JIPMAT 2023 · Question Paper & Solutions

Statement (I). Let one boy do $b$ and one girl do $g$ of the work per day.

$$5(4b+8g)=\frac{1}{3},3(7b+11g)=\frac{1}{3}$$

So $20b+40g=21b+33g\Rightarrow b=7g$. Then $5(28g+8g)=180g=\frac{1}{3}\Rightarrow g=\frac{1}{540}$.

The last $\frac{1}{3}$ in 2 days needs a daily rate of $\frac{1}{6}=90g$. The current team $7B+11G$ gives $49g+11g=60g$, so $30g$ more is needed, i.e. 30 girls, not 32. Statement (I) is false.

Statement (II). A does 6% in 6 days, so A takes 100 days; C takes twice as long, 200 days. B is $166\frac{2}{3}\%=\frac{5}{3}$ more efficient than C, so B's efficiency is $\frac{8}{3}$ of C's and B takes $200\times \frac{3}{8}=75$ days.

$$\frac{1}{75}+\frac{1}{200}=\frac{8+3}{600}=\frac{11}{600}\Rightarrow \frac{600}{11}=54\frac{6}{11}\text{ days}$$

Statement (II) is true.

Option D is the 'I false, II true' slot (its printed wording repeats 'true', a typo), matching the key.

Put $t=y+\frac{1}{y}$, so $y^2+\frac{1}{y^2}=t^2-2$.

$$7t-2(t^2-2)=9\Rightarrow 2t^2-7t+5=0\Rightarrow t=1\text{ or }t=\frac{5}{2}$$

• $t=1$: $y^2-y+1=0$ has no real roots.
• $t=\frac{5}{2}$: $2y^2-5y+2=0\Rightarrow y=2$ or $y=\frac{1}{2}$.

Only $y=2$ is an integer, so there is 1 integral solution.

Statement (I). Let $x-y=k$, $x+y=5k$, $xy=12k$. Then $x=3k$, $y=2k$, so $6k^2=12k\Rightarrow k=2$. The numbers are 6 and 4 and their product is 24, not 18. False.

Statement (II). Last digits of powers of 2 cycle $2,4,8,6$ with period 4. $91=4\times 22+3$, so $2^{91}$ ends in 8, not 6. False.

Both are false.

Every match eliminates exactly one player, and the tournament ends when one champion remains. So $300-1=$ 299 players must be eliminated, needing 299 matches.

$$1\text{ week}=604800\text{ s},1\text{ day}=86400\text{ s},1\text{ h}=3600\text{ s},1\text{ min}=60\text{ s}$$

$$P(604800+86400+3600+60+1)=694861P$$

Items cost $15000+13000+35000=63000$, so $90000-63000=27000$ goes to the bank.

Bank after 2 years: $27000\times 1.15^2=27000\times 1.3225=35707.5$.

Items sold at 80%: $0.8\times 63000=50400$.

Total $=86107.5$, a change of $86107.5-90000=-3892.5$.

$$\frac{-3892.5}{90000}\times 100=-4.325\%$$

$7200=2^5\cdot 3^2\cdot 5^2$, so the number of divisors is $(5+1)(2+1)(2+1)=54$.

Excluding 1 and 7200 leaves 52.

$$2\pi r-r=37\Rightarrow r\left(\frac{44}{7}-1\right)=\frac{37}{7}r=37\Rightarrow r=7$$

Circumference $=2\times \frac{22}{7}\times 7=44$ cm.

$$P(A\cup B)=0.25+0.50-0.14=0.61$$

$$P(\text{neither})=1-0.61=0.39$$

• A is false. A tangent touches a circle at exactly one point (a line meeting it at two points is a secant).
• B is true. For any given direction there are exactly two tangents, at the two ends of the diameter perpendicular to that direction, so at most two tangents can be parallel to each other.
• C is true. Every line through an interior point cuts the circle at two points, so no tangent passes through it.

Hence B and C only.

Try $\sqrt{3+\sqrt{5}}=\sqrt{a}+\sqrt{b}$ with $a+b=3$ and $2\sqrt{ab}=\sqrt{5}\Rightarrow ab=\frac{5}{4}$.

$a,b$ are roots of $t^2-3t+\frac{5}{4}=0$, giving $t=\frac{5}{2},\frac{1}{2}$.

Check: ${\left(\sqrt{\frac{5}{2}}+\sqrt{\frac{1}{2}}\right)}^2=\frac{5}{2}+\frac{1}{2}+2\sqrt{\frac{5}{4}}=3+\sqrt{5}$. ✓

Options A, C, D square to $3+2\sqrt{2}$, $4-\sqrt{7}$ and $6-\sqrt{27}$ respectively, none of which equals $3+\sqrt{5}$.

• A: $\frac{x+y+z}{3}=y\Rightarrow x+z=2y$. True.
• B: Sorted: 15, 28, 31, 32, 43, 44, 51, 56, 72. With 9 values the median is the 5th, 43. True.
• C: In 2, 3, 4, 5, 0, 1, 3, 3, 4, 3 the value 3 occurs four times, more than any other. Mode is 3. True.

All three are true.

If $k$ is even, $k^2-1$ is odd and cannot be divisible by 8.

If $k=2m+1$ is odd, $k^2-1=4m(m+1)$, and $m(m+1)$ is even, so $k^2-1$ is divisible by 8.

So $k$ must be an odd integer. 'Integer' and 'natural number' are too broad (e.g. $k=2$ fails), and 'even' is impossible.

Volume dug $=\pi r^{2}h=\frac{22}{7}\times 3.5^2\times 30=1155{\text{m}}^3$.

Platform height $=\frac{1155}{22\times 14}=\frac{1155}{308}=3.75$ m.

$88=8\times 11$.

Divisibility by 8: the last three digits $65b$ must be divisible by 8. $656=8\times 82$, so $b=6$.

Divisibility by 11: digits $5,a,3,2,4,6,5,6$. Sum of odd places $=5+3+4+5=17$; even places $=a+2+6+6=a+14$. Difference $17-(a+14)=3-a$ must be a multiple of 11, so $a=3$.

$$2a+5b=6+30=36$$

Let costs be $5x$ and $6x$. A 10% profit on both: $0.1\times 11x=22\Rightarrow x=20$. Costs are ₹100 and ₹120.

New result: $-0.20\times 100+0.30\times 120=-20+36=16$.

Total profit ₹16.

$\sin \theta =\cos \theta$ with ${\sin }^2\theta +{\cos }^2\theta =1$ gives ${\sin }^2\theta ={\cos }^2\theta =\frac{1}{2}$.

$${\sin }^4\theta +{\cos }^4\theta =\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$

Total $=25\times 36=900$.

First 13 sum $=13\times 32=416$; last 13 sum $=13\times 39=507$.

The 13th observation is counted in both: $416+507-900=23$.

The zeros of $y=f(x)$ are the points where the graph meets the x-axis.

Reading the figure: the curve rises across the axis on the left, comes back down across it (still left of the origin), stays below the axis near O, and then rises across it again on the right. That is 3 intersection points, so 3 zeros.

(I) $V=\pi r^{2}h=\frac{22}{7}\times 49\times 10=1540{\text{cm}}^3$. True.

(II) $\text{TSA}=2\pi r(r+h)=2\times \frac{22}{7}\times 14\times 44=88\times 44=3872{\text{cm}}^2$. True.

For infinitely many solutions the coefficients are proportional:

$$\frac{2}{2p}=\frac{3}{p+q}=\frac{7}{28}=\frac{1}{4}$$

$\frac{1}{p}=\frac{1}{4}\Rightarrow p=4$. Then $p+q=12\Rightarrow q=8$.

$(p,q)=(4,8)$.

By Pythagoras, $A{C}^2-A{B}^2=B{C}^2=25$, i.e. $(AC-AB)(AC+AB)=25$, so $AC+AB=25$.

With $AC-AB=1$: $AC=13$, $AB=12$.

At angle C: opposite side $AB=12$, adjacent $BC=5$, hypotenuse 13. So $\sin C=\frac{12}{13}$, $\cos C=\frac{5}{13}$.

$$\frac{1+\frac{12}{13}}{1+\frac{5}{13}}=\frac{25}{18}$$

Let $|E|=26$, $|G|=48$, $|E\cap G|=8$, $|G\cap H|=8$, none $=24$, so $|E\cup G\cup H|=76$.

• English but not Hindi $=23\Rightarrow |E\cap H|=26-23=3$. B true.
• English only $=18$, so $|E\cap (G\cup H)|=8=|E\cap G|+|E\cap H|-|E\cap G\cap H|=8+3-x\Rightarrow x=3$. C true.
• Inclusion–exclusion: $76=26+48+|H|-8-8-3+3\Rightarrow |H|=18$. A true.

All three statements hold.

Proper divisors of 220: 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110. Their sum is 284.

Check: proper divisors of $284=2^2\cdot 71$ are 1, 2, 4, 71, 142, summing to 220. ✓ So $y=284$.

(I) The sum is even when both are even or both odd: $\frac{9+9}{36}=\frac{1}{2}=0.50$. True.

(II) The product is odd only when both are odd: $\frac{9}{36}=\frac{1}{4}$. So $P(\text{even product})=1-\frac{1}{4}=0.75$. True.

Factor theorem: $f(1)=2+a+b+3=0\Rightarrow a+b=-5$.

Remainder theorem: $f(-2)=-16+4a-2b+3=15\Rightarrow 4a-2b=28\Rightarrow 2a-b=14$.

Adding: $3a=9\Rightarrow a=3$, $b=-8$.

For the same distance, half the time means double the speed: $b+c=2(b-c)\Rightarrow b=3c$.

(I) The ratio is $3:1$, not $2:1$. False.

(II) $b=6\Rightarrow c=2$ km/h. True.

$$P=\underset{A,B}{\underbrace{\frac{1}{3}\cdot \frac{1}{3}\cdot \frac{3}{4}}}+\underset{A,D}{\underbrace{\frac{1}{3}\cdot \frac{2}{3}\cdot \frac{1}{4}}}+\underset{B,D}{\underbrace{\frac{2}{3}\cdot \frac{1}{3}\cdot \frac{1}{4}}}=\frac{3}{36}+\frac{2}{36}+\frac{2}{36}=\frac{7}{36}$$

(I) $x^2+3x+1=x^2-4x+4\Rightarrow 7x-3=0$, which is linear. So it is not a quadratic equation. True.

(II) Discriminant $=(2\sqrt{3}{)}^2-4\cdot 1\cdot 3=12-12=0$, so the roots are real and equal. True.

The cuboid is $10\times 5\times 5$ cm.

$$\text{TSA}=2(10\cdot 5+5\cdot 5+10\cdot 5)=2(50+25+50)=250{\text{cm}}^2$$

(Equivalently, $2\times 150-2\times 25=250$, removing the two hidden faces.)

Even numbers divisible by 5 are multiples of 10: $10,20,\dots ,1000$ (100 terms).

$$S=\frac{100}{2}(10+1000)=50\times 1010=50500$$

So the Assertion (45050) is false. The Reason is the standard, correct formula for the sum of an AP, so it is true.

The inscribed circle's diameter equals the side, so $r=3$ cm and area $=\pi r^2=9\pi {\text{cm}}^2$.

$V=\frac{1}{3}\pi r^{2}h$, so $\frac{h_1}{h_2}=\frac{V_1/V_2}{(r_1/r_2{)}^2}=\frac{1/3}{4/9}=\frac{3}{4}$.

Ratio of heights $=3:4$.

Table values are percentages (maximum 120 marks per subject):

• A: 70, 75, 70, 65, 60, 70
• B: 80, 70, 75, 60, 50, 75
• C: 75, 72, 71, 70, 60, 65
• D: 60, 70, 65, 55, 60, 70
• E: 70, 75, 65, 50, 55, 60
• F: 65, 70, 60, 65, 50, 60

(order: Physics, Biology, Maths, Geology, Economics, English). All subjects have the same maximum, so a student's overall percentage is the simple average of the six percentages.

• A: $\frac{70+75+70+65+60+70}{6}=\frac{410}{6}=68.33\%$
• B: $\frac{80+70+75+60+50+75}{6}=\frac{410}{6}=68.33\%$
• C: $\frac{75+72+71+70+60+65}{6}=\frac{413}{6}=68.83\%$

No option lists 68.33, 68.33, 68.83 (options A–C use 68.44/69.29/68.94/68.14/68.29), so the answer is No Answer.

Average percentage across the six students:

• Economics: $\frac{60+50+60+60+55+50}{6}=\frac{335}{6}=55.83\%$, i.e. $0.5583\times 120=67$ marks.
• Physics: $\frac{70+80+75+60+70+65}{6}=\frac{420}{6}=70\%$, i.e. $0.70\times 120=84$ marks.

So the averages are Economics 67, Physics 84. Option A has these two values, but lists Physics first (84, 67), the reverse of the order the question asks for.

Option D (56, 70) gives the average *percentages* (55.83 rounded, and 70) in the asked order, not marks out of 120. Options B and C match neither reading.

The official key takes A, i.e. the correct values with the order ignored.

Overall percentages (simple average of the six values):

• A $=68.33\%$, B $=68.33\%$, C $=68.83\%$
• D $=\frac{380}{6}=63.33\%$, E $=\frac{375}{6}=62.5\%$, F $=\frac{370}{6}=61.67\%$

Only A, B and C exceed 65%, so 3 students. Option D wrongly includes D (63.33%), and options B and C leave out students who qualify.

Average across the six students for each subject:

• Physics $=\frac{420}{6}=70\%$
• Biology $=\frac{75+70+72+70+75+70}{6}=\frac{432}{6}=72\%$
• Maths $=\frac{406}{6}=67.67\%$
• Geology $=\frac{365}{6}=60.83\%$
• Economics $=55.83\%$
• English $=\frac{400}{6}=66.67\%$

The highest is Biology, 72%. Options C and D state the wrong figures for Maths (67.67%) and English (66.67%). Physics really is 70% but is not the highest.

Take 2005 total sales as 100.

• A: $45\to 45\times 1.5=67.5$ (gain 22.5)
• E: $22\to 22\times 1.1=24.2$ (gain 2.2)
• Others unchanged.

Total gain $=22.5+2.2=24.7$ on a base of 100, i.e. growth of 24.70%.

From Q38, 2006 total $=1.247\times$ (2005 total).

$$\text{2005 total}=\frac{216000}{1.247}\approx 173215$$

A's 2005 sale $=45\%$ of this $=0.45\times 173215\approx 77,947$.

On a base of 100:

• B: $15\times 10\%=+1.5$
• A: $45\times 10\%=-4.5$

Net change $=1.5-4.5=-3$, i.e. a decrease of 3%.

From the clues: $B>D>E>C$ (in speed).

C is not the slowest, so someone is slower than C, and the only one left is A: $B>D>E>C>A$.

B is the fastest. A is the slowest, and C and D are in the middle.

Each letter moves forward 3 places: B→E, J→M, M→P, S→V.

Apply the same to AWPL: A→D, W→Z, P→S, L→O, giving DZSO.

The other options break the +3 rule, e.g. DYRO has W→Y (+2).

Weekly totals (in hundreds):

• I: $28.3+15.4+40.2=83.9$
• II: $24.5+12.3+36.1=72.9$
• III: $20.4+14.2+42.9=77.5$
• IV: $24.4+24.4+39.8=88.6$

The highest is week IV and the second highest is week I.

$$\frac{28.3+24.5+20.4+24.4}{4}=\frac{97.6}{4}=24.4\text{ hundred}=2440$$

Letter positions: C=3, A=1, T=20. Adding 3 to each gives 6, 4, 23, i.e. 6423.

LION: L=12, I=9, O=15, N=14, and adding 3 gives 15, 12, 18, 17, i.e. 15121817.

Option B (1291514) is just LION's plain positions 12, 9, 15, 14 with no shift, option C (961211) subtracts 3 instead of adding it, and option D (14131716) follows no consistent shift.

Two years share a calendar if they start on the same weekday and are both leap or both non-leap.

2008 is a leap year (366 days, 2 odd days), so 2009 starts on Thursday. 2009 is non-leap.

Odd days year by year: 2009 (1), 2010 (1), 2011 (1), 2012 (2, leap), 2013 (1), 2014 (1). Total $=7\equiv 0$, so 2015 starts on Thursday too and is non-leap.

The other options fail: 2011 starts on Saturday, 2013 on Tuesday, and 2016 is a leap year.

Car A: 30 km along the highway, a right turn and 15 km away from the highway, a left turn and 30 km parallel to it, then 15 km back to the highway.

A's progress along the highway $=30+30=60$ km from X.

Car B is 35 km from Y, i.e. $120-35=85$ km from X.

Distance $=85-60=$ 25 km.

Number the seats 1–6 in order so that for a person facing the centre, 'right' is the next seat (+1) and 'left' is the previous seat (−1).

• Put P at 1, so U (opposite) is at 4.
• P is immediately left of R, so R's left neighbour is P, which puts R at 2.
• S is immediately right of U, so S is at 5.
• T is exactly between P (1) and S (5), so T is at 6.
• Q takes the remaining seat, 3.

Q (seat 3) sits between R (2) and U (4).

Checking the other options: R is opposite S, not Q. U, not Q, is opposite P. P's immediate left is T.

Differences: $1,2,4,8,6,2,?$

These are the units digits of successive powers of 2: $1,2,4,8,1\underline{6},3\underline{2},6\underline{4}$.

The next difference is 4, so the term is $33+4=$ 37.

• EIWNTR = WINTER
• LCUOD = CLOUD
• UMRSME = SUMMER
• PIGRSN = SPRING

Winter, summer and spring are seasons; CLOUD is not. It is also the only 5-letter word among 6-letter ones.

The pattern alternates $+3,-1,+4,-1,+5,-1,+6,-1,\dots$:

$15\stackrel{+3}{\to }18\stackrel{-1}{\to }17\stackrel{+4}{\to }21\stackrel{-1}{\to }20\stackrel{+5}{\to }25\stackrel{-1}{\to }24\stackrel{+6}{\to }30\stackrel{-1}{\to }29\stackrel{+7}{\to }$ 36

$1\times 2,3\times 4,5\times 6,7\times 8,9\times 10$, so the next term is $11\times 12=$ 132.

(Equivalently, the differences 10, 18, 26, 34 grow by 8, so the next difference is 42 and $90+42=132$.)

Angle between the hands at $h$:$m$ is $|30h-5.5m|$.

• (I) 7:15: $|210-82.5|=127.5^{\circ }$, so C–I
• (II) 9:15: $|270-82.5|=187.5^{\circ }$ (the reflex angle; the smaller angle is $172.5^{\circ }$), so D–II
• (III) 4:30: $|120-165|=45^{\circ }$, so B–III
• (IV) 2:30: $|60-165|=105^{\circ }$, so A–IV

A–IV, B–III, C–I, D–II is option A. Every other option pairs at least A or C wrongly.

The hour hand moves $30^{\circ }$ per hour, so $135^{\circ }$ takes $\frac{135}{30}=4.5$ hours.

3:00 + 4 h 30 min = 7:30.

Take the house as $(0,0)$, with east as $+x$ and north as $+y$.

• 150 m south: $(0,-150)$
• Facing south, left is east; 300 m: $(300,-150)$
• Facing east, left is north; 350 m: $(300,200)$
• Facing north, right is east; 50 m: $(350,200)$

The stadium is 350 m east and 200 m north of the house, i.e. North-East.

Read 'between' as immediately between, as these stacking puzzles intend.

• Electronics is directly above Chemistry.
• Botany sits between Electronics and Commerce. It cannot be below Electronics, because Chemistry is there, so the order upward is Chemistry, Electronics, Botany, Commerce.
• Statistics sits between Commerce and Environmental Science, so it goes above Commerce, with Environmental Science above it.

Bottom to top: Chemistry, Electronics, Botany, Commerce, Statistics, Environmental Science. That uses all six books, so Environmental Science is at the top.

• PWSORAR = SPARROW (bird)
• GPNEIO = PIGEON (bird)
• TPORAR = PARROT (bird)
• JRABLNI = BRINJAL (vegetable)

BRINJAL is the only one that is not a bird.

Use reverse alphabet positions (A=26, …, Z=1) and multiply them.

• H=19, E=22: $19\times 22=418$ ✓
• G=20, O=12: $20\times 12=240$ ✓

SHE: S=8, H=19, E=22, so $8\times 19\times 22=$ 3344.

• P is from Bombay, and the Tennis player is from Jaipur, so P does not play Tennis. From clue 4, P plays Chess.
• Q plays Football or Chess, and Chess is taken, so Q plays Football.
• R and S share Tennis and Cricket. S does not play Cricket, so S plays Tennis (and is from Jaipur) and R plays Cricket.
• The Cricket player is not from Calcutta, and Bombay and Jaipur are taken, so R is from Ahmedabad. Q is from Calcutta.

The Ahmedabad person (R) plays Cricket.

Number the seats 0–7 in order so that for someone facing the centre, 'right' is $+1$ and 'left' is $-1$.

• F = 0. E is 4th to the right of F, so E = 4.
• E is 2nd to the left of A, so $A-2=4$ and A = 6.
• D is not next to A, so D ≠ 5, 7. D is in {1, 2, 3}.
• B is 3rd to the right of D: D = 1 would put B on E's seat (4), and D = 3 would put B on A's seat (6). So D = 2 and B = 5.
• D is not next to C, so C ≠ 1, 3. Hence C = 7.
• G and H take 1 and 3. F never sits next to H, so H ≠ 1. Hence H = 3 and G = 1.

The circle is F0, G1, D2, H3, E4, B5, A6, C7. Going from G (1) to A (6) the short way passes seats 0 and 7, i.e. C and F.

Options C and D name people on the longer arc along with others (D, H, E, B), and option B includes A itself.

Ecology is the study of organisms in relation to their environment. Histology is the study of tissues.

History is a different discipline, hormones are studied in endocrinology, and fossils in palaeontology.

Ram's son is Gita's real brother, so Gita (and her real sister Sita) are Ram's daughters.

Sham is Ram's brother, so he is Sita's (paternal) uncle. He cannot be her brother, father or grandfather.

The statement has the form $I\Rightarrow P$. Its logically equivalent contrapositive is $\neg P\Rightarrow \neg I$: 'I did not play, so the game was not interesting.' That is option D.

• A is the converse ($P\Rightarrow I$), which does not follow.
• B and C describe an interesting game not being played, which contradicts the statement.

CONSULTATION contains C, O×2, N×2, S, U, L, T×2, A, I.

• NATION (N×2, A, T, I, O) can be formed.
• STATION (S, T×2, A, I, O, N) can be formed.
• SOLUTION (S, O×2, L, U, T, I, N) can be formed.
• SALUTE needs an E, which CONSULTATION does not have.

2 October is a Monday, so the Wednesdays are 4, 11, 18 and 25 October.

Strictly between 14th and 21st means 15–20, and strictly between 15th and 20th means 16–19. The only Wednesday in both ranges is 18th October.

16th, 17th and 19th are Monday, Tuesday and Thursday.

Going clockwise from 4: $4,7,12,21,38,?$

Each term is twice the previous minus an increasing number:

$4\times 2-1=7,7\times 2-2=12,12\times 2-3=21,21\times 2-4=38,38\times 2-5=$ 71.

(Equivalently, the differences $3,5,9,17,33$ are $2^n+1$.)

A polite imperative ('Please …') is reported with requested + object + to-infinitive. 'Here' becomes there, and the present 'return' is back-shifted to the past, returned.

• A changes 'him' to 'her', the wrong object.
• C keeps the present tense 'returns' after a past reporting verb.
• D uses the present 'requests' but then a past perfect, an inconsistent mix.

The passage says the monkeys had no warm home on a cold, stormy night and were looking everywhere for shelter. That supports option C, 'were in search of a warm shelter'.

• A (food) and D (killing birds) are never mentioned as their reason.
• B, 'liked running about in dark', has no support. The monkeys are described as worried, not enjoying themselves.

This question is word-for-word identical to Q69, where the official key gives C. The key's B for Q68 looks like a key error. It is kept here as printed.

The passage states the monkeys had no warm home in the cold and 'looked everywhere for shelter'. They even went towards what they thought was fire to warm themselves. So they were searching for a warm shelter.

Food (A) is never mentioned. Nothing suggests they enjoyed running in the dark (B). Their attack on the bird came later, out of anger, and was not why they were running (D).

The passage says one bird 'had a kind heart' and 'felt sorry for the monkeys', so B gives its motive directly.

• A describes what the monkeys *called* her advice, not her reason for giving it.
• C and D have no support; she only wanted to spare them a futile chase.

The bird's well-meant advice made the angry monkey pull it out of its nest. Its final lament shows that helping a fool brought harm on the helper, so C fits.

• A and B are the opposite of what happened (no thanks, no friendship).
• D is wrong because the harm fell on the helper, not on the fool.

Both events are completed past actions in sequence, and 'after' already marks the order, so the simple past in both clauses is correct: 'Who replaced the player after she was injured?'

• A puts the *later* action in the past perfect ('had replaced'), which reverses the time logic.
• B uses 'she were', a subject–verb agreement error.
• C uses the past continuous 'was replacing' for a single completed act, which is unidiomatic.

The sentence is a general comparison between two ways of communicating. A general statement needs a generic noun phrase, 'Writing a letter' (or 'Writing letters'). The definite article 'the letter' wrongly points to one particular letter.

Parts B ('is often cheaper': singular verb with a gerund subject) and C ('than telephoning': parallel gerund) are correct. So the error is in part A.

Considerate means thoughtful about and caring for the feelings and needs of others. Its opposite is lacking concern for others, i.e. indifferent (uncaring, unconcerned).

• Comprehensive means all-inclusive, which is unrelated.
• Atrocious means extremely bad or wicked, and malignant means harmful or evil. Both describe cruelty or harm, not the absence of care, so neither is the direct opposite.

'Let us …' expresses a proposal that includes the speaker, so it is reported as proposed/suggested that they should + verb.

• A ('proposed him') is ungrammatical and drops the speaker from the group.
• C ('agreed … with him') changes the meaning from a proposal to agreement.
• D ('proposed me that I') is ungrammatical and wrongly narrows the group to the listener.

Dormitory is spelt correctly. The correct forms of the others are disaster, documentation and deviation.

The proverb is 'Time and tide wait for none'. Two subjects joined by 'and' take a plural verb, and a general truth takes the simple present. So the answer is 'wait'.

'Waits' is singular, 'waiting' is not a finite verb, and 'had waited' is the wrong tense for a timeless truth.

Reproach means to express disapproval or disappointment to someone, i.e. to rebuke or reprimand.

'Appreciate' is nearly the opposite, and 'understand' is unrelated. 'To approach again' is a false guess from the word's parts ('re-' + 'approach').

An impatient client wants things done faster, so expedite means to speed up, i.e. accelerate.

Evaluating, reversing or justifying the process would not answer the client's impatience.

'When I was young, (Q) shopkeepers would cycle (S) around the town (P) delivering food to customers (R).'

The time clause Q opens the sentence and needs the main clause S with its subject and verb. P completes 'cycle' with a place, and the participle phrase R describes the purpose. The other orders leave S or R without a proper link (e.g. RSPQ starts with a dangling participle).

The standard negative infinitive is not to + verb: 'decided not to perform'.

• A splits the infinitive ('to not perform'), which is avoided in formal usage.
• C wrongly uses 'did not decided' (after 'did' the base form 'decide' is needed), and it also changes the meaning.
• D uses the present 'decide' with a singular subject in a past context.

'This card entitles (Q) you to take (S) an extra person (P) with you free (R).'

'Entitles' needs an object plus a to-infinitive ('entitles you to take'), 'take' needs its object 'an extra person', and 'with you free' completes the idea. No other order gives a grammatical sentence.

Break down is the phrasal verb for a vehicle or machine that stops working.

'Break apart' means to split into pieces, 'break up' means to end a relationship or disperse, and 'break into' means to enter by force. None of these fits a car failing on a journey.

To put one's foot down is to take a firm stand and refuse to change, i.e. refused to yield.

The context confirms it: she was asked to reconsider, and she stood firm. Agreeing to consider (A), ignoring (B) or listening carefully (C) do not convey that firmness.

In the future simple, will finish becomes will be finished in the passive. The object 'the work' becomes the subject, and the agent follows as 'by him'.

• A and B wrongly change the tense to 'would' and 'in a fortnight' to 'that fortnight' (A also keeps 'he' as the subject).
• C changes the tense to the past ('was finished').

Palpable means able to be touched or felt, and so clearly noticeable. The closest synonym is perceivable (perceptible), which covers perception by any sense.

'Visible' and 'audible' are limited to single senses (sight, hearing), and 'presentable' is unrelated.

The paragraph says critical reading is demanding: slow down, read with a pencil, mark reactions and questions, be an active participant. That adds up to thoughtful, careful attention (D).

• A adds 'dull', which the author never says.
• B narrows the point to questioning truth, which is only one of the activities mentioned.
• C plays on the word 'critical' ('critical times'), which is unrelated to the paragraph.

Someone who believes even the most outlandish excuses is easily deceived, i.e. naive.

'Intelligent' and 'critical' are the opposite of what the context shows. 'Dishonest' describes the employees, not Mr. Powers.

A comparison with 'more … ' is completed by than, as in Shakespeare's line from *Hamlet*.

'Then' (time), 'thin' (not a connector) and 'where' do not complete a comparison.

The passive 'was … written' is simple past, so the active question uses did + base verb: 'Why did your father write such a letter?'

• A shifts the tense to the past perfect.
• B wrongly uses 'did … wrote' (after 'did' the base form is needed).
• D shifts to the past continuous.

Famished means extremely hungry.

'Agricultural' may tempt through its association with food and famine, but it is unrelated in meaning. 'Prestigious' and 'feasible' are also unrelated.

Ostensible means stated or appearing to be true but not necessarily so, i.e. apparent (as in 'the ostensible reason').

'Strong', 'desirable' and 'fateful' do not carry the sense of outward appearance.

Put off means to postpone, which fits delaying a departure 'for a week'.

'Put down' means to suppress or write down, and 'put on' means to wear or stage something. 'Put of' is not a phrasal verb.

This is from Francis Bacon's *Of Studies*: read not to contradict, nor to believe, but to weigh and consider. The parallel structure 'not to …; nor to …; but to …' needs the negative infinitive not to + verb.

• A splits the infinitive ('to not').
• B scrambles the word order.
• C uses the adjective 'contradictory' where a verb is needed.

Communicable (as in a communicable disease) means able to be passed from one person to another, i.e. contagious.

'Communication' and 'commutation' are nouns that only look similar. 'Understandable' confuses communicable with 'comprehensible'.

The correct spelling is un-equi-voc-al-ly: 'unequivocal' + '-ly' keeps both l's.

Option A drops one l, C starts with 'en-', and D replaces 'equi' with 'iqui'.

After the auxiliary 'doesn't' the main verb takes its base form (like), and after 'to' the infinitive is also base form (get).

• A has 'to gets'.
• B and D have 'doesn't likes'.

The author presents mediation as an efficient, successful, cheaper alternative that seeks mutually acceptable solutions. The hedged claim in D, that it can be effective, matches.

• A (society relies too much on litigation) is not discussed.
• B makes claims about litigation's limits beyond what is stated.
• C ('the best way') overstates: the author says litigation is not *always* the best, not that mediation always is.

'Hardly' at the start of a sentence requires inversion with the past perfect: Hardly had + subject + past participle. Option A has this ('Hardly had the function got over').

• B uses 'have … getting', which is ungrammatical.
• C garbles both the meaning and the grammar ('was not got over', 'the crowd were').
• D has no subject after 'had' and uses the present 'disperse'.

Strictly, the standard pattern also needs when before the second clause ('… got over when the crowd dispersed'). A is the only option with the correct inverted structure, so it is the best choice.

The subject is the singular 'a bird', and a proverb states a timeless truth, so the simple present singular is is needed.

'Are' and 'were' are plural, and 'was' is the past tense.