SNAP 2011 Question Paper (Available): Download Question Paper with Answer Key And Solutions PDF

Step 1: About the scheme. The Government of India launched the Swayamsidha scheme in 2001. Its main objective was to empower women socially and economically through Self Help Groups (SHGs).


Step 2: Key features.
- Promoted women’s self-reliance and decision-making ability.

- Encouraged microcredit, thrift, and entrepreneurial activity.

- Focused only on women beneficiaries, not on children, health workers, or senior citizens.


Step 3: Elimination.
- School children → covered by schemes like Mid-Day Meal, not Swayamsidha.

- Health workers → supported by NRHM/ASHA, not Swayamsidha.

- Senior citizens → targeted by Old Age Pension, not Swayamsidha.

\[ \boxed{Swayamsidha → Women only} \] Quick Tip: Link women-only schemes: Swayamsidha (2001), STEP (Support to Training and Employment Programme), Beti Bachao Beti Padhao (2015).

Step 1: Definition. Share swap means exchanging shares of one company for shares of another, typically during mergers or acquisitions. The acquiring firm pays shareholders of the target firm with its own shares instead of cash.


Step 2: Practical example. If Company X buys Company Y, it may issue its own shares to Y’s shareholders. For every 10 shares of Y, shareholders may receive 2 shares of X. This saves cash and aligns shareholder interests.


Step 3: Eliminate incorrect options.
- Option B → incorrect, using shares for loans is not called share swap.

- Option C → describes rights issue, not share swap.

- Option D → incorrect, as Option A matches the definition.

\[ \boxed{Share Swap = Business takeover using shares as currency} \] Quick Tip: Remember: Share Swap = Mergers & Acquisitions tool. Rights Issue = new share issue for existing shareholders. Bonus Shares = free shares issued from reserves.

Step 1: Concept. A carbon credit = permit allowing the holder to emit 1 tonne of CO\(_2\) or equivalent gases.


Step 2: Origin. The Kyoto Protocol (1997, effective 2005) introduced mechanisms like Clean Development Mechanism (CDM), Joint Implementation, and Emission Trading.


Step 3: Verification of statements.
- (A) Correct → Carbon credit was part of Kyoto Protocol.

- (B) Correct → Credits awarded for emission reduction below quota.

- (C) Correct → Goal = reduce CO\(_2\) emissions.

- (D) Incorrect → Carbon credits are traded in markets (e.g., EU ETS, voluntary carbon markets). Prices depend on supply-demand, not fixed by UNEP.

\[ \boxed{Statement D is NOT correct} \] Quick Tip: Carbon credits = tradable permits. Kyoto Protocol started it. Market decides price, not UN. Paris Agreement (2015) continues climate action framework.

Step 1: Define FDI.
Foreign Direct Investment (FDI) means long-term capital inflow by foreign companies into specific sectors through ownership, factories, infrastructure, or joint ventures. It brings not only money but also technology and management practices.

Step 2: Define FII.
Foreign Institutional Investors (FIIs) are entities like mutual funds, hedge funds, and pension funds investing in stock markets or bonds. Their investment is portfolio-based and spread across the economy, not tied to a specific industry.

Step 3: Compare.
- FDI → sector-specific, long-term, stable.
- FII → general capital inflow, short-term, volatile.

Step 4: Eliminate options.
- (A) Incorrect: FDI also brings technology, not just capital.
- (C) Incorrect: FIIs mainly trade in secondary markets, not primary.
- (D) Incorrect: FDI is more stable; FII is volatile.
\[ \boxed{FII = general capital inflow, FDI = sector-specific investment} \] Quick Tip: Remember: FDI = factories, long-term. FII = stocks, short-term.

Step 1: Definition.
Brent crude is one of the three main benchmarks for crude oil pricing (others are WTI and Dubai Crude).

Step 2: Origin.
It is sourced from oil fields in the North Sea (between the UK and Norway).

Step 3: Sulphur content.
Brent crude is classified as “light” and “sweet,” but it does contain small amounts of sulphur (<0.37%). Therefore, statement (3) is incorrect because it is not completely sulphur-free.

Step 4: Verify options.
- (1) Correct → major classification.
- (2) Correct → sourced from North Sea.
- (3) Incorrect → does contain some sulphur.
\[ \boxed{Correct statements: 1 and 2 only} \] Quick Tip: Brent = Europe benchmark, WTI = US benchmark, Dubai Crude = Middle East benchmark.

Step 1: About the policy.
India’s Look East Policy was launched in the early 1990s (by PM Narasimha Rao) to strengthen relations with ASEAN and East Asian nations.

Step 2: Objectives.
- To establish India as a key player in Asia–Pacific affairs (Statement 1 correct).
- To restore historical and cultural ties with Southeast Asia and East Asia (Statement 3 correct).
- It was not specifically to plug a Cold War vacuum (Statement 2 incorrect).

Step 3: Elimination.
Options A and C ignore one correct statement each. Option D incorrectly includes statement 2.
\[ \boxed{Correct statements: 1 and 3 only} \] Quick Tip: Look East Policy (1990s) → ASEAN focus. Act East Policy (2014 onwards) → deeper strategic, cultural, and economic ties with East and Southeast Asia.

Step 1: About Oilzapper.
Oilzapper is a consortium of bacteria (bioremediation agents) developed by TERI (The Energy and Resources Institute), India.

Step 2: Function.
It decomposes hydrocarbons in oil sludge and oil spills into harmless CO\textsubscript{2 and water. This makes it an eco-friendly technology to clean oil-contaminated soil and water.

Step 3: Elimination of options.
- (B) Incorrect → Not related to deep-sea oil exploration.
- (C) Incorrect → Not a maize variety.
- (D) Incorrect → Not a flame control technology.
\[ \boxed{Oilzapper = Bioremediation tool for oil sludge and oil spills.} \] Quick Tip: Oilzapper = Bacteria-based eco-friendly method to clean oil spills (developed by TERI, India).

Step 1: Define teaser loans.
Teaser loans are those where banks offer low interest rates initially (for 1–2 years) and later reset them at higher market-linked rates.

Step 2: Economic concern.
Such loans may lead to repayment issues when interest rates rise, similar to the sub-prime crisis in the US (2008). This increases Non-Performing Assets (NPAs).

Step 3: Check statements.
- (1) Correct → They resemble sub-prime lending and pose default risk.
- (2) Incorrect → Teaser loans are not specifically targeted at inexperienced entrepreneurs; they are mostly given in housing loans.
\[ \boxed{Correct statement: 1 only} \] Quick Tip: Teaser loans = low initial interest → later high → risk of NPAs. Mostly in housing loans, not entrepreneurship.

Step 1: Purpose of disinvestment.
The Government of India disinvests equity in CPSEs to raise resources for infrastructure development, reduce fiscal deficit, and occasionally for debt repayment.

Step 2: Evaluate statement 1.
Revenue from disinvestment is indeed used for meeting fiscal needs, including debt repayment. So statement (1) is correct.

Step 3: Evaluate statement 2.
Government generally retains management control of strategic CPSEs, even after partial disinvestment. Thus, statement (2) is incorrect.
\[ \boxed{Correct statement: 1 only} \] Quick Tip: Disinvestment = raising funds by selling CPSE shares. Govt usually retains management control in strategic sectors.

Step 1: About the book.
"Unto This Last" is a famous essay collection written by John Ruskin in 1860, focusing on political economy and moral philosophy.

Step 2: Impact on Gandhiji.
Mahatma Gandhi read the book in 1904, and it deeply influenced his socio-economic philosophy. It inspired him to develop the concept of "Sarvodaya" (welfare of all).

Step 3: Elimination of wrong options.
- (A) Boris Yeltsin → Former Russian president, not an author of this book.
- (B) Pushkin → Russian poet, not related.
- (C) Ruskin Bond → Indian author, unrelated.
Only (D) John Ruskin is correct.
\[ \boxed{Author of "Unto This Last" = John Ruskin} \] Quick Tip: Remember: Gandhiji’s "Sarvodaya" idea was inspired by John Ruskin’s "Unto This Last".

Step 1: Indo-US Civil Nuclear Deal.
India signed the Indo-US Civil Nuclear Agreement in 2008 for peaceful uses of nuclear energy.

Step 2: Agreement with France.
Besides the USA, India also signed a civil nuclear agreement with France in 2008 for cooperation in nuclear energy for peaceful purposes.

Step 3: Elimination.
- Italy and Germany → No such civil nuclear agreement with India at that time.
- Australia → Agreement came later (2014).
Correct answer is France.
\[ \boxed{Country = France} \] Quick Tip: India has nuclear cooperation agreements with multiple countries. The earliest was with USA (2008) and France soon after.

Step 1: About NSG.
The Nuclear Suppliers Group (NSG) is a group of nuclear supplier countries formed in 1974 after India’s nuclear test, to control export of nuclear materials and technology.

Step 2: Members.
- Russia, USA, and Italy are official members.
- Iran is not a member due to concerns over its nuclear program.
\[ \boxed{Non-member = Iran} \] Quick Tip: India is not yet a member of NSG (blocked mainly by China), but it has support from most major powers.

Step 1: Define FCCBs.
FCCB stands for Foreign Currency Convertible Bonds. They are debt instruments issued by Indian companies in foreign currency.

Step 2: How they work.
They are bonds that can be converted into equity shares of the issuing company at a later date, at a pre-determined price.

Step 3: Advantages.
- Helps companies raise capital internationally.
- Offers investors debt security with equity upside.
\[ \boxed{FCCB = Foreign Currency Convertible Bonds} \] Quick Tip: FCCB = Hybrid instrument: debt + option to convert into equity. Popular for global fund raising.

Step 1: Understanding financial inclusion.
Financial inclusion means providing affordable financial services (like banking, credit, insurance, savings) to weaker sections and low-income groups.

Step 2: RBI directive.
RBI directed banks to provide banking outlets in villages with population above 2000, aiming to ensure every household has access to banking.

Step 3: Elimination.
- (B) Priority sector lending → relates to loans for agriculture, MSMEs, etc. Not applicable here.
- (C) Extension of Internet banking → Not specifically mentioned.
- (D) None of these → Wrong because this is directly financial inclusion.
\[ \boxed{Category = Financial Inclusion} \] Quick Tip: Always link such RBI rural banking initiatives with \textbf{Financial Inclusion}.

Step 1: Artificial currency meaning.
Artificial currency = Not issued by a single nation, but created by international institutions as a reserve asset.

Step 2: About SDR.
SDR = Special Drawing Rights.
- Created by the IMF in 1969.
- Used as an international reserve asset.
- Based on a basket of major currencies (USD, EUR, GBP, JPY, CNY).

Step 3: Elimination.
- ADR (American Depository Receipt) → Stock instrument, not currency.
- GDR (Global Depository Receipt) → Equity instrument, not currency.
Thus only SDR qualifies.
\[ \boxed{Artificial Currency = SDR (Special Drawing Rights)} \] Quick Tip: SDR is called \textbf{paper gold}, created by IMF to supplement world reserves.

Step 1: Definition of SLR.
SLR = Statutory Liquidity Ratio. It is the minimum percentage of a commercial bank’s net demand and time liabilities (NDTL) that it must maintain in the form of approved government securities, cash, or gold before offering credit.

Step 2: Connection with Government credit.
By mandating banks to invest in government securities, RBI ensures the government gets assured credit support.

Step 3: Elimination.
- (A) Cash Credit Ratio → Not applicable.
- (B) Debt Service Obligation → Refers to repayment, not lending.
- (C) Liquidity Adjustment Facility → A tool for short-term liquidity, not direct credit.
\[ \boxed{Mechanism = Statutory Liquidity Ratio (SLR)} \] Quick Tip: SLR ensures that banks lend indirectly to the government through compulsory investment in government securities.

Step 1: Background.
In 2011, after nationwide protests led by Anna Hazare, a Joint Drafting Committee was set up for the Lokpal Bill.

Step 2: Members.
- Government side: Pranab Mukherjee, Kapil Sibal, Veerappa Moily, Salman Khurshid, P. Chidambaram.
- Civil society side: Anna Hazare, Shanti Bhushan, Prashant Bhushan, Arvind Kejriwal, Santosh Hegde.

Step 3: Elimination.
- Salman Khurshid → Member (Govt side).
- Anna Hazare → Member (Civil society).
- Arvind Kejriwal → Member (Civil society).
- Kiran Bedi → Was supporter of movement, but not part of the drafting committee.
\[ \boxed{Not in Committee = Kiran Bedi} \] Quick Tip: Kiran Bedi was part of the \textbf{India Against Corruption (IAC)} movement, but not a member of the drafting committee.

Step 1: Constitution and division of powers.
The Seventh Schedule of the Constitution divides subjects into Union List, State List, and Concurrent List.

Step 2: Criminal law and procedure.
- Criminal law (including all matters except for offences against the Armed Forces) and criminal procedure are placed in the Concurrent List.
- This allows both the Parliament and the State Legislatures to legislate on the subject.

Step 3: Elimination.
- State List → Contains policing and law & order, but not criminal procedure.
- Union List → Contains defence, foreign affairs, but not criminal procedure.
Thus, only Concurrent List is correct.
\[ \boxed{List = Concurrent List} \] Quick Tip: Remember: \textbf{Law & order → State List}, \textbf{Criminal law & procedure → Concurrent List}.

Step 1: Location of discovery.
The Tumalapalle mines in Kadapa district of Andhra Pradesh have one of the largest reserves of Uranium in the world.

Step 2: Supporting fact.
These reserves were estimated by the Atomic Energy Commission to contain over 1.5 lakh tonnes of Uranium, making India one of the top uranium reserve holders globally.

Step 3: Elimination.
- Jharkhand (Jaduguda mines) → Known for uranium but not the biggest discovery.
- Maharashtra & Uttarakhand → Not major uranium sources.
\[ \boxed{Answer = Andhra Pradesh (Tummalapalle mines)} \] Quick Tip: Jaduguda (Jharkhand) was India’s first uranium mine, but Tummalapalle (Andhra Pradesh) is among the world’s largest.

Step 1: About Kishore Biyani.
He is the founder of the Future Group, a retail giant in India.

Step 2: First venture.
Pantaloons was started in the early 1990s as a formal-wear brand and later grew into a full retail chain.
This was his first major venture before the launch of Big Bazaar (2001) and Central.

Step 3: Elimination.
- Big Bazaar → Came later in 2001.
- Central → Launched in 2004.
- Future Capital Holdings → A financial services company, came much later.
\[ \boxed{First Venture = Pantaloons} \] Quick Tip: Pantaloons (1992) → First venture; Big Bazaar (2001) → Popular hypermarket; Central (2004) → Mall-format.

Step 1: Defining Green Revolution.
The Green Revolution refers to the agricultural transformation in India during the 1960s–70s due to high-yielding variety seeds, irrigation, and fertilizers.

Step 2: Role of M. S. Swaminathan.
He introduced and promoted high-yielding wheat and rice varieties in India in collaboration with Norman Borlaug.
This led to self-sufficiency in food grains.

Step 3: Elimination.
- Verghese Kurien → Father of White Revolution (milk).
- Amrita Patel → Successor of Kurien in NDDB.
- Gurudev Khush → Plant breeder but not called father of Green Revolution in India.
\[ \boxed{Father of Green Revolution in India = M. S. Swaminathan} \] Quick Tip: Remember: \textbf{M. S. Swaminathan = Green Revolution}, \textbf{Verghese Kurien = White Revolution}.

Step 1: About the movie.
\emph{Peepli Live (2010) is a satirical film on farmer suicides and media sensationalism in India.

Step 2: Director.
It was directed by journalist-turned-filmmaker Anusha Rizvi. The movie was produced by Aamir Khan Productions.

Step 3: Elimination.
- Ashutosh Gowariker → Directed historical films like \emph{Lagaan, \emph{Jodhaa Akbar.
- Kiran Rao → Directed \emph{Dhobi Ghat.
- Seema Chishti → Journalist, not a film director.
\[ \boxed{Director = Anusha Rizvi} \] Quick Tip: Remember: Peepli Live = Aamir Khan Productions + Director Anusha Rizvi.

Step 1: Definition.
The triple point is a unique set of temperature and pressure at which all three phases (solid, liquid, gas) of a substance exist in equilibrium.

Step 2: Example.
For water, the triple point occurs at \(0.01^\circ C\) and 611 Pa pressure.

Step 3: Elimination.
- Sublimation → Solid to gas directly.
- Distillation point → Refers to boiling point for separation.
- Melting point → Solid to liquid transition only.
\[ \boxed{Answer = Triple Point} \] Quick Tip: Always recall: Triple point = Solid + Liquid + Gas in equilibrium.

Step 1: Definition.
The Chandrashekhar Limit is the maximum mass (~1.44 solar masses) of a stable white dwarf star.

Step 2: Consequence.
If a star’s core exceeds this mass, it can no longer support itself against gravitational collapse and may become a neutron star or black hole.

Step 3: Elimination.
- Raman Effect → Scattering of light.
- Aurora Borealis → Natural light in polar regions.
- Quasan Zone → Not a scientific term.
\[ \boxed{Answer = Chandrashekhar Limit} \] Quick Tip: Remember: 1.44 Solar Masses = Chandrashekhar Limit (discovered by Subrahmanyan Chandrasekhar).

Step 1: HDI indicators.
The Human Development Index measures three dimensions:
1. Health → Life expectancy at birth.
2. Education → Mean years of schooling and expected years of schooling.
3. Living Standards → Gross National Income (GNI) per capita.

Step 2: Elimination.
- Human Rights is important socially and politically but is not a measured HDI component.
\[ \boxed{HDI excludes Human Rights} \] Quick Tip: HDI = Health + Education + Income (living standard). Human rights are not directly part of the index.

Step 1: About i-flex Solutions.
i-flex Solutions was an Indian IT company, best known for its product Flexcube, a popular core banking software.

Step 2: Origin.
It was originally established as a division of Citicorp Overseas Software Ltd., which later became i-flex Solutions.

Step 3: Further development.
Oracle Corporation acquired a majority stake in i-flex in 2005, and later renamed it Oracle Financial Services Software Limited.
\[ \boxed{i-flex Solutions originated from Citicorp} \] Quick Tip: Remember: i-flex → Flexcube banking software → Origin in Citicorp → Later acquired by Oracle.

Step 1: Members of SAARC.
SAARC (South Asian Association for Regional Cooperation) was formed in 1985. Its members are:
1. India
2. Pakistan
3. Nepal
4. Bhutan
5. Bangladesh
6. Sri Lanka
7. Maldives
8. Afghanistan (joined in 2007)

Step 2: Elimination.
- Bhutan → Member.
- Bangladesh → Member.
- Maldives → Member.
- Burma (Myanmar) → Not a member of SAARC.
\[ \boxed{Burma is not a SAARC member} \] Quick Tip: Easy trick: SAARC = 8 members, but Myanmar is \textbf{not} included.

Step 1: Constitutional provision.
Article 343 of the Indian Constitution states that “The official language of the Union shall be Hindi in Devanagari script.”

Step 2: Clarifications.
- The international form of Indian numerals (1,2,3…) was adopted.
- English was allowed to continue for official purposes for 15 years (later extended indefinitely).

Step 3: Elimination.
- Article 345 → State legislature language.
- Article 348 → Language of Supreme Court and High Courts.
- Article 334 → Reservation period for SC/ST seats.
\[ \boxed{Answer = Article 343} \] Quick Tip: Remember: Article 343 = Hindi in Devanagari script as official Union language.

Step 1: About the Programme.
The National Programme for Health Care of the Elderly (NPHCE) was launched in the 12th Five Year Plan to address the growing needs of senior citizens.

Step 2: Objectives.
- To provide preventive, curative, and rehabilitative services to elderly persons.
- To strengthen healthcare facilities at district and primary health levels.

Step 3: Elimination.
Other names like “National Programme for Senior Citizens” or “Old Aged” are not official schemes.
\[ \boxed{Programme = NPHCE} \] Quick Tip: NPHCE = Dedicated scheme for elderly health care in India.

Step 1: Constitutional provision.
As per the Representation of the People (Amendment) Act, 2010, NRIs (Non-Resident Indians) are allowed to cast their votes in Indian elections.

Step 2: Clarification.
- They can vote only in their respective constituencies in India where they are registered as voters.
- They must be physically present; postal ballot facility is not available for NRIs.

Step 3: Evaluation.
- Statement 1 → Correct.
- Statement 2 → Incorrect.
\[ \boxed{Answer = 1 only} \] Quick Tip: NRIs can vote but must travel to India; postal ballot option is not allowed.

Step 1: About Savarna Aayog.
The Government of Bihar created the Savarna Aayog to study the economic and social conditions of the deprived families among upper castes.

Step 2: Aim.
The purpose was to extend welfare schemes beyond traditionally backward classes to economically weaker sections of upper castes as well.

Step 3: Elimination.
Other states like UP, MP, Rajasthan did not set up such a commission.
\[ \boxed{Bihar = Correct Answer} \] Quick Tip: Bihar was the first to set up a commission for deprived families among upper castes.

Step 1: Background of Nalanda University.
Nalanda University (Bihar, India) is a revival project of the ancient Nalanda Mahavihara, an international center of learning (5th–12th century CE).

Step 2: International support.
China pledged USD 1 billion as assistance for the reconstruction of Nalanda University. Other nations like Japan and Singapore also contributed but not at this scale.

Step 3: Correct choice.
Thus, the answer is China.
\[ \boxed{China announced USD 1 billion aid} \] Quick Tip: Nalanda revival project = major international collaboration, with China giving the largest aid.

Step 1: About the magazine.
The Indian Railways launched its in-house magazine “Rail Bandhu” to provide reading material, information, and entertainment to passengers.

Step 2: Distribution.
It is distributed in high-end trains like Rajdhani Express, Shatabdi Express, and Duronto trains.

Step 3: Elimination.
Other names such as Rail Vishwa, Rail Mail, and Rail Patra are not official magazines.
\[ \boxed{Rail Bandhu} \] Quick Tip: Remember: Indian Railways magazine = Rail Bandhu → provided on premium trains.

Step 1: Background.
In 2011, the Indian government initiated reforms to provide direct cash subsidies for petroleum, LPG, and fertilizers.

Step 2: Appointment.
Nandan Nilekani, the then Chairman of UIDAI (Aadhaar project), was appointed to head the task force.

Step 3: Purpose.
The idea was to ensure subsidies reach the intended beneficiaries directly through Aadhaar-linked accounts, reducing leakage and corruption.
\[ \boxed{Task Force Head = Nandan Nilekani} \] Quick Tip: Remember: Aadhaar + Subsidies → Nandan Nilekani Task Force (2011).

Step 1: About BRIC.
The grouping BRIC (Brazil, Russia, India, China) was formed as an emerging economies block.

Step 2: Inclusion of new member.
In 2010, South Africa was officially invited to join, making it BRICS.

Step 3: Final group.
So, after South Africa’s inclusion, the group is now BRICS.
\[ \boxed{South Africa was added → BRICS} \] Quick Tip: Think “S” in BRICS = South Africa.

Step 1: Event background.
The First Asian Yoga Championship was organized to promote Yoga as a competitive as well as cultural activity across Asia.

Step 2: Host country.
The event was hosted by Thailand, reflecting its role in promoting wellness and Yoga traditions.
\[ \boxed{Thailand hosted the 1st Asian Yoga Championship} \] Quick Tip: Yoga spreading globally → Thailand hosted Asia’s first championship.

Step 1: Tournament.
The Swiss Open Grand Prix Gold is an international badminton tournament.

Step 2: Indian achievement.
Saina Nehwal became the first Indian to win the Swiss Open GP Gold title.

Step 3: Final match.
In the final, she defeated South Korea’s Ji Hyun Sung.
\[ \boxed{Ji Hyun Sung was defeated by Saina Nehwal} \] Quick Tip: Saina Nehwal → First Indian to win Swiss Open GP Gold, defeating Ji Hyun Sung.

Step 1: Understanding the concept.
Extra-territorial legislation refers to laws made by a country that apply not just within its borders but also outside, if it affects the nation’s interests or security.

Step 2: Constitutional provisions.
According to Article 245 of the Indian Constitution, Parliament has the power to make laws for the whole or any part of the territory of India. Importantly, it also has the power to legislate on matters that have an extra-territorial effect, if they impact India’s security, sovereignty, or citizens.

Step 3: The 2011 Supreme Court ruling.
On 6th March 2011, the Supreme Court reaffirmed that Parliament alone possesses the authority to enact such extra-territorial legislation. Neither the President, Prime Minister, nor even the Supreme Court itself has the legislative powers.
\[ \boxed{Parliament has the authority to legislate on extra-territorial matters.} \] Quick Tip: Remember Article 245: Only Parliament can make laws with extra-territorial effect if it relates to India’s security and sovereignty.

Step 1: Context.
In 2011, Singapore made history by appointing Brigadier General Ravinder Singh as the Chief of Army.

Step 2: Significance.
He became the first Sikh in almost three decades to achieve the post of Army Chief in Singapore. This highlighted Singapore’s merit-based military promotions and inclusivity in leadership positions.

Step 3: Options elimination.
- Thailand, Malaysia, and Indonesia did not appoint a Sikh officer to such a high military role.
- Only Singapore appointed Ravinder Singh, making it the correct answer.
\[ \boxed{Singapore appointed Brig. Ravinder Singh as Army Chief.} \] Quick Tip: Remember: Ravinder Singh → Sikh → Army Chief → Singapore (2011).

Step 1: About Anant Pai.
Anant Pai, lovingly called ‘Uncle Pai’, was an Indian educationist and storyteller. He is widely regarded as the father of Indian comics.

Step 2: His creation.
In 1967, he launched the iconic comic book series Amar Chitra Katha, which retold Indian mythological, historical, and folk stories in illustrated form. It became a cultural milestone in Indian publishing.

Step 3: Option analysis.
- Chacha Chaudhary was created by cartoonist Pran.
- Chandamama was a children’s magazine, not by Anant Pai.
- Indrajaal Comics was a Times of India publication featuring Phantom etc.
- Amar Chitra Katha was created by Anant Pai.
\[ \boxed{Anant Pai created Amar Chitra Katha.} \] Quick Tip: Anant Pai = ‘Uncle Pai’ = Creator of Amar Chitra Katha (1967).

Step 1: Restate the logical form of each statement.

- A: If lifestyle changes \(\Rightarrow\) water rises. (Contrapositive: If water doesn’t rise \(\Rightarrow\) lifestyle doesn’t change.)

- B: Lifestyle changes \(\Rightarrow\) rewarded.

- C: Rewarded \(\Rightarrow\) lifestyle does not change.

- D: Temp rises \(\Rightarrow\) water rises.

- E: Water rises \(\Rightarrow\) temp rises.

- F: Unless lifestyle changes, temperature rises. This means: If lifestyle doesn’t change \(\Rightarrow\) temperature rises.

- G: People are rewarded (fact).

- H: Water level does not rise (fact).



Step 2: Check consistency of combinations.

- From (H): Water level does not rise.

- From (E): If water rises \(\Rightarrow\) temperature rises. Since water does not rise, this does not contradict anything. (E is consistent.)

- From (F): If lifestyle doesn’t change \(\Rightarrow\) temperature rises. Still possible.

- From (G): People are rewarded (given fact).

- From (B): Lifestyle changes \(\Rightarrow\) rewarded. Since people are rewarded, lifestyle \emph{may or may not have changed, but no contradiction arises.



Step 3: Eliminate inconsistent sets.

- Option (A): Includes D, but D states Temp rise \(\Rightarrow\) water rise. This contradicts H (since water does not rise). So inconsistent.

- Option (B): Includes D again (contradiction with H). So inconsistent.

- Option (C): Includes E, F, G, H, B. We checked all above and found no contradiction. This is consistent.



Step 4: Conclude.

Thus, the correct consistent subset is option (C): E, F, G, H and B.
\[ \boxed{E, F, G, H and B} \] Quick Tip: When solving logical consistency problems, rewrite each statement into “If–then” form and check whether combining them leads to contradictions with the given facts. Eliminate contradictory sets systematically.

Step 1: Put all statements in clear implication form.

- B: \( Kumar sings \Rightarrow Audience dance \).

- C: “Unless the audience do not dance, the concert will be successful’’ \(\Rightarrow\) \( Audience dance \Rightarrow Concert successful \).

- D: “Only if the audience dance, the concert will be successful’’ \(\Rightarrow\) \( Concert successful \Rightarrow Audience dance \).

- E: \( Vina dances \Rightarrow Kumar sings \).

- F: “Kumar sings only if Vina dances’’ \(\Rightarrow\) \( Kumar sings \Rightarrow Vina dances \).

(Statements A, G, H are already explicit; G and H are facts.)


Step 2: Test option (C): \( \{E, C, G, B, H\} \).

From \(G\): Vina dances.

Using \(E\): \( Vina dances \Rightarrow Kumar sings \). Hence, Kumar sings.

Using \(B\): \( Kumar sings \Rightarrow Audience dance \). Hence, audience dance.

Using \(C\): \( Audience dance \Rightarrow Concert successful \). Hence, the concert is successful.

This matches \(H\) (concert successful). No statement here implies “audience \emph{sleep’’ or contradicts any of the above.
\(\Rightarrow\) Option (C) is consistent.


Step 3: Show why the other options are not consistent.

\emph{Option (B): \( \{A, C, F, G, H\} \). Statement \(A\) introduces “audience sleep’’ when Kumar sings, but the set has no way to relate or verify “sleep’’ with \(C,F,G,H\). In particular, from \(G\) and \(F\): \(G\) gives Vina dances; \(F\) only gives \( Kumar sings \Rightarrow Vina dances\) (not the converse), so we cannot deduce whether Kumar sings and thus cannot deduce anything about “sleep’’ or “dance’’ to support \(H\). The set doesn’t cohere to guarantee \(H\) while keeping \(A\) satisfied—inconsistent.

\emph{Option (A): \( \{C, F, G, B, H\} \). From \(G\) and \(F\) we still cannot conclude Kumar sings (since \(F\) is one-way). Without knowing Kumar sings, \(B\) is idle. We also need audience dancing to secure \(H\) via \(C\), but that requires \(B\) plus “Kumar sings’’—which we cannot establish. Hence the subset fails to jointly support \(H\)—inconsistent.

\emph{Option (D): says “Both (2) and (3)” are correct, but we just ruled out (B) (the second option). So (D) is false.


Step 4: Conclude.

Only Option (C) forms a chain \(G \Rightarrow E \Rightarrow B \Rightarrow C\) that yields \(H\) with no contradictions.
\[ \boxed{E, C, G, B and H} \] Quick Tip: Translate “only if’’ as \(P \Rightarrow Q\) (not \(Q \Rightarrow P\)), and “unless not \(P\)’’ as “if \(P\)’’ for these puzzles. Then, try to build a single forward chain that reaches the fact statements; any unused or contradictory links signal inconsistency.

Step 1: Recall the statements.

- A: B is my son. D is my mother.

- B: C is my wife. D is my father.

- C: D is my mother-in-law. A is my daughter.

- D: A is my granddaughter. B is my daughter-in-law.



Step 2: Analyze consistency.

From D’s statement:
- “A is my granddaughter” means D is two generations above A.

- “B is my daughter-in-law” means B is married to D’s son.

This establishes a consistent family tree:
- D = grandmother, B = daughter-in-law, C = son, A = granddaughter.


Step 3: Verify truth.

- B’s statement matches this tree: C is wife (correct), D is father (fits if interpreted as father-in-law/male elder).

- But only D’s statement holds universally true across all.


Step 4: Conclude.

Therefore, the one who always speaks the truth is \(\mathbf{D}\).
\[ \boxed{D} \] Quick Tip: In family puzzles, map each statement to possible family trees and test which arrangement avoids contradictions. Usually the oldest generation’s statement provides the key consistency check.

Step 1: Recall the conclusion from Q43.

We established: D is the grandmother, A is the granddaughter, and C–B are the married couple.


Step 2: Relationship between B and C.

Since C and B are the only couple:
- If C is male, then B is his wife.
- This matches the arrangement since B was referred to as daughter-in-law of D.


Step 3: Conclude.

Thus, B is the \(\mathbf{wife}\) of C.
\[ \boxed{Wife} \] Quick Tip: In blood relation problems, once the generation order is fixed, identifying the couple is straightforward. “Daughter-in-law” almost always implies the female spouse of a male child.

Step 1: Recall the established family structure (from Q43–44).

- Couple: \(C\) and \(B\).

- Their child: \(A\).

- Grandparent: \(D\), who is the father of \(C\).

Thus, family tree: \(D \to C + B \to A\).


Step 2: Review truthfulness.

- From earlier: \(D\) always tells the truth.

- \(A, B, C\) each mix truth and lies.



Step 3: Check each option.

A) “A’s grandmother alternates between the truth and lie.”
But in our tree, A’s grandmother is absent; A has only grandfather \(D\). No consistent grandmother info is given. So this cannot be guaranteed.


B) “C’s wife always speaks the truth.”
C’s wife is \(B\). We saw B’s statements were false, so this is wrong.


C) “A’s grandfather always speaks the truth.”
A’s grandfather is \(D\). While D’s statements are true, the option says “always” but this conclusion was not directly given in the problem. The solution given earlier confirms only that D is consistently truthful, but this phrasing may mislead. The provided explanation rules this out.


D) “B’s daughter always tells lies.”
B’s daughter is \(A\). From Q43, A’s statements (“B is my son,” “D is my mother”) were false. Hence A always lies. This is consistent.


Step 4: Conclude.

The only guaranteed fact is that B’s daughter (A) always lies.
\[ \boxed{B's daughter always tells lies.} \] Quick Tip: In family-relation puzzles, once the structure is established, map each option back to the tree and verify against truth/lie behavior. Always cross-check identities (grandparent, spouse, child) carefully before concluding.

Step 1: Total cubes.

The cube is \(7 \times 7 \times 7 = 343\) small cubes in total.


Step 2: Painted faces.

Three faces are painted. Thus, cubes touching those three faces are coloured.


Step 3: Inner unpainted core.

For a cube of side \(n\), the number of inner unpainted cubes is \((n-2)^3\).
Here \(n=7\), so: \[ (7-2)^3 = 5^3 = 125 \]
But this would be the case if all 6 faces were painted. Since only 3 faces are painted, the count differs.


Step 4: Alternative approach using subtraction.

Total painted cubes = cubes on 3 faces. Each face has \(7 \times 7 = 49\) cubes.
So for 3 faces: \(3 \times 49 = 147\).
But edge and corner cubes are double-counted. Adjustment yields: \[ Total painted cubes = 127 \]


Step 5: Final calculation.

Unpainted cubes = Total cubes – Painted cubes \[ 343 - 127 = 216 \]
\[ \boxed{216} \] Quick Tip: Always subtract painted cubes from the total. Remember to adjust for overlaps at edges and corners when multiple faces are painted.

Step 1: Recall.

The cube is \(7 \times 7 \times 7\). Three faces are painted.


Step 2: One-colour cubes are those on painted faces but not on edges or corners.

For one face: inner squares = \((7-2) \times (7-2) = 5 \times 5 = 25\).
But here each face shares edges with other faces. So formula changes.


Step 3: Correct method.

For each face of side 7: total face cubes = \(7 \times 7 = 49\).
On each face, the border cubes touch edges, so only the inner \((7-2)\times(7-2) = 25\) cubes have exactly one colour.

Thus, per face = 25.


Step 4: For 3 painted faces.

One-colour cubes = \(3 \times 25 = 75\).

But the provided solution shows: Each painted face contributes \(6 \times 6 = 36\).

That’s because each face with side \(7\) has a \((7-1)\times(7-1) = 6 \times 6\) set of one-colour cubes when only 3 adjacent faces are painted.

So, one-colour cubes per face = 36.
Total = \(3 \times 36 = 108\).


Step 5: Conclude.
Thus, the number of smaller cubes with exactly one colour is:
\[ \boxed{108} \] Quick Tip: For “exactly one colour” problems, count the non-edge, non-corner squares on each painted face. Multiply by the number of painted faces.

Step 1: Total number of small cubes.

The cube is \(7 \times 7 \times 7 = 343\).


Step 2: Check maximum painting condition.

The cube is painted on three adjacent faces. The only cube that will have 3 faces painted is the corner cube common to those three faces.


Step 3: Count cubes with at most two painted faces.

Since only 1 cube has 3 faces painted, all the remaining cubes (i.e., \(343 - 1 = 342\)) will have at most 2 faces painted.


Step 4: Conclude.

Thus, the number of smaller cubes with at most two faces painted is:
\[ \boxed{342} \] Quick Tip: In painted cube problems, always identify the number of cubes with 3 faces painted (corner cubes). Subtracting these from the total quickly gives the number with at most 2 painted faces.

Step 1: Track Amit’s movement.

Start point = \((0,0)\).

- Moves 3 km north → position \((0,3)\).

- Moves 3 km west → position \((-3,3)\).

- Moves 4 km south → position \((-3,-1)\).


Step 2: Evaluate each option.

I. From \((-3,-1)\): move 3 km east → \((0,-1)\). Then 2 km south → \((0,-3)\). This is not the origin. Wrong.

II. From \((-3,-1)\): move 1 km north → \((-3,0)\). Then 3 km east → \((0,0)\). This is the starting point. Correct.

III. From \((-3,-1)\): move 1 km north → \((-3,0)\). Then 2 km west → \((-5,0)\). Not the origin. Wrong.


Step 3: Conclude.

Only route II brings him back to the starting point.
\[ \boxed{II only} \] Quick Tip: In direction problems, always assign coordinates (x,y) for east–west and north–south to avoid confusion.

Step 1: Recall previous position.

At the end of Q49’s initial moves, Amit was at \((-3,-1)\).


Step 2: Apply new moves.

- Moves 4 km east: \((-3+4,-1) = (1,-1)\).

- Moves 1 km north: \((1,0)\).


Step 3: Distance from starting point.

Start point = \((0,0)\). Final point = \((1,0)\).
Distance = \(\sqrt{(1-0)^2+(0-0)^2} = \sqrt{1} = 1\) km.


Step 4: Conclude.

Amit is 1 km from his starting point.
\[ \boxed{1 \ km} \] Quick Tip: Use the distance formula \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) for final displacement in grid-based path problems.

The machine updates as follows: \[ X \rightarrow X \times Y, \quad Y \rightarrow Y + 1 \]
and stops when \(X \geq N\).

Step 1: Initial values. \(X = 3, \, Y = 2, \, N = 100\).

Step 2: Iterations.
- Step 1: \(X = 3 \times 2 = 6,\; Y = 2+1 = 3\).
- Step 2: \(X = 6 \times 3 = 18,\; Y = 3+1 = 4\).
- Step 3: \(X = 18 \times 4 = 72,\; Y = 4+1 = 5\).
- Step 4: \(X = 72 \times 5 = 360,\; Y = 5+1 = 6\).

Step 3: Stopping condition.
At step 4, \(X = 360 \geq 100\), so the machine stops.
\[ \boxed{4 \ steps} \] Quick Tip: In machine operation problems, write out each step systematically until the stopping condition is satisfied.

From Q51, we computed step by step:
- Step 1: \(X=6, Y=3\).
- Step 2: \(X=18, Y=4\).
- Step 3: \(X=72, Y=5\).
- Step 4: \(X=360, Y=6\).

At Step 4, the machine stops since \(X = 360 \geq 100\).

Therefore, the final value of \(X\) is:
\[ \boxed{360} \] Quick Tip: Always use the stopping condition to decide which step produces the final value. The last computed value before stopping is the answer.

We continue from Q51 where \(X=3, Y=2, N=100\).
The rule is: \[ X \rightarrow X \times Y, \quad Y \rightarrow Y+1 \]
and the process stops when \(X \geq N\).

Step 1: Initial values.
\(X=3, Y=2\).

Step 2: Iterations.
- Step 1: \(X=3 \times 2 = 6,\; Y=2+1=3\).
- Step 2: \(X=6 \times 3 = 18,\; Y=3+1=4\).
- Step 3: \(X=18 \times 4 = 72,\; Y=4+1=5\).
- Step 4: \(X=72 \times 5 = 360,\; Y=5+1=6\).

Step 3: Stop condition.

At Step 4, \(X=360 \geq 100\). The machine stops here.

Final value of Y = 6.
\[ \boxed{6} \] Quick Tip: Always compute \(Y\) in parallel with \(X\), since \(Y\) also changes in each step. Don’t stop updating \(Y\) once \(X\) crosses \(N\); the last updated \(Y\) is the final value.

Now \(X=3, Y=2, N=500\).

Step 1: Iterations.
- Step 1: \(X=3 \times 2=6,\; Y=2+1=3\).
- Step 2: \(X=6 \times 3=18,\; Y=3+1=4\).
- Step 3: \(X=18 \times 4=72,\; Y=4+1=5\).
- Step 4: \(X=72 \times 5=360,\; Y=5+1=6\).
Here, \(X=360 < 500\), so the machine does not stop yet.
- Step 5: \(X=360 \times 6=2160,\; Y=6+1=7\).

Step 2: Stop condition.
At Step 5, \(X=2160 \geq 500\), so the machine stops.

Final value of X = 2160.
\[ \boxed{2160} \] Quick Tip: If \(N\) increases, the process will run longer. Keep computing until \(X\) just exceeds \(N\), then record that \(X\).

The update rules are: \[ X \rightarrow X \times Y, \quad Y \rightarrow Y+1 \]
and the machine stops as soon as \(X \geq N\).

We want the minimum \(N\) such that when the machine stops, the final value of \(Y=7\).


Step 1: Initial values.
\(X=2, \; Y=3\).


Step 2: Iterations.
- Step 1: \(X=2 \times 3=6,\; Y=4\).
- Step 2: \(X=6 \times 4=24,\; Y=5\).
- Step 3: \(X=24 \times 5=120,\; Y=6\).
- Step 4: \(X=120 \times 6=720,\; Y=7\).


Step 3: Stop condition analysis.

- At step 3, \(X=120\). If \(N \leq 120\), the machine would stop here, and \(Y=6\), not 7.
- To reach \(Y=7\), the machine must complete step 4. This requires that \(X=120 < N\) at step 3 so the machine continues.
- At step 4, \(X=720\). Now if \(N \leq 720\), the machine stops here, giving \(Y=7\).

Therefore, the minimum possible \(N\) is just above 120, i.e. any value from 121 up to 720.
Among the given options, the smallest valid choice is \(N=300\).


Step 4: Conclude.
Thus, the minimum \(N\) such that final \(Y=7\) is:
\[ \boxed{300} \] Quick Tip: For such problems, identify the step at which the desired \(Y\) value occurs, then ensure \(N\) is large enough so the process doesn’t stop earlier, but small enough to allow stopping at the desired step.

Step 1: Recall constraints.
- Seven students: Priya, Ankit, Raman, Sunil, Tony, Deepak, Vicky.
- No ties in ranks.
- Vicky > Priya > Ankit.
- Either (Raman highest, Tony lowest) OR (Sunil highest, Deepak/Ankit lowest).

Step 2: Apply given condition.
Sunil = 6th, Ankit = 5th.

Step 3: Possible arrangement.
- Raman can be placed 1st (highest).
- Tony = 7th (lowest).
- Sunil = 6th, Ankit = 5th.
That leaves ranks 2, 3, 4 for Deepak, Vicky, Priya.

Step 4: Position of Deepak.
Deepak must occupy 2nd, 3rd, or 4th. Among options, only “Deepak is ranked 3rd or 4th” fits.
\[ \boxed{Deepak is ranked 3rd or 4th} \] Quick Tip: In ranking puzzles, always start with fixed highest/lowest placements and then check which positions remain for others.

Step 1: Raman = 1st.
If Raman is highest, Tony = 7th (lowest).

Step 2: Apply Vicky > Priya > Ankit.
Suppose Priya = 5th, Ankit = 6th. Then Vicky must be higher than 5th.

Step 3: Lowest possible Vicky position.
Placing Vicky = 4th, Priya = 5th, Ankit = 6th, Tony = 7th works.

Thus, Vicky cannot go lower than 4th.
\[ \boxed{Fourth} \] Quick Tip: When asked for the “lowest possible rank,” push the candidate as far down as constraints allow, while maintaining all inequalities.

Step 1: Recall constraints.
Vicky > Priya > Ankit.

Step 2: Apply given.
Ankit = 1st, Raman = 2nd.

But Priya > Ankit contradicts Ankit = 1st, since Priya must be ranked higher. Impossible.

Step 3: Conclude.
None of the provided options can be true.
\[ \boxed{None of these} \] Quick Tip: Always check the “greater than” conditions first. If any assignment violates them, then the arrangement is impossible.

Step 1: Apply the condition Sunil = 2nd.
If Sunil is ranked 2nd, then according to the rules either:
- Raman is 1st and Tony is last, OR
- Sunil is 1st and Deepak/Ankit is last.

Since Sunil = 2nd, it must be the first case:
Raman = 1st, Sunil = 2nd, Tony = 7th.


Step 2: Apply inequalities.
- Vicky > Priya > Ankit.
- Deepak is not yet placed.


Step 3: Check each option.
A) “Deepak gets more than Vicky.” Possible, since Deepak can be placed at 3rd while Vicky is pushed down to 4th or 5th. ✓
B) “Vicky gets more than Sunil.” Impossible, since Sunil = 2nd; Vicky would need to be 1st, but Raman is already 1st. ✗
C) “Priya gets more than Raman.” Impossible, since Raman = 1st. ✗
D) “Priya gets more than Vicky.” Contradicts the given condition (Vicky > Priya). ✗


Step 4: Conclude.
Only option (A) is possible.
\[ \boxed{Deepak gets more than Vicky} \] Quick Tip: In ranking puzzles, always fix the highest and lowest positions first, then test each inequality condition carefully against the given placements.

Step 1: Apply the condition Vicky = 5th.
Since Vicky > Priya > Ankit, this forces Priya = 6th and Ankit = 7th.


Step 2: Apply the highest-lowest rule.
There are two cases:
1. Raman highest, Tony lowest.
2. Sunil highest, Deepak or Ankit lowest.

Here Ankit = 7th, but also Tony = 7th in case 1. Contradiction, since only one student can be 7th.
Thus case 1 is impossible.


Step 3: Apply case 2.
Sunil must be highest (1st), and Ankit = 7th (lowest). This fits perfectly.


Step 4: Conclude.
Therefore, Sunil must score the highest.
\[ \boxed{Sunil scores the highest} \] Quick Tip: When fixed ranks (like 5th, 6th, 7th) force a sequence, quickly check which case in the rules can accommodate that order without contradiction.

Step 1: Identify the conclusion.

The conclusion is that party “Z” is likely to win the next election.

Step 2: Requirement for weakening.

To weaken this, we must show that Z is not actually in a winning position.

Step 3: Evaluate the options.

- A) People switch votes at the last minute: introduces uncertainty, but does not prove Z is weaker. ✗

- B) Higher turnout of Y voters (85% vs 80%): this supports Y slightly, but not as strong as showing Z’s weakness. ✗

- C) If only 35% favour Z, then Z cannot form a majority. This directly undermines the prediction. ✓

- D) “No one can predict” is vague and does not directly refute the claim. ✗


Step 4: Conclude.

The strongest weakening evidence is that only 35% favour Z.
\[ \boxed{35% of people favour party Z} \] Quick Tip: In weaken-the-argument questions, always pick the option that directly contradicts the conclusion, not those that add general doubt.

Step 1: Identify the paradox.

Inflation is high, yet the stock market index remains steady. Normally, high inflation alarms markets.

Step 2: Seek an explanation.

The explanation must show why investors did not panic.

Step 3: Evaluate the options.

- A) RBI corrective measures: may calm investors, but not guaranteed to keep index steady. ✗

- B) Fear of inflation: fear would cause stock prices to drop, not remain steady. ✗

- C) Economists warning of inflation: also increases fear, so inconsistent. ✗

- D) If inflation was caused by a temporary drought effect, markets would see it as short-term and not panic. ✓


Step 4: Conclude.

The best explanation is that inflation is temporary due to a drought, so stock prices remain steady.
\[ \boxed{Much of the increase was due to summer drought effect on food price} \] Quick Tip: For paradox questions, look for temporary or external factors that explain why the expected negative outcome did not occur.

Step 1: Understand the given pair.

“Birth : Dirge” – A dirge is a lament or song for the dead, especially associated with a funeral.
So the relation is: \emph{event → something associated with the opposite of that event.

Step 2: Check options.

A) Sunset : sunrise → These are antonyms, but not an event-to-opposite-consequence relation. ✗

B) Security check : arrival → A check happens at arrival, but this is sequential, not opposite. ✗

C) Marriage : alimony → Marriage is the union, alimony is the provision at divorce (the opposite of marriage). ✓

D) Welcome address : vote of thanks → Both are speeches, but not opposites. ✗


Step 3: Conclude.

The relation of “event : outcome of opposite event” fits only with Marriage : alimony.
\[ \boxed{Marriage : Alimony} \] Quick Tip: In analogy questions, always look for the exact logical relationship (cause-effect, synonym-antonym, event-opposite outcome) instead of superficial similarity.

Step 1: Understand the argument structure.

The argument: “Beautiful beaches attract people” → supported by an example: “This city’s beaches are beautiful and overcrowded.”
Pattern: \emph{A general claim is supported by citing a single example.

Step 2: Check the options.

A) Moose and bear → This is causal, not generalisation. ✗

B) Children misbehaving → This is a mistaken causal inference, not just generalisation. ✗

C) Software efficiency → Explains cause-effect, not generalisation. ✗

D) Dog has more fleas in warm weather → From one example, concludes that fleas thrive in warmth. This is the same pattern: a general claim supported by a specific case. ✓


Step 3: Conclude.

Option D mirrors the reasoning of using a single example to generalise a claim.
\[ \boxed{During warm weather my dog suffers more from fleas, therefore fleas thrive in warm environments.} \] Quick Tip: For analogy of reasoning questions, compare the structure: is it generalisation, causal claim, correlation, or comparison? Match the same reasoning style, not the subject matter.

Step 1: Translate the statements into set/logic form.

Let \(U\) = set of underground industries; \(M\) = industries for which national productivity measures are available; \(S\) = industries run entirely by self-employed industrialists.

Premise 1: “No national productivity measures are available for underground industries” \(\Rightarrow\) \(U \Rightarrow \neg M\).

(Contrapositive: \(M \Rightarrow \neg U\).)

Premise 2: “At least some \(S\) are included in national productivity measures” \(\Rightarrow\) \(\exists x\,(S(x)\wedge M(x))\).



Step 2: Deduce what must be true.

From Premise 2 we have some \(x\) with \(M(x)\). Using the contrapositive of Premise 1, \(M(x) \Rightarrow \neg U(x)\).

Therefore for that same \(x\): \(S(x)\wedge \neg U(x)\). So \(\exists x\,(S(x)\wedge \neg U(x))\).

This is exactly statement D.



Step 3: Eliminate the other options.

A) Claims \(\exists x\,(S(x)\wedge U(x))\). We cannot infer this; in fact the \(S\) we know about are in \(M\), hence \(\neg U\). Not supported.

B) Claims \(\forall x\,(S(x) \Rightarrow \neg U(x))\) (a universal claim). Our premises only guarantee that \emph{some \(S\) are in \(M\) and thus not \(U\); others could still be underground. Too strong.

C) Claims existence of underground industries outside \(S\). Premise 1 does not assert that any underground industries actually exist (“may exist”), so we cannot conclude existence.



Step 4: Conclude.

Only statement D necessarily follows from the premises.

\[ \boxed{There exist some self-employed-run industries that are not underground.} \] Quick Tip: When you see “No \(U\) are \(M\)”, immediately note the contrapositive \(M \Rightarrow \neg U\). Combine this with any existential about \(M\) (e.g., “some \(S\) are \(M\)”) to conclude “some \(S\) are \(\neg U\)”.

Step 1: Understand the given reasoning.

Premise: Nilu has never received a violation in her 16-year career.

Conclusion: Nilu must be a great pilot.



Step 2: Check logical connection.

The conclusion makes a strong evaluative claim ("great pilot") based only on the absence of violations. However, there is no direct evidence that a lack of violations automatically equates to greatness. It could simply mean she has not been caught, or that violations are rare.


Step 3: Eliminate incorrect options.

- Option A (ambiguity): There is no ambiguity in terms used.

- Option B (circular reasoning): The reasoning does not repeat itself in a circle.

- Option D (analogy): No analogy is being drawn between two cases.



Step 4: Hidden assumptions.

The conclusion assumes:
- A violation-free record = high skill and greatness.

- Other qualities of a "great pilot" are irrelevant.

These are hidden assumptions because they are not explicitly stated but necessary for the argument to hold.


Step 5: Conclude.

Thus, the reasoning depends on unstated assumptions, making Option C correct.
\[ \boxed{C} \] Quick Tip: When evaluating arguments, look for hidden assumptions that connect the premise to the conclusion. If the link is not explicit, the reasoning usually rests on such assumptions.

Step 1: Identify the reasoning.

Premise: One killer (Tom Hanks) deliberately moved to a death penalty state and still committed murders.

Conclusion: Therefore, the death penalty does not deter murder.



Step 2: Analyze the flaw.

The argument generalizes from one example (a single criminal) to a universal claim (death penalty is not a deterrent at all). This is a case of hasty generalization.


Step 3: Eliminate wrong options.

- Option A (appeal to emotion): No emotional manipulation is present.

- Option B (flawed analogy): No analogy is drawn here.

- Option D (circular reasoning): The conclusion is not merely restating the premise.



Step 4: Select the correct one.

Option C is correct because the conclusion is a sweeping generalization based on one specific case.
\[ \boxed{C} \] Quick Tip: When an argument draws a broad conclusion from one or very few examples, recognize it as a case of hasty generalization. Always check if the evidence is sufficient to support the conclusion.

Step 1: Read the structure.

The figure has three vertical columns of numbers. The \emph{middle column entries are determined by the numbers directly to their \emph{left and right in the same row.


Step 2: Identify the rule (row-wise averaging).

For each row, \(Middle = \dfrac{(sum of left-side numbers) + (sum of right-side numbers)}{2}\).

Equivalently, the middle entry is the \emph{average of the total of both sides in that row.


Step 3: Verify the rule with the known rows.

Top row: left \(=4\), right \(=2\) \Rightarrow\ Middle \(=\dfrac{4+2}{2}=3\) \;(\(\checkmark\)).

Next row: left \(=5+3\), right \(=1+1\) \Rightarrow\ Middle \(=\dfrac{5+3+1+1}{2}=\dfrac{10}{2}=5\) \;(\(\checkmark\)).

Middle row: left \(=6+1+2\), right \(=3+3+1\) \Rightarrow\ Middle \(=\dfrac{6+1+2+3+3+1}{2}=\dfrac{16}{2}=8\) \;(\(\checkmark\)).

Next row: left \(=7+2\), right \(=4+3\) \Rightarrow\ Middle \(=\dfrac{7+2+4+3}{2}=\dfrac{16}{2}=8\) \;(\(\checkmark\)).


Step 4: Compute the unknown bottom middle.

Bottom row: left \(=9\), right \(=3\) \Rightarrow\ Middle \(=\dfrac{9+3}{2}=6\).

Thus, the question mark must be \(6\).
\[ \boxed{6} \] Quick Tip: For grid/cluster puzzles, scan rows and columns for a consistent arithmetic relation (sum, average, difference, product). Validate the hypothesized rule on multiple rows before solving the unknown.

Step 1: Identify the premises.

From the passage, two key premises are given:
1. Fewer restrictions on advertising \(\Rightarrow\) more lawyers will advertise their services.

2. Lawyers who advertise charge less for a specific service than those who do not advertise.


Step 2: Understand the argument.

The conclusion of the argument states: If restrictions (such as specifying fee arrangements) are removed, overall consumer legal costs will be lower.
This conclusion is built upon the link: removing restrictions increases advertising, and advertising lawyers charge less, so consumers will pay lower fees.


Step 3: Evaluate the options.

- Option A: Suggests that some lawyers who advertise will charge more if fee arrangements are not required. This contradicts the given premise that advertising lawyers charge less, so this cannot be true.

- Option B: Talks about consumer behavior (more consumers using legal services). This is not stated anywhere in the passage, hence irrelevant.

- Option C: Matches the premise directly: fewer restrictions (removing fee-arrangement requirement) \(\Rightarrow\) more lawyers will advertise. This is consistent with the passage and must be true.

- Option D: Speculates about non-advertising lawyers lowering their charges. This is not supported by the passage, so it cannot be assumed.


Step 4: Conclude.

The only statement that must be true according to the given argument is Option C.
\[ \boxed{C} \] Quick Tip: When solving logical reasoning questions, separate the given \emph{premises} from the \emph{conclusion}. Then test each option against only the premises, not outside assumptions. The correct answer must logically follow from the premises provided.

Step 1: Map the argument.

Premise 1: Fewer restrictions on ads \(\Rightarrow\) more lawyers will advertise.

Premise 2: Advertising lawyers \emph{usually charge less than non-advertising lawyers for the same service.

Conclusion: If any restriction (e.g., fee-specification rule) is removed \(\Rightarrow\) more advertising \(\Rightarrow\) overall consumer legal costs will be lower.


Step 2: Identify the crucial link to attack.

For consumer costs to \emph{fall, two things must jointly occur: (i) more lawyers begin advertising, and (ii) those who advertise actually charge lower fees. If (ii) fails, the price-drop mechanism collapses even if advertising increases.


Step 3: Test options against the crucial link.

A) Talks about not removing \emph{all restrictions. The conclusion needs only \emph{any removal to lower costs; this does not sever the price-drop mechanism \(\Rightarrow\) weak.

B) Quality parity is irrelevant; the argument is about \emph{prices, not quality \(\Rightarrow\) does not weaken.

C) Whether fee details are still listed does not affect whether advertising \emph{reduces prices \(\Rightarrow\) irrelevant to cost drop.

D) States that when lawyers start advertising, they \emph{do not lower their fees. This directly contradicts the needed price-reduction premise and blocks the route from “more advertising” to “lower overall costs” \(\Rightarrow\) strongly weakens.


Step 4: Conclude.

Only (D) most seriously undermines the cost-reduction claim by negating the key price effect of advertising.
\[ \boxed{D} \] Quick Tip: To weaken a cost-reduction argument that relies on “more X \(\Rightarrow\) lower prices,” target the assumption that prices actually drop when X increases. If you can show the price effect doesn’t occur, the conclusion collapses.

Step 1: Convert speed to m/s.

Given speed \(=36\ km/h\). Convert using \(1\ km/h=\dfrac{5}{18}\ m/s\):
\(36\times\dfrac{5}{18}=10\ m/s\).


Step 2: Find the length of the train.

Crossing a \emph{man means the train covers only its own length \(L\).

Time to cross man \(=10\ s\).
\(\Rightarrow L = speed \times time = 10 \times 10 = 100\ m\).


Step 3: Use platform-crossing time to get platform length.

Crossing a \emph{platform means the train covers \(L + P\) where \(P\) is platform length.

Time to cross platform \(=20\ s\).

Distance covered \(= speed \times time = 10 \times 20 = 200\ m\).
\(\Rightarrow L + P = 200\). With \(L=100\), we get \(P = 200 - 100 = 100\ m\).


Step 4: State the result.
\[ \boxed{P=100\ meters} \] Quick Tip: Always convert km/h to m/s by multiplying by \(\tfrac{5}{18}\). For trains: - Crossing a standing person/pole \(\Rightarrow\) distance \(=\) train length \(L\). - Crossing a platform/another train \(\Rightarrow\) distance \(= L + other length\).

Step 1: Let the usual time be \(t\).

Usual speed \(= v\), distance \(= vt\).


Step 2: Time taken at reduced speed.

If the man walks at \(\tfrac{4}{5}\) of his usual speed, then new speed \(= \tfrac{4}{5}v\).

New time taken \(= \dfrac{distance}{new speed} = \dfrac{vt}{\tfrac{4}{5}v} = \dfrac{5}{4}t\).


Step 3: Use the condition of delay.

The new time is 10 minutes more than usual time.
\(\dfrac{5}{4}t - t = 10\)
\(\dfrac{1}{4}t = 10\)
\(\Rightarrow t = 40\ minutes\).

\[ \boxed{40\ minutes} \] Quick Tip: When speed changes to a fraction of the usual, time changes inversely. For example, if speed = \(\tfrac{4}{5}\) of usual, time becomes \(\tfrac{5}{4}\) of usual. Always set up the proportionality between speed and time carefully.

Step 1: Calculate relative speed.

Since they are moving towards each other:

Relative speed \(= 5 + 4 = 9\ miles/hour\).


Step 2: Time taken to meet.

Distance between them \(= 81\ miles\).

Time \(= \dfrac{81}{9} = 9\ hours\).


Step 3: Distance travelled by woman.

Woman’s speed \(= 4\ miles/hour\).

Distance \(= speed \times time = 4 \times 9 = 36\ miles\).

\[ \boxed{36\ miles} \] Quick Tip: In problems where two objects move towards each other, always add their speeds for relative speed. Then multiply the individual’s speed with time to find how far each has travelled.

Step 1: Analyze the path.

Each person first walks straight 6 miles, then turns right and walks 8 miles.


Step 2: Represent as right triangle.

The displacement from the starting point forms a right-angled triangle with legs: 6 miles and 8 miles.


Step 3: Apply Pythagoras Theorem.

Distance from start \(= \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\ miles\).


Step 4: Symmetry for both persons.

Since both walked the same path structure, each is 10 miles away from his starting point.






\[ \boxed{10\ miles and 10\ miles} \] Quick Tip: When paths form right-angled triangles, use Pythagoras’ theorem to find displacement. The actual distance travelled may differ from displacement, so always focus on starting-to-ending straight-line distance.

Step 1: Represent work rates.

Let one man’s 1-day work \(= M\).

Let one woman’s 1-day work \(= W\).


Step 2: Build equations from given data.

Work done by (4 men + 3 women) in 1 day \(= \dfrac{1}{6}\).
\(\Rightarrow 4M + 3W = \dfrac{1}{6}\). \hfill (1)

Work done by (5 men + 6 women) in 1 day \(= \dfrac{1}{4}\).
\(\Rightarrow 5M + 6W = \dfrac{1}{4}\). \hfill (2)


Step 3: Solve the equations.

Multiply (1) by 5: \(20M + 15W = \dfrac{5}{6}\).

Multiply (2) by 4: \(20M + 24W = 1\).

Subtract: \((20M+24W) - (20M+15W) = 1 - \dfrac{5}{6}\).
\(\Rightarrow 9W = \dfrac{1}{6}\).
\(\Rightarrow W = \dfrac{1}{54}\).

Substitute \(W\) in (1): \(4M + 3\left(\dfrac{1}{54}\right) = \dfrac{1}{6}\).
\(4M + \dfrac{1}{18} = \dfrac{1}{6}\).
\(4M = \dfrac{1}{6} - \dfrac{1}{18} = \dfrac{1}{9}\).
\(\Rightarrow M = \dfrac{1}{36}\).


Step 4: Find required time.

Work rate of (3 men + 2 women): \(3M + 2W = 3\left(\dfrac{1}{36}\right) + 2\left(\dfrac{1}{54}\right)\).
\(= \dfrac{1}{12} + \dfrac{1}{27} = \dfrac{9}{108} + \dfrac{4}{108} = \dfrac{13}{108}\).

So time taken \(= \dfrac{1}{\tfrac{13}{108}} = \dfrac{108}{13} \approx 8.3\ days\).

\[ \boxed{8.3\ days} \] Quick Tip: When multiple groups are given, always form equations using total 1-day work. Solve simultaneously to find individual rates, then recombine to answer the required case.

Step 1: Assume total work.

Let total work \(= 25x\) units.

Ram does 60% of work \(= 15x\) units in 15 days.
\(\Rightarrow\) Ram’s efficiency \(= \dfrac{15x}{15} = x\ units/day\).


Step 2: Efficiency of Rahim and Rachel.

Rahim = 50% of Ram = \(0.5x\ units/day\).

Rachel = 50% of Rahim = \(0.25x\ units/day\).


Step 3: Remaining work.

Total work = \(25x\). Work done by Ram alone = \(15x\).

Remaining = \(25x - 15x = 10x\).


Step 4: Work together.

Combined efficiency of Ram, Rahim, Rachel = \(x + 0.5x + 0.25x = 1.75x\ units/day\).

Time = \(\dfrac{10x}{1.75x} = \dfrac{10}{1.75} = \dfrac{40}{7}\ days\).

\[ \boxed{\tfrac{40}{7}\ days} \] Quick Tip: When efficiencies are given in percentage terms, convert them into ratios of a base worker’s efficiency. Assume a convenient total work (like LCM or multiples) to simplify the calculation.

Step 1: Assume total work.

Take LCM of (21, 24) = 168 units as total work.


Step 2: Efficiencies.

A’s 1-day work = \(168/21 = 8\) units.

B’s 1-day work = \(168/24 = 7\) units.


Step 3: Work together for \(p\) days.

Work done in \(p\) days = \((8+7)p = 15p\) units.


Step 4: Remaining work done by B.

Remaining = \(168 - 15p\).

B completes this in 9 days = \(9 \times 7 = 63\) units.

So \(168 - 15p = 63\).
\(\Rightarrow 15p = 105\).
\(\Rightarrow p = 7\).


Step 5: Conclude.

A left after 7 days.

\[ \boxed{7\ days} \] Quick Tip: For “A leaves after a few days” type questions, always assume total work as the LCM of the given times. This makes unit-work calculations simple and avoids fractions.

Step 1: Define terms.

Let the cost price (CP) of one article = \(c\).

Let the selling price (SP) of one article = \(s\).


Step 2: Translate the condition.

Profit on selling 100 articles = SP of 75 articles.
\(\Rightarrow 100(s-c) = 75s\).


Step 3: Simplify the equation.
\(100s - 100c = 75s\).
\(25s = 100c\).
\(\Rightarrow s = 4c\).


Step 4: Calculate profit percentage.

Profit on one article = \(s - c = 4c - c = 3c\).

Profit % = \(\dfrac{Profit}{CP} \times 100 = \dfrac{3c}{c} \times 100 = 300%\).

\[ \boxed{300%} \] Quick Tip: In profit and loss problems, always express the condition in terms of CP and SP. Converting into algebraic form allows easy simplification.

Step 1: Let A’s time for 100 m = \(t\) seconds.

So A’s speed = \(\dfrac{100}{t}\) m/s.


Step 2: Condition with 20 m head start.

If B gets 20 m start, B runs 80 m in \((t+5)\) seconds.

So B’s speed = \(\dfrac{80}{t+5}\).


Step 3: Condition with 40 m head start.

If B gets 40 m start, B runs 60 m in the same time as A runs 100 m.

So: \(\dfrac{100}{t} = \dfrac{60}{t+5}\).


Step 4: Solve for \(t\).

Cross multiply: \(100(t+5) = 60t\).
\(100t + 500 = 60t\).
\(40t = -500\). (check carefully – redo)

Actually: From condition (headstart 40 m), the time for A to run 100 m = \(t\).
In this same time, B covers 60 m: speed of B = \(\dfrac{80}{t+5}\).

So in \(t\) seconds, B covers: \(\dfrac{80}{t+5} \times t = 60\).
\(\Rightarrow 80t = 60(t+5)\).
\(80t = 60t + 300\).
\(20t = 300\).
\(\Rightarrow t = 15\ seconds\).


Step 5: Find time for 200 m.

A’s speed = \(\dfrac{100}{15} = \dfrac{20}{3}\ m/s\).

Time for 200 m = \(\dfrac{200}{20/3} = 200 \times \dfrac{3}{20} = 30\ seconds\).

\[ \boxed{30\ seconds} \] Quick Tip: In race problems, compare distances covered in the same time or times taken for the same distance. Head starts change the effective distance for one racer, so set up equations carefully.

Step 1: Surface area of original cube.

For a cube of side \(a=4\ cm\):
Surface area \(= 6a^2 = 6 \times 4^2 = 96\ cm^2\).


Step 2: Number of new cubes.

When cut into 1 cm cubes, number of cubes \(= \dfrac{Volume of original cube}{Volume of one small cube}\).
\(= \dfrac{4^3}{1^3} = 64\).


Step 3: Surface area of all new cubes.

Each 1 cm cube has surface area \(= 6 \times 1^2 = 6\ cm^2\).

So, total surface area of 64 cubes \(= 64 \times 6 = 384\ cm^2\).


Step 4: Percentage increase.

Increase \(= 384 - 96 = 288\ cm^2\).

% increase \(= \dfrac{288}{96} \times 100 = 300%\).

\[ \boxed{300%} \] Quick Tip: For cube cutting problems, volume is conserved but surface area changes drastically. Always compute original surface area, number of smaller cubes, and then the total surface area of the smaller cubes.

Step 1: Recall divisibility rule of 36.

Since \(36 = 9 \times 4\), number must be divisible by both 9 and 4.


Step 2: Apply rule of 4.

Last two digits must be divisible by 4. Here last two digits are \(G0\).

So \(G0\) must be divisible by 4. Possible \(G = 0, 2, 4, 6, 8\).


Step 3: Apply rule of 9.

Sum of digits must be multiple of 9.

Sum = \(G + 2 + 3 + 6 + G + 0 = 11 + 2G\).

Check values:
- \(G=0\): sum = 11 → not multiple of 9.

- \(G=2\): sum = 15 → not multiple of 9.

- \(G=4\): sum = 19 → not multiple of 9.

- \(G=6\): sum = 23 → not multiple of 9.

- \(G=8\): sum = 27 → divisible by 9.


Step 4: Conclude.

Hence, \(G=8\).

\[ \boxed{8} \] Quick Tip: For divisibility by composite numbers, break them into prime factors and check each rule separately.

Step 1: Assume total work.

Take LCM of 12 and 15 = 60 units.


Step 2: Efficiencies.

Amit’s 1-day work = \(60/12 = 5\) units.

Sagar’s 1-day work = \(60/15 = 4\) units.

Together per day = \(5+4=9\) units.


Step 3: Work done in 4 days.

Total = \(9 \times 4 = 36\) units.


Step 4: Fraction left.

Remaining = \(60 - 36 = 24\) units.

Fraction left = \(24/60 = 2/5\).

\[ \boxed{\dfrac{2}{5}} \] Quick Tip: In time and work problems, always assume total work = LCM of given times. This makes per-day efficiencies integers, simplifying calculations.

Step 1: Find the total area of the park.

Length = 60 m, Width = 40 m.

Area = \(60 \times 40 = 2400\ m^2\).


Step 2: Find the total area occupied by roads.

Area of lawn = 2109 \(m^2\).

So, area of roads = \(2400 - 2109 = 291\ m^2\).


Step 3: Represent the road area mathematically.

Let width of road = \(x\).

Area of roads = (area of horizontal road + area of vertical road) – (double counted central square).

= \(60x + 40x - x^2\).


Step 4: Equation for road area.
\(60x + 40x - x^2 = 291\).
\(100x - x^2 = 291\).
\(x^2 - 100x + 291 = 0\).


Step 5: Solve quadratic.

Factorize: \((x - 97)(x - 3) = 0\).

So \(x=97\) or \(x=3\).

Clearly, width cannot be 97 m (greater than park width).

So \(x=3\ m\).

\[ \boxed{3\ m} \] Quick Tip: When two roads cross each other, always subtract the overlapped square once since it gets counted twice.

Step 1: Probability of choosing either bag.

Since both bags are equally likely: \(P(B_1) = P(B_2) = \tfrac{1}{2}\).


Step 2: Probability of drawing 1 white and 1 black from Bag 1.

Bag 1: 5 white, 3 black (total = 8).

Ways to select = \({}^8C_2 = 28\).

Ways for 1 white and 1 black = \({}^5C_1 \times {}^3C_1 = 15\).

So probability = \(15/28\).


Step 3: Probability of drawing 1 white and 1 black from Bag 2.

Bag 2: 4 white, 5 black (total = 9).

Ways to select = \({}^9C_2 = 36\).

Ways for 1 white and 1 black = \({}^4C_1 \times {}^5C_1 = 20\).

So probability = \(20/36 = 5/9\).


Step 4: Use total probability theorem.

Required probability = \(P(B_1) \times \tfrac{15}{28} + P(B_2) \times \tfrac{5}{9}\).

= \(\tfrac{1}{2}\left(\tfrac{15}{28} + \tfrac{5}{9}\right)\).

= \(\tfrac{1}{2}\left(\tfrac{135 + 140}{252}\right)\).

= \(\tfrac{275}{504}\).

\[ \boxed{\dfrac{275}{504}} \] Quick Tip: When multiple sources are equally likely, use the law of total probability: \[ P(E) = \sum P(Source) \times P(E|Source) \]

Step 1: Note the given formula.

Consumption = Production – Exports.


Step 2: Consumption values (from table).

2006 → 72.5, \quad 2007 → 88, \quad 2008 → 108, \quad 2009 → 105, \quad 2010 → 127.


Step 3: Compute percentage increase.

2007 over 2006 = \(\dfrac{88-72.5}{72.5} \times 100 = 21.38%\).

2008 over 2007 = \(\dfrac{108-88}{88} \times 100 = 22.73%\).

2009 over 2008 = \(\dfrac{105-108}{108} \times 100 = -2.78%\).

2010 over 2009 = \(\dfrac{127-105}{105} \times 100 = 20.95%\).


Step 4: Conclude.

The highest increase occurred in 2008 (22.73%).

\[ \boxed{2008} \] Quick Tip: Always compute year-on-year changes using the formula \(\dfrac{New - Old}{Old} \times 100\). Negative results imply a decrease.

Step 1: Recall formula.

Population = \(\dfrac{Consumption}{Per capita consumption}\).


Step 2: Use 2008 data.

Consumption = Production – Exports = \(230 - 138 = 92\) million kg.

Per capita consumption = 38.7 kg.


Step 3: Calculate.

Population = \(\dfrac{92 \times 10^6}{38.7}\).

= 2.79 million (approx).

\[ \boxed{2.79\ million} \] Quick Tip: When per capita consumption is given, total consumption ÷ per capita gives the population directly.

Step 1: Use the formula.

Ratio = \(\dfrac{Exports}{Consumption}\).


Step 2: Ratios from table.

2006 → \(120/80 = 1.5\).

2007 → \(130/110 = 1.18\).

2008 → \(120/105 = 1.14\).

2009 → \(120/103 = 1.17\).

2010 → \(131/114 = 1.15\).


Step 3: Identify maximum.

The highest ratio = 1.5 in 2006.

\[ \boxed{2006} \] Quick Tip: For “highest ratio” questions, always compute for each year and compare values, instead of relying on visual inspection of graphs.

Step 1: Formula for population.

Population = \(\dfrac{Consumption}{Per capita consumption}\).


Step 2: Use the given data table.
\[ \begin{array}{|c|c|c|c|} \hline Year & Consumption (million kg) & Per capita (kg) & Population (millions)
\hline 2006 & 72.5 & 36.25 & 2.00
2007 & 88 & 35.2 & 2.5
2008 & 108 & 38.7 & 2.79
2009 & 105 & 40.5 & 2.59
2010 & 127 & 42 & 3.02
\hline \end{array} \]


Step 3: Compare populations.

Highest value of population = 3.02 million in 2010.

\[ \boxed{2010} \] Quick Tip: When population = \(\dfrac{Consumption}{Per capita consumption}\), higher consumption and lower per capita values yield higher population.

Step 1: Recall the concept.

A pie chart represents the whole as a circle of \(360^\circ\). Here, the “whole” = 24 hours of a day.
Therefore: \[ 1^\circ \; in pie chart = \frac{24}{360} = \frac{1}{15} \; hours \]


Step 2: Angle for school sector.

The chart shows school corresponds to \(105^\circ\).


Step 3: Convert angle into hours.
\[ Time in school = 105 \times \frac{1}{15} = 7 \; hours \]


Step 4: Calculate percentage of the day.

Since the day = 24 hours, \[ % of time in school = \frac{7}{24} \times 100 \approx 29.16% \approx 30% \]

\[ \boxed{30%} \] Quick Tip: Always convert the angle to hours when the pie chart is based on 24 hours. Then use the ratio of hours spent to total hours to calculate the percentage.

Step 1: Angles for games and sleeping.

Games = \(30^\circ\).

Sleeping = \(120^\circ\).


Step 2: Convert angles to hours.
\[ Games time = 30 \times \frac{1}{15} = 2 \; hours \] \[ Sleeping time = 120 \times \frac{1}{15} = 8 \; hours \]


Step 3: Compare.

The ratio of time spent on games to sleeping = \(\dfrac{2}{8} = \dfrac{1}{4}\).


Step 4: Express in percentage.
\[ \dfrac{1}{4} \times 100 = 25% \]

\[ \boxed{25%} \] Quick Tip: When comparing two categories in a pie chart, convert both to hours (or keep them as angles since ratio is the same). Then calculate the ratio and convert into percentage.

Step 1: Note current distribution.

Games = \(30^\circ\), Homework = \(45^\circ\), Sleeping = \(120^\circ\).


Step 2: New condition.

Games time must become equal to Homework time = \(45^\circ\).

So, Games angle increases from \(30^\circ\) to \(45^\circ\).


Step 3: Source of adjustment.

The extra \(15^\circ\) (i.e., \(45 - 30 = 15^\circ\)) is taken from Sleeping since other activities are constant.


Step 4: Calculate decrease in sleeping.

Original sleeping angle = \(120^\circ\).

New sleeping angle = \(120 - 15 = 105^\circ\).

Decrease = \(15^\circ\).


Step 5: Percentage decrease.
\[ % decrease = \frac{15}{120} \times 100 = 12.5% \]

\[ \boxed{12.5%} \] Quick Tip: When one activity’s time is increased, always identify which sector decreases. Then calculate the decrease relative to the original value of that sector.

Step 1: Angles from the pie chart.

School sector = \(105^\circ\), Homework sector = \(45^\circ\).


Step 2: Convert angles into hours.

Since \(360^\circ\) = 24 hours, \[ 1^\circ = \frac{24}{360} = \frac{1}{15}\ hours \]


Step 3: Hours for school and homework.

School time = \(105 \times \frac{1}{15} = 7\ hours\).

Homework time = \(45 \times \frac{1}{15} = 3\ hours\).


Step 4: Find the difference.

Difference = \(7 - 3 = 4\ hours\).

\[ \boxed{4\ hours} \] Quick Tip: When comparing times in pie charts, always convert each sector angle into hours, then subtract.

Step 1: Time spent on homework.

Homework sector angle = \(45^\circ\).

Convert into hours: \[ \frac{45}{360} \times 24 = 3\ hours \]


Step 2: Time for Mathematics.

Mathematics = \(\tfrac{1}{3}\) of homework = \(\tfrac{1}{3} \times 3 = 1\ hour\).


Step 3: Remaining homework subjects.

Total homework = 3 hours.

Maths = 1 hour.

Other subjects = \(3 - 1 = 2\ hours\).

\[ \boxed{2\ hours} \] Quick Tip: When a fraction of a total is allocated to one activity, multiply the fraction by the total. Subtract this from the total to find the remainder for other activities.

Step 1: Identify the figure.

We have three parallel lines cut by two transversals. The segments AB, BC are on one transversal, and DE, EF are on the other transversal.


Step 2: Apply Basic Proportionality Theorem (Thales’ theorem).

According to the theorem, when three parallel lines cut two transversals, the corresponding segments are proportional: \[ \frac{AB}{BC} = \frac{DE}{EF} \]


Step 3: Substitute values.
\[ \frac{AB}{BC} = \frac{DE}{EF} \] \[ \frac{2}{4} = \frac{1.5}{EF} \]


Step 4: Solve for EF.

Cross multiply: \[ EF = \frac{1.5 \times 4}{2} \] \[ EF = \frac{6}{2} = 3 \ cm \]

\[ \boxed{3\ cm} \] Quick Tip: In problems involving parallel lines and transversals, always apply the Basic Proportionality Theorem (Thales' theorem). Ratios of corresponding segments remain equal.

Step 1: Use the power rule of logarithms.

For any \(a>0\), \(\log_{10}(10^{a})=a\).

\Rightarrow \(\log_{10} 10 = 1,\; \log_{10} 10^{2} = 2,\; \ldots,\; \log_{10} 10^{n} = n\).


Step 2: Convert the expression to an arithmetic sum.

Thus, \[ \log_{10} 10 + \log_{10} 10^{2} + \cdots + \log_{10} 10^{n} = 1 + 2 + \cdots + n. \]


Step 3: Apply the sum of first \(n\) natural numbers.
\[ 1+2+\cdots+n = \frac{n(n+1)}{2} \;=\; \frac{n^{2}+n}{2}. \]
\[ \boxed{\dfrac{n^{2}+n}{2}} \] Quick Tip: Whenever you see \(\log_{b}(b^{k})\), immediately reduce it to \(k\). Long-looking sums of logs often telescope into simple arithmetic progressions.

Step 1: Translate the statement into an equation.

Let the number be \(x \neq 0\). Given: \[ x + \frac{1}{x} = 3\!\left(x - \frac{1}{x}\right). \]


Step 2: Clear denominators and simplify.

Multiply both sides by \(x\) (\(x\neq 0\)): \[ x^{2} + 1 = 3\left(x^{2} - 1\right) \Rightarrow x^{2} + 1 = 3x^{2} - 3. \]
Bring all terms to one side: \[ 0 = 3x^{2} - 3 - x^{2} - 1 = 2x^{2} - 4 \Rightarrow 2x^{2} = 4 \Rightarrow x^{2} = 2. \]


Step 3: Conclude the values of \(x\).
\[ x = \pm \sqrt{2}. \]
\[ \boxed{\pm \sqrt{2}} \] Quick Tip: Equations involving \(x\) and \(\tfrac{1}{x}\) are easiest after multiplying through by \(x\) (or \(x^{2}\)) to clear fractions—just remember to keep \(x\neq 0\).

Step 1: Identify the first and last multiples of 9 in the range.

Numbers strictly between \(10\) and \(300\) that are divisible by \(9\) start at \(9\times 2=18\) and go up to \(9\times 33=297\) (since \(9\times 34=306>300\)).


Step 2: Treat as an arithmetic progression (AP).

The AP is \(18, 27, 36, \ldots, 297\) with first term \(a=18\), common difference \(d=9\), last term \(l=297\).


Step 3: Count the terms.

For an AP, \(n=\dfrac{l-a}{d}+1 \Rightarrow n=\dfrac{297-18}{9}+1=\dfrac{279}{9}+1=31+1=32\).
\[ \boxed{32} \] Quick Tip: To count multiples of \(k\) in \((A,B)\), use the first multiple \(\ge A+1\) and the last multiple \(\le B-1\), then apply \(n=\dfrac{l-a}{k}+1\).

Step 1: Use the relation between permutations and combinations.

Recall \({}^{n}P_{x} = {}^{n}C_{x}\,x!\).

Given \({}^{n}P_{x}=336\) and \({}^{n}C_{x}=56\), \[ 336 = 56 \cdot x! \Rightarrow x! = \frac{336}{56}=6 \Rightarrow x=3. \]


Step 2: Substitute \(x=3\) into the combination value.
\[ {}^{n}C_{3} = 56 \Rightarrow \frac{n(n-1)(n-2)}{3!}=56 \Rightarrow \frac{n(n-1)(n-2)}{6}=56 \Rightarrow n(n-1)(n-2)=336. \]


Step 3: Solve for \(n\).

Test \(n=8\): \(8\cdot7\cdot6=336\) \Rightarrow true. Hence \(n=8\).
\[ \boxed{n=8,\; x=3} \] Quick Tip: When both \({}^{n}C_{x}\) and \({}^{n}P_{x}\) are given, first use \({}^{n}P_{x}={}^nC_x\,x!\) to get \(x\), then plug into the combination to find \(n\).

Step 1: First triangle.

Side of the equilateral triangle = 24 cm.

Perimeter = \( 3 \times 24 = 72 \, cm \).


Step 2: Second triangle.

Joining the midpoints halves each side: new side = \( \frac{24}{2} = 12 \, cm \).

Perimeter = \( 3 \times 12 = 36 \, cm \).


Step 3: Third triangle.

Again, sides halve: new side = \( \frac{12}{2} = 6 \, cm \).

Perimeter = \( 3 \times 6 = 18 \, cm \).


Step 4: General pattern.

Perimeters form a geometric progression: \[ 72, 36, 18, 9, \dots \]
First term \(a = 72\), common ratio \(r = \frac{1}{2}\).


Step 5: Infinite sum.

The sum of an infinite GP is \[ S = \frac{a}{1-r} = \frac{72}{1 - \tfrac{1}{2}} = \frac{72}{\tfrac{1}{2}} = 144. \]


Final Answer: \[ \boxed{144 \, cm} \] Quick Tip: Whenever a process involves repeatedly halving or doubling, look for a geometric progression. The formula \( S = \frac{a}{1-r} \) is key for infinite sums when \(|r|<1\).

Step 1: Total days in a leap year.

A leap year has 366 days = 52 weeks + 2 extra days.


Step 2: Nature of the extra days.

These 2 extra days can be:
(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun).

Thus, there are 7 equally likely cases.


Step 3: Condition for 53 Sundays.

If one of the extra days is Sunday, then total Sundays = 53.

This happens in 2 cases: (Sat, Sun) and (Sun, Mon).

So, probability = \(\tfrac{2}{7}\).


Step 4: Condition for 53 Mondays.

If one of the extra days is Monday, then total Mondays = 53.

This happens in 2 cases: (Sun, Mon) and (Mon, Tue).

So, probability = \(\tfrac{2}{7}\).


Step 5: Combined condition.

We need either 53 Sundays or 53 Mondays.

But note that (Sun, Mon) counts both, so it is common.

By addition rule: \[ P = \tfrac{2}{7} + \tfrac{2}{7} - \tfrac{1}{7} = \tfrac{3}{7}. \]


Final Answer: \[ \boxed{\tfrac{3}{7}} \] Quick Tip: In leap-year probability problems, always check the 2 extra days beyond 52 weeks. The overlapping case must be subtracted when events are not mutually exclusive.

Step 1: Recall intercept form.

The intercept form of a line is \[ \frac{x}{a} + \frac{y}{b} = 1 \]
where \(a\) is the x-intercept and \(b\) is the y-intercept.


Step 2: Rewrite the given equation.

Given: \(3x + 4y - 12 = 0\).

Rearranging: \[ \frac{3x}{12} + \frac{4y}{12} = 1 \] \[ \frac{x}{4} + \frac{y}{3} = 1 \]


Step 3: Identify the intercepts.

From the equation, \(a = 4\) and \(b = 3\).


Final Answer: \[ \boxed{x-intercept = 4, y-intercept = 3} \] Quick Tip: To find intercepts quickly, set \(y=0\) for x-intercept and \(x=0\) for y-intercept. Alternatively, reduce the line equation into intercept form.

Step 1: Use the average condition.

Average of 4 primes = 35.
\[ \frac{a+b+c+d}{4} = 35 \quad \Rightarrow \quad a+b+c+d = 140 \]


Step 2: Use the equidistant conditions.

- \(a\) and \(d\) are equidistant from 36. So: \[ a + d = 72 \]
- \(b\) and \(c\) are equidistant from 34. So: \[ b + c = 68 \]


Step 3: Verify consistency.

From Step 1: \(a+b+c+d = 140\).

From Step 2: \((a+d) + (b+c) = 72 + 68 = 140\). ✓ Correct.


Step 4: Check prime pairs.

We need primes satisfying: \[ a+d=72, \quad b+c=68, \quad aTesting prime combinations: \[ a=29, \quad b=31, \quad c=37, \quad d=43 \]
All are prime and satisfy conditions.


Step 5: Find difference.
\[ d - a = 43 - 29 = 14 \]


Final Answer: \[ \boxed{14} \] Quick Tip: In prime-based average/equidistant problems, always translate conditions into equations like \(a+d\) and \(b+c\). Then test small prime values systematically.

Step 1: Selling Price and Profit.

Selling Price (SP) of rasgulla = Rs. 15 per kg.

Profit = \(66 \tfrac{2}{3}%\) = \(\tfrac{2}{3}\) of cost price.
\[ SP = CP \times \left(1 + \tfrac{2}{3}\right) = \tfrac{5}{3} \times CP \]

So, \[ 15 = \tfrac{5}{3} \times CP \quad \Rightarrow \quad CP = 9 \]


Step 2: Express cost in terms of flour and sugar.

1 kg rasgulla is made of flour and sugar in ratio 5:3.
Let CP of flour = 3x and CP of sugar = 7x (since sugar:flour = 7:3).


Step 3: Compute weighted average CP.

Per kg rasgulla = \(\tfrac{5}{8} \, kg flour + \tfrac{3}{8} \, kg sugar\).

So CP = \[ \frac{5}{8} \times (3x) + \frac{3}{8} \times (7x) = \frac{15x + 21x}{8} = \frac{36x}{8} = \frac{9x}{2} \]


Step 4: Equating with known CP.
\[ \frac{9x}{2} = 9 \quad \Rightarrow \quad x = 2 \]


Step 5: Find sugar cost.

CP of sugar = \(7x = 7 \times 2 = 14\).


Final Answer: \[ \boxed{Rs.\;14 \; per kg} \] Quick Tip: In mixture problems, always reduce profit percentage into fraction form. Then use weighted averages of components (in ratio form) to match with cost price.

Step 1: Assume original price.

Let original price of sugar = Rs. \(x\) per kg.


Step 2: Quantity with original price.

With Rs. 240, sugar bought = \(\tfrac{240}{x}\) kg.


Step 3: Quantity after 20% reduction.

New price = \(0.8x\).

Sugar bought with Rs. 240 = \(\tfrac{240}{0.8x} = \tfrac{300}{x}\) kg.


Step 4: Condition of 6 kg more.
\[ \frac{300}{x} - \frac{240}{x} = 6 \] \[ \frac{60}{x} = 6 \quad \Rightarrow \quad x = 10 \]


Final Answer: \[ \boxed{Rs.\;10 \; per kg} \] Quick Tip: In price-reduction problems, always compare quantities bought with the same money. The difference gives the relation to solve for original price.

Step 1: Volume of sphere.

Let the radius of sphere = \(r\).

Volume of sphere: \[ V_s = \frac{4}{3}\pi r^3 \]


Step 2: Volume of cone formed.

When recast, cone radius = \(r\), cone height = \(h\).

Volume of cone: \[ V_c = \frac{1}{3}\pi r^2 h \]


Step 3: Equating volumes.

Since melted volume = recast volume: \[ \frac{4}{3}\pi r^3 = \frac{1}{3}\pi r^2 h \] \[ 4r = h \]


Step 4: Ratio of height to radius.
\[ \frac{h}{r} = \frac{4r}{r} = 4 : 1 \]


Final Answer: \[ \boxed{4 : 1} \] Quick Tip: In melting-and-recasting problems, always equate the volumes of the original and new solids. Then solve for the required ratio or dimension.

Step 1: Recall relation.

Time taken = \(\tfrac{Distance}{Speed}\).

We need ratio of \(\tfrac{Time}{Speed}\).


Step 2: Assume distance.

Let distance covered = 16 km (for easy calculation).


Step 3: Compute times.

- Scooter speed = 1 km/h → Time = 16 h.

- Car speed = 4 km/h → Time = 4 h.

- Train speed = 16 km/h → Time = 1 h.


Step 4: Compute time/velocity for each.

- Scooter: \(\tfrac{16}{1} = 16\).

- Car: \(\tfrac{4}{4} = 1\).

- Train: \(\tfrac{1}{16} = \tfrac{1}{16}\).


Step 5: Ratio.
\[ 16 : 1 : \tfrac{1}{16} = 256 : 16 : 1 \]


Final Answer: \[ \boxed{256 : 16 : 1} \] Quick Tip: When equal distances are covered, times are inversely proportional to speeds. Ratios involving time/speed must be carefully scaled to avoid fractions.

Step 1: Work rates.

A alone completes in 15 days → Work rate of A = \(\tfrac{1}{15}\).

B is twice efficient → Work rate of B = \(\tfrac{2}{15}\).


Step 2: Let B join after \(x\) days.

Then:
- A works alone for \(x\) days.

- A and B together work for \((11 - x)\) days.


Step 3: Write total work equation.
\[ x \times \frac{1}{15} + (11 - x) \times \left(\frac{1}{15} + \frac{2}{15}\right) = 1 \] \[ \frac{x}{15} + (11 - x)\times \frac{3}{15} = 1 \] \[ \frac{x}{15} + \frac{33 - 3x}{15} = 1 \] \[ \frac{33 - 2x}{15} = 1 \quad \Rightarrow \quad 33 - 2x = 15 \quad \Rightarrow \quad x = 9 \]


Step 4: Find days of working together.

They worked together for \(11 - x = 11 - 9 = 2\) days.


Final Answer: \[ \boxed{2 \; days} \] Quick Tip: In work problems, convert efficiency into work rates. Always set up an equation based on total work = 1, and solve systematically.

Step 1: Write total.
\[ A + B + C + D = 56 \quad (Eq 1) \]


Step 2: B, C, D together.
\[ B + C + D = 4.6A \quad (Eq 2) \]


Step 3: A, C, D together.
\[ A + C + D = \frac{11}{3} B \quad (Eq 3) \quad since 366.66% = \tfrac{11}{3} \]


Step 4: C condition.
\[ C = 0.4(A + B + D) = \tfrac{2}{5}(A+B+D) \quad (Eq 4) \]


Step 5: Solve equations.

From Eq (1) and (2): \[ A + (4.6A) = 56 \quad \Rightarrow \quad 5.6A = 56 \quad \Rightarrow \quad A = 10 \]

From Eq (1) and (3): \[ 10 + C + D = \tfrac{11}{3}B \]

From Eq (2): \[ B + C + D = 46 \]

From Eq (1): \[ 10 + B + C + D = 56 \quad \Rightarrow \quad B + C + D = 46 \]

Now using Eq (4): \[ C = 0.4(10 + B + D) = 0.4(B + D + 10) \]

Substituting values, solving gives: \[ B = 12, \quad C = 16, \quad D = 18 \]


Final Answer: \[ \boxed{18 \; lakhs} \] Quick Tip: When multiple contribution ratios are given, form equations step by step. Then substitute systematically to eliminate variables.

Step 1: Assume Saroj’s salary.

Let Saroj’s salary = 100x.


Step 2: Raju’s and Ram’s salaries.

Raju’s salary = \(100x - 20x = 80x\).

Ram’s salary = \(100x - 30x = 70x\).


Step 3: Compare difference.

Difference = \(80x - 70x = 10x\).


Step 4: Percentage comparison.
\[ Percent more = \frac{10x}{70x} \times 100 = \frac{10}{70}\times 100 = 14.28% \]


Final Answer: \[ \boxed{14.28%} \] Quick Tip: When comparing “how much more,” always take the difference over the smaller quantity.

Step 1: Recall cylinder volume.

Wire is cylindrical: \[ V = \pi r^2 h \]


Step 2: New radius.

New radius = \(\tfrac{r}{3}\). Let new height = \(h_2\), original height = \(h_1\).


Step 3: Equating volumes.
\[ \pi r^2 h_1 = \pi \left(\tfrac{r}{3}\right)^2 h_2 \] \[ r^2 h_1 = \tfrac{r^2}{9} h_2 \] \[ h_2 = 9h_1 \]


Step 4: Ratio of lengths.
\[ \frac{h_2}{h_1} = 9 : 1 \]


Final Answer: \[ \boxed{9 \; times} \] Quick Tip: If radius is scaled down by factor \(k\), then length increases by factor \(k^2\) to keep volume constant.

Step 1: Recall passage context.

The passage begins with: “For man, the vast marvel is to be alive. For man as for flower and beast and bird, the supreme triumph is to be most vividly, most perfectly alive.”


Step 2: Interpret usage of ‘triumph’.

Here, “triumph” refers to the ultimate achievement or success of life itself. It is not about sin, loss, or sorrow, but about the **victory of life**.


Step 3: Conclude.

Thus, triumph in the author’s usage means **victory**.


Final Answer: \[ \boxed{victory} \] Quick Tip: When interpreting words in RC passages, always check the contextual meaning rather than the dictionary meaning alone.

Step 1: Refer to the text.

The author writes: “The dead may look after the afterwards. But the magnificent here and now of life in the flesh is ours, and ours only for a time.”


Step 2: Meaning of ‘only for a time’.

The author stresses that life in the flesh is temporary, lasting only for a short while, not forever or months.


Step 3: Conclusion.

Thus, the living should cherish life only for the short while it exists.


Final Answer: \[ \boxed{for only a short while} \] Quick Tip: Look for explicit phrases in the passage (like “only for a time”) when answering inference-based questions.

Step 1: Passage reference.

The author says: “We ought to dance with rapture, that we should be alive and in the flesh.”


Step 2: Meaning of ‘rapture’.

Rapture denotes a feeling of intense ecstasy, happiness, or joy. In this context, it reflects the joy of being alive and part of existence.


Step 3: Eliminate wrong options.

- Trepidation = fear, does not fit.

- Thrill = excitement, partial but not exact.

- Fear = opposite meaning.

- Joy = perfect match with ecstasy.


Final Answer: \[ \boxed{joy} \] Quick Tip: Words like “rapture” or “ecstasy” often denote extreme joy in literature passages. Contextual clues (e.g., “dance with rapture”) confirm the correct meaning.

Step 1: Passage reference.

In the last line of the passage, the author says: “the mind has no existence by itself, it is only the glitter of the sun on the surface of the waters.”


Step 2: Interpretation.

The comparison here shows that the mind is not an independent entity but merely a reflection, much like sunlight glitters on the water’s surface.


Step 3: Eliminate wrong options.

- Mirage = optical illusion, not correct here.

- Illusion = false perception, weaker meaning.

- Magic = not implied.

- Reflection = best fit.


Final Answer: \[ \boxed{a reflection} \] Quick Tip: Always look at similes/metaphors in passages carefully — they often directly guide the meaning of abstract words.

Step 1: Identify style of writing.

The author is not describing society (social) or prescribing right/wrong (moral). While reflective elements exist, the dominant tone is philosophical.


Step 2: Justify.

The passage discusses the essence of life, triumph of being alive, the relation of man with nature, and the concept of the mind’s existence — all abstract and philosophical ideas.


Final Answer: \[ \boxed{philosophical} \] Quick Tip: Tone questions often ask for the dominant style — focus on the overall theme rather than isolated lines.

Step 1: Recall central theme.

The author emphasizes the supreme triumph of being alive, the joy of existence, and the importance of living in the “here and now.”


Step 2: Analyze options.

- \emph{The Surface of the Waters → focuses only on the last metaphor, too narrow.

- \emph{My Mind → does not capture the broader theme of life.

- \emph{The Human Race → partially relevant, but not central.

- \emph{Alive and Kicking → perfectly conveys the essence: the celebration of life itself.


Step 3: Conclude.

Hence, the most suitable title is “Alive and Kicking.”


Final Answer: \[ \boxed{Alive and Kicking} \] Quick Tip: When choosing titles, always pick the one that reflects the main theme of the passage, not just one small part.

Step 1: Recall definitions.

- Redundancy = unnecessary repetition of words or ideas.

- Hyperbole = deliberate exaggeration.

- Alliteration = repetition of consonant sounds at the beginning of words.

- Allegory = symbolic story or representation.


Step 2: Apply to question.

Since the question refers to repetition of messages or superfluous use of expressions, the correct term is **redundancy**.


Final Answer: \[ \boxed{redundancy} \] Quick Tip: Redundancy means unnecessary repetition. Always distinguish it from hyperbole (exaggeration) and alliteration (sound repetition).

Step 1: Analyze given relation.

Pen is dipped in ink and used to write on paper.


Step 2: Match with options.

- (A) Watch : dial : strap → parts of a watch, unrelated.

- (B) Book : paper : words → composition of book, unrelated to dipping/using.

- (C) Farmer : plough : field → person-tool-field relation, not analogous.

- (D) Colour : brush : canvas → Brush is dipped in colour/paint and used to paint on canvas. Perfect analogy.


Final Answer: \[ \boxed{colour : brush : canvas} \] Quick Tip: In analogy questions, carefully identify the functional relationship between the first set of terms, then find the exact parallel.

Step 1: Meaning of ‘realia’.

In educational context, “realia” refers to real-life objects used as teaching aids to make learning more effective.


Step 2: Evaluate options.

- Theoretical constructs → abstract, opposite of realia.

- Fabricated examples → artificial, not correct.

- Objects from real life → exact definition of realia.

- Based on reality → vague, not precise.


Step 3: Conclude.

Hence, the correct option is **objects from real life**.


Final Answer: \[ \boxed{objects from real life} \] Quick Tip: Realia = real-life teaching aids. For example: using currency notes in a math class or fruits in a language lesson.

Step 1: Analyze subject-verb agreement.

The subject = “gloves” (plural). So, verb should be “do” and “belong” (not belongs).


Step 2: Check options.

- (A) Incorrect → “Are belonging” is wrong tense form.

- (B) Incorrect → “Does this gloves” is wrong; “gloves” is plural, should be “do”.

- (C) Incorrect → “Do … belongs” is wrong; plural subject takes base verb “belong”.

- (D) Correct → “Do these gloves belong” is correct.


Final Answer: \[ \boxed{‘Do these gloves belong to you?’ she asked.} \] Quick Tip: With plural subjects, always use “do” and the base form of the verb (not -s form).

Step 1: Identify correct article usage.

We must use: “a house”, “a street”, “the countryside” → correct article sequence.


Step 2: Check options.

- (A) Correct → all articles used properly.

- (B) Incorrect → “the house in the street countryside” is grammatically wrong.

- (C) Incorrect → “the street in the countryside” creates confusion, needs article adjustment.

- (D) Incorrect → “a house is old” is wrong; it should be “the house is old”.


Final Answer: \[ \boxed{Option (A)} \] Quick Tip: Always check article usage carefully (a/an/the) when multiple nouns are listed in a sentence.

Step 1: Rule of comparison.

In English grammar, “compare” is followed by “with” when comparing two or more things.


Step 2: Check options.

- (A) Incorrect → “compare … to” suggests similarity, not proper here.

- (B) Incorrect → “for their own” is ungrammatical.

- (C) Correct → “compare … with their own” is accurate.

- (D) Incorrect → “teaching method” should be plural; also “with their own” makes sense only in plural.


Final Answer: \[ \boxed{Option (C)} \] Quick Tip: Use “compare with” when examining similarities and differences. Use “compare to” only when highlighting resemblance.

Step 1: Analyze each option.

- (A) Correct → Proper clause structure, use of “that” and “which” is grammatically valid.

- (B) Incorrect → “what” is wrongly used in place of “that.”

- (C) Incorrect → “for which” construction is incorrect in this context.

- (D) Incorrect → “wherein” is misapplied and ungrammatical here.


Step 2: Conclude.

Only (A) gives a grammatically correct and meaningful sentence.


Final Answer: \[ \boxed{Option (A)} \] Quick Tip: Relative pronouns like “that” and “which” must be used carefully for defining clauses. Avoid misuse of “what” or “wherein.”

Step 1: Identify correct tense.

The word “yesterday” clearly refers to a past action. Therefore, the correct tense must be **simple past**.


Step 2: Eliminate wrong options.

- (A) Wrong → “have completed” is present perfect, which doesn’t fit with “yesterday.”

- (B) Wrong → “did completed” is grammatically incorrect (double past).

- (C) Wrong → “have had completed” is clumsy and incorrect.

- (D) Correct → “I completed the work yesterday” is simple past and accurate.


Final Answer: \[ \boxed{I completed the work yesterday.} \] Quick Tip: When a definite past time (e.g., yesterday, last week) is given, always use the simple past tense.

Step 1: Rule of preposition with “crash.”

The verb “crash” is normally followed by the preposition **into** (crash into something).


Step 2: Eliminate wrong options.

- (A) Wrong → “crashed with” is incorrect usage.

- (B) Correct → “crashed into” is the proper idiomatic usage.

- (C) Wrong → “crashed against” is uncommon and incorrect here.

- (D) Wrong → “crashed before the truck” changes meaning entirely.


Final Answer: \[ \boxed{The train couldn’t stop in time and crashed into the truck.} \] Quick Tip: Always use “crash into” to indicate direct collision.

Step 1: Understand the meaning of “referred to”.

The phrase “referred to” means “called” or “described as” in the context of identifying something by a certain name.


Step 2: Eliminate incorrect options.

- (B) “included in” means being a part of something, which does not match the sense here.

- (C) “supported for” means to be in favor of something, which is unrelated.

- (D) “reformed as” means changed or improved, which is not the same.


Step 3: Confirm the correct option.

Only (A) “described as” conveys the exact meaning of “referred to.”

\[ \boxed{described as} \] Quick Tip: When a word is “referred to” in a passage, it usually means it is “called” or “described as” something. Look for synonymous expressions of naming/description.

Step 1: Understand the meaning of “independent”.

In this passage, “independent” is used in the sense of being free from outside influence or control, especially when describing a legal body.


Step 2: Analyze each option.

- (A) “self-centered” means selfish, which is not the same.

- (C) “self-seeking” means pursuing selfish goals, not correct.

- (D) “self-possessed” means calm and confident, unrelated here.


Step 3: Confirm the correct option.

Only (B) “impartial” matches the idea of being free from external control or bias.

\[ \boxed{impartial} \] Quick Tip: Always match the synonym with the context in which the word is used. “Independent” in political/legal contexts often refers to impartiality, not personal traits.

Step 1: Meaning of “deter”.

The word “deter” means to discourage, stop, or prevent someone from doing something.


Step 2: Analyze the options.

- (A) “swell” means to increase in size, not related.

- (C) “propel” means to push or drive forward, opposite in sense.

- (D) “lucubrate” means to study diligently, completely unrelated.


Step 3: Confirm correct option.

Only (B) “prevent” matches the meaning of “deter.”

\[ \boxed{prevent} \] Quick Tip: “Deter” is almost always used in contexts of discouraging or preventing an action, especially in law and policy.

Step 1: Meaning of “grievances”.

“Grievances” refers to problems, injustices, or complaints that people raise, usually against authority or unfair treatment.


Step 2: Eliminate incorrect options.

- (B) “dishonesty” means lack of truthfulness, not the same.

- (C) “committees” are groups of people, irrelevant here.

- (D) “opinions” are beliefs or views, not necessarily complaints.


Step 3: Confirm correct option.

Only (A) “complaints” correctly matches “grievances.”

\[ \boxed{complaints} \] Quick Tip: “Grievances” are formal or informal complaints raised by citizens or employees. Always link it with dissatisfaction or injustice.

Step 1: Meaning of “ombudsman”.

An ombudsman is an independent official appointed to investigate complaints, especially those made by individuals against government agencies, bureaucrats, or politicians.


Step 2: Analyze options.

- (B) Complaints by government against citizens → incorrect, opposite direction.

- (C) Complaints by citizens against citizens → not the role of ombudsman.

- (D) Complaints by government against officials → not the function of ombudsman.


Step 3: Confirm correct option.

Only (A) fits the accurate definition of “ombudsman.”

\[ \boxed{a government appointee who investigates complaints by private persons against bureaucrats and/or politicians} \] Quick Tip: Remember: an ombudsman protects ordinary citizens by addressing their complaints against government functionaries, ensuring accountability.

Step 1: Identify the head verb and its complements.

The core phrase is \emph{to discuss (in detail) assorted problems. The verb \emph{discuss already means “talk \emph{about.” Therefore, it does not take the preposition \emph{about.


Step 2: Apply the rule of redundancy.

If a verb semantically includes a preposition, adding that preposition is redundant and ungrammatical in standard usage.

Examples: \emph{discuss the plan (not \emph{discuss about the plan); \emph{enter the room (not \emph{enter into the room) in formal style.


Step 3: Check the other parts for correctness.

(A) \emph{came to (a decision) — idiomatic collocation.

(B) \emph{to discuss — infinitive of purpose is correct.

(D) \emph{a long time — correct adverbial noun phrase.


Step 4: Correct the sentence.

Remove \emph{about.

\Rightarrow\ Corrected: \emph{The committee came to a decision to discuss in detail assorted problems that people have been facing for a long time.

\[ \boxed{about} \] Quick Tip: Avoid redundancy: \textbf{discuss} \(\Rightarrow\) never use \emph{about}; \textbf{repeat} \(\Rightarrow\) avoid \emph{again}; \textbf{return} \(\Rightarrow\) avoid \emph{back}.

Step 1: Recall the tense rule for Past Perfect.

Past Perfect (had + V\textsubscript{3}) is required only to show an action completed \emph{before another past action/time point. Without a second past reference, use the Simple Past.


Step 2: Locate the time framework in the sentence.

The clause \emph{you must admit refers to a present assertion about a past fact; inside that, \emph{he had had tremendous influence merely states a past condition, not one prior to another specific past event. Hence \emph{had (simple past of \emph{have) suffices.


Step 3: Evaluate other choices.

(A) \emph{see eye to eye — correct idiom meaning “agree.”

(C) \emph{influence over — correct preposition with \emph{influence.


Step 4: Correct the sentence.

Replace \emph{had had with \emph{had.

\Rightarrow\ Corrected: \emph{…you must admit that he had tremendous influence over a great many followers.

\[ \boxed{had had} \] Quick Tip: Use \textbf{Past Perfect} only with a clear “\emph{past-before-past}” timeline (e.g., \emph{By 2010, he had published five papers}). Otherwise, prefer \textbf{Simple Past}.

Step 1: Check for redundant adverbs.

The verb repeat already carries the meaning “do again.” Therefore, adding \emph{again duplicates the meaning \Rightarrow redundancy/error.


Step 2: Confirm remaining parts.

(A) \emph{inquired with — acceptable collocation in contemporary usage (formal alternative: \emph{inquired of).

(B) \emph{to repeat — correct infinitive complement after “would like.”


Step 3: Correct the sentence.

Delete \emph{again.

\Rightarrow\ Corrected: \emph{The Principal inquired with the students if they would like their teacher to repeat the lesson.

\[ \boxed{again} \] Quick Tip: Watch for classic redundancies: \emph{repeat again}, \emph{combine together}, \emph{each and every} (in formal writing).

Step 1: Subject–verb agreement in “one of + plural noun.”

Structure: \emph{One of the + plural noun + singular verb. Here, \emph{One of the security guards (singular subject) \(\Rightarrow\) \emph{rushed (singular past form) — correct.


Step 2: Parallel verbs in a series.

\emph{rushed forward, unlocked the gate and asked… — three coordinated simple past verbs; parallelism is correct.


Step 3: Word choice and prepositions.

\emph{forward (adverb) — apt with \emph{rushed. \emph{objectionable (adjective) correctly modifies the understood noun (e.g., \emph{items). The clause \emph{asked whether I had… uses \emph{whether properly for a yes/no content question.


Step 4: Conclusion.

All elements are grammatical and idiomatic; hence No error.

\[ \boxed{no error} \] Quick Tip: With “\textbf{one of the + plural noun},” keep the verb \textbf{singular}. Also ensure parallel verb forms in coordinated actions.

Step 1: Identify the opening phrase.

The sentence starts with “I always told them …” so the next fragment should logically continue with a clause beginning with “How for me she …” (P).


Step 2: Arrange the next logical unit.

P = “How for me she …” needs continuation. The natural next part is “Was like a family’s” (R), forming: \emph{How for me she was like a family’s….


Step 3: Complete the thought.

After R, the correct addition is S = “Elder and how she has,” so the sentence reads: \emph{How for me she was like a family’s elder and how she has ….


Step 4: Conclude with the remaining phrase.

Finally, Q = “Always exuded warmth” provides the correct ending.


Step 5: Verify full sentence.

“I always told them how for me she was like a family’s elder and how she has always exuded warmth and consideration for me.”
This is grammatically correct and meaningful.

\[ \boxed{PRSQ} \] Quick Tip: When solving rearrangement questions, always look for the most natural starting fragment and check for subject–verb agreement to build the sequence logically.

Step 1: Identify the logical start.

The sentence begins “A year or …” which indicates a time reference. The correct continuation is “so after leaving Bangkok” (R).


Step 2: Describe Bangkok.

After R, the descriptive fragments of Bangkok follow. Q = “the largest urban area and primary” fits as the next part.


Step 3: Complete the descriptive noun phrase.

Q must be completed with S = “city of Thailand and a place of unique beauty.” Together, they describe Bangkok fully: “the largest urban area and primary city of Thailand and a place of unique beauty.”


Step 4: Conclude with the main action.

Finally, P = “I picked up a liking for learning foreign languages.” This links naturally after the descriptive setting.


Step 5: Verify full sentence.

“A year or so after leaving Bangkok, the largest urban area and primary city of Thailand and a place of unique beauty, I picked up a liking for learning foreign languages.”
This is grammatically complete and meaningful.

\[ \boxed{RQSP} \] Quick Tip: In sentence rearrangement, descriptive clauses (like place descriptions) usually come together as one block before the main clause. Always group related ideas.

Step 1: Understand usage of “comprise.”

The verb “comprise” means “to consist of” or “to be made up of.” It is a transitive verb and does not take an additional preposition like “of.”


Step 2: Check each option.

- (A) “comprises of” — incorrect; redundancy (same as saying “consists of of”).

- (C) “consists” — incomplete; correct usage is “consists of,” so this is wrong here.

- (D) “comprises to” — ungrammatical, does not exist in English usage.


Step 3: Confirm correct answer.

Thus, the correct choice is (B) “comprises.”
Sentence: \emph{This book comprises five sections.

\[ \boxed{comprises} \] Quick Tip: Remember: “comprise” = “consist of.” Avoid “comprises of.” Example: \emph{The team comprises ten players.}

Step 1: Identify the grammatical case required.

Here, the pronoun is the object of the preposition “to.” Objects require the objective case, not the subjective one.


Step 2: Check each option.

- (A) “I” — subjective case, incorrect in this context.

- (B) “me” — correct objective case.

- (C) “myself” — reflexive pronoun, but no reflexive action is implied.

- (D) “myself too” — grammatically awkward and incorrect.


Step 3: Confirm correct answer.

Correct sentence: \emph{…the rest went directly to my daughter and me.

\[ \boxed{me} \] Quick Tip: Use “I” for subject and “me” for object. Example: \emph{She and I went to the market.} / \emph{This gift is for you and me.}

Step 1: Recall the meaning of “hinterland.”

Hinterland = an area lying inland, away from the coast or riverbanks, often less developed compared to coastal/urban regions.


Step 2: Eliminate incorrect options.

- (A) “Isthmus” = narrow strip of land joining two larger land areas, wrong.

- (B) “Archipelago” = group of islands, not related.

- (D) “Swamps” = wetland area, not inland territory.


Step 3: Confirm correct answer.

Thus, the correct word is (C) “Hinterland.”

\[ \boxed{Hinterland} \] Quick Tip: “Hinterland” = the inland area behind the coast. Think of “back country.” Opposite of coastal/port regions.

Step 1: Identify voice and tense.

The given sentence “Who is creating this mess?” is in the active voice, present continuous tense (\emph{is creating).


Step 2: Change from active to passive.

Active: \emph{Who is creating this mess?
Passive: \emph{By whom is this mess being created?


Step 3: Check all options.

- (A) Incorrect tense and construction.

- (B) Uses “has been created” (present perfect continuous passive), wrong tense.

- (C) Word order error → not grammatically correct.

- (D) Correct tense + correct passive structure.


Step 4: Conclude.

The only grammatically correct passive transformation is (D).

\[ \boxed{By whom is this mess being created?} \] Quick Tip: When changing questions into passive voice, always maintain the tense and ensure subject–auxiliary inversion is correct (e.g., “Who is creating” → “By whom is … being created”).

Step 1: Identify the original structure.

The sentence is in active voice: \emph{You should open the wine…. It has two clauses:
- Main clause: “You should open the wine”
- Time clause: “before you use it”


Step 2: Convert to passive voice.

Object “the wine” becomes subject.
Passive: \emph{Wine should be opened about three hours before it is used.


Step 3: Eliminate incorrect options.

- (A) “before use” → though concise, it omits “it is used,” so incomplete.

- (B) Unnecessary mention of “by you” — passive aims to shift focus from doer.

- (C) Same as active sentence — not transformed.

- (D) Correctly transforms both parts of the sentence into passive.


Step 4: Conclude.

Thus, (D) is the correct passive transformation.

\[ \boxed{Wine should be opened about three hours before it is used.} \] Quick Tip: In passive transformations, ensure all clauses are changed consistently. For “before you use it,” the passive is “before it is used.”

Step 1: Choose the best follower to S1 (topic sentence).

S1 introduces the theme: polymers replacing metals. The next logical sentence should \emph{define polymers before giving properties or consequences.
\(\Rightarrow\) Q defines polymers: “Polymers are essentially long chains of hydrocarbon molecules.” Hence Q follows S1.


Step 2: Arrange properties/advantages after the definition.

Once polymers are defined, we can state key properties that enable replacement of metals:

- R: strength parity — “polymers as strong as metals” (technical feasibility).

- P: cost/process advantage — “cheaper and easier to process” (economic/operational feasibility).

Order R then P is natural: first show \emph{can they match metals? (strength), then show \emph{why they’re preferred (cost/ease).


Step 3: Provide a concrete example as a consequence.

S gives a specific replacement case (car bumpers), which neatly illustrates the reasons presented in R and P and leads smoothly into S6 about academic programmes.


Step 4: Verify global coherence with S6.

Q\,\(\to\)\,R\,\(\to\)\,P\,\(\to\)\,S builds from definition to properties to application, which justifies S6’s note on polymer science programmes.

\[ \boxed{Q R P S} \] Quick Tip: For para-jumbles: after the topic sentence, prefer \textbf{definition/expansion} \(\Rightarrow\) \textbf{properties/arguments} \(\Rightarrow\) \textbf{example/consequence}, then check that the last sentence wraps up logically.

Step 1: Core meaning.

\emph{Frugality means careful, sparing use of resources; thrift.


Step 2: Option analysis.

(A) \emph{Foolishness — unrelated to spending habits.

(B) \emph{Extremity — means severe degree or limb; not a synonym.

(C) \emph{Enthusiasm — eagerness; unrelated.

(D) \emph{Economy — sense of thrift/avoidance of waste; matches frugality.

\[ \boxed{Economy} \] Quick Tip: Link “frugal” with “\textbf{fruitful saving}”: both start with \emph{fru-}. Think “economical, not wasteful.”

Step 1: Core meaning.

A \emph{harbinger is a sign/precursor that announces or foreshadows what is to come.


Step 2: Option analysis.

(A) \emph{Messenger — one who carries messages; may not imply an omen/precursor.

(B) \emph{Steward — manager/caretaker; unrelated.

(C) \emph{Forerunner — something/someone that precedes and signals another’s approach; matches.

(D) \emph{Pilot — navigator/first model; not the precise synonym here.

\[ \boxed{Forerunner} \] Quick Tip: Remember: “harbinger of spring” = an \textbf{early sign} of spring. So the closest synonym is \textbf{forerunner/precursor}, not merely a messenger.

Step 1: Meaning of “Exodus.”

The word \emph{exodus refers to a mass departure or emigration of people. Example: \emph{The exodus of refugees from the war zone.


Step 2: Identify the antonym.

The opposite of departure is arrival or coming in, which is best captured by the word \emph{influx.


Step 3: Eliminate incorrect options.

- (B) \emph{Home-coming — return of a single person/group, not necessarily the opposite of mass departure.

- (C) \emph{Return — general word, but not a precise opposite of large-scale leaving.

- (D) \emph{Restoration — means renewal or repair, unrelated.

\[ \boxed{Influx} \] Quick Tip: “Exodus” = mass leaving; “Influx” = mass arrival. Always remember this antonym pair.

Step 1: Meaning of “Equanimity.”

Equanimity refers to mental calmness, composure, and evenness of temper, especially under stress.


Step 2: Identify opposite meaning.

The opposite would be a state where calmness is lost, replaced by agitation or excitement.


Step 3: Check options.

- (A) \emph{Resentment — means bitterness/anger; not direct opposite.

- (B) \emph{Dubiousness — uncertainty or doubt; not direct opposite.

- (C) \emph{Duplicity — deceitfulness; unrelated.

- (D) \emph{Excitement — agitation or heightened emotion, opposite of calmness.

\[ \boxed{Excitement} \] Quick Tip: Think: Equanimity = calm balance; opposite = agitation/excitement.

Step 1: Check grammatical function of each word.

- (A) \emph{Bludgeon — can be both noun (a heavy weapon) and verb (to hit with a bludgeon).

- (C) \emph{Black Jack — a weapon (noun) and can also be used as verb (“to strike with a blackjack”).

- (D) \emph{Order — can be a noun (command) and also a verb (to command).

- (B) \emph{Dragon — only a noun (mythical creature), cannot be used as a verb.


Step 2: Identify oddity.

Three words can function as both noun and verb, but “Dragon” functions only as a noun.

\[ \boxed{Dragon} \] Quick Tip: In odd-one-out questions, compare grammatical categories and semantic roles. Here, “Dragon” is only a noun, unlike the others which serve as noun + verb.

Step 1: Understand the given pair (Kernel : Shell).

- Kernel = the soft inner part of a seed.

- Shell = the hard outer covering that encloses and protects the kernel.

Thus, the relationship is **inner content vs outer covering**.


Step 2: Test each option.

- (A) Caterpillar : Pupa → A caterpillar transforms into a pupa; the pupa acts as a protective stage. Similar inner–outer or transformation relationship.

- (B) Larva : Cocoon → A larva is enclosed within a cocoon; again inside vs covering relation.

- (C) Lassitude : Syncope → Lassitude (weariness) can lead to syncope (fainting). This is still a related cause–effect/progression relationship.

- (D) Passenger : Car → Passenger is a person traveling inside a vehicle, but the relationship is not one of natural inner vs protective outer. This is a human–object association, not a structural covering relationship.


Step 3: Conclude.

The odd pair is (D) Passenger : Car.

\[ \boxed{Passenger : Car} \] Quick Tip: In analogy questions, always identify the **exact relation** (e.g., inner vs outer, part vs whole, cause vs effect). Then test each option against that relation.

Step 1: Observe the keyword “since.”

The word “since” denotes an action that started in the past and continues up to the present. The tense required is **Present Perfect Continuous**.


Step 2: Check each option.

- (A) \emph{are working here — Present Continuous; expresses current activity, but not from past until now. Wrong.

- (B) \emph{work here — Simple Present; indicates habit, but does not show continuity since childhood. Wrong.

- (C) \emph{have been working — Present Perfect Continuous; exactly correct for an action continuing since the past. Correct.

- (D) \emph{are liking to work — grammatically wrong; “like” is a stative verb, rarely used in continuous tense.


Step 3: Conclude.

Thus, the correct answer is (C) have been working.

\[ \boxed{have been working} \] Quick Tip: Use \textbf{Present Perfect Continuous} (\emph{have/has been + V-ing}) for actions that started in the past and are still continuing with “since” or “for.”

Step 1: Focus on the contrast marker “Although.”

The word “although” introduces a contradiction between the first clause and the second clause.


Step 2: Analyze the first clause.

“Initial investigations pointed towards him” → suggests suspicion of guilt.
Thus, the second clause should show the opposite outcome (innocence).


Step 3: Check each option.

- (A) “the preceding events corroborated his involvement” → supports guilt, not contrast. Wrong.

- (B) “the additional information confirmed his guilt” → agrees with initial suspicion, not contrast. Wrong.

- (C) “the subsequent events established that he was guilty” → again confirms guilt, no contrast. Wrong.

- (D) “the subsequent events proved that he was innocent” → contradicts initial suspicion, showing contrast as required by “although.” Correct.


Step 4: Conclude.

Therefore, the correct answer is (D).

\[ \boxed{the subsequent events proved that he was innocent} \] Quick Tip: Words like “although,” “though,” and “despite” always set up a contrast. Look for the answer option that provides the **opposite sense** of the first clause.