IPMAT Indore 2019 · Question Paper & Solutions

Interior-angle sum is $(n-2)\times 180^{\circ }$.

$$(n-2)\times 180<2019⟹n-2<11.21⟹n-2\le 11$$

So $n_{\max }=13$ (sum $=1980^{\circ }<2019^{\circ }$; $n=14$ gives $2160^{\circ }$).

Note on the printed stem: the PDF renders the first base as $a^{2}c$, but the other two bases are $b^{2}c$ and $c^{2}a$, so the cyclic pattern shows the intended expression is
$$\frac{1}{1+{\log }_{a^{2}b}\frac{c}{a}}+\frac{1}{1+{\log }_{b^{2}c}\frac{a}{b}}+\frac{1}{1+{\log }_{c^{2}a}\frac{b}{c}}$$

Write $A=\log a,B=\log b,C=\log c$ and $S=A+B+C$. Then
$$1+{\log }_{a^{2}b}\frac{c}{a}=\frac{(2A+B)+(C-A)}{2A+B}=\frac{S}{2A+B}$$
so the first term is $\frac{2A+B}{S}$. Similarly the others are $\frac{2B+C}{S}$ and $\frac{2C+A}{S}$.

$$\text{Sum}=\frac{3(A+B+C)}{S}=3$$

(With the literal base $a^{2}c$ the sum is not a constant, so the printed base is a typo.)

Put $y=0$: $f(x)f(0)=f(3)+3f(x)-3f(0)$, i.e. $f(x)(f(0)-3)=f(3)-3f(0)$.

If $f(0)\ne 3$, $f$ would be a constant $c$; then the original relation gives $c^2=c+6y$ for all $y$, impossible. So $f(0)=3$ and hence $f(3)=9$.

Put $x=0$: $3f(y)=f(3)+3f(y)-3f(y)+6y=9+6y$, so $f(y)=2y+3$.

Check: $(2x+3)(2y+3)=4xy+6x+6y+9$ and the right side is $(4xy+9)+3(2x+2y+3)-3(2y+3)+6y=4xy+6x+6y+9$. ✓

$$f(8)=2(8)+3=19$$

For $4\times 4$ matrices $\det (kM)=k^4\det M$, $\det B^{-1}=1/\det B$, $\det B^T=\det B$.

$$\det (2A{B}^{-1}{C}^3{B}^T)=2^4\cdot 5\cdot \frac{1}{-3}\cdot {\left(\frac{1}{2}\right)}^3\cdot (-3)=16\cdot 5\cdot \frac{1}{8}=10$$

$$(I+A{)}^2-50A=I+2A+A^2-50A=I-48A$$

Since $A^2=0$, every eigenvalue of $A$ is $0$, so every eigenvalue of $I-48A$ is $1$. Hence $\det (I-48A)=1$.

(Alternatively, $(I-48A)(I+48A)=I-48^2{A}^2=I$, and $\det (I-48A)=\det (I+48A)$ by the same eigenvalue argument, so the determinant squared is 1 and it is $1$.)

Shares are $3k,5k,7k$, total $15k$.

$$15k-16=7k⟹8k=16⟹k=2$$

Total $=15\times 2=30$. Check: $30-16=14=7\times 2$. ✓

New revenue $=0.75\times 1.20=0.90$ of the old revenue.

So revenue falls by $10\%$: $x=10$.

Let $y=\frac{3x^2-6x+12}{x^2+2x+4}$ (denominator is always positive). Then
$$(y-3)x^2+(2y+6)x+(4y-12)=0$$

For real $x$, the discriminant is $\ge 0$:
$$(2y+6{)}^2-4(y-3)(4y-12)\ge 0⟺(y+3{)}^2\ge 4(y-3{)}^2⟺|y+3|\ge 2|y-3|$$

This gives $1\le y\le 9$. The maximum $9$ is attained at $x=-2$: $\frac{12+12+12}{4-4+4}=9$.

So $k=9$.

Centre $(0,0)$, radius $2$. Distance from $(-5,0)$ to the centre is $5$.

The farthest point is on the line through the centre, on the far side: $5+2=7$.

Let $x^{12}=y^{16}=z^{24}=t$ and $L=\log t$. Then $\log x=\frac{L}{12},\log y=\frac{L}{16},\log z=\frac{L}{24}$.

• $3{\log }_{y}x=3\cdot \frac{16}{12}=4$
• $4{\log }_{z}y=4\cdot \frac{24}{16}=6$
• $n{\log }_{x}z=n\cdot \frac{12}{24}=\frac{n}{2}$

AP: $4,6,\frac{n}{2}$ needs $\frac{n}{2}=8$, so $n=16$.

Use the identity
$$\sin x+\sin y-\sin (x+y)=4\sin \frac{x}{2}\sin \frac{y}{2}\sin \frac{x+y}{2}$$

So the equation holds iff $x=2k\pi$, $y=2k\pi$, or $x+y=2k\pi$.

With $|x|+|y|=1$ we have $|x|,|y|,|x+y|\le 1<2\pi$, so the only options are $x=0$, $y=0$ or $x+y=0$:
- $x=0$: $y=\pm 1$ (2 pairs)
- $y=0$: $x=\pm 1$ (2 pairs)
- $y=-x$: $2|x|=1$, giving $(\frac{1}{2},-\frac{1}{2}),(-\frac{1}{2},\frac{1}{2})$ (2 pairs)

Total $=6$.

Centre $(3,5)$, $r^2=9+25-k=34-k$.

• Does not touch/intersect the axes: $r$ is less than the distance to the nearer axis (the $y$-axis, distance $3$): $34-k<9\Rightarrow k>25$.
• $(1,4)$ not outside: $(1-3{)}^2+(4-5{)}^2\le r^2\Rightarrow 5\le 34-k\Rightarrow k\le 29$.

So $k\in (25,29]$ and $a+b=54$.

A row of three $\pm 1$ entries sums to $1$ only as two $+1$s and one $-1$. The same holds for columns, so the $-1$s sit one per row and one per column: they form a permutation matrix $P$.

Hence $M=J-2P$, where $J$ is the all-ones matrix. Since $P^{T}J=J$,
$$\det M=\det P\cdot \det (P^{T}J-2I)=\pm \det (J-2I)$$

$J$ has eigenvalues $3,0,0$, so $J-2I$ has eigenvalues $1,-2,-2$ and determinant $4$.

$|\det M|=4$ (a brute-force check over all $2^9$ fillings confirms this).

$x\in {\bigcup }_{k=2}^{25}Q(k)$ exactly when $x$ has a divisor $k$ with $2\le k<x$, i.e. when $x$ is composite.

So $P-\bigcup Q(k)$ is the set of primes up to 25: $2,3,5,7,11,13,17,19,23$, which has $9$ elements.

Let the tile's legs be $a$, so its hypotenuse is $a\sqrt{2}=R$ and its area is $\frac{a^2}{2}=\frac{R^2}{4}$.

$$\frac{\pi R^2}{R^2/4}=4\pi \approx 12.57$$

So $12$ whole tiles.

One solution is $(x,y)=(-1,1)$. The general solution is $x=-1+7t,y=1-4t$.

• $|x|<100$: $-100<-1+7t<100\Rightarrow -14.1<t<14.4\Rightarrow t\in [-14,14]$
• $|y|<100$: $-100<1-4t<100\Rightarrow -24.75<t<25.25$ (always true for the above)

So $t=-14,\dots ,14$: $29$ solutions (e.g. $t=-14$ gives $(-99,57)$, $t=14$ gives $(97,-55)$). Brute force confirms 29.

Sum $=5\times 110=550$. To maximise the largest, make the middle three as small as possible: $101,102,103$.

$$\text{Largest}=550-100-101-102-103=144$$

• 1-digit numbers: $9$ digits
• 2-digit numbers: $90\times 2=180$ (running total $189$)
• 3-digit numbers: $900\times 3=2700$ (running total $2889$)

Remaining: $6389-2889=3500=4\times 875$ digits of 4-digit numbers, ending exactly at the last digit of the 875th 4-digit number, $1000+874=1874$.

The digit is $4$.

$$x+y=xy⟺xy-x-y+1=1⟺(x-1)(y-1)=1$$

So $x-1=y-1=1$ or $x-1=y-1=-1$, giving $(2,2)$ and $(0,0)$.

Number of pairs $=2$.

• Multiples of 4: $\lfloor 999/4\rfloor =249$
• Multiples of 6: $166$
• Multiples of both (of 12): $83$

Multiples of 4 or 6: $249+166-83=332$.

Odd multiples of 9: $9m$ with $m$ odd and $9m\le 999$, so $m=1,3,\dots ,111$: $56$ numbers. These are odd, so none of them is a multiple of 4 or 6.

Total $=332+56=388$.

With principal $P$ and rate $r$ (as a decimal):
- CI for 2 years: $P[(1+r{)}^2-1]=P(2r+r^2)$
- SI for 2 years: $2Pr$

$$2r+r^2=2(2r)⟹r^2=2r⟹r=2$$

So the rate is $200\%$ per annum.

$21=3\times 7$, so $n$ is limited by the smaller of the exponents of 3 and 7 in $50!$.

• Power of 7: $\lfloor 50/7\rfloor +\lfloor 50/49\rfloor =7+1=8$
• Power of 3: $16+5+1=22$

So $n_{\max }=8$.

$29\equiv 2(\mathrm{mod}9)$, so $(29^{29}{)}^{29}\equiv 2^{841}(\mathrm{mod}9)$.

$2^6=64\equiv 1(\mathrm{mod}9)$ and $841=6\times 140+1$, so $2^{841}\equiv 2^1=2$.

Remainder $=2$.

Divisible by 6, 7 and 9 means divisible by $\text{lcm}=126$.

$$453096=126\times 3596$$

Checking the others: $453078$, $453042$ and $453054$ leave remainders $108$, $72$ and $84$ on division by 126. Only 96 works.

(Quick filter: the digit sum must be a multiple of 9. $4+5+3+0=12$, so the two digits must sum to 6 or 15, which leaves only 96 and 78. 78 fails divisibility by 7.)

• Minimum overlap: $x\ge 70+50-100=20$.
• Maximum overlap: the overlap cannot exceed the smaller group, so $x\le 50$.

Hence $20\le x\le 50$.

By inclusion–exclusion:
$$65=40+25+20-10-8-|C\cap H|+5=72-|C\cap H|$$

So $|C\cap H|=7$.

Note on the printed stem: the PDF shows only the expression $\frac{x-3}{x^2+3x+2}$ with no inequality sign or right-hand side. The standard version of this question is $\frac{x-3}{x^2+3x+2}\ge 1$.

$$\frac{x-3}{x^2+3x+2}-1\ge 0⟺\frac{-(x^2+2x+5)}{(x+1)(x+2)}\ge 0$$

$x^2+2x+5=(x+1{)}^2+4>0$ always, so we need $(x+1)(x+2)<0$, i.e. $x\in (-2,-1)$.

Largest $b-a=-1-(-2)=1$.

(With "$\le 1$" or "$<1$" instead, the solution set would be unbounded, so no finite answer would exist. That supports the $\ge 1$ reading.)

Upper bound: $a^2+b^2+c^2-(ab+bc+ca)=\frac{1}{2}[(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]\ge 0$, so $ab+bc+ca\le 1$. Equality holds at $a=b=c=\frac{1}{\sqrt{3}}$.

Lower bound: $(a+b+c{)}^2=1+2(ab+bc+ca)\ge 0$, so $ab+bc+ca\ge -\frac{1}{2}$. Equality holds when $a+b+c=0$, e.g. $a=\frac{1}{\sqrt{2}},b=-\frac{1}{\sqrt{2}},c=0$.

The expression is continuous on the sphere, so its range is $[-\frac{1}{2},1]$. (The PDF drops the brackets in options B and D.)

Since ${\log }_2$ is increasing, we need $0<\frac{3x-1}{2-x}<2$.

Positive: $\frac{3x-1}{2-x}>0⟺\frac{1}{3}<x<2$.

Less than 2: $\frac{3x-1}{2-x}-2=\frac{5x-5}{2-x}<0⟺x<1$ or $x>2$.

Intersection: $x\in (\frac{1}{3},1)$.

$0.343=0.7^3$. The base $0.7$ is less than 1, so the inequality reverses when we compare exponents:
$$2x^2-3x+4>3⟺2x^2-3x+1>0⟺(2x-1)(x-1)>0$$

So $x<\frac{1}{2}$ or $x>1$.

Let the radius be $R$. A chord of length $R$ lies at distance $d=\sqrt{R^2-R^2/4}=\frac{\sqrt{3}}{2}R$ from the centre.

Each inscribed circle is centred on the perpendicular bisector of the chord. Its diameter spans from the chord's midpoint to the big circle, so the diameters are $R-d$ (minor side) and $R+d$ (major side).

Ratio of radii:
$$\frac{R+d}{R-d}=\frac{2+\sqrt{3}}{2-\sqrt{3}}=(2+\sqrt{3}{)}^2=7+4\sqrt{3}$$

Ratio of areas:
$$k=(7+4\sqrt{3}{)}^2=49+48+56\sqrt{3}=97+56\sqrt{3}$$

Take side $1+n$, so $AP=1$ and $PB=n$ (and similarly on every side).

Each corner triangle has legs $1$ and $n$, area $\frac{n}{2}$. The four together have area $2n$.

$$[PQRS]=(1+n{)}^2-2n=1+n^2$$

Ratio $=(1+n^2):(1+n{)}^2$. (Equivalently, $P{Q}^2=1+n^2$ and $PQRS$ is a square.)

A chord of length $\ell$ in a circle of radius 6 subtends a central angle $2\theta$ with $\ell =12\sin \theta$.

• $AB=3$: $\sin \alpha =\frac{1}{4}$, $\cos \alpha =\frac{\sqrt{15}}{4}$
• $BC=2$: $\sin \beta =\frac{1}{6}$, $\cos \beta =\frac{\sqrt{35}}{6}$

B lies on minor arc AC, so the central angles add and $AC=12\sin (\alpha +\beta )$:
$$\sin (\alpha +\beta )=\frac{1}{4}\cdot \frac{\sqrt{35}}{6}+\frac{\sqrt{15}}{4}\cdot \frac{1}{6}=\frac{\sqrt{35}+\sqrt{15}}{24}$$

$$AC=\frac{\sqrt{15}+\sqrt{35}}{2}\text{ m}\approx 4.89\text{ m}$$

The curve is a rhombus with vertices $(\pm 3,0)$ and $(0,\pm 2)$, so its diagonals are $6$ and $4$.

$$\text{Area}=\frac{6\times 4}{2}=12\text{ square units}$$

Check: the first-quadrant piece is a triangle with legs 3 and 2, area $3$. Four such triangles give $12$.

This is option A. The official key's "3 square units" is the area of just one quadrant's triangle.

Place the points at $(0,0)$ and $(1,0)$. Then $\frac{\sqrt{x^2+y^2}}{\sqrt{(x-1{)}^2+y^2}}=\frac{3}{2}$ gives
$$4(x^2+y^2)=9[(x-1{)}^2+y^2]⟺5x^2+5y^2-18x+9=0$$

Equal coefficients of $x^2$ and $y^2$ and no $xy$ term: a circle, the Apollonius circle, centre $(\frac{9}{5},0)$, radius $\frac{6}{5}$.

A straight line would occur only for ratio $1:1$.

Let $s=\sin x-\sin y$. Then
$$s^2+1=(\sin x-\sin y{)}^2+(\cos x+\cos y{)}^2=2+2\cos (x+y)$$

So $s^2=1+2\cos (x+y)\le 3$, giving $|s|\le \sqrt{3}$.

Attained: $x=\frac{\pi }{3},y=-\frac{\pi }{3}$ gives $\cos x+\cos y=1$ and $s=\sqrt{3}$. Swapping gives $-\sqrt{3}$. By continuity every value between is attained.

Range $=[-\sqrt{3},\sqrt{3}]$.

$${\sin }^6\theta +{\cos }^6\theta =({\sin }^2\theta +{\cos }^2\theta {)}^3-3{\sin }^2\theta {\cos }^2\theta =1-3{\sin }^2\theta {\cos }^2\theta$$

Squaring $\sin \theta +\cos \theta =m$ gives $\sin \theta \cos \theta =\frac{m^2-1}{2}$. So
$${\sin }^6\theta +{\cos }^6\theta =1-\frac{3(m^2-1{)}^2}{4}$$

Check at $\theta =0$ ($m=1$): the value is $1$, and the formula gives $1$. Printed option D, $1-\frac{3(m^2+1{)}^2}{4}$, would give $-2$.

So the intended answer is D with $(m^2-1{)}^2$. The PDF prints a $+$ sign, which is a typo; no option is correct exactly as printed.

A matrix times its inverse is $I$:
$$\left[\begin{array}{cc}2& -0.5\\ -1& x\end{array}\right]\left[\begin{array}{cc}1& 1\\ 2& 4\end{array}\right]=\left[\begin{array}{cc}1& 0\\ -1+2x& -1+4x\end{array}\right]$$

This must equal $I$, so $-1+2x=0$ and $-1+4x=1$. Both give $x=0.5$.

Note on the printed stem: the endpoints $\frac{5\pm \sqrt{57}}{2}$ are the roots of $x^2-5x-8$, so the intended function is $f(x)=\frac{x^3-5x^2-8x}{3}=\frac{x(x^2-5x-8)}{3}$. The PDF prints $+5x^2$.

Sign: zeros at $0$ and $\frac{5\pm \sqrt{57}}{2}\approx -1.27,6.27$.
- For $x<-1.27$: $x<0$ and $x^2-5x-8>0$, so $f<0$.
- For $x>6.27$: both factors are positive, so $f>0$.

Monotonicity: $f^{\prime }(x)=\frac{3x^2-10x-8}{3}=\frac{(3x+2)(x-4)}{3}>0$ for $x<-\frac{2}{3}$ and for $x>4$. Both intervals above lie inside these, so $f$ is increasing on both.

This matches option C.

With the literal $+5x^2$, $f(-2)=\frac{28}{3}>0$, so $f$ is not negative on $(-\infty ,-1.27)$ and no option is exactly true. Option C's pattern would also be correct for the literal $+5x^2$ function if the endpoints were $\frac{-5\pm \sqrt{57}}{2}$, so C is the intended answer under either single-sign fix.

A sum of distances to points on a line is minimised at the median point, $x=b$.

$$f(b)=(a-b)+0+(b-c)=a-c$$

(For $c\le x\le a$, $|x-a|+|x-c|=a-c$ is constant, and $|x-b|$ is smallest at $x=b$.)

Let the GP be $\alpha ,\alpha r,\alpha r^2,\alpha r^3$.

• $\alpha +\beta =\alpha (1+r)=1$
• $\gamma +\delta =\alpha r^2(1+r)=4$

Dividing: $r^2=4$, so $r=\pm 2$.

• $r=2$: $\alpha =\frac{1}{3}$, so $p=\alpha \beta =\frac{2}{9}$, not an integer. Rejected.
• $r=-2$: $\alpha =-1$, and the terms are $-1,2,-4,8$. Then $p=(-1)(2)=-2$ and $q=(-4)(8)=-32$.

$p+q=-34$.

Let $f(x)=(1+x-2x^2{)}^6$. The sum of even-index coefficients is $\frac{f(1)+f(-1)}{2}$.

• $f(1)=(1+1-2{)}^6=0$
• $f(-1)=(1-1-2{)}^6=64$

So $A_0+A_2+\cdots +A_{12}=32$. With $A_0=f(0)=1$:
$$A_2+A_4+\cdots +A_{12}=31$$

(Expanding the polynomial directly confirms this.)

The first AP contains numbers $\equiv 2(\mathrm{mod}3)$. The second contains odd numbers from 3 to 101.

Common terms are odd and $\equiv 2(\mathrm{mod}3)$, i.e. $\equiv 5(\mathrm{mod}6)$: $5,11,17,\dots$, with the upper limit set by 101.

$5+6k\le 101\Rightarrow k\le 16$, so there are $17$ common terms (5 to 101).

The answer depends on how "an Ace appears at the $n$th turn" is read.

Reading 1: the first Ace appears at the $n$th draw. This is the reading that gives the key.
- $f(1)=\frac{4}{52}=\frac{1}{13}\approx 0.0769$
- $f(2)=\frac{48}{52}\cdot \frac{4}{51}\approx 0.0724$
- $f(3)=\frac{48}{52}\cdot \frac{47}{51}\cdot \frac{4}{50}\approx 0.0681$

So $\frac{1}{13}>f(2)>f(3)$, which is option B.

Reading 2: the $n$th card drawn is an Ace (not necessarily the first). By symmetry every position is equally likely to hold any given card, so $f(n)=\frac{4}{52}=\frac{1}{13}$ for all $n$. That is option D.

The literal wording fits Reading 2 better; the key assumes Reading 1.

Ordered triples from $\{1,\dots ,6\}$ summing to 15:
- permutations of $(6,6,3)$: 3
- permutations of $(6,5,4)$: 6
- $(5,5,5)$: 1

That is $10$ equally likely outcomes.

First throw $4$: the other two sum to 11, giving $(5,6),(6,5)$, which is 2 outcomes.

$$P=\frac{2}{10}=\frac{1}{5}$$

Persons, not families, are sampled, so larger families are over-represented in proportion to their size.

With $N$ families of each size, the number of people is $2N,4N,6N$. A random person reports size $s$ with probability $\propto s$:
$$E=\frac{2\cdot 2+4\cdot 4+6\cdot 6}{2+4+6}=\frac{56}{12}\approx 4.667$$

The tempting answer 4 is the average per family, not per person.

$({\log }_{a}b{)}^{-1}={\log }_{b}a$, and ${\log }_{900}4={\log }_{30^2}{2}^2={\log }_{30}2$.

$${\log }_{30}3+{\log }_{30}2+{\log }_{30}5={\log }_{30}30=1$$

Both logs must be defined, so $f,g>0$.

• $a>1$: ${\log }_a$ is increasing, so $f<g$, i.e. $g>f>0$.
• $0<a<1$: ${\log }_a$ is decreasing, so $f>g>0$.

This is option A. C and D ignore that the direction depends on the base, and B swaps the cases.

$6(a^2+b^2+c^2)=564\Rightarrow a^2+b^2+c^2=94$ with positive integers.

Checking squares up to 81:
- $81+9+4$: edges $9,3,2$, volume $729+27+8=764$
- $49+36+9$: edges $7,6,3$, volume $343+216+27=586$

No other decomposition exists (e.g. $94-64=30$, $94-25=69$, $94-16=78$ and $94-1=93$ are not sums of two positive squares). An exhaustive search confirms only 764 and 586.

Write each with exponent 100:
- $2^{300}=8^{100}$
- $3^{200}=9^{100}$
- $4^{100}$
- $2^{100}+3^{100}<2\cdot 3^{100}<9^{100}$

So $3^{200}$ is the greatest.

Interior points satisfy $x\ge 1$, $y\ge 1$, $x+y\le 30$.

For $x=k$ ($1\le k\le 29$) there are $30-k$ values of $y$:
$$\sum _{k=1}^{29}(30-k)=29+28+\cdots +1=\frac{29\cdot 30}{2}=435$$

(Pick's theorem agrees: area $\frac{961}{2}$, boundary points $93$, so $I=\frac{961}{2}-\frac{93}{2}+1=435$.)

Let the speeds be $a>b$.

• Towards each other: $a+b=\frac{x}{u}$
• Same direction: $a-b=\frac{x}{v}$

$$\frac{b}{a}=\frac{\frac{x}{u}-\frac{x}{v}}{\frac{x}{u}+\frac{x}{v}}=\frac{v-u}{v+u}$$

Copper fractions: P has $\frac{5}{7}$, Q has $\frac{3}{7}$.

Mix $a:b$ with copper $=\frac{1}{2}$:
$$\frac{5a+3b}{7(a+b)}=\frac{1}{2}\Rightarrow 10a+6b=7a+7b\Rightarrow b=3a$$

So $a:b=1:3$. Mixing $b:a=3:1$ (3 parts P, 1 part Q), copper $=\frac{3\cdot \frac{5}{7}+\frac{3}{7}}{4}=\frac{18}{28}$ and zinc $=\frac{10}{28}$.

Cu : Zn $=18:10=9:5$.

• Numbers with 1 to 6 digits are all below 3000000: $2+4+8+16+32+64=126$.
• 7-digit numbers must start with 1 (a leading 3 gives at least 3111111): $2^6=64$.

Total $=126+64=190$.

Each product $a_i{a}_{i+1}$ is $\pm 1$. They sum to 0, so $n$ is even, with $\frac{n}{2}$ of them equal to $-1$.

The product of all $n$ terms is $(a_1{a}_2\cdots a_n{)}^2=1$, so the number of $-1$ terms, $\frac{n}{2}$, is even. Hence $4\mid n$.

Every multiple of 4 works: repeat the block $1,1,1,-1$, whose cyclic products are $1,1,-1,-1$.

Option A is impossible and D is too restrictive, so the answer is C.

Exports read from the chart (Rs. Cr), 2006–2017: $280,240,230,210,200,220,210,200,200,200,220,200$.

$$\text{Sum}=2610,\text{Average}=\frac{2610}{12}=217.5\approx 220$$

• 2006→2011: $280\to 220$, decline $\frac{60}{280}\times 100=21.43\%$
• 2012→2017: $210\to 200$, decline $\frac{10}{210}\times 100=4.76\%$

Difference $=16.67$ percentage points, closest to $16.5$.

Imports minus exports, 2006–2017:
$-60,40,30,70,110,70,110,100,120,110,30,40$

The maximum is $320-200=120$ (2014).

Balance (exports − imports) for 2007–2017:
$-40,-30,-70,-110,-70,-110,-100,-120,-110,-30,-40$

Three-year averages:
- 2007–09: $-140/3\approx -46.7$
- 2015–17: $-180/3=-60$
- 2014–16: $-260/3\approx -86.7$
- 2010–12: $-290/3\approx -96.7$

The most negative is 2010–2012.

• 2009→2010: $280\to 310$, $+10.7\%$
• 2010→2011: $310\to 290$, a decrease
• 2013→2014: $300\to 320$, $+6.7\%$
• 2008→2009: $260\to 280$, $+7.7\%$

Among the options the maximum is 2009 to 2010. (2006→2007, $+27\%$, is larger but is not an option.)

The passage says each castaway would be set to a task according to his capacity: the strongest dig, the most dexterous make shoes, the best educated prospect and plan irrigation. So he does it because he is suited to it.

• A (weapons) and B (unexploited resources) are never mentioned.
• C (finding manual work beneath him) adds a motive the passage does not suggest.

"Justly" means the suspicion is warranted. The examples (the man altering the sluices at night, the shoemaker hiding his bark) show the others would *naturally, and rightly*, suspect secrecy. So suspicion of secrecy is natural.

• A and B split hairs over the two words; the author uses them together, not in contrast.
• D misreads "instantly" as describing how fast relationships break down.

The passage says he would want more in exchange than the trouble of making the sandals *deserved*. He hides the supply to overcharge, which is an unfair practice.

• A (a strong bargain) is neutral or approving, but the author disapproves.
• C (barter) describes the medium of exchange, not the point of the example.
• D (intent to make trouble) is not his motive; his motive is profit.

Near the end the author says that, whatever the work, there would certainly be difficulties about it. So hardships are part of any work.

• B (not complaining) is not stated.
• C and D distort the later idea of mutual help: the author says difficulties can be eased by others' help, not that help must be expected or traded.

The author says the fastest progress comes from helping each other, but that help can be given *only* so long as the men are frank and open. The rest of the passage attacks secrecy and ends by praising having no secrets. Openness is the condition everything else depends on, so D captures the author's central belief.

• B (co-operation) is tempting, but the passage treats co-operation as something transparency makes possible. The distinctive argument is about openness.
• A (multi-talented members) contradicts the division of labour described.
• C (dealing with secretive people) is a side remark, not the belief.

The shipwreck is a thought experiment. Its lessons (division of labour, openness, mutual help) apply to any community.

• B: the island is called uninhabited, not rich in resources.
• C: the castaways are ordinary men with different capacities, not an ideal world.
• D: the setting is primitive, but the point is not about primitive worlds.

Her findings: no brain differences are solely gender-related, and brains are individual, adaptable, with unbounded potential. So a brain's potential is not tied to gender.

• A ("alike") overstates: the passage says brains are highly individualised, not identical, only that differences are not gender-defined.
• B and C are recommendations the passage makes, not what the research established.

"Gendered" introduces how we still assume gender determines skills and preferences, from toys and colours to careers (Barbie for girls, Lego for boys). That is a society stuck on stereotypes.

• A and C state the stereotypes as if they were true, but the writer rejects them.
• D restates the opening line about the debate, not the meaning of "gendered".

The last paragraph lists the harms: it can be used to say girls need not do science and boys must, to condemn boys for being emotional because that is seen as a feminine trait, and to proclaim that boys are meant to lead.

• A matches the "feminine trait" example: a binary view promotes labelling traits as masculine or feminine, and the other harms flow from that.
• B (determining and limiting academic choices) matches the science examples.
• D (male dominance) matches the leadership example.
• C (distorting scientific data) is described as how science has been misused, not as a danger of the binary view.

Since the question asks for *one of* the dangers, B and D are also directly supported by the text.

The final paragraph says that, given the findings, we should drop the myths and see ourselves afresh. That means using new scientific insight to understand human abilities better.

• B: the writer calls the debate a distraction, not something to continue.
• C: the writer accepts the findings rather than questioning them.
• D: the passage says gendered roles were survival measures in *earlier* ages, and rejects them today.

This question is printed identically to Q70 (same stem, same options), yet the key gives D here and A for Q70.

On the passage, D cannot be right. The writer says gendered roles were survival measures only in *earlier ages*, and argues that the male–female brain debate is a distraction to be discarded. The best answer to this stem is A, as in Q70.

The original Q71 was most likely a different question that was lost when the paper was reproduced; its key entry (D) no longer matches the printed text.

"Unbounded potential" means potential without limits. The opposite is limited, i.e. restricted.

• A (imprisoned) is about physical confinement of a person, not the limits of a quality.
• B (aggressive) and D (fearful) are unrelated.

"Cut someone some slack" means to be lenient with them. The sentence says: don't pity him and relax his diet rules, since he must avoid junk food.

• A and D are literal phrases that don't fit "do not … and".
• C ("be slack with his eating") is not idiomatic.

"Hang in there" is the idiom for persisting through difficulty.

"Hang in up there", "hang on there" and "hang on to that" are either not idioms or mean something else (holding an object).

"Pass muster" means to meet the required standard, which fits a high bar that poorly prepared students may not reach.

• B ("past muster") is a misspelling of the idiom.
• A and C are not idioms, and errors are not what one "passes".

"Scrape together" means to assemble something with difficulty from scattered bits. Here the party gathers enough seats through alliances to make a majority.

• A: "scrape through" means to barely succeed and does not take "a majority" as its object in this sense.
• C and D are not idioms.

"Play it by ear" means to decide as things unfold, without a fixed plan. That fits not knowing how the parents will react.

• B ("face the music") means to accept punishment.
• C ("break the ice") means to ease initial awkwardness.
• A (counting on fingers) is literal.

"Slip through one's fingers" is the idiom for losing a chance through carelessness.

• A, B and C are non-idiomatic ("escaped away" is also redundant).

"The reason why" is redundant, since "reason" already means "why". Drop "why": "the reason he did".

"The place where" is standard and should stay.

• A and D keep the redundant "why".
• C drops "where", giving "the place her father disappeared", which is informal. B is the correct formal version.

Only two teams are compared ("yours or theirs"), so the comparative better is needed, not the superlative "best".

• A and B keep "best".
• C ("Which one team") is awkward and unidiomatic.

The pronoun is the object of "contacted" (the leaders contacted *him/them*), so the object form whom is required. The past perfect "had contacted" fits the past "demanded".

• B doubles "had".
• C changes the tense to "have contacted", which clashes with "demanded".
• D keeps "who" and drops "the".

A negative adverbial ("not once", "never") at the start of a clause requires subject–auxiliary inversion: not once did he apologise.

• B and D lack inversion.
• C inverts but says "give apology" without an article.

The idiom is "wash one's hands of" something, not "off".

"The Municipal Council" is a singular collective body, so "its" is right.

• A uses "from".
• B and C switch to "their", and C also has "hand".

"Avail" needs "of" before its object. Strictly, standard British/American usage is "avail *oneself* of", and "avail of" is accepted in Indian English. No option includes "oneself", so C is the best choice.

"Offered by the company" is the correct preposition.

• A and B omit "of".
• D uses "offered from", which is unidiomatic.

The paragraph's topic is how infrastructure (paved surfaces) worsens flooding, ending with a call for mindful land use. C states that topic and is expanded by the next sentence. "Actually" signals the surprising point that infrastructure can make things worse.

• A (climate change is not the only cause) is a tempting link, but it brings in climate change, which the paragraph never develops.
• B (flooding after heavy or low rainfall) is off-topic and odd.
• D contradicts the paragraph, which treats reduced water retention as harmful.

The paragraph gives a sequence of coping strategies: first accept old age, then adopt a positive attitude, [blank], and above all peace of mind.

• A (humour and fun keep one's mind off grim realities) directly develops "positive attitude" and fits the "spice of life" tone before "above all". Its one flaw is the agreement slip "the role … are".
• D (nostalgia helps one accept old age) goes back to the *first* point, acceptance. Dwelling on a radically different past is not obviously a positive attitude.
• B (living beyond 90) and C (stoically bearing stiff joints) drift back to physical decline.

On coherence A is the better fit, but the official key gives D.

The paragraph is entirely about the *causes* of the Arab Spring, listing them tentatively ("widely believed", "some have speculated", "may have been responsible as well").

• B continues that pattern with another cited cause ("Some also cite …"), matching the hedged, additive tone.
• D starts "To sum up" but then defines *what* the Arab Spring was instead of summing up its causes, so it does not follow from the paragraph.
• A and C pick one cause as "the main reason", which contradicts the hedged, multi-cause discussion.

On logic B is the best continuation; the official key gives D.

• Blank 1: someone who persists despite obstacles is assiduous (diligent).
• Blank 2: the contrast is a complaining person, querulous.
• Blank 3: such a person spends time whining about troubles, which matches querulous.

A (artful/doubtful) and D (hotheaded/scared) break the contrast. In C, "dishonest" does not explain lack of progress or link to "gossiping about his troubles".

Tribalism is the opposite of multiculturalism. It surfaces in times of war or economic distress, and it leads to ghettoising people of different ethnicities.

• C: "antithesis" would fit the first blank, but "prosperity" does not pair with war.
• A: "rise … of multiculturalism" would not cause ghettoisation.
• B: "tyranny of multiculturalism" and "growth" are illogical.

The sentence is about textiles and dance. Traditional motifs (design patterns) belong to textile culture, and the artiste created a textile culture for dance.

• A: "processed" a culture is unidiomatic.
• C: totems are not brought into textile mainstream, and one does not "evolve" a culture in this transitive sense.
• D: "described" a textile culture does not fit an achievement "to be celebrated".

• Blank 1: the letter serves as a guide or map of sorts for the writer's own thinking. Both "map" and "guide" work.
• Blank 2: the book is based on the letter "with some details changed", which describes a version of the letter. A *précis* is a condensed summary, and a book is not a summary of a letter. "Details changed" signals a variant, not a condensation.

So A (map, version) fits both blanks best. The sentence also closely follows the introduction to Chimamanda Ngozi Adichie's *Dear Ijeawele*, which uses "map" and "version". The key's D fails on "precis".

The idea is the observer effect: the mind takes part in, and shapes, what it observes.

• C (entering, shaping, perceived): "mind entering into it" is natural; "shaping the very thing that is being perceived" states the observer effect exactly.
• B (intruding, making, construed): "intruding" wrongly implies an unwelcome presence, and "construed" means interpreted, which does not fit the physical sense of shaping observed reality.
• A (getting into / acknowledged) and D (penetrating / forming / seen) are awkward.

This line also matches a well-known quotation from physicist Fred Alan Wolf, which reads with the words of option C. On both logic and source, C is best; the official key gives B.

• C is a clear error. "Critic" is a *person*, and a body of critical argument is a critique. "Gandhi's critique of the West" is correct.
• D uses "critique" as a verb (to evaluate critically). That is standard, though "criticise" would be more natural for a person.
• A and B use "critic" correctly for people.

The most clearly incorrect usage is C. The key's D is at most a matter of style.

"Teamed" means paired or combined, so "the wedding was teamed in Bhojpuri style" makes no sense. It should be something like "themed" or "conducted".

• A: "teemed with" (was full of) is correct.
• C: "team" as a noun is correct.
• D: "teamed with" (paired with) is correct.

The slogan is "Get it right the first time". "Write" is a homophone error.

• A ("the right way") and D ("rite", a ritual) are correct.
• C is a deliberate pun: driving on the right side is wrong in India, which drives on the left. The usage is correct.

To destroy a building completely is to raze it to the ground. "Raise" means to lift or build, which is the opposite.

• B (raising children) and C (raising a toast) are standard.
• D uses "rose" (past tense of "rise") grammatically, even if "the sun rose" would be more natural than "the rays rose".

To reveal one's innermost feelings is to bare one's soul, so it should be "bared his soul". "Barred" means blocked.

• A ("bard", a poet) is correct.
• C ("barred from entering") is correct.
• D is wordplay but valid: the Bar (lawyers' body) barred (banned) all bards (poets).

Anchors:
- 2 introduces the hunter-gatherer wandering for food, so it opens.
- 3 ("As the crops began to give better yields … reduced his need to go in search") is the outcome of farming, so it must close.
- 4 ("This was followed by an attempt to grow food by scattering the spare grains") marks the step into cultivation.
- 1 ("He just harvested the wild grains") describes simple harvesting without tending.

Among the options only D (2-4-1-3) starts with 2 and ends with 3:
- A (2431) and C (2341) end on 1, an anticlimax after yields have improved.
- B (2314) mentions "the crops" before any crops exist.

In D, after scattering grains (4) he still merely harvested what grew (1), and as those crops improved (3) his need to forage fell.

• 3 opens: life seems idyllic *at first*, compared with Mumbai.
• 2 turns: "Soon, however, she notices" the lack of medical care and education.
• 1 elaborates on 2: people are "one injury away from starvation", which follows from no medical care, and also at risk for a misspoken word.
• 4 adds, with "Moreover", the political disruptions and closes the frame: far from a "Shangri-La utopia", echoing the "idyllic" of sentence 3.

B (3241) separates 1 from the medical-care point it explains and ends on a fragmentary line. C and D do not start with the "at first" sentence, or break the "however … moreover" progression.

Fixed links:
- 5 opens, introducing "a study spanning 170 countries".
- 4 and 1 refer back to it as "The study".
- 3 (dictators lack voter pressure to fund healthcare) explains the mechanism, so it comes late.

Key order A (5-4-1-2-3): study (5), its broad suggestion (4), its specific findings (1), the author's reaction "Incredible as it may sound … good for the body too" (2), then the reason (3), which follows 2's contrast of democracy with dictatorship.

Option C (5-4-2-1-3) is equally coherent: the two general claims 4 and 2 sit together, then the specific evidence (1), then 3. Here 1's mention of heart disease leads straight into 3's mention of heart disease.

B (54231) ends on the study's publication details, and D (45132) opens with "The study" before any study is introduced. A and C both read well.