JIPMAT 2022 · Question Paper & Solutions

Doubling means the interest earned equals the principal: $SI=P$.

$$P=\frac{P\times R\times 10}{100}\Rightarrow R=\frac{100}{10}=10\%$$

All three invest for the same period, so profit is shared in the ratio of capital:

$$25000:50000:25000=1:2:1$$

Q's share $=\frac{2}{4}\times 48000=24000$.

• A: $6x^2+x-12=(3x-4)(2x+3)=0\Rightarrow x=\frac{4}{3},-\frac{3}{2}$ → IV.
• D: $2x^2-5x-3=(2x+1)(x-3)=0\Rightarrow x=3,-\frac{1}{2}$ → III.

Only option B contains both A-IV and D-III, so the answer is B.

The remaining pairings have printing slips in the paper: $8x^2+16x-10=182$ (not 202) gives $x^2+2x-24=0$, roots $-6,4$ (I); and $x^2+45x+324=0$ has roots $-9,-36$, which matches II, $(9,36)$, apart from the sign. Neither slip changes the answer, because A and D already pin it down.

$M+F=5000$ and the increase is $0.05M+0.03F=202$.

Substitute $F=5000-M$: $0.05M+150-0.03M=202\Rightarrow 0.02M=52\Rightarrow M=2600$, so $F=2400$.

Check: $130+72=202$.

Let $B=3$. Then $A=3+1=4$.

B is less than A by $\frac{4-3}{4}\times 100=25\%$.

• A: $\sin 49^{\circ }=\cos 41^{\circ }$, so ${\sin }^{2}41^{\circ }+{\cos }^{2}41^{\circ }=1$. True.
• B: ${\sin }^{2}60^{\circ }-2\tan 45^{\circ }-{\cos }^{2}30^{\circ }=\frac{3}{4}-2-\frac{3}{4}=-2\ne -1$. False.
• C: $\frac{1}{1+{\tan }^2\theta }=\frac{1}{{\sec }^2\theta }={\cos }^2\theta$, so the sum is ${\sin }^2\theta +{\cos }^2\theta =1$. True.

So A and C only.

The first term is printed as $x^2$, but a polynomial can't have two $x^2$ terms. The intended polynomial is $x^4+10x^3+25x^2+15x+m$.

By the factor theorem, $f(-7)=0$:

$$2401-3430+1225-105+m=0\Rightarrow 91+m=0\Rightarrow m=-91$$

$\pi r^2=22176\Rightarrow r^2=22176\times \frac{7}{22}=7056\Rightarrow r=84$ cm.

Circumference $=2\times \frac{22}{7}\times 84=528$ cm $=5.28$ m.

Cost $=5.28\times 50=₹264$.

Each term splits: $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$.

The sum telescopes:

$$\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots +\left(\frac{1}{n}-\frac{1}{n+1}\right)=1-\frac{1}{n+1}=\frac{n}{n+1}$$

Check with $n=1$: $\frac{1}{2}$. Correct.

A finishes the whole work in $3\times 4=12$ days. B takes $4\times 6=24$ days.

Their daily work rates are $\frac{1}{12}:\frac{1}{24}=2:1$. Working together, pay is split in this ratio.

A's share $=\frac{2}{3}\times 180=120$.

The tops differ in height by $11-6=5$ m and are 12 m apart horizontally.

$$d=\sqrt{12^2+5^2}=\sqrt{169}=13\text{ m}$$

$500000\times (0.9{)}^n=364500\Rightarrow (0.9{)}^n=0.729=(0.9{)}^3$, so $n=3$ years.

Milk $=21$ L and water $=9$ L. Milk doesn't change, and it must make up 60% of the new mixture.

New total $=\frac{21}{0.6}=35$ L, so add $35-30=5$ L of water.

Check: $\frac{14}{35}=40\%$.

• I: $392=2^3\times 7^2$, so the exponents sum to $3+2=5$. True.
• II: $\frac{13}{2^3\times 5}=\frac{13}{40}=0.325$, which ends after 3 decimal places. In general, a denominator $2^m{5}^n$ terminates after $\max (m,n)=3$ places. False.

$$P(\text{exactly one})=p(1-2p)+2p(1-p)=3p-4p^2=\frac{5}{9}$$

$$36p^2-27p+5=0\Rightarrow p=\frac{27\pm 3}{72}=\frac{1}{3}\text{ or }\frac{5}{12}$$

Both are valid probabilities, since $2p\le 1$ in each case ($\frac{2}{3}$ and $\frac{5}{6}$). So both statements are true.

Let the upstream speed be $u$ and the downstream speed be $v$:

$$\frac{30}{u}+\frac{28}{v}=7,\frac{21}{u}+\frac{21}{v}=5$$

From the second equation, $\frac{1}{u}+\frac{1}{v}=\frac{5}{21}$. The first can be rewritten as $28\left(\frac{1}{u}+\frac{1}{v}\right)+\frac{2}{u}=7$, which gives $\frac{20}{3}+\frac{2}{u}=7$. So $u=6$ and $v=14$.

Still water $=\frac{6+14}{2}=10$ km/h, and stream $=\frac{14-6}{2}=4$ km/h.

Check: $5+2=7$ h and $3.5+1.5=5$ h. I is false and II is true.

The dimensions are misprinted; they should read 15 m 17 cm = 1517 cm and 9 m 2 cm = 902 cm.

The largest square tile has side $\text{HCF}(1517,902)$. Since $1517=37\times 41$ and $902=22\times 41$, the HCF is 41 cm.

Number of tiles $=\frac{1517}{41}\times \frac{902}{41}=37\times 22=814$.

When they meet, A has covered $60-12=48$ km. B has covered $60+12=72$ km in the same time.

So $\frac{v_A}{v_B}=\frac{48}{72}=\frac{2}{3}$. With $v_B-v_A=4$, one ratio part is 4 km/h, giving $v_A=8$ and $v_B=12$ km/h.

$12H+H=403\Rightarrow H=31$, so $L=372$.

Other number $=\frac{H\times L}{93}=\frac{31\times 372}{93}=124$.

Check: $124=31\times 4$ and $93=31\times 3$, so HCF $=31$ and LCM $=31\times 12=372$.

Let the son be $s$, so the father is $6s$. Then $6s+4=4(s+4)\Rightarrow 2s=12\Rightarrow s=6$.

The ages are 6 and 36. Check: in 4 years, $40=4\times 10$.

Three years ago their ages were $8k$ and $9k$, so in 3 years they will be $8k+6$ and $9k+6$.

$$12(8k+6)=11(9k+6)\Rightarrow 96k+72=99k+66\Rightarrow k=2$$

Amisha now $=16+3=19$ years. Check: $22:24=11:12$.

Joining the four centres gives a square of side $2r$. Its diagonal is $2\sqrt{2}r$.

The inner circle sits at the centre of the square, so each outer centre is $\sqrt{2}r$ away. The inner circle touches each outer circle, so its radius is

$$\sqrt{2}r-r=(\sqrt{2}-1)r$$

The roots satisfy $\sin \alpha +\cos \alpha =-\frac{b}{a}$ and $\sin \alpha \cos \alpha =\frac{c}{a}$.

Square the sum:

$$\frac{b^2}{a^2}=1+2\cdot \frac{c}{a}\Rightarrow b^2=a^2+2ac$$

$a+b+c=3M$, so

$$a^2+b^2+c^2=(a+b+c{)}^2-2(ab+bc+ca)=9M^2-0=9M^2$$

The mean of the squares is $\frac{9M^2}{3}=3M^2$, so $K=3$.

• A: $\frac{2+3+5+7+11}{5}=\frac{28}{5}=5.6$ → III.
• B: The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. Their sum is 60, so the mean is $\frac{60}{8}=7.5$ → II.
• C: $\frac{4+8+12+16+20+24}{6}=\frac{84}{6}=14$ → IV.
• D: The $y$-terms cancel, so the mean is $\frac{5x}{5}=x=12$ → I.

Simplify the nested roots: $\sqrt{8-2\sqrt{15}}=\sqrt{5}-\sqrt{3}$ and $\sqrt{11+2\sqrt{30}}=\sqrt{6}+\sqrt{5}$.

First term: $\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{(\sqrt{5}+\sqrt{3}{)}^2}{2}=\frac{8+2\sqrt{15}}{2}=4+\sqrt{15}$.

Second term: $\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}-\sqrt{5}}=\frac{(\sqrt{6}+\sqrt{5}{)}^2}{1}=11+2\sqrt{30}$.

Sum $=15+\sqrt{15}+2\sqrt{30}$. A numeric check gives about 29.83 both ways.

$3\times 5=6-15+15=6$ and $5\times 3=10-9+15=16$.

Sum $=22$.

The numerator is misprinted as $325\times 325\times 325+175+175+175$. The intended term is $175\times 175\times 175$, which matches the denominator's pattern.

With $a=325$ and $b=175$:

$$\frac{a^3+b^3}{a^2-ab+b^2}=\frac{(a+b)(a^2-ab+b^2)}{a^2-ab+b^2}=a+b=500$$

• I: $A{B}^2+B{C}^2=108+36=144=A{C}^2$. By the converse of Pythagoras, the right angle is opposite AC, which is $\angle B$. True.
• II: $A{B}^2=2A{C}^2=A{C}^2+B{C}^2$, so the right angle is opposite AB, which is $\angle C$. True.

Take $\theta$ acute.

• I: $\sin \theta =\frac{5}{13}\Rightarrow \cos \theta =\frac{12}{13}\Rightarrow \tan \theta =\frac{5}{12}$. True.
• II: $\cot \theta =\frac{12}{5}$ gives a 5-12-13 triangle, so $\sin \theta =\frac{5}{13}$, not $\frac{5}{12}$. False.

Let the number be $10a+b$. Then $10a+b=7(a+b)\Rightarrow 3a=6b\Rightarrow a=2b$.

Reversing lowers it by 18: $9(a-b)=18\Rightarrow a-b=2$. So $b=2$, $a=4$, and the number is 42.

Check: $7\times 6=42$ and $42-24=18$.

Case 1: $a+b+c\ne 0$. Adding numerators and denominators of equal ratios gives

$$k=\frac{a+b+c}{2(a+b+c)}=\frac{1}{2}$$

Case 2: $a+b+c=0$. Then $b+c=-a$, so $k=\frac{a}{-a}=-1$.

Both values are possible.

$r^2=(4-2{)}^2+(5-2{)}^2=4+9=13$.

The circle is $(x-2{)}^2+(y-2{)}^2=13$.

Total taxes $=32.0+34.6+38.8+42.4+42.6=190.4$.

Total cost of components $=296+332+382+410+428=1848$.

Ratio $=\frac{190.4}{1848}\approx 0.103$. Compare the options: $\frac{1}{117}\approx 0.009$, $\frac{3}{70}\approx 0.043$, $\frac{14}{138}\approx 0.101$, $\frac{12}{141}\approx 0.085$.

The closest is 14 : 138.

2000: $288+296+26.4+12.2+32.0=654.6$.

2003: $305+410+28.6+15.2+42.4=801.2$.

$\frac{654.6}{801.2}\times 100\approx 81.70\%$.

• 2001: $\frac{299-288}{288}\approx 3.82\%$
• 2002: $298<299$, a decrease
• 2003: $\frac{305-298}{298}\approx 2.35\%$
• 2004: $\frac{316-305}{305}\approx 3.61\%$

The maximum is in 2001.

Complete the table first (used for Q37–40):

• 40 girls are 50% of the students, so there are 80 students and 40 boys.
• Half the students (40) have grade A or B. With 30 grade B in total, 10 have grade A.
• Boys with A $=5$, so girls with A $=5$.
• $\frac{1}{4}$ of 40 boys $=10$ boys have grade C. So boys with B $=40-5-10=25$.
• Girls with B $=30-25=5$, and girls with C $=40-5-5=30$.

Girls with B = 5 and girls with C = 30.

Girls are 40 and form 50% of the class, so boys $=40$ and total $=80$. See the completed table in Q37.

A or B together $=\frac{80}{2}=40$, and B $=30$, so A $=10$ in total.

Boys with A $=5$, so girls with A $=5$. The answer is 5, 10.

Boys with C $=\frac{1}{4}\times 40=10$.

Boys with B $=40-5\text{ (A)}-10\text{ (C)}=25$. The answer is 25, 10.

Mita is Deepak's child. Sheela is Deepak's sister, so she is Mita's aunt (father's sister).

Arun is Sheela's husband, which makes him Mita's uncle.

The chain is Sparrows ⊂ Birds ⊂ Mountains ⊂ Skies.

• I (All sparrows are skies): follows along the chain.
• IV (All birds are skies): follows, since Birds ⊂ Mountains ⊂ Skies.
• II (All skies are birds) and III (All mountains are sparrows) reverse the inclusions. "All X are Y" does not imply "All Y are X", so they don't follow.

Each term moves the first letter to the end: AGDLX → GDLXA → DLXAG → LXAGD → XAGDL → AGDLX.

The cycle closes back at AGDLX, which confirms the rule.

Count subjects for each person:

• A: History, Biology, Maths, Chemistry = 4
• B: Physics = 1
• C: History, Biology, Chemistry = 3
• D: Physics, Chemistry = 2

Only A has four.

The pie chart gives Red 21%, Blue 14%, Green 18%, Pink 12%, Yellow 21%, Orange 14% of 1000 balls.

Red $=210$. Light : dark $=3:4$, so dark red $=210\times \frac{4}{7}=120$.

Green $=18\%$ of $1000=180$.

Orange + Yellow + Pink $=14\%+21\%+12\%=47\%$.

Light red $=210\times \frac{3}{7}=90$.

Yellow $=210$ with light : dark $=9:5$, so dark yellow $=210\times \frac{5}{14}=75$.

Difference $=90-75=15$.

Dark pink $=120\times \frac{7}{8}=105$.

Dark green $=180\times \frac{3}{4}=135$.

$105:135=7:9$.

Rohit likes badminton and is in neither the theatre club nor the dance club. Everyone belongs to exactly one club, so Rohit is in the Eco club.

Blessy is also in the Eco club but does not like badminton, so she is ruled out.

Within each group the letter gaps are +2, +1, +2: C(3) E(5) F(6) H(8).

The first letters go C(3), I(9), ?, U(21), rising by 6 each time. So the missing group starts with O(15).

O(15), Q(17), R(18), T(20) gives OQRT.

The pie chart gives Toy Making 30%, Paper Quilling 20%, Flower Making 25%, Mask Making 15%, Clay Modelling 10%.

Mask making + paper quilling $=15\%+20\%=35\%$ of $40=14$.

Toy making $=30\%$, so students not in toy making $=70\%$ of $40=28$.

None of the options (16, 12, 10, 20) equals 28. The official key gives "E", which is not one of the printed options, confirming the question is flawed.

The tempting option is 12, but that is the number who did take part in toy making ($30\%$ of 40).

Flower making $=25\%$ of $40=10$.

Rani's sister's brother is Rani's brother. His father is Rani's father.

The only son of Rani's father is Rani's brother.

Positions are 1, 3, 6, 10, with gaps +2, +3, +4.

The next gaps are +5 and +6, giving 15 = O and 21 = U.

The operations alternate between $+2$ and $÷2$:

$$6\stackrel{+2}{\to }8\stackrel{÷2}{\to }4\stackrel{+2}{\to }6\stackrel{÷2}{\to }3\stackrel{+2}{\to }5\stackrel{÷2}{\to }\frac{5}{2}$$

The next term would be $\frac{5}{2}+2=\frac{9}{2}$. The paper prints it as $\frac{9}{12}$, most likely a misprint of $\frac{9}{2}$.

A second check: the odd-position terms 6, 4, 3, ? have differences −2, −1, so the next difference is −½, giving $\frac{5}{2}$.

Going clockwise, the sectors hold consecutive cubes: $2^3,3^3,4^3,5^3,6^3$.

The missing number is $7^3=343$.

Each code is the sum of the squares of the digits:

• $6^2+2^2+3^2=49$
• $5^2+1^2+6^2=62$
• $4^2+3^2+8^2=89$

So $583\to 25+64+9=98$.

The rule is $\times 3+6$: $1\to 9\to 33\to 105\to 321$.

The differences confirm it: 8, 24, 72 triple each time, so the next is 216 and $105+216=321$. The term 158 is wrong.

Net east–west: $3-6=3$ km west. Net north–south: 4 km south.

Distance $=\sqrt{3^2+4^2}=5$ km.

Going clockwise, each number is one-third of the previous: $594÷3=198$ and $198÷3=66$.

The missing number is $66÷3=22$.

Deer and lions are both animals, so both circles sit inside the "animals" circle. No deer is a lion, so the two inner circles don't overlap. That is diagram A.

• B (nested circles) would mean every deer is a lion.
• C puts one class outside animals.
• D shows overlaps that don't exist.

Botany is the study of plants, and entomology is the study of insects.

Birds belong to ornithology, and germs to microbiology or bacteriology.

After substituting the symbols, the expression becomes

$$5\times (13\times 4÷2-6+15-5)$$

Inside the brackets: $52÷2=26$, then $26-6+15-5=30$.

Result $=5\times 30=150$.

The letter shifts are +1, +2, +3, +4: L→M, C→E, Q→T, J→N.

Applying them to ABDF: A→B, B→D, D→G, F→J gives BDGJ.

A contemporary is someone who lives or works in the same period as another person.

• A predecessor came before.
• A successor comes after.
• Contemptuous means scornful and has nothing to do with time.

The fixed expression from Shakespeare's *Hamlet* is "the soul of wit". "Soul of X" means the essence of X.

"With" and "for" don't fit the idiom. "Off" is a spelling trap for "of".

Break down means to lose emotional control, such as starting to cry. That fits a speech interrupted midway.

• "Broke apart" is used for objects or groups.
• "Broke of" and "broke for" are not phrasal verbs with a sensible meaning here.

A doctor administers a course of medicine, and antibiotics combat (fight) a disease. Both words fit the medical context.

• A: you relieve symptoms or pain, not a disease.
• B: "fake out" makes no sense here.
• C: "prescribed" could work, but you don't "prove" a disease.

The blank needs an emphatic pronoun that refers back to "fear", a thing, so itself is correct. This is F. D. Roosevelt's well-known line.

Ourselves, myself and yourself refer to people and don't agree with "fear".

To sift evidence or information is to go through it closely and sort out what matters, which means to examine it.

Calamity (disaster), worthiness and problematic are unrelated. Two of them aren't even verbs.

The standard pattern is prefer X to Y.

• "Then" is a spelling confusion with "than", and "prefer" doesn't take "than" anyway.
• "Prefer … over" is common in casual speech but is not the accepted formal idiom tested here.
• "Against" is wrong.

The active verb is simple past ("compelled"), so the passive is was + past participle: "The enemy was compelled to surrender by us."

• A: changes the tense to present ("is compelled").
• B: is still active, and in present tense.
• C: "was compelling" is active continuous, not passive.

"Best" is already a superlative, so D is the clean, correct form.

• B ("more best") and C ("more better") are double comparisons, which are clearly wrong.
• A ("the very best") is the tempting option. "Very" used as an intensifier before a superlative is standard, idiomatic English (e.g. "the very best"), so A is also grammatical.

The examiners intend D, the plain superlative.

A negative infinitive puts "not" before "to": order someone not to do something.

• A splits the infinitive ("to not leave"), which traditional grammar treats as incorrect.
• B ("for not leave") is ungrammatical.
• D uses "order" without the past-tense ending.

Occasional has double c and single s.

• "Ocasional" is missing a c.
• "Occasionaly" should be "occasionally", with double l.
• "Gratitus" is not a word; the intended word is "gratuitous".

The idiom is time off work, meaning leave from work. Part B wrongly uses "of" instead of "off".

Parts A and C are correct.

The active verb is past continuous ("were helping"), so the passive is was/were being + past participle: "was being helped".

• D ("was helped") changes the tense to simple past.
• C changes it to present.
• A reverses the meaning.

Part A has the error. When a whole species is meant, English uses the ("The white rhinoceros is close to extinction"), not "a". The word is also misspelt: it should be "rhinoceros".

Parts B and C ("is close to extinction") are correct.

"Although/though" already signals contrast, so it must not be paired with "but". D does this correctly.

• A uses both "though" and "but", a double conjunction.
• B and C are awkward or ungrammatical and change the meaning.

• Simile: the beloved is compared to a rose using like.
• Alliteration: the repeated r sound in "red red rose".

A metaphor would state the comparison directly without "like" or "as", so options with metaphor fail. There is no exaggeration here, so it isn't hyperbole.

After "are", the blank needs a passive participle describing people's condition, so enslaved fits. It also matches the image of a "prison". The sentence paraphrases a well-known line by Alphonse Daudet.

• "Enslave" is the base verb and can't follow "are".
• "Unslaved" is not a word.
• "Prisoners" is grammatical but repeats "prison" awkwardly and loses the idea of a people losing their freedom.

As written the sentence has no main verb, because "entitling" is only a participle. Part B should read entitled her to: "The admission ticket entitled her to win a prize."

Parts A and C are fine.

People are grateful to those who made something possible. This is the natural phrasing of a book's acknowledgements.

• A: "plausible" means believable, which doesn't describe a book being made.
• B: "feasible" describes plans, and "honoured to" someone is unidiomatic.
• D: "flexible" makes no sense here.

Ebullient means cheerful and full of energy or enthusiasm, which is closest to energetic.

• Indolent and lethargic are opposites.
• Athletic refers to physical fitness, not temperament.

For a question, "said to" becomes asked, and the wh-word stays as the link word without "that". The clause takes statement order, and the tense shifts back ("are doing" becomes "was doing"). "You" becomes "I" because it refers to the listener, "me".

• A uses the wrong pronoun ("he").
• B adds "that" before a wh-question.
• D changes "what" to "why".

Neither of + plural noun takes a singular verb in formal grammar, because it means "not either one". So "is" is correct.

"Are" and "were" are plural, and "am" is only used with "I".

After the auxiliary did, the verb takes its base form: "No sooner did the chairman arrive than …".

• "Had arrived" needs the pattern "No sooner had the chairman arrived".
• "Arrived" can't follow "did".
• "Will arrive" clashes with the past tense of "put up".

• Q → P: "when I decided to go abroad" is completed by "for five years". P then goes on to the main clause, "I was sure".
• P → R: "I was sure" needs its that-clause: "that my grandmother would be upset".
• R → S: "and at her age one could never tell" adds the reason for the worry, so it comes last.

The order is QPRS.

The passage takes mutual co-operation and sacrifice for the common good as the test of civilization, which amounts to developing qualities of mutual help.

Wealth, power and education are never offered as the measure. The passage even suggests that humans, despite their advances, lag behind bees and ants on this test.

The passage says bees and ants have learnt co-operation and sacrifice for the common good better than man. By that test they are superior, so the reason is their instinct for co-operation.

Physical strength and a carefree life are never mentioned. "Rude and blind behaviour" would count against them, not for them.

The passage asks whether humans have really progressed and measures that progress by moral qualities: co-operation, sacrifice and serving a larger cause. The closing saying ends with sacrifice "for the soul". So the theme is man's moral and spiritual development.

• A: the passage is not about empires.
• B and D: truth and idealism are too narrow or not discussed.

The passage links exploitation to selfishness and a lack of co-operation, which is a moral failing. Among the options, moral turpitude (moral corruption) fits best.

• Lack of education is never mentioned.
• High ethics would prevent exploitation, not cause it.
• "Weakness of some people" blames the victims, which the passage does not do.

An exclamation becomes exclaimed that. "How wise" turns into "very wise", "I" becomes "he" (the speaker), and "am" shifts back to "was".

• A and C keep "I", and C also keeps the present tense.
• D changes the feeling to "regretted".

Dissimulate means to hide one's true feelings or intentions, or to pretend. Its opposite is reveal.

• "Concealed" is close to a synonym, a trap.
• "Circulate" merely sounds similar.
• "Concrete" is unrelated.

After the auxiliary did (not), the main verb takes its base form: "didn't like".

• A ("didn't liked") marks the past tense twice.
• B ("hadn't like") needs the participle "liked".
• C ("He haven't") breaks subject–verb agreement.

The present perfect needs the past participle: "As I haven't seen all the evidence". Part A has the error.

Parts B and C are correct.

Jocund means cheerful and light-hearted, so its antonym is gloomy.

• "Cheerful" is a synonym, a trap.
• "Optimistic" is close in spirit.
• "Jealous" is unrelated.

• S opens the sentence and needs a clause after "that".
• Q follows: "we would be held".
• P completes the phrase "held responsible for anything".
• R describes "anything" as "that went wrong".

The sentence reads: *We were told that we would be held responsible for anything that went wrong.* So the order is SQPR.