Step 1: Start from S1 and look for a definition-expansion.

S1 gives a general statement about a library. A natural continuation is to specify \emph{what it contains. Sentence Q does exactly this by listing “books, journals, and digital resources.”
\(\Rightarrow\) \underline{Q should come immediately after S1.


Step 2: Resolve the anaphora “these resources.”

Sentence S mentions “\emph{these resources,” which must refer to something introduced earlier. Only Q introduces resources explicitly, so S must follow Q.
\(\Rightarrow\) partial order: S1 \,\(\to\)\, Q \,\(\to\)\, \underline{S.


Step 3: Bring in the users, then the effect on them.

After resources and their management (Q, S), the next logical idea is \emph{who uses them. Sentence P introduces “Students and researchers … visit to find useful information,” which connects naturally to the resources.

Sentence R talks about “visitors” focusing due to the peaceful environment—this refers back to the visitors in P, so R must follow P.
\(\Rightarrow\) order of the middle four becomes: Q \,\(\to\)\, S \,\(\to\)\, P \,\(\to\)\, R.


Step 4: Conclude with S6.

With contents (Q), management (S), users (P), and environment (R) established, S6’s “Thus” provides a coherent summary.



\boxed{\text{Proper sequence of the four sentences: QSPR. Quick Tip: For para-jumbles, track pronoun references (e.g., “these resources”) and noun continuity (e.g., “visitors”) to lock relative positions before checking global coherence.

The idiom “to burn the midnight oil” originated from the days when people used oil lamps for light. Working or studying until late night meant literally burning oil.
\(\Rightarrow\) Therefore, it refers to working hard late into the night, especially on important tasks or studies.


\boxed{\text{To burn the midnight oil = To work late into the night Quick Tip: Think of “midnight oil” as lamp oil burned during late-night study or work sessions. The literal image helps recall the figurative meaning.

The phrase “hit the nail on the head” comes from carpentry: hitting the nail at its head is the correct and precise action. Figuratively, it means to describe or do something with perfect accuracy.
\(\Rightarrow\) Hence, the idiom means “to be very precise or accurate.”


\boxed{\text{Hit the nail on the head = To be very precise or accurate Quick Tip: For idioms, imagine the literal scenario—here, hitting a nail at the head is exact and correct, which translates into precision in meaning.

Step 1: Meaning of "Ardent"

The word ardent means intensely passionate, eager, or enthusiastic about something. It conveys strong emotional involvement and zeal.


Step 2: Check each option

- (a) Indifferent: means having no particular interest or concern; apathetic. This is the opposite of passionate. ✔

- (b) Zealous: means full of zeal, strongly devoted, or enthusiastic. This is actually a synonym, not an opposite. ✘

- (c) Fervent: means showing passionate intensity, again a synonym of ardent. ✘

- (d) Enthusiastic: means excited or keen interest. This is again a synonym. ✘


Step 3: Conclusion

The only word that is the exact opposite of Ardent is Indifferent.

\[ \boxed{Answer = (a) Indifferent} \] Quick Tip: When dealing with synonym–antonym questions, first identify whether the given word carries a \emph{positive intensity} (like ardent) or a \emph{negative/neutral lack of intensity} (like indifferent). Then eliminate the synonyms to focus on the true opposite.

The original sentence is in Direct Speech with the present perfect tense: “I have completed…”.
When converting to Indirect Speech, the tense usually shifts one step back:
- Present Perfect (\textit{have completed) \(\Rightarrow\) Past Perfect (\textit{had completed).
Also, the pronoun “I” changes to “she” (since the speaker is “she”).

Thus, the correct transformation is: \(\Rightarrow\) “She said that she had completed her assignment on time.”


\boxed{\text{Correct Indirect Speech: She said that she had completed her assignment on time. Quick Tip: In reported speech, remember to shift the tense back: Present \(\Rightarrow\) Past, Present Perfect \(\Rightarrow\) Past Perfect, etc., unless the sentence expresses a universal truth.

The given sentence is in Indirect Speech. It expresses a universal truth: “The Earth revolves around the Sun.”
Rule: Universal truths and facts remain in the present tense, even when reported.

Therefore, the correct direct form remains: \(\Rightarrow\) “The teacher said, ‘The Earth revolves around the Sun.’”


\boxed{\text{Correct Direct Speech: The teacher said, “The Earth revolves around the Sun.” Quick Tip: Universal truths and scientific facts always remain in the present tense in reported speech, regardless of the reporting verb’s tense.

The phrase “Even though she was unwell” implies a contrast is coming. Instead of stopping or avoiding work, the logical continuation is that she showed persistence and determination. \(\Rightarrow\) Hence, option (a) is the most meaningful completion.


\boxed{\text{Even though she was unwell, she continued working with determination. Quick Tip: When a sentence begins with “Even though…,” look for a contrasting but positive action in the completion.

When a fire alarm rings, the natural and logical action is evacuation. Other options either contradict the urgency (a, d) or assume unrealistic immediate response (c). \(\Rightarrow\) Thus, option (b) makes the most sense.


\boxed{\text{The moment the fire alarm rang, people evacuated the building immediately. Quick Tip: Always choose the option that reflects logical, real-life action in emergencies for such sentence completions.

The clause “If we don’t leave now” indicates a warning of negative consequence. Among the given options, missing the train is the logical and direct outcome. Other options contradict the urgency. \(\Rightarrow\) Therefore, option (b) is correct.


\boxed{\text{If we don’t leave now, we might miss the train. Quick Tip: In conditional sentences starting with “If we don’t…,” usually the result highlights a loss, risk, or negative outcome.

Step 1: Meaning of "Benevolent"

The word benevolent means well-meaning, kind, charitable, or showing goodwill towards others. It implies helpfulness and compassion.


Step 2: Check each option

- (a) Kind: directly matches the meaning of benevolent. ✔

- (b) Cruel: opposite meaning, not correct. ✘

- (c) Selfish: opposite of being generous, not correct. ✘

- (d) Hostile: unfriendly, again the opposite, not correct. ✘


Step 3: Conclusion

Thus, the synonym of benevolent is Kind.

\[ \boxed{Answer = (a) Kind} \] Quick Tip: Remember: “Benevolent” = “Good + Willing.” Break it into Latin roots: \emph{bene} (good) + \emph{volens} (wishing).

Step 1: Meaning of the word

An artifact is an object made by a human being, often of cultural or historical interest.


Step 2: Check each option

- (a) Artifact: Correct spelling. ✔

- (b) Artifect: Incorrect spelling. ✘

- (c) Arttefect: Extra “t” makes it wrong. ✘

- (d) Arrtefact: Extra “r” makes it wrong. ✘


Step 3: Conclusion

The correctly spelt word is Artifact.

\[ \boxed{Answer = (a) Artifact} \] Quick Tip: Spelling questions often insert double letters or replace vowels. Focus on the root word (\emph{arti} + \emph{fact} = “something made with skill”).

The sentence highlights that the act was one of revenge.
- “Vindictive” means having a strong desire for revenge. \(\Rightarrow\) Fits perfectly.

- “Malicious” means intending to harm, but not necessarily for revenge.

- “Spiteful” is similar, but weaker and less formal than “vindictive.”

- “Callous” means emotionally insensitive, which doesn’t imply revenge.


Thus, the most appropriate word is vindictive.


\boxed{\text{Answer: Vindictive Quick Tip: When the clue word is “revenge,” the closest synonym is always “vindictive.”

The context shows persistence and firmness of purpose despite repeated failures.
- “Resolute” means firmly determined. \(\Rightarrow\) Fits perfectly.

- “Obstinate” means stubborn, often in a negative sense.

- “Relentless” means not giving up, but is usually harsher and doesn’t fit as smoothly.

- “Unwavering” is close, but “resolute” is the best single-word fit here.


Thus, the correct answer is resolute.


\boxed{\text{Answer: Resolute Quick Tip: Look for positive determination in contexts of repeated failures—“resolute” captures firmness with a positive tone.

Step 1: Identify the relation in the stem

A florist is a professional who deals in/sells works with flowers.

Relation type: \emph{profession/person \(\Rightarrow\) \emph{thing they handle or provide.


Step 2: Test each option

(a) Chef : Food — A chef’s profession involves preparing/providing food. \(\Rightarrow\) Matches.

(b) Author : Books — An author creates books (creator \(\Rightarrow\) product), not primarily “deals in/sells” them. Relation differs.

(c) Painter : Canvas — A painter uses a canvas as a medium; canvas is not the good provided. Relation differs.

(d) None of these — Not applicable since (a) fits well.


Step 3: Conclusion

Option (a) mirrors the \emph{profession \(\Rightarrow\) item handled/provided relation most closely.

\[ \boxed{Answer = (a) Chef : Food} \] Quick Tip: For analogies, first \textbf{name the relation} (e.g., profession \(\Rightarrow\) item provided, creator \(\Rightarrow\) creation, tool \(\Rightarrow\) user). Then eliminate choices that shift to a different relation.

Step 1: Interpret the stem

A sentinel is one who keeps watch/guards. Relation: \emph{agent (person) \(\Rightarrow\) core duty/purpose (verb).


Step 2: Evaluate options

(a) Artist : Paint — Agent \(\Rightarrow\) action, but “paint” is the \emph{medium/action of creating art; not a \emph{core protective duty like “watch.” Partially similar but weaker fit.

(b) Soldier : Battle — Agent \(\Rightarrow\) event/occasion; “battle” is not the soldier’s \emph{constant duty but an occurrence. Relation drifts from purpose.

(c) Guard : Protect — Agent \(\Rightarrow\) \emph{core duty/purpose. This exactly parallels sentinel : watch. \(\Rightarrow\) Best fit.

(d) None of these — Not applicable since (c) fits perfectly.


Step 3: Conclusion

Option (c) mirrors the \emph{agent \(\Rightarrow\) essential duty relation exactly.

\[ \boxed{Answer = (c) Guard : Protect} \] Quick Tip: When two nouns appear (e.g., sentinel : watch), check if the second is a \textbf{duty/purpose}. Prefer choices where the second word states the agent’s \textbf{function}, not a tool or an occasional event.

Step 1: Use formula for SI for 2 years. \[ SI = \frac{P \times R \times T}{100} \]
Here, \( SI = 14400, \, T = 2 \). \[ 14400 = \frac{P \times R \times 2}{100} \] \[ 14400 = \frac{2PR}{100} \quad \(\Rightarrow\) \quad 14400 = \frac{PR}{50} \] \[ PR = 14400 \times 50 = 720000 \]

Step 2: Use formula for CI for 2 years. \[ CI = P \left( \left(1+\frac{R}{100}\right)^2 - 1 \right) \] \[ 16560 = P \left( \frac{R}{100} + \frac{R^2}{10000} \right) \times 2 \]
But more directly: \[ CI - SI = \frac{P \times (R/100)^2 \times 2}{2} \]
Actually formula: For 2 years, \[ CI - SI = \frac{P \times R^2}{100^2} \]

Step 3: Calculate CI - SI. \[ CI - SI = 16560 - 14400 = 2160 \]
So, \[ 2160 = \frac{P \times R^2}{100^2} \] \[ P \times R^2 = 2160 \times 10000 = 21600000 \]

Step 4: Divide two relations.
From Step 1: \( PR = 720000 \).
From Step 3: \( PR^2 = 21600000 \).
\[ \frac{PR^2}{PR} = \frac{21600000}{720000} \] \[ R = 30% \]

Step 5: Find principal. \[ PR = 720000 \quad \(\Rightarrow\) \quad P \times 30 = 720000 \] \[ P = \frac{720000}{30} = 24000 \]


\boxed{Principal = Rs. 24,000, Rate = 30% per annum Quick Tip: For 2 years, use the relation \( CI - SI = \frac{P \times R^2{100^2} \). This shortcut saves time compared to expanding the full CI formula.

Step 1: Convert ratios to algebra.

Let \(x-100=1k,\; y-25=2k \(\Rightarrow\) x=k+100,\; y=2k+25.\)

From \((y-25):(z-50)=5:6\), let \(y-25=5m,\; z-50=6m\). But \(y-25=2k\), so \(2k=5m \(\Rightarrow\) m=\frac{2k{5\). Hence \(z=50+6m=50+\frac{12k}{5}\).


Step 2: Use \(x+y+z=850\).
\((k+100)+(2k+25)+\left(50+\frac{12k}{5}\right)=850\)
\(\Rightarrow \frac{27k}{5}+175=850 \(\Rightarrow\) \frac{27k{5=675 \(\Rightarrow\) 27k=3375 \(\Rightarrow\) k=125.\)


Step 3: Find \(x\) and \(y\) and sum.
\(x=125+100=225,\; y=2\cdot125+25=275 \(\Rightarrow\) x+y=225+275=500.\)



\boxed{\text{Original x+y=500 Quick Tip: When ratios involve shifted values (like \(x-100\)), set them equal to \(k\)-multiples, back-substitute in the sum, and solve.

\(\log 87600=\log(8.76\times10^4)=\log 8.76+4\);
\(\log 23100=\log(2.31\times10^4)=\log 2.31+4\).

So, \(\log 87600+\log 23100-8=(\log 8.76+4)+(\log 2.31+4)-8
\Rightarrow \log 8.76+\log 2.31.\)


\boxed{\log 87600+\log 23100-8=\log 8.76+\log 2.31 Quick Tip: Write numbers in scientific form \(a\times10^n\) and use \(\log(ab)=\log a+\log b\) to cancel powers of \(10\) quickly.

Step 1: Compute required totals.

Total for \(12\) months \(=12\timesRs. 3200=Rs. 38400.\)

Total Jan--Jun \(=6\timesRs. 3200=Rs. 19200.\)

Total Jul--Oct \(=Rs.(3000+3600+3900+4200)=Rs. 14700.\)

Spent till Oct \(=Rs.(19200+14700)=Rs. 33900.\)


Step 2: Amount left for Nov \& Dec.

Remaining \(=Rs.(38400-33900)=Rs. 4500.\)
Let Nov \(=N\) and Dec \(=D\) with \(N=\tfrac{1}{2}D \(\Rightarrow\) D=2N.\)

Then \(N+D=3N=Rs. 4500 \(\Rightarrow\) N=Rs. 1500.\)



\boxed{November’s expenditure =\ Rs. 1500 Quick Tip: With averages across periods, convert to totals first; then apply given ratios (here, \(N=\tfrac{1{2}D\)) to split the remainder.

Step 1: Model the situation

Let the observers be \(A\) and \(B\) with \(AB=10\) km, and let the balloon be vertically above point \(P\) on the same line. Suppose \(\angle APB=30^\circ\) at the farther observer and \(\angle BP A=60^\circ\) at the nearer observer (larger angle \(\Rightarrow\) nearer).

Let \(AP=d\) so \(BP=d-10\), and let the height be \(h\).


Step 2: Use \(\tan\theta=\dfrac{opposite}{adjacent}\)

From \(A\): \(\tan 30^\circ=\dfrac{h}{d}\Rightarrow h=\dfrac{d}{\sqrt{3}}\).

From \(B\): \(\tan 60^\circ=\dfrac{h}{d-10}\Rightarrow h=(d-10)\sqrt{3}\).


Step 3: Equate the two expressions for \(h\)
\(\dfrac{d}{\sqrt{3}}=(d-10)\sqrt{3}\ \Rightarrow\ d=3(d-10)\ \Rightarrow\ 2d=30\ \Rightarrow\ d=15\).

Hence \(h=\dfrac{15}{\sqrt{3}}=5\sqrt{3}\ km\).

\[ \boxed{h=5\sqrt{3}\ km} \] Quick Tip: In height–distance problems with two angles from points on a line, the \textbf{larger angle corresponds to the nearer observer}. Set up distances accordingly and use \(\tan\theta\).

Step 1: Compute total outlay

Cost per article \(=\) 120 paisa \(=\) \(Rs. 1.20\).

Total cost \(C = 1500\times 1.20=Rs. 1800\).


Step 2: Fix the marked selling price per article

Price chosen so that selling only 1200 items gives \(80%\) profit on \emph{total outlay.

So required revenue \(= C + 0.8C = 1.8C = 1.8\times 1800=Rs. 3240\).

Hence price per article \(p=\dfrac{3240}{1200}=Rs. 2.70\).


Step 3: Actual sale and profit

Spoilt \(=240\Rightarrow\) sold \(=1500-240=1260\) articles.

Actual revenue \(R=1260\times 2.70=Rs. 3402\).

Profit \(=R-C=3402-1800=Rs. 1602\).

Profit % on outlay \(=\dfrac{1602}{1800}\times 100=89%\).

\[ \boxed{Actual Profit =89%} \] Quick Tip: When a price is set for a target profit on \emph{total outlay}, compute the per-unit price from that target revenue first, then apply real quantities sold.

Observation: Focusing on the right block that produces \(102\): it is formed from the numbers just below it — \(8\) (to its immediate left in the middle row), and \(6\) and \(4\) in the bottom row — together with the missing value \(x\) in the middle row of that block.

Rule (consistent with this block):
\[ 102 \;=\; 8^2 \;+\; (6\times 4) \;+\; x \] \(\Rightarrow\ 102 \;=\; 64 \;+\; 24 \;+\; x \ \Rightarrow\ x=102-88=14.\)

\[ \boxed{x=14} \] Quick Tip: In grid puzzles, totals often combine a \textbf{square term} with a \textbf{product term}. Try decomposing the target into recognizable chunks (e.g., \(a^2+b\times c\)) and solve for the unknown.

Non-green marbles \(=3+3=6\); Total marbles \(=3+3+7=13\).

Required probability \(=\dfrac{\binom{6}{2}}{\binom{13}{2}}=\dfrac{15}{78}=\dfrac{5}{26}\).
\(\Rightarrow\) Both chosen are from the 6 non-green marbles.


\boxed{Probability=\dfrac{5{26 Quick Tip: When picking without replacement, use combinations: favourable \(\binom{\text{wanted}{2}\) over total \(\binom{all}{2}\).

Amount \(=P(1+0.06)(1+0.07)(1+0.08)(1+0.09)\).

So, \(A=45000\times1.06\times1.07\times1.08\times1.09 \approx 45000\times1.335 \approx Rs. 6.01\times10^4\).
\(\Rightarrow\) Approximately Rs. 60000.


\boxed{\text{Amount \approx Rs. 60000 Quick Tip: For varying yearly rates, multiply sequential growth factors \((1+r_i)\); rounding the final factor gives a quick estimate.

Assume cost price (CP) per kg \(=Rs. 100\).

Marked price \(=100\times1.30=Rs. 130\). After 15% discount, billed SP per “kg” \(=130\times0.85=Rs. 110.5\).

But he supplies only \(0.8\) kg, whose cost to him \(=0.8\times100=Rs. 80\).

Profit \(=Rs.(110.5-80)=Rs. 30.5\).

Profit % \(=\dfrac{30.5}{80}\times100=38.125%\). \(\Rightarrow\) Includes hidden gain from short-weight (\(1/0.8=1.25\), i.e., \(25%\) extra).


\boxed{\text{Actual Profit =38.125% Quick Tip: For false weights, compute SP on the billed quantity but CP on the actual quantity delivered.

Step 1: Height of the trapezium

Since \(\angle ABC=90^\circ\) and \(AB\parallel CD\), \(BC\) is perpendicular to both bases \(\Rightarrow\) height \(h=BC=4\ \mathrm{cm}\).


Step 2: Find the other base \(AB\)

Place \(A(0,0)\), let \(AB=x\) so \(B(x,0)\). Since \(CD\parallel AB\) and \(BC=4\), take \(C(x,-4)\) and \(D(x-5,-4)\) (because \(CD=5\)).

Vector \(\overrightarrow{AD}=(x-5,-4)\). Given \(\angle BAD=45^\circ\), slope magnitude of \(AD\) is \(|-4/(x-5)|=\tan 45^\circ=1 \(\Rightarrow\) |x-5|=4\).

Choose \(x-5=4\) (keeps vertices in order) \(\Rightarrow x=9 \(\Rightarrow\) AB=9\ \mathrm{cm\).


Step 3: Area
\(\displaystyle Area=\frac{(AB+CD)}{2}\times h=\frac{(9+5)}{2}\times 4=7\times 4=28\ \mathrm{cm}^2.\)

\[ \boxed{28\ \mathrm{cm}^2} \] Quick Tip: When one leg is perpendicular to the bases in a trapezium, that leg is the \textbf{height}. Use coordinates or projections with the given angle to recover the unknown base.

Step 1: Let the numbers be in \(3:2\)

Let selected \(=3k\), unselected \(=2k\) \(\Rightarrow\) applicants \(A=5k\).


Step 2: Apply the hypothetical change

Applied \(A-100\), selected \(3k-50\). Then unselected becomes \((A-100)-(3k-50)=(2k-50)\).

Given ratio \((3k-50):(2k-50)=7:4\).


Step 3: Solve for \(k\)
\(\displaystyle \frac{3k-50}{2k-50}=\frac{7}{4}\Rightarrow 4(3k-50)=7(2k-50)\Rightarrow 12k-200=14k-350\Rightarrow 2k=150\Rightarrow k=75.\)

Hence \(A=5k=375\).

\[ \boxed{375} \] Quick Tip: Translate ratio statements into variables first (\(3k,2k\)). For “if less/more” scenarios, adjust both selected and unselected consistently before forming the new ratio.

Step 1: Compute LCM
\(21=3\cdot 7,\ 24=2^3\cdot 3,\ 28=2^2\cdot 7 \(\Rightarrow\) LCM=2^3\cdot 3\cdot 7=168.\)


Step 2: Find the first multiple in \([2000,3000]\)
\(168\times 11=1848<2000\), \(168\times 12=2016\) \(\Rightarrow\) smallest required number \(=2016\).

\[ \boxed{2016 \] Quick Tip: For “exactly divisible by several numbers,” take the \textbf{LCM} and scan multiples within the interval. Multiplying once more than the floor of \(\frac{lower bound}{LCM}\) gives the first valid multiple.

Step 1: Check the pattern of growth.

Observe each term:
\(5 \to 6\): \(5 \times 1 + 1 = 6\)
\(6 \to 16\): \(6 \times 2 + 4 = 16\)
\(16 \to 57\): \(16 \times 3 + 9 = 57\)
\(57 \to 244\): \(57 \times 4 + 16 = 244\)
\(244 \to 1245\): \(244 \times 5 + 25 = 1245\)


Step 2: Generalize the rule.

Next term \(=\) (Previous term \(\times n\)) \(+ n^2\), where \(n\) is the step number.


Step 3: Apply for the next term.

Here \(n=6\):
\(1245 \times 6 + 36 = 7470 + 36 = 7506\)



\boxed{\text{Next term of the series is 7506 Quick Tip: In number series, look for operations of the form “multiply by \(n\) then add \(n^2\)” or similar compound patterns.

Step 1: Largest possible bag size

Equal bag size must divide each heap \(\Rightarrow\) use \(\gcd(435,493,551)\).
\(\gcd(435,493)=\gcd(435,58)=\gcd(58,29)=29\). Also \(551=29\times 19\) \(\Rightarrow\) \(\gcd=29\ kg\).

Step 2: Number of bags

Bags \(=\dfrac{435}{29}+\dfrac{493}{29}+\dfrac{551}{29}=15+17+19=51\).

\[ \boxed{51} \] Quick Tip: “Least number of equal bags’’ \(\Rightarrow\) take the \(\gcd\) as bag size, then sum the quotients.

Let boys \(=B\), girls \(=G\). Then \(B+G=300\) and \(2.4B+2.8G=776\).

Multiply by 10: \(24B+28G=7760\Rightarrow 6B+7G=1940\).

Using \(B=300-G\): \(6(300-G)+7G=1940\Rightarrow 1800-G=1940\Rightarrow G=140\).

\[ \boxed{140} \] Quick Tip: Two linear conditions (headcount and total money) \(\Rightarrow\) solve by substitution after clearing decimals.

\(\sqrt{4^{6x^{2}}}=4^{3x^{2}}\), \quad \(\sqrt{25^{\,y/2}}=25^{y/4}=(5^{2})^{y/4}=5^{y/2}\),
\(\sqrt{9}=3\), \quad \(\sqrt{z^{4}}=z^{2}\) (assuming \(z\ge 0\)).

Multiply: \(4^{3x^{2}}\cdot 5^{y/2}\cdot 3\cdot z^{2}\).

\[ \boxed{4^{3x^{2}}\cdot 5^{y/2}\cdot 3\,z^{2}} \] Quick Tip: \(\sqrt{a^{m}}=a^{m/2}\). For bases like \(25\) and \(9\), rewrite as powers of primes (\(5^2, 3^2\)) before halving the exponent.

Step 1: Extract import values from the bar graph

2018: 30,\ 2019: 45,\ 2020: 40,\ 2021: 35,\ 2022: 50,\ 2023: 45,\ 2024: 48.


Step 2: Compute year-on-year falls

- 2020 vs 2019: fall \(=45-40=5\). Percentage fall \(=\tfrac{5}{45}\times100 \approx 11.1%\).

- 2021 vs 2020: fall \(=40-35=5\). Percentage fall \(=\tfrac{5}{40}\times100=12.5%\).

- 2023 vs 2022: fall \(=50-45=5\). Percentage fall \(=\tfrac{5}{50}\times100=10%\).

- 2024 vs 2023: \emph{increase, not a fall.


Step 3: Conclusion

Among actual falls, the lowest % fall is \(10%\) in 2023. But since 2024 shows an \emph{increase (no fall at all), its effective fall is the lowest (zero).

\[ \boxed{2024} \] Quick Tip: Always check if the data shows a rise instead of a fall. An increase means fall \(=0%\), which is lower than any positive fall percentage.

Step 1: Form equations

Let orange \(=O\), banana \(=B\), apple \(=A\).

Equation (1): \(2O+3B+4A=15\).

Equation (2): \(3O+2B+A=10\).


Step 2: Required expression

We need \(3O+3B+3A\).


Step 3: Manipulate equations

Multiply (2) by 3: \(9O+6B+3A=30\).

Multiply (1) by 3: \(6O+9B+12A=45\).

Subtract second from first: \((9O-6O)+(6B-9B)+(3A-12A)=30-45\).
\(\Rightarrow 3O-3B-9A=-15\ \(\Rightarrow\) O-B-3A=-5\).

This single relation plus the originals allows solving for \(3O+3B+3A\). Add Eqn (1) and Eqn (2): \((2O+3O)+(3B+2B)+(4A+A)=15+10\).
\(\Rightarrow 5O+5B+5A=25 \(\Rightarrow\) O+B+A=5\).

So \(3O+3B+3A=15\).

\[ \boxed{Rs. 15} \] Quick Tip: In linear system word problems, sometimes the exact values are not needed. Combine equations cleverly to form the target expression directly.

Step 1: Assume equal quantities

Let each glass contain \(300\) ml (for easy calculation).


Step 2: First glass (ratio \(2:1\))

Alcohol \(=\frac{2}{3}\times 300=200\) ml,\quad Water \(=100\) ml.


Step 3: Second glass (ratio \(1:1\))

Alcohol \(=\frac{1}{2}\times 300=150\) ml,\quad Water \(=150\) ml.


Step 4: Combine into third glass

Total Alcohol \(=200+150=350\) ml.

Total Water \(=100+150=250\) ml.


Step 5: Simplify ratio

Alcohol : Water \(=350:250=7:5\).

\[ \boxed{7:5} \] Quick Tip: When mixing two solutions of equal volume, simply take the average contribution of each component. Always assume an easy number like 100 or 300 ml for quick calculations.

October has \(31\) days. From 2 Oct to 2 Nov is a shift of \(31\) days.

Day shift \(=31 \bmod 7 = 3\) days ahead.

Monday \(\Rightarrow\) Tuesday (1) \(\Rightarrow\) Wednesday (2) \(\Rightarrow\) Thursday (3).
\(\Rightarrow\) \boxed{\text{Thursday.

Assume M is innocent \(\Rightarrow\) M’s statement is true. Then for “only one true”, both B and V must be false.

B false \(\Rightarrow\) B is involved; V false then says “B is involved” is false — contradiction (two truths).

Hence M cannot be innocent \(\Rightarrow\) M is involved (M’s statement is false).

Now either B is innocent (B true, V false) or B is involved (B false, V true) — both satisfy “only one true”.
\(\Rightarrow\) The definite culprit is \boxed{M.

Elapsed actual time: Mon 3{:00 AM \(\to\) Tue 7{:30 PM \(= 24 + 16.5 = 40.5\) hours.

Gain \(= 2\) min/hour \(\Rightarrow 40.5\times 2 = 81\) minutes \(= 1\) h \(21\) m.

Shown time \(=\) actual time \(+\) gain \(= 7{:}30 PM + 1{:}21 = 8{:}51 PM\).
\(\Rightarrow\) \boxed{8{:51\ \text{PM.

Interpret the first two symbols as letter–number: L1, J2, _, F4, D5 (letters go \(-2\) each step: L, J, H, F, D; numbers go \(+1\)).

After “\#” the same digit repeats; the next letter increases by \(+4\) more each step from the first letter:
\(L\to O (+3)\), \(J\to Q (+7)\), so next must be \(H\to S (+11)\), then \(F\to U (+15)\), \(D\to W (+19)\).

Final part “\(\sim\)” repeats the second digit incremented by \(1\).

Therefore the missing code should be H3\#3S4\(\sim\)4, which is not among the options.
\(\Rightarrow\) \boxed{\text{None of these.

“Father of my father” \(\Rightarrow\) A’s grandfather.

“Grandson of A’s grandfather” \(\Rightarrow\) A or A’s brother (male of A’s generation in that lineage).

“Wife of that grandson” \(\Rightarrow\) either A’s own wife or his brother’s wife.

Among standard relations listed, the certain relation w.r.t. A is a brother’s wife \(\Rightarrow\) \boxed{\text{Sister-in-law.

Step 1: Decode the relation.

“Narmada Bachao Andolan” is associated with its leader \(\Rightarrow\) Medha Patkar.

Step 2: Apply to the second pair.

“Bhudan Andolan” (land-gift movement) was led by \(\Rightarrow\) Vinoba Bhave.
\(\Rightarrow\) The correct pair is (a).



\boxed{Answer: Vinoba Bhave Quick Tip: In analogy questions on movements, map \emph{movement : leader. Remember key pairs like Bhudan–Vinoba Bhave, Chipko–Sunderlal Bahuguna, etc.

Step 1: Check differences.
\(26-14=12,\ 36-26=10,\ 44-36=8,\ 50-44=6\)

Step 2: Spot the pattern.

Differences decrease by \(2\): \(12,10,8,6,4,2\).

Step 3: Fill the missing term.

Next number \(=50+4=54\); then \(54+2=56\) \(\Rightarrow\) consistent.



\boxed{\text{Missing term = 54 Quick Tip: When terms look irregular, examine first-level differences; if they form an arithmetic pattern, extend it to fill the blank.

Step 1: Assumption 1.

Issuing a notice presumes employees will read it \(\Rightarrow\) implicit.

Step 2: Assumption 2.

Declaring a new compulsory system presumes it helps the company (accuracy, compliance, transparency) \(\Rightarrow\) implicit.
\(\Rightarrow\) Both assumptions are taken for granted.



\boxed{\text{Both 1 and 2 are implicit Quick Tip: For “notice/compulsory” statements, typical hidden assumptions are: (i) people will read/follow it, (ii) the action benefits the issuer’s objective.

Step 1: Separate causes from effects.

Cause: Earthquake (in ocean) \(\Rightarrow\) Tsunami.

Step 2: Effects reported.

Human losses (casualties/deaths) and property loss \(\Rightarrow\) effects.

Step 3: Choose option listing only effects.

Option (c) lists consequences (casualties, property loss, death) without mixing in causes.



\boxed{\text{Effects: casualties and property loss (including deaths) Quick Tip: In cause–effect questions, remove the initiating events; keep only the outcome phrases like casualties, damage, losses.

Check I: From “Some \(W\) are \(M\)” we only know \(W\cap M\neq \varnothing\). This does \emph{not imply \(M\subseteq W\). So I does \emph{not logically follow.

Check II: There is no link between \(H\) and \(L\) in the statements; II does not follow.

Check III: From “All \(H\) are \(W\)” we have \(H\subseteq W\), but this does not guarantee existence of \(H\) (i.e., “Some \(W\) are \(H\)”). Without existential import, III doesn’t necessarily follow.
\(\Rightarrow\) Under standard syllogism rules, none of I/II/III follows, so (a) would be logically correct. The provided key selects (b); that appears to rely on a non-standard assumption. Quick Tip: “Some” claims existence; “All \(A\) are \(B\)” does not guarantee that \(A\) exists. Be careful not to convert or add existence where it isn’t stated.

Minute hand at 55 min \(\Rightarrow 55\times 6=330^\circ\).

Hour hand at \(5+\frac{55}{60}\) hours \(\Rightarrow (5+\tfrac{55}{60})\times 30=177.5^\circ\).

Angle \(=|330-177.5|=152.5^\circ\).

\[ \boxed{152.5^\circ} \] Quick Tip: Hour hand moves \(0.5^\circ\) per minute. Use: angle \(=\lvert 30H-5.5M\rvert\).

Days between 8 Jul 2007 and 9 Dec 2007 \(=\) 154 \(=\) \(22\) weeks \(\Rightarrow\) same weekday.

Since 9 Dec 2007 is Sunday, 8 Jul 2007 is also Sunday.
\(\Rightarrow\) None of the given options match; the official key’s “Thursday” conflicts with the calendar. Quick Tip: If the day gap is a multiple of 7, the weekday is unchanged.

Two common interpretations:

(A) Equal \emph{interval} between chimes. At 6 o’clock there are 6 chimes \(\Rightarrow\) 5 intervals \(=10\) s \(\Rightarrow\) interval \(=2\) s. For 10 o’clock: 9 intervals \(\Rightarrow 18\) s. (Not listed.)

(B) Equal time per \emph{chime}. Time per chime \(=\frac{10}{6}\) s \(\Rightarrow\) for 10 chimes \(=\frac{10}{6}\times 10=16.\overline{6}\) s \(\approx 16.7\) s.

Most exam keys assume model (B), giving option (c). Quick Tip: Read wording carefully: “regular time intervals” usually means equal gaps between chimes (model A). Some keys, however, treat it as equal time per chime.

Since Roma is the daughter of Mr. \& Mrs. Mohan and Roma is Sumit’s wife,
\(\Rightarrow\) Mrs. Mohan is the mother of Sumit’s wife.
\(\Rightarrow\) \boxed{\text{Mrs. Mohan is Sumit’s mother-in-law.
Quick Tip: When X marries Y (daughter of A \& B), then A is father-in-law of X and B is mother-in-law of X. Track via “of” links carefully.

Surname in the passage follows the paternal line: Adhir Mishra \(\Rightarrow\) his son Sumit Mishra.

Sohan is Sumit’s son \(\Rightarrow\) Sohan’s surname is also Mishra.
\(\Rightarrow\) \boxed{\text{Sohan Mishra
Quick Tip: Unless stated otherwise, competitive questions assume children carry the father’s surname. Verify lineage before concluding.

Assume X at \((0,0)\) and first move is east.

1) X \(\to\) Y: \(80\) m east \(\Rightarrow\) position \((80,0)\).

2) Turn right (towards south) \(50\) m \(\Rightarrow\) \((80,-50)\).

3) Turn right (towards west) \(70\) m \(\Rightarrow\) \((10,-50)\).

4) Turn right (towards north) \(50\) m \(\Rightarrow\) \((10,0)\).

Distance from start \(=\) distance between \((0,0)\) and \((10,0)\) \(=\) \(\boxed{10\ metres}\).
Quick Tip: Fix an axis (east as +x, north as +y) and track coordinates with each turn. Parallel opposite legs often cancel out.

From the given clues:
- Keshav is second in height but younger than Rekha.

- Parul is taller than Megha but younger in age.

- Rekha and Megha are of the same age, but Rekha is the tallest.

- Nikku is taller than Parul and elder to Rekha.


The age ranking in descending order is:
- First: Rekha (tallest, so assumed oldest).

- Second: Keshav (second in height, younger than Rekha).

- Third: Nikku (elder to Rekha, and taller than Parul).

- Fourth: Parul (younger than Megha, and younger than Keshav).

- Fifth: Megha (youngest).


Hence, \boxed{\text{None of these are in the third position by age. Quick Tip: When arranging based on age, first check the height clues to establish the correct order. Then apply age-related clues.

From the clues:
- Nikku is taller than Parul and elder to Rekha.

- Keshav is second in height, younger than Rekha.

- Rekha is tallest between her and Megha.


The height order is:
- First: Parul (shortest).

- Second: Megha.

- Third: Keshav (second tallest).

- Fourth: Rekha (tallest among those mentioned).

- Fifth: Nikku (taller than Parul).


\boxed{\text{Keshav is in the fourth position based on height. Quick Tip: For arranging in ascending order, track both height and age relationships to place them correctly.

The second, fifth, and eighth letters of "CARETAKER" are: A, T, and E.

Now, check for possible meaningful words using these letters:

- "ATE" (meaning "to consume").

- "EAT" (meaning "to consume food").


Hence, the first letter of the word formed is M since multiple words can be formed.
\(\Rightarrow\) \boxed{\text{Answer: M Quick Tip: When asked about forming meaningful words from specific letters, check if more than one valid word can be formed. If so, answer with M.

- Graduates or teachers but not politicians are represented in the regions where the circle and small triangle intersect but do not overlap with the large triangle (politicians).

- The regions A and E fit this description.
\(\Rightarrow\) \boxed{\text{A and E Quick Tip: Always check for the regions that fall within the desired shapes and exclude overlaps with other conditions.

- Graduate politicians but not members of Parliament are represented in the regions that overlap between the big triangle (politicians) and the circle (graduates) but do not overlap with the rectangle (members of Parliament).

- The regions B and C fit this description.
\(\Rightarrow\) \boxed{\text{B and C Quick Tip: For graduate politicians, look for regions where both the politician triangle and the graduate circle overlap but exclude the Parliament rectangle.

- "All sweet things are fluids" implies that all sweet things belong to the fluid category.

- "Some fluids are coloured things" implies that some but not all fluids belong to the coloured category.

- Combining these, we can conclude:
- Some sweet things are coloured things (B)

- Some sweet things are fluids (C)

- Some fluids are sweet things (D)
\(\Rightarrow\) \boxed{\text{B, C and D only Quick Tip: When analyzing logical implications, follow the inclusion and intersection rules to deduce the valid conclusions.

- The gate is at the center of the plot's side. Lemon and Coconut trees are on either side of the gate.

- The mango tree is not at the corner where the cabin is located, and the well and cabin are at opposite corners.

- Thus, the diagonally opposite trees in the plot are the Lemon tree and the Neem tree.
\(\Rightarrow\) \boxed{\text{Neem tree and Lemon tree Quick Tip: For problems involving placement on a plot, use the corner and side relationships to deduce opposite corners based on restrictions given.

Observe the pattern:

- The first part of the sequence increases by 1 (B, D, H, F...).

- The second part of the sequence (number) increases by 1 each time (2, 3, 5...).

- The last part is a 3-letter sequence that keeps moving alphabetically (CD, E3F, etc.).

So, the next term is: G3HI.

\(\Rightarrow\) \boxed{\text{G3HI Quick Tip: In letter-number sequence problems, analyze each part (letter, number, sequence) separately to identify incremental patterns.

The given letters are: R, A, C, E, T.

- Arrange them in the order 5, 1, 2, 3, 4.

So, we get the word: REACT.

\(\Rightarrow\) \boxed{\text{5, 1, 2, 3, 4 Quick Tip: When asked to form a meaningful word from a set of letters, check each option’s number sequence to see if the letters form a valid word.