about changes".

• Brought on is usually used for something bad, such as an illness or attack.
• Brought along means brought with you.
• Brought up means raised a topic or raised a child.

There are two errors:

1. "Hopes to + verb": the infinitive is needed, so "hopes to continue", not "to continuing".
2. The subject "hopes" is plural, so the verb must be "seem".

Only A fixes both.

• B keeps "continuing" and "seems".
• C uses the unidiomatic "hopes on continued talks".
• D has "to continued" and "seems".

• Tense: the sentence is in the past ("streets *were* filled", "time *seemed*"), so the verb should be "provided", which also fixes agreement with the singular "countryside".
• Pronoun: "for their seeking" should be "for those seeking solace".
• "A welcome retreat" (a retreat that is welcome) is the natural collocation.

A uses "for who seek", which is ungrammatical. B and C use present-tense "provides" against the past-tense context, and B has "them seeking". D is correct.

The consequence "I will not attend" is about the present and future, so the cause must be a present state: "I have a headache".

"Had been having" (past perfect continuous) and "was having" (B) put the headache in the past, and "used to have" (D) describes a past habit; none of these justifies skipping a meeting now. A drops the article ("have headache").

"All the time" marks a habitual action, which takes the simple present, matching "He is very careless". The verb is lose, not "loose" (to set free or slacken).

A uses "loosing" (wrong word). B has "has loses" (ungrammatical). C's "was losing" is past tense, which clashes with "is". D is correct.

The original compares *thrillers* with a *filmmaker*, which is a faulty comparison. It should compare thrillers with those of other filmmakers.

• The subject "thrillers" is plural, so the verb is "have".
• Impact built up over history calls for the present perfect "have had".
• Hitchcock is himself a filmmaker, so the comparison needs "any other filmmaker".

• A keeps the faulty comparison and drops "other".
• B uses "are having", which is unidiomatic for impact.
• C uses singular "has" with plural "thrillers".

The next sentence starts with "Athletes must therefore undergo rigorous training to maintain peak physical condition", so the blank must give a physical reason for training. A does this with physical demands and injury risk.

• B (higher education) and C (other career paths) don't lead to training.
• D is about psychological pressure, which fits the later sentence on mental resilience, not "therefore … physical condition".

"This happened after a lorry … went missing" needs an earlier situation that was caused by the missing lorry. Panic and confusion among traders fits, and the rest (lost contact, unsure whether hijacked or stolen) explains it.

• B says the vehicle was hijacked and the tomatoes sold, but the traders later "were not sure" if it was hijacked, a contradiction.
• C is unrelated to the theft.
• D happened *before* the lorry went missing, so "this happened after" makes no sense.

The paragraph ends on whether AI can be creative *in the same sense as humans*. C continues that debate by naming what AI-made work may lack compared with human creativity.

• A (fear of machines) changes the subject.
• B (calculation speed) is about computing power, not creativity.
• D (ethics, privacy, bias) introduces a different issue.

The opening contrasts an era that prizes privacy and individualism with Yamamoto blurring public and private. D picks this up directly: he stands out by advocating a return to more interconnected living. It also leads into the next sentence about breaking down barriers and building community.

• A is generic and doesn't link to the privacy/individualism contrast.
• B is award news that breaks the flow of ideas.
• C (traditional and modern aesthetics) shifts to style, away from public vs private.

• Blank 1: the outcome of enemies' *intentional* ___ calls for a deliberate scheme, such as plotting or machinations. "Errors" and "inaction" aren't intentional schemes.
• Blank 2: the example is the planet turning into an oven, so "catastrophic events". "Outstanding", "advantageous" and "indescribable" don't fit.
• Blank 3: "a matter of power and prowess" means capability, matching the claim that events can be controlled.

B fails at "outstanding events", so C is the only option that fits all three blanks.

• Blank 1: critical thinking lets people navigate information discerningly, telling good from bad. "Discreetly" means secretly, which is wrong.
• Blank 2: fake news is widespread, a pervasive challenge. "Holistic" makes no sense, and "systematic" is off.
• Blank 3: media literacy is vital for fostering informed citizenship. "Inhibiting" reverses the meaning.

Only A works in all three blanks.

• Blank 1: "without precedent in history" is the standard phrase. "Without parallel" also works, but "without similarity" or "without equivalent" do not.
• Blank 2: "containing and mitigating the spread" works. "Eradicating the spread" is odd, since you eradicate a disease, not its spread.
• Blank 3: "reducing the incidence of the disease" (how many new cases occur) fits with limiting pressure on hospitals. "Reducing the outbreak" (D) is unidiomatic.

B fits all three.

• Blank 1: the paragraph is about populists escaping restraints, so the ones who benefit are aspiring autocrats. Entrepreneurs are off-topic, and "democrats" clashes with escaping restraints.
• Blank 2: connecting directly with people means bypassing the checks parties provided. "Enforcing" reverses the meaning, and "unscrupulous" is an adjective where a verb is needed.
• Blank 3: echo chambers create the impression that the whole people uniformly backs the leader.

C fits all three.

The story is a probe finding that a race was fixed.

• Blank 1: an investigation "found" wrongdoing. An appraisal or overview doesn't uncover facts like this.
• Blank 2: he slowed down deliberately (on purpose). "Systematically" and "particularly" don't fit.
• Blank 3: "after signalling to the one behind him". "Beckoning to" is close, but C's "overview" fails blank 1, and "speaking" and "waving" are weaker.

D fits all three.

• Blank 1: a pressing (urgent) challenge fits rising xenophobia. "Sophisticated" is wrong.
• Blank 2: nurturing mutual understanding and respect is the natural collocation. "Expediting" and "stimulating" are odd with "understanding".
• Blank 3: challenging stereotypes goes with fighting intolerance. Archetypes, propositions and truisms aren't what fuels xenophobia.

B fits all three.

D should read "trail of blood", a track left behind. A "train" is a sequence or a vehicle, not a track.

A ("train of thought"), B (train a dog, meaning teach) and C (travel by train) are all correct.

Neglected cattle that wander into fields have strayed, not "stayed away"; the intended word is "stray".

A (stay awake), C (a stay order) and D (stay with grandparents) are correct.

*Proscribe* means forbid; *prescribe* means lay down or recommend. A manual prescribes instructions, it doesn't forbid them, so C is wrong.

A, B and D all describe forbidding something, which is correct. (A also has an agreement slip, "rules … proscribes", but the word itself is used correctly.)

Calming a tense situation is defusing it (removing the fuse). To *diffuse* is to spread out, so C should say "defused".

A (aroma spreads), B (glass scatters light) and D (fan disperses smoke) all correctly use "diffuse" to mean spread out.

A *callus* (noun) is hardened skin. The adjective meaning unfeeling or cruel is callous, so D should say "callous indifference".

A, B and C correctly use "callus" for the patch of thickened skin.

The notice of an intended marriage that lets the community raise objections is the banns. A ban (a prohibition) doesn't invite objections, so C should use "banns".

A, B and D all use "bans" to mean prohibits, which is correct.

Suggested order: 3241

• 3 opens with the facts: the Minister announces over 2800 lightning deaths last year.
• 2 gives the public reaction: people demand compensation for victims' families.
• 4 "The Minister responded": a response needs something to respond to, which is the demand in 2. His response is that pamphlets on precautions are being distributed.
• 1 "Farmers, however, … ignore the warnings": the contrast shows why awareness pamphlets may not be enough, so it follows 4.

Suggested order: 25143

• 2 states the main idea: the new social media are no longer very social.
• 5 explains it: apps serve AI-selected clips based on viewing behaviour, *not social connections*.
• 1 "Meanwhile, people are posting less": a parallel trend on the user side.
• 4 Debate is moving to closed platforms like WhatsApp and Telegram, continuing the retreat from public posting.
• 3 "Some of the consequences of this…": "this" is the move to closed groups, and campaigners now target private groups. It needs 4 before it.

The question is misprinted: sentences 2 and 3 are word-for-word the same, so no unique order exists.

The logical flow is still clear:

• 4 introduces space's weaker gravity: less pressure on the spine, so you grow up to two inches taller.
• 1 "In this microgravity environment, your blood also…": "this" and "also" need 4 before it.
• 2/3 "Both effects are only short-lived…": this sums up the spine and heart effects, so it comes last.

So the order is 4, 1, then the duplicated sentence (4123 or 4132). The original probably had a sentence about the effect on the heart in one of those slots.

Suggested order: 3142

• 3 introduces kava by its regional names and its sacred role in Pacific countries.
• 1 "The muddy concoction": describes how the drink is made and its cultural role in sharing, which needs kava already introduced.
• 4 "A swig (or two) taken from a bowl or shell": moves from how it's shared to what drinking it does, profound relaxation.
• 2 "this blissful state": refers back to the relaxation and numbness in 4.

Suggested order: 32415

• 3 introduces the topic: strategic giving is emerging and "differs … between countries".
• 2 "There are no comparable data on different countries. But one report…": picks up "countries" and gives India's figure (USD 13 billion).
• 4 adds a second Asian figure: China, USD 21 billion.
• 1 "By contrast, in America … almost USD 500 billion": contrasts with the Asian figures just given.
• 5 "By any measure, however…": comes back to the data caveat in 2. Even though the data aren't comparable, Asian philanthropy is clearly far smaller than the West's.

(Placing 5 before 1 is possible, but then "By contrast" would follow a sentence already about the West, so 32415 reads better.)