JIPMAT 2024 · Question Paper & Solutions

Required number $=\text{LCM}(12,21,35)+6$.

$12=2^2\cdot 3,21=3\cdot 7,35=5\cdot 7\Rightarrow \text{LCM}=2^2\cdot 3\cdot 5\cdot 7=420$.

Number $=420+6=426$.

Let maximum marks be $M$ and pass mark $P$.

$P=\frac{xM}{100}+a$ and $P=\frac{yM}{100}-b$.

Equating: $\frac{(y-x)M}{100}=a+b\Rightarrow M=\frac{100(a+b)}{y-x}$.

In an equilateral triangle the centroid coincides with the circumcentre.

So the distance from the centroid to any vertex equals the circumradius: $d=R=20\sqrt{3}$ cm.

Each 5 years multiplies the amount by 2. $16=2^4$, so it needs $4$ doubling periods.

Time $=4\times 5=20$ years.

Final price as a fraction of marked price:

• (A) $0.75\times 0.85=0.6375$
• (B) $0.70\times 0.90=0.630$
• (C) $0.65\times 0.95=0.6175$
• (D) $0.85\times 0.80=0.680$

Ascending: $0.6175<0.630<0.6375<0.680$, i.e. (C), (B), (A), (D).

Let 5 articles cost ₹100 each (total ₹500).

3 articles at 20% gain give profit $3\times 20=₹60$; the other 2 give no profit.

Gain $=\frac{60}{500}=12\%$.

Combine ratios: $a:b=5:3=10:6$ and $b:c=2:5=6:15$, so $a:b:c=10:6:15$.

Take $a=10k,b=6k,c=15k$. Then $c-a=5k$.

• Statement I: $10(a+c)=10(25k)=250k\ne 5k$ — false.
• Statement II: $10a+25b=100k+150k=250k\ne 5k$ — false.

Both statements are false.

$xy=\text{HCF}\times \text{LCM}=6\times 210=1260$.

$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{72}{1260}=\frac{2}{35}$.

Check: $x=6m,y=6n$ with $m,n$ coprime, $mn=35$, $m+n=12\Rightarrow (m,n)=(5,7)$, so the numbers are 30 and 42 (sum 72 ✓). $\frac{1}{30}+\frac{1}{42}=\frac{7+5}{210}=\frac{12}{210}=\frac{2}{35}$.

The correct value is $\frac{2}{35}$. In this printed paper option (D) reads $\frac{12}{35}$, which is not equal to it; other transcriptions of the JIPMAT 2024 paper show option (D) as $\frac{2}{35}$, so (D) is the intended answer and the printed "12" is a typo.

Five consecutive numbers with average $a$: $a-2,a-1,a,a+1,a+2$ (sum $5a$).

Next two: $a+3,a+4$. New sum $=5a+2a+7=7a+7$.

New average $=\frac{7a+7}{7}=a+1$ — it increases by 1.

• (A) $\sqrt{15\times 163÷5-89}=\sqrt{489-89}=\sqrt{400}=20$ → (II)
• (B) $\sqrt{15^2+11\times 3^2}=\sqrt{225+99}=\sqrt{324}=18$ → (IV)
• (C) $\sqrt{8^2\times 7\times 5^2-175}=\sqrt{11200-175}=\sqrt{11025}=105$ → (I)
• (D) $\sqrt{91+\sqrt{70+\sqrt{121}}}=\sqrt{91+\sqrt{81}}=\sqrt{100}=10$ → (III)

So (A)-(II), (B)-(IV), (C)-(I), (D)-(III).

$72=8\times 9$.

Divisibility by 8: last three digits $\overline{9b6}$ must be divisible by 8. Checking $b=0\dots 9$: only $936=8\times 117$ and $976=8\times 122$ work, so $b=3$ or $7$.

Divisibility by 9: digit sum $9+4+a+2+9+b+6=30+a+b$ must be a multiple of 9, so $a+b=6$ or $15$.

• $b=3$: $a=3$ (rejected since $a\ne b$) or $a=12$ (not a digit).
• $b=7$: $a=8$ (sum 15) ✓.

$2a+3b=16+21=37$.

For an equilateral triangle, circumradius $R=\frac{s}{\sqrt{3}}$, so $s=R\sqrt{3}$.

$s=20\sqrt{3}\times \sqrt{3}=60$ cm.

Class X: $\frac{24}{30}=80\%$. Class Y: $\frac{28}{35}=80\%$.

The percentages are equal, so Statement I (X is higher) is false and Statement II (equal) is true.

Decimal values:

• (A) $\frac{5}{9}\approx 0.556$
• (B) $\frac{7}{11}\approx 0.636$
• (C) $\frac{8}{15}\approx 0.533$
• (D) $\frac{11}{17}\approx 0.647$

Descending: (D) > (B) > (A) > (C).

Numerator: $\sqrt[4]{0.0625}=0.5$ (since $0.5^4=0.0625$), $\sqrt[3]{0.008}=0.2$, $\sqrt{0.09}=0.3$.

$0.5+0.2+0.3-1=0$.

Denominator: $\sqrt[5]{32}=2$, so $\sqrt[3]{62.5\times 2}=\sqrt[3]{125}=5\ne 0$.

Value $=\frac{0}{5}=0$.

Measure time $t$ in hours after 9 AM and position from A.

• Ram: reaches B at $t=1$ (10:00), then returns: position $10-5t$ for $1\le t\le 2$.
• Shyam: position $10(t-0.75)$ from $t=0.75$, reaching B at $t=1.25$ (10:15).

Before 10:00, Shyam cannot catch Ram ($5t=10t-7.5$ gives $t=1.5>1$). After Ram turns back, they approach each other:

$10t-7.5=10-5t\Rightarrow 15t=17.5\Rightarrow t=\frac{7}{6}$ h $=1$ h 10 min.

They meet at 10:10 AM, about $4\frac{1}{6}$ km from A.

Shyam reaches B at 10:15 ($t=1.25$) and returns: position $5-10(t-1.25)=17.5-10t$.

Ram (returning since 10:00): position $10-5t$.

Both moving towards A; Shyam catches Ram when $17.5-10t=10-5t\Rightarrow 5t=7.5\Rightarrow t=1.5$ h.

Shyam overtakes Ram at 10:30 AM (2.5 km from A).

Let roots be $2k$ and $3k$.

Product: $6k^2=54\Rightarrow k=\pm 3$. Sum: $p=5k=\pm 15$.

Only $15$ is among the options.

Surface area $\propto$ side${}^2$.

New area factor $=0.81^2=0.6561$.

Decrease $=1-0.6561=0.3439=34.39\%$.

Statement I: $\cot 30^{\circ }=\sqrt{3}$.

LHS $=\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{(\sqrt{3}+1{)}^2}{2}=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}$.

RHS $=2\left(\frac{\sqrt{3}}{2}+1\right)=\sqrt{3}+2$. True.

Statement II: $2\cdot \frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}-1\cdot 1=1-1=0$. True.

Both are true.

Reason: $S_n=\frac{a(1-r^n)}{1-r}$ ($r\ne 1$) is the standard GP sum formula — true.

Assertion: with $a=1,r=\frac{1}{2}$:

$S_n=\frac{1-(1/2{)}^n}{1/2}=2\left(1-\frac{1}{2^n}\right)=\frac{2^n-1}{2^{n-1}}$.

The assertion's numerator $2^{n-1}-1$ is wrong; e.g. for $n=1$ it gives $0$, but the sum is $1$. So (A) is false and (R) is true.

Let C's efficiency be 1 unit/day. B takes $0.8$ of C's time, so B's efficiency $=\frac{1}{0.8}=1.25$.

A is 40% less efficient than B: $A=0.6\times 1.25=0.75$.

A + B $=2$ units/day. 80% of the work $=2\times 12=24$ units, so total work $=30$ units.

60% of the work $=18$ units. B + C $=2.25$ units/day.

Days $=\frac{18}{2.25}=8$.

They contradict when exactly one tells the truth:

$P=0.75\times 0.20+0.25\times 0.80=0.15+0.20=0.35$.

So 35% of cases.

Let cost price of one article be $c$.

SP of 2 $-$ CP of 2 $=$ CP of 3: $800-2c=3c\Rightarrow c=160$.

CP of 2 $=₹320$, profit $=₹480$.

Profit % $=\frac{480}{320}\times 100=150\%$.

Let each distance be $d$. Total time $=\frac{d}{a}+\frac{d}{b}+\frac{d}{c}=d\cdot \frac{bc+ac+ab}{abc}$.

Average speed $=\frac{3d}{\text{total time}}=\frac{3abc}{ab+bc+ac}$ (harmonic mean of $a,b,c$).

Since $\alpha$ is a root, $a{\alpha }^2+b\alpha +c=0\Rightarrow \alpha (a\alpha +b)=-c\Rightarrow \frac{1}{a\alpha +b}=-\frac{\alpha }{c}$.

Similarly $\frac{1}{a\beta +b}=-\frac{\beta }{c}$.

Sum $=-\frac{\alpha +\beta }{c}=-\frac{-b/a}{c}=\frac{b}{ac}$.

Check whether the square of the longest side equals the sum of squares of the other two:

• (A) $20^2+22^2=884\ne 576$ — no.
• (B) $15^2+32^2=1249\ne 1369$ — no.
• (C) $11^2+60^2=121+3600=3721=61^2$ — yes.
• (D) $6^2+8^2=100=10^2$ — yes.

So (C) and (D) only.

$f(1)=1-6+11-6=0$, so $(x-1)$ is a factor.

Dividing: $x^3-6x^2+11x-6=(x-1)(x^2-5x+6)=(x-1)(x-2)(x-3)$.

For a fixed area, the more sides a regular figure has (the closer it is to a circle), the smaller its perimeter. With area $A$:

• Circle: $P=2\sqrt{\pi A}\approx 3.54\sqrt{A}$
• Regular pentagon: $P\approx 3.81\sqrt{A}$
• Square: $P=4\sqrt{A}$
• Equilateral triangle: $P=3\sqrt{\frac{4A}{\sqrt{3}}}\approx 4.56\sqrt{A}$

Ascending: Circle, Pentagon, Square, Triangle = (A), (D), (B), (C).

$x=3$ is a root of both.

$9-3p-q=0\Rightarrow 3p+q=9$.

$45-9p-15q=0\Rightarrow 3p+5q=15$.

Subtracting: $4q=6\Rightarrow q=\frac{3}{2}$; then $3p=9-\frac{3}{2}=\frac{15}{2}\Rightarrow p=\frac{5}{2}$.

Square: $1+2\sin \theta \cos \theta =\frac{7}{4}\Rightarrow 2\sin \theta \cos \theta =\frac{3}{4}$.

$(\sin \theta -\cos \theta {)}^2=1-2\sin \theta \cos \theta =\frac{1}{4}\Rightarrow \sin \theta -\cos \theta =\pm \frac{1}{2}$.

Among the options, $\frac{1}{2}$.

Hugs: one per pair $=\left(\genfrac{}{}{0ex}{}{15}{2}\right)=105$.

Gifts: each of 15 people gives 2 gifts to each of the other 14: $15\times 14\times 2=420$.

Difference $=420-105=315$.

Let $P(2,2),Q(6,3),R(4,11)$.

$P{Q}^2=4^2+1^2=17$, $Q{R}^2=2^2+8^2=68$, $P{R}^2=2^2+9^2=85$.

$P{Q}^2+Q{R}^2=17+68=85=P{R}^2$, so the triangle is right-angled at Q.

(The sides are also unequal, but "right-angled" is the specific property the question targets; it is not isosceles or equilateral.)

Table: 8→₹, 3→H, 0→R, 7→A, 4→M, 6→%, 9→L, 2→K, 5→E, 1→#.

First letter Y and last symbol \* occur only under Exception I, so the number must begin and end with a non-zero odd digit.

Middle letters H, ₹, M, E decode to 3, 8, 4, 5. So the number is (odd) 3 8 4 5 (odd).

• (A) 138450 — ends in 0 (even), so Exception I does not apply. ✗
• (B) 338457 — starts 3, ends 7. ✓
• (C) 538459 — starts 5, ends 9. ✓
• (D) 438458 — starts and ends even; Exception II would give β…X. ✗

Answer: (B) and (C) only.

Let there be $n$ notes of each kind: $2n+10n+50n=62n=620\Rightarrow n=10$.

Total notes $=3\times 10=30$.

Given directly: C → Bengaluru by car; B → Kolkata by aeroplane; D → boat; E → train.

The only mode left is bus, so A → bus.

Cities left for A, D, E: Pune, Delhi, Mumbai. Bus does not go to Delhi or Pune, so A → Mumbai. Delhi was not reached by boat, so D → Pune and E → Delhi.

Pairs: A–Bus ✓, B–Aeroplane ✓, D–Boat ✓, but E travelled by train, not car. So "E – Car" is not correct.

Alphabet positions of GREEN: G=7, R=18, E=5, E=5, N=14. Add 2 to each: 9, 20, 7, 7, 16. Write them in reverse order: 16 7 7 20 9 → 1677209 ✓.

BLUE: B=2, L=12, U=21, E=5 → add 2: 4, 14, 23, 7 → reverse: 7 23 14 4 → 723144.

• (A) 8, 13, 21, 32, 46, 63, ?: differences 5, 8, 11, 14, 17 (+3 each), next +20 → 83 (III).
• (B) 20, 80, 30, 60, 40, ?, 50, 20: two interleaved series — 20, 30, 40, 50 and 80, 60, ?, 20 (−20 each) → 40 (I).
• (C) 0, 2, 3, 5, 8, 10, 15, 17, 24, 26, ?: pairs $(n^2-1,n^2+1)$: (0,2), (3,5), (8,10), (15,17), (24,26); next is $6^2-1=$ 35 (IV).
• (D) 6, 13, 25, 51, ?: $\times 2+1,\times 2-1$ alternately: 13, 25, 51, then $51\times 2-1=$ 101 (II).

So (A)-(III), (B)-(I), (C)-(IV), (D)-(II).

Reading intended by the key:

• The statement presents non-conventional sources as the answer to the energy crisis, implying a shift, driven by modern technology, away from conventional sources — taken to support I.
• An "energy crisis" that needs new sources presupposes that conventional sources are running out through over-use — taken to support II.

Caution: strictly, the statement is a prediction about what non-conventional energy *will* do. It says nothing about replacement already happening "gradually" (I) or about the *cause* of depletion being environmental over-exploitation (II). On a strict inference standard, neither conclusion is forced, which would point to option (C).

When people face the centre, a person's left is the next seat clockwise and their right is the next seat anticlockwise.

Place Radhika at seat 0 and number the seats anticlockwise:

• Sonia is opposite Radhika → seat 3.
• Purnima is to Radhika's right → seat 1; she is to Deepti's left → Deepti at seat 2.
• Monika is to Radhika's left → seat 5.
• Kamini is to Sonia's right (seat 4) and to Monika's left (seat 4) ✓.

Order anticlockwise: Radhika, Purnima, Deepti, Sonia, Kamini, Monika.

After the swaps Deepti ↔ Kamini and Monika ↔ Radhika: Deepti is at seat 4, and her left (clockwise) neighbour is seat 3, Sonia.

The flash and the sound are produced together. Light travels at about $3\times 10^8$ m/s and sound at about 340 m/s in air, so the light reaches us almost instantly while the sound arrives later.

Both statements are true, and R is exactly why A happens.

From 10 AM Wednesday to 1 PM Thursday is 27 hours.

In that time the watch goes from 2 min slow to 2.5 min fast, gaining 4.5 min.

To be correct it must gain 2 min: time needed $=\frac{2}{4.5}\times 27=12$ hours.

10 AM Wednesday + 12 h = 10:00 PM, Wednesday.

Position shifts from AFHO to GBDJ: A→G (+6), F→B (−4), H→D (−4), O→J (−5).

Apply the same to CHFM: C+6=I, H−4=D, F−4=B, M−5=H → IDBH.

Tuberculosis, diphtheria and typhoid are all caused by bacteria. Measles is caused by a virus, so it is the odd one out.

Read each circle clockwise.

• (A) 149, 175, 203, ?: differences +26, +28, next +30 → 233 (IV).
• (B) 3, ?, 81, 6561: each number is the square of the previous: $3^2=9,9^2=81,81^2=6561$ → 9 (III).
• (C) 0, 7, 26, ?: $1^3-1,2^3-1,3^3-1$, next $4^3-1=$ 63 (I).
• (D) 9, 13, 22, ?: differences $2^2,3^2$, next $4^2$: $22+16=$ 38 (II).

So (A)-(IV), (B)-(III), (C)-(I), (D)-(II).

The first seven terms are products of neighbouring terms from the last eight (3, 50, 4, 25, 2, 6, 30, 3), taken from the end backwards:

• $30\times 3=90$
• $6\times 30=180$
• $2\times 6=12$
• $25\times 2=50$
• $4\times 25=100$
• $50\times 4=$ 200
• $3\times 50=$ 150

So the missing pair is 200, 150.

Conscience stops a person from doing wrong. In the same way, police act to prevent and check crime.

Court and law deal with crime but don't act against it the way police do. Discipline is not something police act against.

A $4\times 4\times 4$ cube ($n=4$):

• Three faces painted (corners): 8
• Two faces painted (edges): $12(n-2)=24$
• One face painted: $6(n-2{)}^2=24$
• No face painted: $(n-2{)}^3=8$

Check: (A) 36 ✗ (it is 24). (B) 24 ✓. (C) 8 ✓. (D) at least two faces $=24+8=32$, not 30 ✗.

So (B) and (C) only.

• First letters: A(1), B(2), D(4), ?, K(11) — gaps +1, +2, +3, +4 → G(7).
• Last letters: Y(25), V(22), R(18), ?, G(7) — gaps −3, −4, −5, −6 → M(13).
• Numbers: $2\times 2+2^2=8$, $8\times 3+3^2=33$, $33\times 4+4^2=$ 148, $148\times 5+5^2=765$ ✓.

Missing term: G148M.

The man's son's sister is the man's daughter.

She is the woman's mother-in-law, so she is the mother of the woman's husband.

So the husband is the son of the man's daughter, i.e. the man's grandson.

He cannot be the man's son-in-law: the man's son-in-law would be married to the man's daughter, but here she is his mother. The printed key (D, son-in-law) does not match this; the logically correct option is (A) Grandson.

Education: ₹150 thousand (2011) → ₹300 thousand (2015).

Increase $=\frac{300-150}{150}\times 100=100\%$.

Totals (₹ thousand): 1500, 1700, 2000, 2400, 2900.

Year-on-year increases: 200, 300, 400, 500, rising by 100 each year. The next increase should be 600.

2016 estimate $=2900+600=3500$ thousand $=₹35,00,000$.

Total house rent over five years $=₹1050$ thousand; total expenditure $=₹10500$ thousand.

Percentage $=\frac{1050}{10500}\times 100=10\%$.

With PR north–south and Q west of S: P is north, Q west, R south, S east. Seen from above, clockwise order is P → S → R → Q → P.

• Radha: from Q, clockwise: Q → P → S → R → Q.
• Mohan: from R, anticlockwise: R → S → P → Q.

Along Radha's path, Mohan starts $90\times 3=270$ m ahead, walking towards her. Their speeds are in the ratio $8:10$, so Radha covers $\frac{8}{18}$ of any closing distance.

• 1st crossing: closing distance 270 m → Radha walks 120 m: Q→P (90) then 30 m along PS.
• 2nd crossing: a further full lap of 360 m → Radha walks another 160 m, 280 m in all: Q→P (90), P→S (180), S→R (270), then 10 m along RQ.

They cross the second time on QR, 10 m from R.

*The question line is missing from this print; the original JIPMAT 2024 question asks which number is on the face opposite 4.*

Faces seen next to each number across the views:

• Next to 1: 6, 5, 2, 4 → the only number never next to 1 is 3, so 1 is opposite 3.
• Next to 2: 3, 5, 1, 4 → the only number never next to 2 is 6, so 2 is opposite 6.
• That leaves 4 opposite 5. (5 appears next to 6, 1, 3 and 2, never next to 4.)

So the face opposite 4 shows 5.

Label by region.

Whole triangle and its halves (3): the big triangle, and its left and right halves cut by the vertical median.

Triangle above the horizontal line (3): the triangle from the apex down to the horizontal line (whose ends lie on the sides), and its two halves.

Triangle above the rectangle's top edge (3): the apex triangle above the rectangle's top, and its two halves.

Corner triangles (4):

• two small triangles at the bottom corners, outside the rectangle's vertical sides and below the horizontal line;
• two small triangles inside the rectangle's upper corners, between the top edge, the vertical side and the triangle's slanted side.

Diamond (8): its two diagonals cut it into 4 small triangles. Pairs of them give 4 more: top half, bottom half, left half and right half.

Total $=3+3+3+4+8=21$.

Unscrambled: POROT → TROOP, ROPUG → GROUP, BMO → MOB, SALCS → CLASS.

All four are collective nouns for people. Troop, group and class are organised or orderly groups. A mob is a disorderly, unruly crowd, so it is the odd one out. (It is also the only three-letter word; the rest have five letters.)

2016 total expenditure $=₹5800$.

The key treats profit as 10% of receipts: $R-5800=0.10R\Rightarrow R=\frac{5800}{0.9}\approx 6444$.

(If profit is measured on expenditure instead, $R=1.1\times 5800=6380$. The nearest option is still 6444, so the answer is (C) either way.)

The region needed is:

• inside the square (women),
• inside the triangle (government service),
• inside the rectangle (private sector),
• outside the circle (educated persons).

Region 9 meets all four conditions. Region 6 is in the square, triangle and rectangle too, but it lies inside the circle (educated), so it does not fit.

Printing cost = 35%, Advertisement charges = 18%.

Difference $=17\%$ of $360^{\circ }=0.17\times 360^{\circ }=61.2^{\circ }$.

Miscellaneous = 4% = ₹6,000, so 1% = ₹1,500.

Binding and cutting = 18% (the one sector not in the legend); Royalty = 15%.

Difference $=3\%=3\times 1500=₹4,500$.

• Argument I is strong. Unscheduled holidays stop work across the country and cause a real economic and administrative loss, which is a direct and serious reason against the practice.
• Argument II is weak. People can pay homage without a national holiday (condolence meetings, a moment of silence, flags at half-mast), so the holiday isn't necessary for that purpose.

Only argument I is strong.

Middle East & North Africa: 12.5%; Southern Continental Africa: 3%.

Percent greater $=\frac{12.5-3}{3}\times 100=\frac{9.5}{3}\times 100\approx 316.7\%\approx 317\%$.

Positions: I(1) N(2) S(3) T(4) R(5) U(6) C(7) T(8) I(9) O(10) N(11). Check every pair, in either alphabetical direction:

• I(1)–C(7): 5 letters between in the word; C to I in the alphabet has D, E, F, G, H = 5 ✓
• N(2)–T(8): 5 between in the word; N to T has O, P, Q, R, S = 5 ✓
• S(3)–T(4): adjacent in both ✓
• O(10)–N(11): adjacent in both ✓

No other pair works, so there are four pairs.

INFRASTRUCTURE contains: I, N, F, R×3, A, S, T×2, U×2, C, E.

• NATURE — N, A, T, U, R, E ✓
• CHARTER — needs H, which is not present ✗
• FRACTURE — F, R, A, C, T, U, R, E (two R's available) ✓
• RESTRAIN — R, E, S, T, R, A, I, N ✓

Only "Charter" cannot be formed.

Prakash's range: 21–27 May. Jyoti's range: 27–29 May.

The only common date is 27th May.

The sentence is a sudden exclamation of surprise or urgency at the bell ringing, not a plain statement, so it takes an exclamation mark: "There goes the bell!"

It is not a question, so (?) is wrong. A colon needs something to follow, and a full stop loses the exclamatory tone.

The action (waiting) starts before a future point (six o'clock) and continues up to it, and a duration is given ("for an hour"). This calls for the future perfect continuous: "I *shall have been waiting* here for an hour."

• "Shall wait" / "shall be waiting on" do not show the duration up to that future point.
• "Would be waiting" is conditional, so it doesn't fit either.

*Hackneyed* means overused and therefore stale or unoriginal, the same as clichéd.

Important, horrible and heightened do not mean "overused".

• The exclamation "Well done!" becomes a reporting verb: *applauded*.
• Reporting verb in the past, so the present perfect *have made* shifts back to the past perfect *had made*.
• Pronouns change: *you* → *he*, *me* → *him*.

(D) does all three.

• (A) keeps "has".
• (B) and (C) mix tenses ("applauds/says") and keep "well done" in reported speech.

"Is being sung" is the present continuous passive. Its active form is also present continuous: "The players are singing the national anthem."

The other options change the tense: past continuous, simple present and present perfect.

"A gentleman at large" is a (humorous) term for a man with no regular occupation or employment, i.e. a man without a job.

It doesn't mean someone in a crowd, someone who runs around, or someone generous.

The subject "all the students and teachers" is plural (two nouns joined by *and*), so the verb takes its plural present form: like.

• "Likes" is singular.
• "Liking" is not a finite verb.
• "Are like" changes the meaning to "resemble".

The *public purse* is the money held and spent by the government (collected through taxes). The sentence asks whether arts spending should be funded by the government.

• It does not refer to donations or to enthusiasts' own money.
• It does not refer to funds meant for the poor.

Three of these are standard phrasal verbs:

• plan for — prepare for a possible event
• plan on — intend or expect to do something
• plan out — arrange in detail

"Plan in" is not an accepted phrasal verb, so it is the odd one out.

A philistine is a person who is indifferent or hostile to culture, literature and the arts.

• A dictator is an absolute ruler.
• A hypocrite pretends to have virtues.
• "Primitive" means undeveloped or early.

Simple future active "will drop" → passive "will be dropped". The object *you* becomes the subject and the agent goes into a *by*-phrase: "You will be dropped at the airport by the taxi."

• (A) and (C) are still active and change the tense.
• (B) is awkward and not the proper passive transformation.

*Ultra vires* (Latin, "beyond the powers") describes an act done beyond one's legal authority or jurisdiction.

Its opposite, *intra vires*, means "within authority" (option C). It has nothing to do with viruses or disease.

The correct spelling is disease (dis- + ease).

"Dicease", "Desease" and "Discease" are misspellings.

*Dilate* means to widen or expand, so its opposite is contract (become narrower or smaller).

• "Expand" is a synonym.
• "Dilute" means to weaken a liquid.
• "Differentiate" is unrelated.

Two fixes are needed:

1. A period of time (five years) takes for; *since* is used with a point in time (e.g. since 1990).
2. The action ran up to a past moment (1995), so the past perfect continuous *had been teaching* is correct. "Have/has" would wrongly make it present perfect.

Hence "had already been teaching there for five years".

Purpose after *come/came* is expressed with the infinitive: "when I came to see you last evening". "Came for seeing" is incorrect.

Parts (i) and (iii) are grammatically fine.

The blank comes before the noun *choices* and after the adverb *fully*, so it needs an adjective: "fully *rational* choices".

• "Rationale" is a noun meaning reasoning.
• "Rationally" is an adverb.
• "Rationality" is a noun.

The correlative pair is whether … or: "whether through explicit reasoning *or* by implicit presumption".

"And", "along" and "instead" cannot complete a *whether* construction.

The blank is parallel to the adverb *typically* ("___, or even typically, guided") and modifies the verb phrase *guided by reason*, so it needs an adverb: invariably (always).

• "Invariable" is an adjective.
• "Invariability" is a noun.
• "Invariabily" is misspelt.

After the passive participle *presented*, the agent is introduced by by: "presented … *by* Kahneman, Slovic and Tversky".

The passage says a sastraic exposition involves inquiry into three things:

• the nature of substance — (B)
• the nature of language — (D)
• the validity of critical statements — (E)

(A) derivatives and compounds are what Panini's sutras discuss. (C) stylistic patterns belong to formal meaning in poetry. Neither is part of a sastraic inquiry.

Answer: (B), (D) and (E) only.

The passage says Panini's sutras discuss the influence of the concept of language on suffixes, primary and secondary derivatives and compounds, which matches (B).

• (A) The terms jati, kriya, guna and dravya are ones that *poetics* borrows from grammar, not Panini's main subject.
• (C) and (D) Philosophical issues and critical questions are linked to sastraic exposition, not to Panini.

• (A) False. The passage says criticism does *not necessarily* have to choose between textual and contextual study.
• (B) True. The distinction between the two types of linguistic meaning is seen at two levels, formal and textual.
• (C) True. The passage calls grammatical and philosophical understanding a *prerequisite* for critical questions.
• (D) Not supported. The passage only says Panini's sutras are scattered over his works; it says nothing about a compact format.
• (E) True. The passage refers to literary criticism "in the Sanskrit tradition", treated as a sastra, i.e. an established, long-standing tradition.

Answer: (B), (C) and (E) only.

Validity (as in "validity of critical statements") means the quality of being logically sound and true.

• "Inquiry" means investigation.
• "Formulated" means systematically expressed.
• "Compound" is a grammatical term.

• (A) Supported. Only there, among kindred spirits, can they live harmoniously and gain the serenity of ideal companionship.
• (B) Supported. The passage says the habit of intercourse with "the gods" makes one impatient of mere men.
• (C) Supported. It notes that spiritual or intellectual pride is engendered on that high plane and becomes inevitable.
• (D) Not supported. The passage never calls them apathetic to commoners or *solely* eager for lofty ideals; it even says those ideals get shattered by life's realities. This is an overstatement.

Answer: (A), (B) and (C) only.

The phrase (borrowed from *Hamlet*) refers to the habit of deep, sombre reflection. In the passage it follows "a habitual meditation on the vast problems that underlie human life", so it means serious and prolonged thinking, which gives the thinkers' views their sombre setting.

• It is not illness.
• It is not a sudden flash of thought.
• The unhappy mood is the *result* of the thinking, not the meaning of the phrase.

Assertion is correct. The passage explicitly says "It is easy to understand, therefore, why such thinkers fly to the solitude of their own thoughts…", giving a rationale.

Reason is not correct. The passage traces the retreat to their habit of living on a high plane of thought, which makes them impatient of ordinary men and discontented when ideals meet reality. It never says they are *naturally* melancholic, or that they feel *disgust* for the masses. Their melancholy is a result of their thinking, not an innate trait.

Statement I: The author says he has "always thought" the strongest argument for the Baconian theory is that no man, however great his genius, could have written those plays and sonnets without Bacon's liberal education. So the author endorses the theory on the ground that Shakespeare, genius though he was, could not have written them. Correct.

Statement II: The same argument rests on Bacon having the liberal education that the works required. So the author sees Bacon as better equipped, i.e. more resourceful, because of that education. Correct.

Both statements are correct.

• (A) "All men are mortal": *all* qualifies the noun *men* → adjective (IV).
• (B) "He was all alone": *all* modifies the adjective *alone* (= completely) → adverb (III).
• (C) "All spoke in his favour": *all* stands in place of a noun as the subject → pronoun (II).
• (D) "He lost his all": *all* follows the possessive *his* and means "everything he owned" → noun (I).

So (A)-(IV), (B)-(III), (C)-(II), (D)-(I).

• (C) opens the topic: what kind of education should we invest in?
• (D) states the aim: not only an affluent but a just, humane society.
• (A) "For that" refers back to the aim in (D): we need people with high skills *and* values.
• (B) "So" draws the conclusion: we need skill-based and value-based education.

Order: C, D, A, B.

• (A) cannot come first, since "For that" needs something before it.
• (B) must come last, since "So" signals the conclusion.

• (A) I shall speak: shall + base verb → Simple Future (II)
• (B) I shall have spoken: shall have + past participle → Future Perfect (IV)
• (C) I shall have been speaking: shall have been + -ing → Future Perfect Continuous (I)
• (D) I shall be speaking: shall be + -ing → Future Continuous (III)

So (A)-(II), (B)-(IV), (C)-(I), (D)-(III).

• (B) "almonds are commonly called nuts" is the main clause and has the subject, so it comes first.
• (C) "but they are actually seeds" corrects that popular view.
• (D) "and they come from a tree" adds information.
• (A) "that is closely related to the peach tree" is a relative clause that must follow its antecedent, *a tree*.

"Almonds are commonly called nuts but they are actually seeds and they come from a tree that is closely related to the peach tree." Order: B, C, D, A.

• (A) Up for grabs: available for anyone who wants it → (IV)
• (B) Make hay while the sun shines: make use of an opportunity while it lasts → (III)
• (C) Go haywire: go out of control or stop working properly → (II)
• (D) Get off lightly: receive a milder punishment than expected → (I)

So (A)-(IV), (B)-(III), (C)-(II), (D)-(I).

• (A) "The economist were": a singular subject with a plural verb. It should be "The economists were" (or "The economist was"). Error.
• (B) "longer times": in the sense of a period allowed, *time* is uncountable, so it should be "longer time". Error.
• (C) "to pay the loan instead of waiving the entire farm loan": grammatically acceptable.
• (D) "as doing so is not good for credit culture": acceptable.

Errors are in (A) and (B) only.