ICAR Technician 2 March 2022 Shift 3 (Previous Year Paper)

The correct answer is October.

Key Points

• The month of October is globally marked as ‘National Cyber Security Awareness Month (NCSAM)’  with the aim of spreading awareness about cyber security.
• The Ministry of Electronics and Information Technology (MeitY) in association with CERT-IN, NIC and C-DAC observes and organizes unique mass awareness activities.

Additional Information

• Cyber security has been an important issue in the modern age of digital dependency.
• It has a potential for misuse that can lead to substantial damage.
• In these backgrounds, it became essential for securing and protecting data, networks, systems, and privacy from cybercriminals the importance of cyber security has grown.

The correct answer is Prakash Padukone.

Key Points

• Prakash Padukone is a former Indian badminton player.
• He is the first Indian to win the All England Open Badminton Championship.
• He was world No. 1 in 1980.
• He was honoured with the Arjuna Award(1972) & Padma Shri(1982).
• He is the founder of Olympic Gold Quest, a not-for-profit organisation aiming to encourage the best talent in sports.

Additional Information

• Prannoy Kumar
• He is an Indian Badminton player from the state of Kerala.
• He had a career-high rank of World No 8.
• He won the Gold medal in the 2018 Commonwealth games in the mixed team category.
• Chirag Shetty
• He is an Indian badminton player.
• He along with his partner Satwiksairaj is the first men’s double player to be in the top-10 rankings.
• He won silver in the 2018 Commonwealth games in Mens’ doubles category.
• B Sai Praneeth
• He is an Indian Badminton player from Andhra Pradesh.
• His best performance is a Bronze medal in the 2019 World Championships.

The correct answer is Schedule tribes.

Key Points

• Dr Jitendra Singh inaugurated “Tech NEEV@75” as part of Azadi ka Amrit Mahotsav and interacted with successful start-ups including those from the tribal communities on the occasion of Janjatiya Gaurav Diwas.
• Under this program, the Government will be setting up 30 Science Technology and Innovation (STI) Hubs in different parts of the country for Scheduled Tribes (STs) by the end of 2022
• It helps in promoting scientific talent among tribal communities and their overall socio-economic development.

Additional Information

• Tech NEEV/ नीव @ 75”  is a year-long celebration highlighting the impact of Science Technology Innovation (STI) in empowering communities for creating equitable inclusive economic growth.
• Government has decided to celebrate 15th November i.e. the birth anniversary of Bhagwan Birsa Munda as “Janjatiya Gaurav Diwas”.

The correct answer is Goods and Services Tax.

Key Points

• Production cost refers to the expenditure incurred in the process of production of a Good.
• The major components of production cost are:
• Direct materials cost of raw materials involved in the production.
• Labour cost: wages paid to labour for the production.
• Other manufacturing overheads: include many indirect costs like rent, depreciation, utility expenses
• Goods & Services Tax is part of the taxation system by the government and has nothing to do with production.

Additional Information

• The factors of production are:
• Land refers to the expenses related to rent paid for the production
• Labour refers to the expenses involving wages of the labour.
• Capital refers to the expenses in the form of interest on the money
• Entrepreneurship refers to the expenses in the form of payment to the management for taking the risk.
• Utility Expenses
• These are the expenditure incurred for using public utility services like Electricity, Water, Sewage, Telephone bills etc.
• These are the services used by the producer and may not be directly related to the product but are necessary for the line of production.

The correct answer is Devanagari.

Key Points

• Article 343 (1) refers to Hindi as the official language of Government of India and the script used for it is Devanagari script.
• Devanagari script is based on the ancient Brahmi script.
• It uses 47 characters of which 14 are vowels and 33 are consonants.
• Apart from Hindi there are other languages like Marathi, Nepali, Bhojpuri, Rajasthani, Brajbasha etc which use the Devanagari script.
• It is mainly found in the Indian Subcontinent.

Additional Information

• Sharada 
• This script also belongs to the family of Brahmmi scripts.
• It is used in north western parts of the Indian Subcontinent.
• The major languages using the script were Kashmiri and Sanskrit.
• It is written from left to right.
• Kharosthi
• This is an ancient script used by Saks, Yeuzhi and Kushans.
• It can be found in the present day Afghanistan and northern Pakistan regions.
• The languages using the script were Gandhari, Pali.
• It is written from right to left.
• Siddham
• This is a medieval script derived from Gupta script.
• It is the root for languages like Bengali, Odia, Assamese.

The correct answer is Energy.

Key Points

• National Infrastructure Pipeline:
• NIP is launched in August 2020, for the period of five years 2020-2025.
• the National Infrastructure Pipeline (NIP) is a first-of-its-kind, whole-of-government exercise to provide world-class infrastructure across India,
• It will improve the quality of life for all citizens.
• It aims to showcase investment opportunities in India's infrastructure sector, improve project preparation and attract investments into India.
• The NIP aims to capture key greenfield and brownfield projects for investments across all economic and social infrastructure sub-sectors on a best-effort basis
• NIP is a collection of projects and programs totalling INR 102 lakh crore in infrastructure development over the next five years.
• It follows the government's goal of making India a $5 trillion economy by 2024-25.
• NIP covers economic and social infrastructure projects in sectors such as Energy (24%), Roads (19%), Urban (16%), and Railways (13%) that account for almost 70% of India's expected infrastructure capital spending.

The correct answer is Chakma.

Key Points

• Chakma tribe
• The Chakma or Changma is the largest tribal community in Bangladesh.
• The Chakmas also live in Tripura, Assam, Mizoram, Arunachal Pradesh, and Burma (Myanmar).
• In Bangladesh, they are living mostly in the Chittagong Hilly Areas including Cox’s Bazar.
• The origin of Chakma is from the Mongoloid people who are the people of Southwestern Burma.
• They talk in their dialect (Arakanese – Indo-European tongue) as well as Bengali.
• The tribal life, they are very rich in their culture and tradition.
• The Chakma people are very innovative and adaptive than any other tribal people in Bangladesh.
• The people of Chakma believe in the religion of Buddhism.
• They live depending on agriculture.
• They grow different types of crops following the eco-friendly method known as “Jhum cultivation.”
• The women of this community wear homemade Pinon and Khadi.
• The men generally wear the Dhuti.
• They have their own script in their dialect language.
• Even the ancient religious books are written in this language in palm

Additional Information

Important Tribes of the world​

The correct answer is Bharatanatyam.

Key Points

• Bharatanatyam
• Bharatanatyam is a dance form of Tamil Nadu in southern India.
• It has grown out of the art of dancers dedicated to temples.
• It was earlier known as Sadir or Dasi Attam.
• It is the first of India's traditional dances to be refashioned by a theatre artist and to be exhibited widely both at home and abroad.
• Bharatanatyam rests on principles of performance and aesthetics set down in classics such as Bharata's Natyashastra.
• It has a rich repertoire of songs in Telugu, Tamil, and Sanskrit.
• The present-day format of a Bharatanatyam recital, as well as a valuable part of its musical compositions, were created by the famed ‘Tanjore Quartet’ of the nineteenth-century brothers Ponniah, Chinnaiah, Sivanandam and Vadivelu.
• Bharatanatyam has a highly evolved language of Nritta, abstract dance.
• The themes have a wide range spanning human and divine love.
• They are generally classed under the rubric of shringara (romantic love) and Bhakti (devotion).
• The music of Bharatanatyam belongs to the Carnatic system of southern India.
• The musicians accompanying a dance recital include at least one vocalist, a Mridangam (drum)-player, a flutist or violinist, or Veena (lute)-player.
• The group also includes a Nattuvanar, or dance conductor, who recites the dance syllables as he plays a pair of small bronze cymbals.

The correct answer is More members in a nuclear family.

Key Points

• Social Indicators of Poverty
• Illiteracy Level
• Lack of general resistance due to malnutrition
• Lack of access to healthcare
• Lack of Opportunities
• Lack of access to safe drinking water
• Lack of access to safe sanitation facilities

Important Points

• Poverty
• Poverty is a state or condition in which a person or community lacks the financial resources and essentials for a minimum standard of living.
• Poverty means that the income level from employment is so low that basic human needs can't be met.
• According to World Bank 
• Poverty is pronounced deprivation in well-being and comprises many dimensions.
• It includes low incomes and the inability to acquire the basic goods and services necessary for survival with dignity.
• Poverty also encompasses low levels of health and education, poor access to clean water and sanitation, inadequate physical security, lack of voice, and insufficient capacity and opportunity to better one's life.
• In India, 21.9% of the population lives below the national poverty line in 2011.
• Causes of Poverty

1. Population Explosion
2. Low Agriculture Productivity
3. Inefficient resource utilization
4. Low rate of Economic Development
5. Price Rise
6. Unemployment
7. Social Factors
8. Climatic Factors

The correct answer is Convex Mirror.

Key Points

• Convex Mirror
• A Convex Mirror curves outward like the back of a spoon.
• In a convex mirror, light originating from the same point will reflect off the mirror surface and diverge.
• Hence convex mirror is sometimes referred to as a diverging mirror.
• The type of image formed by these mirrors is always imaginary or virtual.
• Uses of Convex Mirror

1. It is used as a rearview mirror in the vehicle because it provides the maximum near field of view.
2. In Sodium reflector lamp.

• Concave Mirror
• A Concave mirror is a mirror that has a slightly inward curve like the inside of a spoon.
• The Curved surfaces focus parallel rays from a great distance into a single point for enhanced intensity.
• Uses of Concave Mirror

1. As a shaving glass
2. As a reflector for headlights of vehicles, and searchlights.
3. In Ophthalmoscope examines the eye, ear, and nose doctors.
4. In Solar cookers.

Confusion Points

• Do not confuse between Concave/Convex Lens and Mirror
• Convex lens
• The convex lens is also called a converging lens.
• Concave lens
• The concave lens is also called a diverging lens.

The correct answer is Estuary.

Key Points

• Estuary:
• An estuary is a coastal area where a river meets the ocean.
• It is strongly affected by tidal actions and is very rich in nutrients and consequently very rich in biodiversity too.
• The organisms here are highly tolerant and many species are endemic. The ecosystem is known to have high food productivity.
• However, this ecosystem has to be protected from pollution.
• Various types of marine ecosystems are:
• Salt marshes
• Mangrove forests
• Coral reefs
• The open ocean
• The deep-sea ocean

The correct answer is Saurav Ghosal.

Key Points

• Saurav Ghosal is currently ranked 15th in the world (PSA Men's World Rankings).
• Saurav has won 5 titles, besides one gold and one silver at the 2014 Asian Games and three bronze medals before that.
• He is the first Indian Squash player to win the Malaysian Open Championships 2021.

Additional Information

• Ritwik Bhattacharya, Ramit Tandon, Mahesh Mangaonkar and Joshna Chinappa are other sports players associated with squash.

The correct answer is Pillars.

Key Points 

• Mauryan art was basically royal art.
• The use of pillars, stupas, caves, sculptures and pottery was a typical example of Mauryan art.
• It was divided into 2 parts:

1. Court art: It flourished with royal initiative.

• E.g: Pillars, Stupas.

2. Popular art: It flourished with individual initiative.

• E.g: Caves, Sculpture, Pottery.

The correct answer is Net Domestic Product index.

Key Points

• The Human Development Index (HDI) is a statistic composite index of life expectancy, education (mean years of schooling completed and expected years of schooling upon entering the education system), and per capita income indicators, which are used to rank countries into four tiers of human development.
• The Human Development Index (HDI) is measured on the basis of the Life Expectancy Index (Long and healthy life), Education Index (Knowledge), and Gross National Income index (Standard of Living).
• The other indices that form the part of the Report are:

1. Inequality-adjusted Human Development Index (IHDI),
2. Gender Development Index (GDI),
3. Gender Inequality Index (GII) and
4. Multidimensional Poverty Index (MPI).

The correct answer is 1865.

Key Points

• Second Anglo-Bhutanese War:
• It was a war fought between British India and Bhutan from 1864 to1865.
• It is also called the Duar War.
• It was won by British India.

The correct answer is 31 March 2023.

Key Points

• The Pradhan Mantri Vaya Vandana Yojana being implemented through LIC (Life Insurance Corporation) of India is open for subscription up to 31 March 2023.
• Pradhan Mantri Vaya Vandana Yojna is a pension scheme that is started by the Indian Government on 4 May 2017.
• This pension scheme is for senior citizens age 60 years and above.
• Assured return of 8% payable monthly for 10 years.
• Exempted from Service Tax or GST.
• Loan of up to 75% of purchase price allowed after 3 policy years.
• Allows for premature exit for treatment of any critical/terminal illness of self or spouse.

The correct answer is Gluten.

Key Points

• Gluten is the sticky substance in wet wheat flour which when combined with yeast helps the dough to rise.
• Gluten is a protein composite found in wheat and related grains.
• Gluten gives the elasticity to dough, helping it rise and keeps its shape, and often gives the final product and **chewy texture.**​

Additional Information

• Secalin is a prolamin glycoprotein found in the grey rye, Secale cereal.
• Secalin is one of the forms of gluten proteins.
• Avenin, a prolamin like wheat gluten, is only minor.
• The content of lipids in the floury oats.
• Hordein is a prolamin glycoprotein, present in barley and some other cereals, together with gladin.

The correct answer is 2006.

Key Points

• The Infrastructure Finance Company Limited (IIFCL) was established in 2006.
• IIFCL is a wholly-owned Government of India company that was set up in 2006 to provide long-term financial assistance to viable infrastructure projects through the Scheme for Financing Viable Infrastructure Projects through a Special Purpose Vehicle called India Infrastructure Finance company Ltd.
• IIFCL has been registered as an NBFC-ND-IFC with the Reserve Bank of India since September 2013 and follows the applicable prudential norms of the RBI.
• This Company comes under the Government of India Ministry of Finance.

The correct answer is Justice Lodha Committee.

Key Points

• On the basis of the recommendation of Justice Lodha Committee was a Differently-abled Committee formed for differently-abled cricketers by the BCCI (Board of Control for Cricket in India).
• A board of three former disabled cricketers was being awaited for the last several years.
• On December 4, 2021, in Kolkata, the proposal was accepted to create a differently-abled committee approved in the BCCI's annual meeting.
• The Differently Abled Cricket Council of India (DCCI) thanked the **BCCI for taking this step.**​

Additional Information

• Justice Chandrachud Committee was related to Indian Match-fixing Investigation (1997).
• Justice Reddy Committee regarding the Distance Education Programs being run in the country by various universities.
• Justice Rastogi Committee regarding the Higher Education and it's role in our society.

The correct answer is Kinetic energy.

Key Points

• The energy that moving particles possess is Kinetic energy.
• Moving Water and Wind are good examples of Kinetic energy.
• Kinetic energy includes electrical, radiant thermal, and **sound energy.**​

Additional Information

• Magnetic Energy that energy is created by the strength of the magnetic field.
• Electrical energy is that energy where's movement of electrically positive charged particles.
• Potential energy is a kind of energy stored in objects.

The correct answer is rectangular.

Key Points

• The towns were planned in a rectangular grid pattern during the Harappan civilisation.
• Town planning was the unique feature of the Indus valley civilisation.
• Harappan cities were planned with chessboard system or grid pattern.
• A grid pattern is a network of intersecting parallel lines.
• A grid pattern, means the streets intersect at right angles and form a pattern of squares when viewed from above.

Important Points

• Common features of Harrapn civilisation Major cities are:
• A town is divided into two parts: one smaller but higher i.e. Citadel and the other much larger but lower i.e Lower Town or walled.
• They used baked bricks in construction, which was a unique feature around the world in that time period.
• The main roads ranged from 9 feet to 30-34 feet wide and were divided into networks of narrow streets.
• The drainage system was so advanced that first roads were laid with drains and then houses were built along with them.

The correct answer is Cholas.

Key Points

• Cholas ruled Tamil Nadu in Medieval India.
• Cholas
• ​The founder of the Cholas dynasty was Vijayalaya.
• The Cholas conquered the Kaveri region by defeating the Pallavas in 850 AD (9th century).
• Vijayalay captured Tanjore in 850 AD.
• Vijayalaya built the town of Thanjavur and a temple for goddess Nishumbhasudini there.
• Rajaraja I (985-1014 AD)
• He built Vridheshwar/ Rajarajeshwar temple at Tanjore.
• Rajendra I (1014-1044 AD)
• ​He conquered Orissa, Bengal, and Sri Lanka.
• He took the title of Gangaikondachola.
• The last ruler of the Chola Dynasty was Rajendra III.
• The Chola dynasty was known for its local Self-government.
• Land revenue and trade tax were the main sources of income for the Chola dynasty.​

Additional Information

The correct answer is Quit India Movement was launched.

Key Points

• The Quit India Movement took place on 8 August 1942 due to growing discontent over the proposal of Cripps.
• Other reasons for the Quit India Movement were high price rise, British pushing India into war, fear of Japanese invasion of India.
• Congress had passed a resolution on 'Quit India' movement at Gowalia Tank, Mumbai.
• Most of the leaders were arrested on 9 August 1942 under Operation Zero Hour, which was launched by British.
• In this movement, Gandhi Ji gave the slogan 'Do or Die'.
• Yusuf Mehrli gave the slogan 'Quit India'. He was a companion of Mahatma Gandhi.
• The parallel government was established in Ballia (UP), Tamulak (Bangal), Satara (Maharashtra), and Talchar (Orissa) during this movement.

Additional Information

• Rowlatt’s Act 
• ​It was passed in 1919.
• Viceroy - Lord Chelmsford
• Aim - To curb sedition activities in India.
• This act authorized the British government to imprison any person without trial who was suspected of terrorist activity for a minimum period of two years.
• In protest against this act, Gandhi started Rowlatt Satyagraha.
• Khilafat Movement 
• ​It was launched by Muslims of British India led by Mohammad Ali and Shaukat Ali.
• The reason behind this movement was to protest the Caliphate (Caliph, Sultan of Turkey) and the restrictions imposed by the Treaty of Svres on the Ottoman Empire after World War I.
• Aim - Restoration of the Khilafat.
• Simon Commission
• Appointed - Nov 1927 by the British Government.
• Arrived in India - on 3 Feb 1928
• ​No. of member- Seven (7)  British (white) Members of Parliament
• No Indian representative in it. so it was known as White Commission.
• Headed by - Lord Simon
• purpose - Suggest to the British Government regarding Indian constitutional reforms.
• Simon Go Back" slogan was coined by Yusuf Meherally.

The correct answer is Odisha.

Key Points

• India successfully launched the Supersonic Missile Assisted Release of Torpedo System from the coast of Odisha in December 2021.
• It was successfully launched from Wheeler Island, at Balasore in Odisha.
• The Supersonic Missile Assisted Torpedo (SMART) was developed by Defence Research and Development Organisation (DRDO) for the Indian Navy.
• The system has been designed to enhance anti-submarine warfare capability.

Important Points

• Stand-off Anti-tank (SANT) Missile 
• ​It is an indigenous Stand-Off Anti-Tank Missile.
• The flight test was done by (DRDO) and Indian Air Force (IAF) at the Pokhran range in Rajasthan.
• It was launched from a Russian-origin Mi-35 helicopter gunship in December 2021.
• It can neutralize targets at a range of up to 10 km.
• It has been designed and developed by Research Center Imran, Hyderabad with DRDO.

Additional Information

• DRDO (As of 2022)
• ​Founded: 1958
• Headquarters: New Delhi
• Chairperson: Samir Kamat

The correct answer is West Bengal.

Key Points

• In West Bengal is the localised thunderstorms associated with violent gust and heavy rain known as ‘Kaal Baisakhi’.
• Kaal Baisakhi
• ​It is a local rainfall and thunderstorm occur in Assam and West Bengal and in India or Bangladesh.
• It occurs in the summer season.
• It harms life and money and generally occurs in the month of Baisakh (April-May), hence it is known as 'Kaal Baisakhi'.
• It originated from Chota Nagpur Plateau.
• Since It is caused by winds coming from the north-west direction, they are also called 'norwesters'.

Additional Information

• During the summer season, strong humid winds start blowing in many places in India, which are known by different names in different places of the country:
• Norwesters in East India
• Cherry Blossoms/Coffee Showers
• It is the local humid wind of Karnataka that is helpful for the coffee plantation and flows in the hot weather season.
• Mango- shower
• It is pre-monsoon rainfall and occurs especially in parts of Kerala, Karnataka and Tamil Nadu.
• It helps in the early ripening of mangoes and hence it is called 'Mango rain' or Mango-shower.

Given: 

PT is a tangent at point C on a circle with centre O and AB is a diameter. BA is produced, it meets PT at point P.

∠BPC = 34°.

Concept used:

The angle subtended by diameter on any point on the circumference of a circle is 90°.

The angle between a tangent and a chord is equal to the angle in the alternate segment.

Calculation:

According to the question, the required figure is:

As we know,

The angle subtended by diameter on any point on the circumference of a circle is 90°.

So, 

∠ACB = 90°

Now, 

∠BCP = ∠BCA + ∠ACP

⇒ ∠BCP = 90° + θ

As per the alternate segment theorem.

∠PBC = ∠ACP = θ

In ΔPCB,

∠CPB + ∠BCP + ∠CBP = 180°

⇒ 34° + 90° + θ + θ = 180°

⇒  124° + 2θ = 180°

⇒ θ = 28°

Therefore,

∠ACP = θ = 28°

∴ 28° is the required answer.

Given:

A can complete a certain work in 35 days.

B is 40% more efficient than A, and C is 25% more efficient than B.

B and C work together for 8 days.

The remaining work is completed by A with the help of D in $3\frac{1}{2}$ days.

Formula used:

Work = Efficiency × Time

Calculation:

The time is taken by A to complete the work = 35 days

The efficiency of A = $\frac{1}{35}$

Now,

The efficiency of B = (100% + 40%) × $\frac{1}{35}$ = $\frac{1.4}{35}$

The efficiency of C = (100% + 25%) × $\frac{1.4}{35}$ = $\frac{1.75}{35}$

The amount of work done by B and C in 8 days = $(\frac{1.4}{35} + \frac{1.75}{35}) × 8$ = 0.72

The remaining amount of work = 1 - 0.72 = 0.28

Let the time taken by D to complete the work = x days

The efficiency of D = $\frac{1}{x}$

Since the remaining work is completed by A with the help of D in $3\frac{1}{2}$ days,

Therefore,

⇒ $(\frac{1}{35} + \frac{1}{x} ) × 3\frac{1}{2} = 0.28$

⇒ $(\frac{1}{35} + \frac{1}{x} )$ × 3.5 = 0.28

⇒ $(\frac{1}{35} + \frac{1}{x} )$ = $\frac{2}{25}$

⇒ $\frac{1}{x}$ = $\frac{9}{175}$

Now, 

The time is taken by D to complete 72% of the original work,

⇒ $\frac{9}{175}$ × t = 0.72

⇒ t = 14 days

∴ 14 days is the required answer.

Given: 

The total surface area of a solid hemisphere is 1848 cm2.

The volume of a solid hemisphere is equal to the volume of a right circular cone, whose height is 28 cm.

(Take $π = \frac{{22}}{7}$)

Formula used:

The total surface area of a solid hemisphere = 3πr2

The volume of the solid hemisphere = $\frac{2}{3}$ × π × r3

The volume of the right circular cone = $\frac{1}{3}$ × π × R2 × h

The base area of the right circular cone = π × R2

Where,

r, be the radius of the hemisphere

R, is the radius of the right circular cone

h, is the height of the right circular cone

Calculation:

The total surface area of a solid hemisphere, 

⇒ 3πr² = 1848 

⇒ 3 × $\frac{22}{7}$ × r2 = 1848

⇒ r2 = 196

⇒ r = 14 cm

The volume of the solid hemisphere,

⇒ V_hemisphere = $\frac{2}{3}$ × π × r³

⇒ Vhemisphere = $\frac{2}{3}$ × π × (14)3 cm³

The volume of right circular cone,

⇒ V_cone = $\frac{1}{3}$ × π × R² × h

⇒ Vcone = $\frac{1}{3}$ × π × (R)2 × (28)

According to the question,

⇒ Vhemisphere = Vcone

⇒ $\frac{2}{3}$ × π × (14)3 = $\frac{1}{3}$ × π × (R)2 × (28)

⇒ 2 × (14)3 = (R)2 × (28)

⇒ R² = 196

Now, 

The base area of right circular cone,

⇒ A_cone = π × R²

⇒ Acone = $\frac{22}{7}$ × 196

⇒ Acone = 616 cm²

∴ The base area of right circular cone is 616 cm².

Shortcut Trick

Put θ = 60° in the given expression.

Left part = tan²60° + cot²60° + 2 = (√3)² + (1/√3)² + 2 = 3 + 1/3 + 2 = 16/3

Right part = 1 / (1 + cos60°) + 1 / (1 − cos60°) = 1 / (1 + 1/2) + 1 / (1 − 1/2) = 2/3 + 2 = 8/3

Expression value = (16/3) ÷ (8/3) = 2

Checking Option 3: sec²60° / 2 = (2)² / 2 = 4 / 2 = 2 (matches exactly).

∴ The correct answer is sec²θ / 2.

Alternate Method

Given:

(tan²θ + cot²θ + 2) ÷ [1 / (1 + cosθ) + 1 / (1 − cosθ)]

Formula Used:

tan²θ + 1 = sec²θ | cot²θ + 1 = cosec²θ | 1 − cos²θ = sin²θ

Calculations:

⇒ First part = tan²θ + cot²θ + 2

⇒ First part = (tan²θ + 1) + (cot²θ + 1) = sec²θ + cosec²θ

⇒ First part = 1 / cos²θ + 1 / sin²θ = (sin²θ + cos²θ) / (sin²θ × cos²θ)

⇒ First part = 1 / (sin²θ × cos²θ) = sec²θ × cosec²θ

⇒ Second part = 1 / (1 + cosθ) + 1 / (1 − cosθ)

⇒ Second part = [(1 − cosθ) + (1 + cosθ)] / [(1 + cosθ) × (1 − cosθ)]

⇒ Second part = 2 / (1 − cos²θ) = 2 / sin²θ = 2cosec²θ

⇒ Substituting both parts back into the given expression:

⇒ Expression = (sec²θ × cosec²θ) ÷ 2cosec²θ

⇒ Expression = sec²θ / 2

∴ The correct answer is sec²θ / 2.

Additional Information

Pythagorean Trigonometric Identities

The three fundamental identities are: sin²θ + cos²θ = 1, sec²θ − tan²θ = 1, and cosec²θ − cot²θ = 1.

Sum to Product Transformation

A very useful simplification trick for competitive exams: sec²θ + cosec²θ = sec²θ × cosec²θ.

Reciprocal Functions

Remember that cosecθ = 1 / sinθ, secθ = 1 / cosθ, and cotθ = cosθ / sinθ.

Given:

The ratio of male and female employees in office D is 9 : 7 and in office C, it is 4 : 5.

The total employee in the office E = 704

Calculation:

The degree-wise distribution for the office E = 79.2^∘

The total number of employees in all departments = $\frac{360^{\circ}}{79.2^{\circ}} \times 704 = 3200$

The total number of employees in the office D = $\frac{57.6^{\circ}}{360^{\circ}} \times 3200 = 512$

The total female employees in offices D = $\frac{7}{9+7} \times 512 = 224$

The total number of employees in the office C = $\frac{64.8^{\circ}}{360^{\circ}} \times 3200 = 576$

The total female employees in offices C = $\frac{5}{4+5} \times 576 = 320$

The total number of employees in offices A and B = $\frac{122.4^{\circ} + 36^{\circ}}{360^{\circ}} \times 3200 = 1408$

The total female employees in offices D and C = 320 + 224 = 544

The required per cent = $\frac{544}{1408} \times 100$ = 38.6%

∴ The total female employees in offices D and C is 38.6% of the total number of employees in offices A and B.

Given:

Seema sells 40% of her goods at a profit of 25%, 20% of the remaining goods at a loss of 35%, and 25% of the remaining goods at a profit of 15%.

Formula used:

Selling price = (1 + profit%) × Cost price

Selling price = (1 - loss%) × Cost price

Calculation:

Let the total quantity of goods = 100

The cost price of 100 units of goods = ₹100

The selling price for the total quantity of goods for 20% profit = (1+ 20%) of ₹100 = ₹120

In the first case:

The unit cost price of good = ₹1

The selling price of 40% of goods at 25% of profit = (1 + 25%) of (40% of 100) of ₹1 = ₹50

The remaining goods = 100 - (40% of 100) = 60

In the second case:

The selling price of 20% of the remaining goods at 35% of loss = (1 - 35%) of (20% of 60) of ₹1 = ₹7.8

The remaining goods = 60 - (20% of 60) = 48

In the third case:

The selling price of 25% of the remaining goods at 15% of profit = (1 + 15%) of (25% of 48) of ₹1 = ₹13.8

The remaining goods = 48 - (25% of 48) = 36

In the last case:

The selling price for the remaining goods = ₹120 - (₹50 + ₹7.8 + ₹13.8) = ₹48.4

The cost price of 36 units of goods = ₹1 × 36 = ₹36

The required profit per cent = $\frac{₹48.4-₹36}{₹36} × 100$ = 34.4%

∴  In order to get 20% profit in the entire transaction, she must sell the remaining goods at a profit 34.4%.

Given:

A T.V is available for ₹1,50,000 cash or ₹17,600 cash down payment followed by three equal half-yearly instalments.

If the rate of interest charged is 20% p.a., compounded half-yearly.

Formula used:

D = $\frac{A}{R} × (1 - \frac{1}{(1+R)^n})$

Where, 

D, is the amount due (or amount on which interest is to be applied)

A, is the value of the instalment

R, is the rate of interest applied

n, is the number of instalments

Calculation:

The price of the T.V = ₹1,50,000

The down payment is done for buying T.V = ₹17,600

The remaining amount (which is paid in three equal instalments) = ₹1,50,000 - ₹17,600 = ₹1,32,400

The rate of interest applied (half-yearly) = 20% ÷ 2 = 10%

The number of instalments, n = 3

Now, 

By substituting the value of the respective variable in the required formula, we get,

⇒ ₹1,32,400 = $\frac{A}{10\text{\%}} × (1 - \frac{1}{(1+\text{10\%})^3})$

⇒ ₹1,32,400 × 10% = A × $(1 - \frac{10^3}{11^3})$

⇒ ₹1,32,400 × 0.1 = A × $(1 - \frac{1000}{1331})$

⇒ ₹13,240 = A × $(\frac{331}{1331})$

⇒ A = ₹53,240

The total amount paid in three instalments = ₹53,240 × 3 = ₹1,59,720

The interest paid = ₹1,59,720 - ₹1,32,400 = ₹27,320

∴ ₹27,320 is the interest charged in the instalment scheme.

Given:

A, B and C enter into a partnership and invest their capitals in the ratio $\frac{1}{3}:\frac{5}{{12}}:\frac{1}{2}$,

A receives one-fourth of the total profit for managing the business.

The remaining profit is divided between them in the ratio of their investments.

At the end of a year, if A gets Rs. 1.26 lakh less than what B and C together.

Calculation:

The ratio in which A, B and C entered into a partnership = $\frac{1}{3}:\frac{5}{{12}}:\frac{1}{2}$

Now, multiply each part of the ratio by 12.

The ratio in which A, B and C entered into a partnership = $\frac{12}{3}:\frac{5× 12}{{12}}:\frac{12}{2} = 4 : 5 : 6$

Let the total profit = 60x

According to the question,

The profit earned by A for managing business = $\frac{1}{4}$ of 60x = 15x

The remaining profit by A, B and C = 60x - 15x = 45x

The remaining profit is shared by A, B and C in the ratio of 4 : 5 : 6.

Therefore,

The profit earned by A = $\frac{4}{15}$ × 45x = 12x

The profit earned by B = $\frac{5}{15}$ × 45x = 15x

The profit earned by C = $\frac{6}{15}$ × 45x = 18x

Therefore, 

The total profit earned by A = 12x + 15x = 27x

Again according to the question,

The profit earned by A gets w.r.t to B and C together,

⇒ (18x + 15x) - 27x = Rs. 1.26 lakh

⇒ 33x - 27x = Rs. 1.26 lakh

⇒ x = Rs. $\frac{1.26}{6}$ lakh = Rs. 0.21 lakhs

The profit earned by C = 18x = 18 × 0.21 lakhs = Rs. 3.78 lakhs

∴ The profit earned by C is Rs. 3.78 lakhs.

Given:

A shopkeeper earns a profit of 20% by selling an article by giving 25% discount on its marked price.

The cost price of the article increases by 10%.

Formula used:

Selling price = (Cost price) × (1 + Profit%)

Selling price = (Marked price) × (1 - Discount%)

Discount% = $\frac{\text{Marked Price} - \text{Selling Price} }{\text{Marked Price}}$ × 100

Calculation:

Let the marked price of the article = ₹100

In the first case:

The selling price after getting 25% discount = (100% - 25%) of ₹100 = ₹75

The cost price of an article on which 20% profit is earned = $\frac{₹75}{100\text {\%} + 20\text {\%}}$ = ₹62.5

In the second case:

The increased cost price = (100% + 10%) of ₹62.5 = ₹68.75

The selling price of the article for earning 20% profit = (100% + 20%) of ₹68.75 = ₹82.5

The required discount percent = $\frac{₹100-₹82.5}{₹100} × 100$ = 17.5% 

∴ The discount per cent should be given now on the same marked price to earning the same percentage of profit as earlier is 17.5%.

Given:

A man travels 25% faster than his usual speed, he would reach Y 15 minutes early.

Formula:

Distance = Speed × Time

Calculation:

Let the usual speed of the man = 100 kmph

Let the usual time to reach the destination = 75 min

In the first case:

The increased speed of the man = (100% + 25%) of 100 = 125 kmph

The time required to reach Y from X = 75 min - 15 min = 60 min = 1 hour

The required distance travelled = 125 × 1 = 125 km

In the second case:

The decreased speed of the man = (100% - 40%) of 100 = 60 kmph

The time required to reach Y from X = t min

The required distance travelled,

⇒ 60 × t = 125

⇒ t = $\frac{125}{60}$ × 60 min = 125 min

Therefore, 

The time (in min) by which man reach the destination = 125 min - 75 min = 50 min

∴ By 50 minutes would he be late if he travels 40% slower than his usual speed.

Given:

The amount received on a sum invested at 8% p.a. simple interest for $3\frac1 2$year exceeds the simple interest on the same sum at the same rate 15 years by Rs. 600. 

Formula used:

Simple Interest = $\frac{\text {Principal} × \text {Rate} × \text {Time}}{100}$

Amount = Principal + Interest

Calculation:

Let 'P' be the principal invested.

The rate of interest applied = 8% p.a.

In the first case:

The time of investment = $3\frac1 2$ = 3.5 years

The interest earned = $\frac{P × 8 × 3.5}{100}$

The final amount earned = P + $\frac{P × 8 × 3.5}{100}$ = 1.28P

In the second case:

The time of investment = 15 years

The interest earned = $\frac{P × 8 × 15}{100}$ = 1.2P

According to the question,

⇒ 1.28P - 1.2P = Rs. 600

⇒ 0.08P = Rs. 600

⇒ P = Rs. 7500

Now, 

The principal invested = Rs. 7500

The time of investment = 5.5 years

The rate of interest applied = 2 × 8 - 16% p.a.

The simple interest earned = $\frac{7500 \times 16 \times 5.5}{100}$ = Rs. 6600

The final amount earned = Rs. 7500 + Rs. 6600 = Rs. 14,100

∴ The final amount earned is Rs. 14,100.

Given:  

X varies directly as √Y.

When Y = 20.25, X = 18

Calculation:

X ∝ √Y

Remove the proportionality sign by the proportionality constant.

It can be written as:

⇒ X = k(√Y)

For the first case:

Substitute the value of the respective variable in the required expression.

⇒ 18 = k × √(20.25)

⇒ 18 = k × 4.5

⇒ k = 4

For the second case:

⇒ X = k(√Y)

Take square on both sides of the equation.

⇒ X² = k² ×(√Y)²

Substitute the value of the respective variable in the required expression.

⇒ 7.22 = 42 × Y

⇒ (8 × 9 × 0.1)2 = 16 × Y

⇒ (64 × 81 × 0.01) = 16 × Y

⇒ Y = (4 × 81 × 0.01)

⇒ Y = 3.24

∴ 3.24 is the required answer.

Given:

The income of A is 75% more than the income of B, and the income of C is 20% less than the combined incomes of A and B.

Calculation:

The income of B = ₹100

The income of A = (100% + 75%) of ₹100 = ₹175

The combined income of A and B = ₹100 + ₹175 = ₹275

The income of C = (100% - 20%) of ₹275 = ₹220

The required percent = $\frac{₹220-₹175}{₹175}$ × 100 = 25.7%

∴ The income of C is 25.7% more than that of A.

Given:

The average weight of some girls in a group is 58.4 kg.

When 15 girls with an average weight 62.2 kg leave the group or 10 girls with an average weight 49.85 kg join the group, the average weight (in kg) of girls in both cases is equal.

Formula used:

Average = $\frac{\text {Sum of 'n' observation}}{\text {n}}$

Calculation:

Let the number of girls in the group = x

The total weight of 'x' girls = 58.4 × x = 58.4x kg

In the first case:

15 girls with an average weight 62.2 kg leave the group.

The total weight of 15 girls = 62.2 × 15 = 933 kg

The new average = $\frac{58.4x - 933}{x - 15}$

In the second case:

10 girls with an average weight 49.85 kg join the group.

The total weight of 10 girls = 49.85 × 10 = 498.5 kg

The new average = $\frac{58.4x + 498.5}{x + 10}$

According to the question,

⇒ $\frac{58.4x - 933}{x - 15} = \frac{58.4x + 498.5}{x + 10}$

⇒ (58.4x - 933)(x + 10) = (58.4x + 498.5)(x - 15)

⇒ 58.4x² + 584x - 933x - 9330 = 58.4x² + 498.5x - 876x - 7477.5

⇒ -349x - 9330 = -377.5x - 7477.5

⇒ 28.5x = 1852.5

⇒ x = 65

∴ Initially, the number of girls in the group is 65.

Alternate MethodLet the number of girls in the group be x.

According to the questions

⇒ (58.4 × x - 62.2 × 15)(x + 10) = (58.4 × x + 49.85 × 10)(x - 15)

⇒ (58.4x - 933)(x + 10) = (58.4x + 498.5)(x - 15)

⇒ 58.4x² - 933x + 584x - 9330 = 58.4x² + 498.5x - 876x - 7477.5

⇒ - 933x + 584x - 498.5x + 876x = 9330 - 7477.5

⇒ 28.5x = 1852.5

⇒ x = 65

∴ The number of girls in the group initially is 65.

Given:

In ΔABC, BE ⊥ AC, CD ⊥ AB and BE and CD intersect at O.

The bisectors of ∠OBC and ∠OCB intersect at P.

∠A = 62°

Formula used:

The intersection of altitude is called orthocenter.

The relationship between the angle made by the orthocenter of a triangle can be seen below figure.

∠BOC = 180^∘ - ∠A

The sum of all three internal angles for a triangle is 180∘.

Calculation:

According to the question, the required figure is:

In ΔABC,

∠BOC = 180∘ - ∠A

⇒ ∠BOC = 180∘ - 62∘

⇒ ∠BOC = 118∘

In ΔBOC,

∠OBC + ∠OCB + ∠BOC = 180∘

⇒ 2x + 2y + 118∘ = 180∘

⇒ 2(x + y) = 62∘

⇒ (x + y) = 31∘

In ΔBPC,

∠PBC + ∠PCB + ∠BPC = 180∘

⇒ x + y + ∠BPC = 180∘

⇒ 31∘ + ∠BPC = 180∘

⇒ ∠BPC = 149∘

∴ The required value of ∠BPC is 149∘.

Given:

$5\frac{2}{3} \div 1\frac{2}{3}$ of $4\frac{1}{4} - 2\frac{4}{5} + \left( {2\frac{1}{3} + 2\frac{1}{2} - \frac{1}{6}} \right) \div 2\frac{1}{3} + 5\frac{1}{3} × \frac{3}{4} \div 2\frac{2}{3}$ 

Concept used:

Calculation:

$5\frac{2}{3} \div 1\frac{2}{3}$ of $4\frac{1}{4} - 2\frac{4}{5} + \left( {2\frac{1}{3} + 2\frac{1}{2} - \frac{1}{6}} \right) \div 2\frac{1}{3} + 5\frac{1}{3} × \frac{3}{4} \div 2\frac{2}{3}$

⇒ $5\frac{2}{3} \div 1\frac{2}{3}$ of $4\frac{1}{4} - 2\frac{4}{5} + \left( {\frac{7}{3} + \frac{5}{2} - \frac{1}{6}} \right) \div 2\frac{1}{3} + 5\frac{1}{3} × \frac{3}{4} \div 2\frac{2}{3}$ 

⇒ $5\frac{2}{3} \div 1\frac{2}{3}$ of $4\frac{1}{4} - 2\frac{4}{5} + \left( { \frac{14+15-1}{6}} \right) \div 2\frac{1}{3} + 5\frac{1}{3} × \frac{3}{4} \div 2\frac{2}{3}$ 

⇒ $5\frac{2}{3} \div 1\frac{2}{3}$ of $4\frac{1}{4} - 2\frac{4}{5} + { \frac{14}{3}}\div 2\frac{1}{3} + 5\frac{1}{3} × \frac{3}{4} \div 2\frac{2}{3}$ 

⇒ $5\frac{2}{3} \div \frac{5}{3}$ of $\frac{17}{4} - 2\frac{4}{5} + { \frac{14}{3}}\div 2\frac{1}{3} + 5\frac{1}{3} × \frac{3}{4} \div 2\frac{2}{3}$

⇒ $5\frac{2}{3} \div \frac{85}{12}- 2\frac{4}{5} + { \frac{14}{3}}\div 2\frac{1}{3} + 5\frac{1}{3} × \frac{3}{4} \div 2\frac{2}{3}$

⇒ $\frac{17}{3} \div \frac{85}{12}- 2\frac{4}{5} + { \frac{14}{3}}\div \frac{7}{3} + 5\frac{1}{3} × \frac{3}{4} \div \frac{8}{3}$

⇒ $\frac{17}{3} \times \frac{12}{85}- 2\frac{4}{5} + { \frac{14}{3}}\times \frac{3}{7} + 5\frac{1}{3} × \frac{3}{4} \times \frac{3}{8}$

⇒ $\frac{4}{5}- \frac{14}{5} + 2 + \frac{16}{3} × \frac{9}{32}$

⇒ $\frac{4}{5}- \frac{14}{5} + 2 + \frac{3}{2}$

⇒ $\frac{8-28+20+15}{10}$

⇒ $\frac{15}{10}$

⇒ $\frac{3}{2}$

∴ Required answer is $\frac{3}{2}$.

Given:

If 1 is subtracted from the numerator and 3 is added to the denominator of a fraction, the fraction becomes $\frac{1}{2}$.

If 1 is added to the numerator and 1 is subtracted from the denominator of the original fraction, it becomes $\frac{5}{6}$. 

Calculation:

Let the original fraction = $\frac{a}{b}$

In the first case:

⇒ $\frac{a-1}{b+3} = \frac{1}{2}$

It can be written as:

⇒ 2 × (a - 1) = b + 3

⇒ 2a - 2 = b + 3

⇒ 2a - b = 5    ----(1)

In the second case:

⇒ $\frac{a+1}{b-1} = \frac{5}{6}$

It can be written as:

⇒ 6 × (a + 1) = 5 × (b - 1)

⇒ 6a + 6 = 5b - 5

⇒ 5b - 6a = 11    ----(2)

Multiply the equation (1) by 5 and add to equation (2).

⇒ 4a = 36

⇒ a = 9

Now, substitute the value of 'a' in equation (1), we get,

⇒ 2 × 9 - b = 5

⇒ 18 - 5 = b

⇒ b = 13

The required fraction = $\frac{a}{b} = \frac{9}{13}$

The sum of the numerator and the denominator of the original fraction = 9 + 13 = 22

∴ 22 is the required answer.

Given:

The height of a right circular cone is trisected by two planes parallel to its base at equal distances.

Formula used:

Cone:

The volume of cone = $\frac{\pi r^2 h}{3}$

Frustum:

The volume of frustum = $\frac{\pi h}{3} (r_1^2 + r_2^2 + r_1r_2)$

Calculation:

According to the question, the required figure is:

The ΔABC ∼ ΔAB'C'

⇒ $\frac{AC}{AC'} = \frac{BC}{B'C'}$

⇒ $\frac{3h}{2h} = \frac{r}{r_1}$

⇒ r₁ = $\frac{2}{3}$r

The ΔAB''C'' ∼ ΔAB'C'

⇒ $\frac{AC''}{AC'} = \frac{B''C''}{B'C'}$

⇒ $\frac{h}{2h} = \frac{r_2}{r_1}$

⇒ $\frac{h}{2h} = \frac{3r_2}{2r}$

⇒ r₂ = $\frac{1}{3}$r

Now, 

The volume of the cone AB''D'',

⇒ V_AB''D'' = $\frac{\pi}{3} r_2^2h$

⇒ VAB''D'' = $\frac{\pi}{3} (\frac{r}{3})^2h$

⇒ VAB''D'' = $\frac{\pi}{27} r^2h$

The volume of frustum cone B''D''B'D',

⇒ V_B''D''B'D' = $\frac{\pi h}{3} (r_1^2 + r_2^2 + r_1r_2)$

⇒ VB''D''B'D' = $\frac{\pi h}{3} [(\frac{2}{3}r)^2 + (\frac{r}{3})^2 + (\frac{r}{3})(\frac{2r}{3})]$

⇒ VB''D''B'D' = $\frac{\pi h}{3} [(\frac{4}{9}r^2) + (\frac{r^2}{9}) + (\frac{2r^2}{9})]$

⇒ VB''D''B'D' = $\frac{\pi h}{3} [\frac{7}{9}r^2]$

The volume of frustum cone B'D'BD,

⇒ VB'D'B'D' = $\frac{\pi h}{3} (r_1^2 + r^2 + r_1r)$

⇒ VB'D'B'D' = $\frac{\pi h}{3} [(\frac{2}{3}r)^2 + (r)^2 + (r)(\frac{2r}{3})]$

⇒ VB'D'B'D' = $\frac{\pi h}{3} [(\frac{4}{9}r^2) + (\frac{9r^2}{9}) + (\frac{6r^2}{9})]$

⇒ VB'D'B'D' = $\frac{\pi h}{3} [\frac{19}{9}r^2]$

The required ratio = $\frac{\pi h}{27} r^2 : \frac{\pi h}{3} [\frac{7}{9}r^2] :\frac{\pi h}{3} [\frac{19}{9}r^2]$ = 1 : 7 : 19

∴ 1 : 7 : 19 is the requried ratio.

Given:

The number of pipes attached = 16

The number of filling pipes = x

The number of emptying pipes = 16 - x

The time is taken by a filling pipe to fill the tank = 8 hours

The time is taken from an emptying pipe to emptying the tank = 10 hours 

When the tank is 65% empty, all the pipes are opened at the same time. It takes one hour to fill the tank completely

Formula used:

Efficiency × TIme = Work

Calculation:

Let the capacity of the tank = 1 unit

The time is taken by a filling pipe to fill the tank = 8 hours

The efficiency for filling pipe = $\frac{1}{8}$

The time is taken from an emptying pipe to emptying the tank = 10 hours 

The efficiency of emptying pipe = $-\frac{1}{10}$

Now, 

The efficiency for 'x' filling pipe = $\frac{x}{8}$

The efficiency for '16 - x' emptying pipes = $-\frac{16-x}{10}$

The remaining part of the tank which is to be filled = 65% of 1 unit

The time taken to fill the 65% of the tank = 1 hours

According to the question,

⇒ ($\frac{x}{8}$ - $\frac{16-x}{10}$) × 1 = 65% of 1 unit

⇒ ($\frac{x}{8}$ - $\frac{16-x}{10}$) =  0.65

⇒ ($\frac{10x - 8(16 - x)}{80}$) =  0.65

⇒ 10x - 128 + 8x =  0.65 × 80

⇒ 18x = 128 + 52

⇒ 18x = 180

⇒ x = 10

The number of emptying pipes = 16 - 10 = 6

∴ The number of emptying pipes is 6.

Given:

In finding the HCF of two numbers by the division method, the last divisor is 28 and the quotients are 1, 30, 1 and 3, respectively.

Formula used:

Dividend = Divisor × Quotient + Remainder

Calculation:

As per the question, we have to solve the question reverse.

The last divisor, d₃ = 28

The respective quotient, q₃ = 3

This is the last step in the divisions' method, the remainder is 0.

Now, substitute the value of the respective variables in the above relation.

(Dividend)₃ = 28 × 3 + 0

(Dividend)3 = 84

As we know, the last step's dividend will be divisor for the next step and the divisor is now taken as the remainder.

The divisor in this step, d₂ = 84

The respective quotient, q₂ = 1

The respective remainder, r₂ = 28

Now, substitute the respective value in the required relation.

(Dividend)2 = 84 × 1 + 28

(Dividend)2 = 112

Similarly the next step,

(Dividend)1 = 112 × 30 + 84

(Dividend)1 = 3444

Similarly the next step,

(Dividend) = 3444 × 1 + 112

(Dividend) = 3556

The required numbers are 3444 and 3556.

The sum of two numbers = 3444 + 3556 = 7000

∴ 7000 is the required answer.

Given: 

If the income of companies A and B in 2017 are in the ratio 3 : 5,

The total expenditure/investment of both the companies was Rs. 350 crores,

The profit per cent earned by company A in 2017 = 35%

The profit per cent earned by company B in 2017 = 50%

Formula used:

Profit% = $\frac{\text{Income} - \text{Investment}}{\text{Investment}} × 100$

It can be written as:

Income = Investment × (Profit%)

Calculation:

Let 'I_A' and 'I_B' are the income of companies A and B in 2017.

Let 'EA' and 'EB' are the expenditure of companies A and B in 2017.

According to the question

The income of company A in 2017, I_A = 3x

The income of company B in 2017, I_B = 5x

Now,

⇒ E_A + E_B = Rs. 350 crores

⇒ $\frac{I_A}{1 + \text{Profit}_A} + \frac{I_B}{1 + \text{Profit}_B}$ = 350

⇒ $\frac{I_A}{1 + 35\text {\%}} + \frac{I_B}{1 + 50\text {\%}}$ = 350

⇒ $\frac{I_A}{1 + 0.35} + \frac{I_B}{1 + 0.50}$ = 350

⇒ $\frac{3x}{1.35} + \frac{5x}{1.50}$ = 350

⇒ $\frac{4.5x + 6.75x}{2.025}$ = 350

⇒ $\frac{11.25x}{2.025}$ = 350

⇒ x = 63

Therefore, 

The income of company A in 2017, IA = 3 × 63 = Rs. 189 crores

The income of company B in 2017, IB = 5 × 63 = Rs. 315 crores

Now, 

The income of companies A and B in 2017, = Rs. 189 + Rs. 315 = Rs. 504 crores

∴  The total income (in Rs. crores) of A and B together in 2017 is 504.

Shortcut Trick

Present Average Age of A, B and C = 21 + 5 = 26 years

Present Average Age of B and C = 25.5 − 3 = 22.5 years

Present Age of A = (Total of A, B, C) − (Total of B, C)

Present Age of A = (26 × 3) − (22.5 × 2) = 78 − 45 = 33 years

∴ The correct answer is 33 years.

Alternate Method

Given:

Average age of A, B, C (5 years ago) = 21 years

Average age of B, C (3 years hence) = 25.5 years

Formula Used:

Total Age = Average Age × Number of Persons

Calculations:

⇒ Sum of ages of A, B, C (5 years ago) = 21 × 3 = 63 years

⇒ Present Sum of ages of A, B, C = 63 + (3 × 5) = 78 years

⇒ Sum of ages of B and C (3 years from now) = 25.5 × 2 = 51 years

⇒ Present Sum of ages of B and C = 51 − (2 × 3) = 51 − 6 = 45 years

⇒ Present age of A = (Present Sum of A, B, C) − (Present Sum of B, C)

⇒ Present age of A = 78 − 45 = 33 years

∴ The correct answer is 33 years.

Additional Information

Average Age Property

If the age of each person in a group increases or decreases by n years, the Average Age of the group also increases or decreases by n years.

Sum and Average Relation

The sum of ages of n people after t years = (Present Sum) + (n × t).

Finding Individual Value

To find the age of a specific person from a group, subtract the sum of the ages of the remaining members from the total sum: A = Sum(All) − Sum(Others).

Given:

$\frac{{\left( {0.\overline {95} \div 1.0\overline 5 } \right) \div 0.4\overline {09} }}{{\left( {0.\overline {84} \div 0.9\overline 3 \div 0.\overline {54} } \right)}}$

**Concept used:**​​

$0.\overline {ab} =\frac{ab}{99}$

$c.\overline {ab} =\frac{cab - c}{99}$

$c.a\overline {b} =\frac{cab - ca}{90}$

Calculation:

$\frac{{\left( {0.\overline {95} \div 1.0\overline 5 } \right) \div 0.4\overline {09} }}{{\left( {0.\overline {84} \div 0.9\overline 3 \div 0.\overline {54} } \right)}}$

⇒ $\frac{{\left( {\frac{95}{99} \times \frac{90}{105 -10} } \right) \times \frac{990}{409 - 4} }}{{\left( {\frac{84}{99} \times \frac{90}{93-9}\times\frac{99}{54} } \right)}}$

⇒ $\frac{{\left( {\frac{95}{99} \times \frac{90}{95} } \right) \times \frac{990}{405} }}{{\left( {\frac{84}{99} \times \frac{90}{84}\times\frac{99}{54} } \right)}}$

⇒ $\frac{{ {\frac{10}{11} } \times \frac{990}{405} }}{{{\frac{5}{3} }}}$

⇒ $\frac{{ {\frac{20}{9} }}}{{ {\frac{5}{3} }}}$

⇒ ${{\left( {\frac{20}{9} } \right)}}\times{{\left( {\frac{3}{5} } \right)}}$

⇒ $\frac{4}{3} = 1\frac{1}{3}$

∴ Required answer is $1\frac{1}{3}$

Given:

If  $5x -\frac{1}{{3x}}=3,x\ne0$

Formula used:

(a - b)² = a² + b² - 2ab

Calculation:

$5x -\frac{1}{{3x}}=3$

Now, multiply the required expression by $\frac{3}{5}$ on both sides of the equation.

⇒ $\frac{3}{5} \times (5x -\frac{1}{{3x}})=\frac{3}{5} \times3$

⇒ $(3x -\frac{1}{{5x}})=\frac{9}{5}$

Take square on both sides of the equation,

⇒ $(3x -\frac{1}{{5x}})^2=\frac{9^2}{5^2}$

⇒ $9x^2 + \frac{1}{{25x^2}} - 2 \times 3x \times \frac{1}{5x}=\frac{9^2}{5^2}$

⇒ $9x^2 + \frac{1}{{25x^2}} - \frac{6}{5}=\frac{81}{25}$

⇒ $9x^2 + \frac{1}{{25x^2}} =\frac{81}{25} + \frac{6}{5}$

⇒ $9x^2 + \frac{1}{{25x^2}} =\frac{81 + 30}{25}$

⇒ $9x^2 + \frac{1}{{25x^2}} =\frac{111}{25}$

⇒ $9x^2 + \frac{1}{{25x^2}} =4\frac{11}{25}$

∴ $4\frac{11}{25}$ is the required answer.

Shortcut Trick

Let 1/D = x and 1/U = y (D = Downstream, U = Upstream speed).

Eq 1: 9x + 4y = 2.5 | Eq 2: 15x + 6y = 4

Multiply Eq 1 by 3 and Eq 2 by 2 to equate y: 27x + 12y = 7.5 and 30x + 12y = 8.

Subtracting gives 3x = 0.5 ⇒ x = 1/6 ⇒ D = 6 kmph.

Substitute x in Eq 1: 9(1/6) + 4y = 2.5 ⇒ 1.5 + 4y = 2.5 ⇒ 4y = 1 ⇒ y = 1/4 ⇒ U = 4 kmph.

Required Time = 8/6 + 10/4 = 4/3 + 5/2 = (8 + 15)/6 = 23/6 hours = 3 hr 50 min.

∴ The correct answer is 3 hr 50 min.

Alternate Method

Given:

Case 1: 9/D + 4/U = 2.5 hours

Case 2: 15/D + 6/U = 4 hours

Formula Used:

Time = Distance ÷ Speed

Calculations:

⇒ Let 1/D = u and 1/U = v

⇒ 9u + 4v = 2.5 --- (i)

⇒ 15u + 6v = 4 --- (ii)

⇒ Multiply (i) by 3: 27u + 12v = 7.5

⇒ Multiply (ii) by 2: 30u + 12v = 8

⇒ Subtracting: (30u − 27u) = 8 − 7.5 ⇒ 3u = 0.5 ⇒ u = 1/6

⇒ Downstream speed (D) = 6 kmph

⇒ Put u = 1/6 in (i): 9(1/6) + 4v = 2.5 ⇒ 1.5 + 4v = 2.5 ⇒ 4v = 1 ⇒ v = 1/4

⇒ Upstream speed (U) = 4 kmph

⇒ Required Time = 8/D + 10/U = 8/6 + 10/4

⇒ Time = 4/3 + 5/2 = (8 + 15)/6 = 23/6 hours

⇒ Time = 3 hours + 5/6 hours = 3 hours + (5/6 × 60) min = 3 hr 50 min

∴ The correct answer is 3 hr 50 min.

Additional Information

Downstream and Upstream Speeds

Downstream speed (D) = Speed of boat in still water (u) + Speed of stream (v). Upstream speed (U) = u − v.

Still Water Speed Formula

Speed of boat in still water (u) = (D + U) ÷ 2.

Stream Speed Formula

Speed of stream (v) = (D − U) ÷ 2.

The correct answer is 53.

Concept:

Lens:

• The transparent curved surface which is used to refract the light and make an image of any object placed in front of it is called a lens.
• A convex lens can converge a beam of parallel rays to a point on the other side of the lens

Power of lens:

• It is equal to the reciprocal of the focal length of the lens/mirror in meters.
• Power (P) = 1/f, where f is the focal length of the lens/mirror.
• P = $\frac{100}{f}$ where f is the focal length in centimeters.
• The unit of power of a lens is Dioptre when the focal length of the lens is taken in meters (m).
• The focal length of the concave lens is negative and hence its power is also negative.
• The focal length of the convex lens is positive and hence its power is also positive.

Lens formula: $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$  where f is focal length, v is image distance and u is object distance.

Explanation:

For first convex lens, L1

Power = $\frac{1}{f}$

f = $\frac{1}{p}$ = $\frac{100}{10}$ = 10 cm

Given: Object distance, u = -15 cm, Focal length, f = 10 cm

 v=?

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{10} = \frac{1}{v} - \frac{1}{-15}$

$\frac{1}{10} = \frac{1}{v} + \frac{1}{15}$

$\frac{1}{10} - \frac{1}{15} = \frac{1}{v}$

$\frac{1}{v} = \frac{3-2}{30}$

$\frac{1}{v} = \frac{1}{30}$

v = 30 cm

For second convex  lens, L2

f = $\frac{1}{p}$ = $\frac{100 × 3}{-10}$ = -10 × 3 = -30 cm

Object distance, u = -50-(-30)= -50 + 30 = -20cm, focal length, f = -30cm

v = ?

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{-30} = \frac{1}{v} - \frac{1}{-20}$

$\frac{1}{-30} = \frac{1}{v} + \frac{1}{20}$

$\frac{1}{-30} - \frac{1}{20} = \frac{1}{v}$

$\frac{1}{v} = \frac{-2-3}{60} = \frac{-5}{60}$

v= -60 / 5 = -12 cm 

Distance between original object and final image = 15 + 50 - 12 = 65 - 12 = 53.

The correct answer is 2.

Concept:

The Modern Periodic Table:

• In 1913, Henry Moseley showed that the atomic number of an element is a more fundamental property than its atomic mass.
• Modern Period Law: The physical and chemical properties of elements are the periodic function of their atomic number.
• The modern periodic table contains 18 vertical columns known as groups and seven horizontal rows known as periods.

Key Points

Period:

• The periods have the same elements that have the same valence shell or the energy shell.
• The number of elements present in a period can be determined by the formula 2n2, where n is the number of the shell from the nucleus.
• For example, the first period has two elements only as they have only 1 valence shell and the second period has 8 elements with 2 shells.

​Groups:

• The group consists of elements that have the same number of electrons in the valence shell or outermost shell of the atom.
• The number of shells increases by one unit as the elements move down in the periodic table in a group.
• Group number indicates the valence electrons of the atom. Group1 and 2 have 1 and 2 valence electrons in the outermost shell respectively.
• But for group no. 13-18 it is calculated by subtracting 10 from the group no. For example, an element belonging to group 13 will have 13-10 = 3 valence electrons.

Electronic configuration:

• The distribution of electrons around the nucleus in different shells.
• There are four shells namely K, L, M, and N.
• Each orbit can hold a maximum of 2n2 electrons, where n is the number of orbits.
• Only 2 electrons can accommodate in the first shell i.e., shell K.
• 8 in shell L
• 18 in shell M
• 32 in shell N

Explanation:

• Period number = 3
• So, Number of shells = Period number = 3
• Group number = 16
• So, Valence electrons in outermost shell = Group number - 10 = 16 -10 = 6

Electronic configuration of Element A:

• Valency refers to the number of electrons lost or gained by an element to complete its duplet or octet.
• Here, Element A has 6 valence electrons, so it will gain 2 electrons to complete its octet.

Thus, it has a valency = 2.

The correct answer is Roasting.

Concept:

• Mineral: A naturally occurring substance with uniform chemical composition is called a mineral.
• Ore: An ore is a mineral from which a metal can be extracted in a profitable way.

According to the reactivity of metals, different methods are employed to extract metals from their ores.

Reactivity series of metals:

• Metals of high reactivity; such as sodium, calcium, magnesium, aluminium, etc. are extracted from their ores by electrolytic reduction.
• Metals of middle reactivity, Iron, zinc, lead, etc. are found in the form of carbonate or sulfide ores. They are either calcinated or roasted.
• Metals of least reactivity, Mercury and copper, which belong to the least reactivity series, are often found in the form of their sulfide ores. They are roasted and then refined.

Explanation:

• Cinnabar: It is an ore of Mercury. Its chemical formula is HgS.
• Mercury is among the least reactive metals in the reactivity series.
• Its sulfide ore is roasted to extract the Mercury metal.
• Roasting: The heating of sulfide ores in the presence of excess air to convert them into oxides is known as Roasting.

Extraction of Mercury Metal:

• Cinnabar (HgS) is first heated in the air.
• This turns HgS (mercury sulfide or cinnabar) into HgO (mercury oxide) by liberation of sulfur dioxide.
• 2HgS + 3O2 → 2HgO + 2SO2
• Mercury oxide so obtained is again heated strongly.
• This reduces mercury oxide to mercury metal.
• 2HgO + 2Hg + O2

So, Extracting a metal from cinnabar ore is through Roasting.

Additional Information

• Electrolytic reduction: Electric current is passed through the molten state of metal ores. Metal being positively charged is deposited over the cathode.
• Thermit process: It is a chemical reaction between metal and metal oxide. A lot of heat energy is liberated in this reaction.
• Calcination: The heating of carbonate ores in the limited supply of air to convert them into oxides is known as Calcination.

The correct answer is 1.26; towards.

Concept:

Refraction of Light: 

• The bending of the ray of light passing from one medium to the other medium is called refraction.

Refractive Index (n):

• n = sin i / sin r, where 'i' =  angle of incidence  and 'r' = angle of refraction.
• Also, n = c/v, where c = speed of light in a vacuum, and v = speed of light in a medium.

Explanation:​

Relative refractive index:

• It refers to the refractive index of one material medium with respect to another one.
• $n_{21}= \frac{n_{2}}{n_{1}}$

Refractive index of glass (dense flint) with respect to turpentine oil:

$n= \frac{Refractive \ index \ of \ glass \ (dense \ flint) }{Refractiev \ index\ of\ water}= \frac{1.65}{1.31}=1.26$

• Greater the refractive index, the denser the medium.
• Thus, glass (1.65) is denser than ice (1.31).
• When the ray of light moves from a denser to a rarer medium, the refracted ray moves away from the normal.
• When the ray of light moves from a rarer to a denser medium, the refracted ray moves towards the normal.
• Thus, when a light ray will move from ice to glass, it will move towards the normal.

The refractive index of glass (dense flint) with respect to ice is 1.26 and the light ray bends towards the normal in the glass.

The correct answer is Lower than 5.5.

Concept:

pH scale:

• A scale developed to measure the concentration of hydrogen ions in a solution is known as the pH scale.
• pH scale ranges from 0 (highly acidic) to 14 (highly alkaline).
• The pH of a neutral solution is 7.
• The basic strength increases as we move from 7 to 14 on the pH scale.
• The acidic strength increases as we move from 7 to 1.

Tooth decay: 

• Tooth decay is damage to a tooth's surface or enamel.
• It happens when bacteria in our mouth make acids that attack the enamel.
• Enamel is the hardest part of the human body, made up of calcium phosphate.

Explanation:

Cause of tooth decay:

• Bacteria in the mouth feed on sugary, starchy foods and drinks (fruit, candy, bread, cereal, sodas, juice, and milk).
• The bacteria convert these carbohydrates into acids.
• Tooth decay starts when the pH of the mouth becomes lower than 5.5.
• Enamel which is composed of calcium phosphate starts to dissolve at such low pH.
• Tooth decay can lead to cavities (dental caries), which are holes in your teeth.
• If tooth decay is not treated, it can cause pain, infection, and even tooth loss.

Additional Information

Prevention of tooth decay: Tooth decay can be prevented  by 

• Eating a healthy diet,
• Drinking plenty of tap water instead of sugary drinks
• Brushing your teeth twice a day
• Using a fluoride toothpaste
• Flossing once a day.

The correct answer is 146g.

Concept:

Mole:

• It is the SI unit of the amount of substance.
• One mole contains exactly 6.022×10²³ atoms/molecules.
• The mass of one mole of a substance in grams is called its molar mass.
• The molar mass in grams is numerically equal to atomic/molecular/formula mass in u.
• So, 1 mole = 6.022 × 10²³ atoms/molecules = Atomic/molecular mass of an element/compound.

​Explanation:

Molar mass of methane (CH4) = 12 + 1 × 4 = 16g

1 mole of methane (CH₄)  = Molar mass of methane (CH4) = 6.022 × 10²³ molecules of methane (CH4)

⇒ 16g CH4 contains 6.022 × 1023 molecules of methane (CH4).

1 molecule of CH₄ has 4 H atoms. 

6.022 × 1023 molecules of methane (CH4) will have H atoms = 4 × 6.022 × 1023 = 2.4088 × 10²⁴

Hydrochloric acid (HCl) has the same number of H atoms as in 16g of CH4.

1 molecule of HCl has 1 H atom. 

2.4088 × 1024 molecules of HCl will have 2.4088 × 1024 H atoms. 

Molar mass of HCl = 1 + 35.5 = 36.5g

1 mole of  HCl = Molar mass of HCl = 6.022 × 1023 molecules of HCl

⇒  6.022 × 1023 molecules of HCl = 36.5 g HCl

1 molecule of HCl = 36.5 / 6.022 × 1023 g

2.4088 × 1024 molecules =  36.5 × 2.4088 × 1024 / 6.022 × 1023 = 36.5 × 4 = 146g 

Thus, 146g of HCl will have the same number of H atoms as are present in 16g of methane.