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ARCHITECT S GUIDEBOOK FOR COMPETITIVE EXAMS Set 1- Solved Question Papers (GATE- 1991 to 2018, UPSC- 2015-2016)

Ar. Swapnil Vidhate B. Arch. (University of Mumbai) M. Plan (NIT, Bhopal)

Published by: Mrs. Rachana S. Vidhate C-801, Sun Satellite, Near Suncity, Anandnagar, Sinhgad Road, Pune- 411051. Phone: +91-9975603199 e-mail: ISBN (13)- 978-9-35-267484-8 ISBN (10)- 9-35-267484-7


Information contained in this book has been obtained by author from sources believed to be reliable and are correct to the best of his knowledge. Every effort has been made to avoid errors and omissions and ensure accuracy. Any error or omission noted may be brought to the notice of the author which shall be taken care of in the forthcoming edition of this book. However, author does not guarantee the accuracy or completeness of any information published herein, and does not take not responsibility or liability for any inconvenience, expenses, losses or damage to any one resulting from the contents of the book. The author of the book has taken all possible care to ensure that the contents of the book do not violate any existing copyright or other intellectual property rights of any person in any manner whatsoever. In the event, the author has been unable to track any source and if any copyright has been infringed, the facts may be brought to the notice of the author in writing for corrective action. Send all correspondence to: Ar. Swapnil S. Vidhate Email:

First Edition

: August 2016.

Second Edition : March 2017. Third Edition

: March 2018.

Price: Rs. 1000/-


Architecture is one of the prestigious careers available to young individuals for work. It carries a level of education superior to that of many other career options. After 5 years of graduation, many students opt to pursue higher education for specialization. At present in India, almost all admissions processes for Architecture and planning specialization courses are based on entrance exams like GATE, CEED, State CET etc. The number of students appearing for these exams are increasing every year which is making the level of competitive exams higher and higher. The recruitment of architects and planners in various government agencies is also done by preliminary examinations. But no standard books are available for students for preparation. Thus, this book is compiled and published as a guidebook to help students and architects for those exams. This book will provide the past trends of questions asked in the exams with their solutions & detailed explanations for better understanding. The book contains all GATE- Architecture & Planning papers from 1991 to 2018 as well as UPSC-CPWD 2015-2016 exam papers which were conducted for recruitment of architects. The book not only provides the questions and answers but also provides the explanations to those. These will be helpful for understanding the basic concepts, formulae and relevant information about the question which students had learnt in the academic years. The Author welcomes suggestions from the readers for improving the book in any manner.

Ar. Swapnil Vidhate (Author)



I express my sincere thanks to my family and friends for their constant help, support and encouragement in compiling the book. encouraged me to compile the book and took all the efforts to publish it. Thank you so much. -Ar. Swapnil Vidhate

Index Question Paper

Page No.

01. GATE 2018


02. GATE 2017


03. GATE 2016


04. GATE 2015


05. GATE 2014


06. GATE 2013


07. GATE 2012


08. GATE 2011


09. GATE 2010


10. GATE 2009


11. GATE 2008


12. GATE 2007


13. GATE 2006


14. GATE 2005


15. GATE 2004


16. GATE 2003 . 17. GATE 2002

353 381

18. GATE 2001


19. GATE 2000


20. GATE 1999


21. GATE 1998


22. GATE 1997


23. GATE 1996


24. GATE 1995


25. GATE 1994


26. GATE 1993


27. GATE 1992


28. GATE 1991


29. UPSC 2015


30. UPSC 2016




Q. 5 carry one mark each.

that best fill the

blanks in the above sentence are. (A) stairs, stares

(C) stares, stairs

(B) stairs, stairs

(D) stares, stares

Ans. A

the blank in the above sentence is. (A) rational

(C) errant

(B) reasonable

(D) good

Ans. C (Errant- Misbehaving / sinful)

Q.3. For 0 x 2 , sin x and cos x are both decreasing functions in the interval ___________. (A) (0, /2) (B) ( /2, ) (C) ( , 3 /2) (D) (3 /2, 2 ) Ans. B Explanation: Draw the curves of sinx and cosx. From the curves, it can be observed that both functions are decreasing in the interval of /2 and

(A) 2

(C) 6

(B) 4

(D) 8

interval respectively.

Ans. C Explanation: Area of equilateral triangle = 2.


Where a = length of each side.

Thus, a= 2 (side of triangle). Thus, perimeter = 3a= 6

Q.5. Arrange the following three-dimensional objects in the descending order of their volumes. (i) A cuboid with dimensions 10 cm, 8 cm and 6 cm. (ii) A cube of side 8 cm. (iii) A cylinder with base radius 7 cm and height 7 cm (iv) A sphere of radius 7 cm. 1


(A) (i), (ii), (iii), (iv)

(C)(iii), (ii), (i), (iv)

(B) (ii), (i), (iv), (iii)

(D) (iv), (iii), (ii), (i)

Ans. D Sr.No. i ii Iii iv Q. 6

Type of Solid Cuboid Cube Cylinder Sphere

Formula to calculate volume Side x Side x Side Side3 x (base radius)2 x height = r2h 4/3 x x radius3

Calculation 10 x 8 x 6 83 22/7x 72x 7 4/3 x 22/7 x 73

Volume 480 cm3 512 cm3 1078 cm3 1437.33 cm3

Q. 10 carry two marks each.

Q.6. An automobile travels from city A to city B and returns to city A by the same route. The speed of the vehicle during the onward and return journeys was constant at 60km/hr. and 90 km/hr. What is the average speed in km/hr. for the entire journey? (A) 72

(C) 74

(B) 73

(D) 75

Ans. A Explanation: Assume the travel distance from city A to B is 180 km. Speed while travelling from city A to B = 60 km /hr. Thus, travel time from A to B = 3 hr. Speed while travelling from city B to A = 90 km /hr. Thus, travel time from B to A = 2 hr. Thus, total travel time for to and fro travel from A to B = 5 hr. Total travel distance covered while to and fro travel from A to B = 180 + 180 = 360 km. Thus, average speed for the entire journey = 360 / 5 = 72 km /hr. Q.7. A set of 4 parallel lines intersect with another set of 5 parallel lines. How many parallelograms are formed? (A) 20

(C) 60

(B) 48

(D) 72

Ans. C Explanation: Total no of parallel lines =5, Intersecting lines =4 he required number of parallelogram = 5C2 Ă— 4C2 =

5!__ x 4!__ = 5 x 4 x 3 x 2 x 1 x 4 x 3 x 2 x 1 = 10 x 6 = 60 3! x 2! 2! x 2! 3x2x1x2x1 2x1x2x1

Q.8. To pass a test, a candidate needs to answer at least 2 out of 3 questions correctly. A total of 6,30,000 candidates appeared for the test. Question A was correctly answered by 3,30,000 candidates. Question B was answered correctly by 2,50,000 candidates. Question C was answered correctly by 2,60,000 candidates. Both questions A and B were answered correctly by 1,00,000 candidates. Both questions B and C were answered correctly by 90,000 candidates. Both questions A and C were answered correctly by 80,000 candidates. If the number of students answering all questions correctly is the same as the number answering none, how many candidates failed to clear the test? (A) 30,000

(C) 3,90,000

(B) 2,70,000

(D) 4,20,000

Ans. D



Total students














Assume in A+B+C



Passed only in A+B



Passed only in B+C



Passed only in C+A



Passed only in A



Passed only in B



Passed only in C


Total Students

570k + 2x

(7) (Adding 1 to 7)


Equating (8) with total no. of students as given. 630k = 570k + 2x. Thus, x = 30k = 30,000 Thus, no. of students who got all A, B & C correct = 30,000 As given, the number of students who answered none of the three questions = Number of students who answered all three correct = 30,000. To pass the test, the students need to answer at least 2 questions. Thus, total no. of failed students

Who got only 1 correct + who got none correct = 150000 + 30000 + 60000 + 30000 + 90000 + 30000 + 30000 = 4,20,000

Q.9. If x2 + x -1 = 0. What is the value of x4 + (_1_) x4 (A) 1

(C) 7

(B) 5

(D) 9

Ans. C



Explanation: x2 + x -1 = 0. Thus, x2 + x = 1. x (x+1) = 1 (x+1) = 1 x Thus, x - 1_ = ( -1) x Squaring both sides. (x - 1_)2 = ( -1)2 x x2 + 1 - 2 = 1 x2 2 x +1=3 x2 4 Squaring both sides. x + 1 + 2 = 9 x4 x4 + 1 = 7 x4 Q.10. In a detailed study of annual crow births in India, it was found that there was relatively no growth during the period 2002 to 2004 and a sudden spike from 2004 to 2005. In another unrelated study, it was found that the revenue from cracker sales in India which remained fairly flat from 2002 to 2004, saw a sudden spike in 2005 before declining again in 2006. The solid line in the graph below refers to annual sale of crackers and the dashed line refers to the annual crow births in India. Choose the most appropriate inference from the above data. (A) There is a strong correlation between crow birth and cracker sales. (B) Cracker usage increase crow birth rate. (C) If cracker sale declines, crow birth will decline. (D) Increased birth rate of crows will cause an increase in the sale of crackers. Ans. A


Q. 25 carry one mark each.

Q.1. In a Color Wheel, Red and Blue colors are (A) Tertiary

(C) Secondary

(B) Complementary

(D) Primary

Ans. D Explanation: Refer Q. 15, GATE-2015 for more information.

maximum number of vanishing points is (A) 1 (B) 2 (C) 3 (D) 6



Ans. C

Q.3. The compressive strength of M-25 concrete is (A) 25 kg/sq.m

(C) 250 N/

(B) 25 N/

(D) 2.5 N/

Ans. B Explanation: Refer Q.20 GATE-1998 for more information.

(A) Late start and early finish time

(C) Late start and late finish time

(B) Early start and early finish time

(D) Early start and late finish time

Ans. B Explanation: Forward pass and Backward pass are the two terms related to Critical Path Method in Project Management. These terms are related to ways of determining the early or late start [forward pass] or early or late finish [backward pass] for an activity. Forward pass is a technique to move forward through a diagram to calculate activity duration. Backward pass is its opposite. The calculation helps to find out the Critical Path along the network which is the path through the project network in which none of the activities have float (total float is zero) i.e. A critical path satisfies following 3 conditions: EST =LST EFT=LFT Ej

E i = Lj

Li = tij

The critical path can be identified by determining the following four parameters for each activity: EST-earliest start time: The earliest time at which the activity can start given that all its precedent activit ies must be completed first = Ei EFT-earliest finish time: Equal to the earliest start time for the activity plus the time required to complete the activity = EST(i-j) + tij LFT-latest finish time: The latest time at which the activity can be completed without delaying (beyond its targeted completion time) the project = L j LST-latest start time: Equal to the latest finish time minus the time required to complete the activity = LFT (i-j) - tij

Q.5. Collapse of the World Trade Center (WTC), New York, in 2001, was due to (A) Wind load failure

(C) Thermal performance failure of reinforcement steel in RCC

(B) Foundation failure

(D) Thermal performance failure of structural steel Ans. D Explanation: The Twin Towers of the World Trade Center (WTC) collapsed on September 11, 2001, as a result of being struck by two jet airliners. The building performance study (BPS) team issued its report in May 2002, finding that the aircraft impacts caused "extensive structural damage, including localized collapse" and that the resulting fires "further weakened the steel-framed structures, eventually leading to total collapse". They also emphasized the role of the fires and found that sagging floors 5


pulled inward on the perimeter columns. This led to the inward bowing of the perimeter columns and failure of the south face of WTC 1 and the east face of WTC 2, initiating the collapse of each of the towers.

Q.6. During the construction of tall buildings, the equipment used for hoisting building materials to the upper floors is a (A) Goods lift

(C) Gantry crane

(B) Capsule lift

(D) Tower crane

Ans. D Type of Crane

Used for

Goods lift

It is used for transfer of goods in Industries, warehouses, etc. It is also used in under construction buildings for transfer of labors and light weight building materials.

Capsule Lift

Gantry Crane

Tower Crane

Type of Crane

Capsule lifts are used for transfer of passengers. They have aesthetically attractive interiors with large glass viewing panel. They are often installed along with the atrium or around courtyard. Such that, they provide view of entire space while using the lift. Gantry Crane is also called as portal crane. The "portal" being the empty space straddled by the gantry. Gantry cranes are used mainly for marine applications. They are also used at container terminals for loading and unloading.

Tower cranes are used in the construction of highrise buildings to lift steel, concrete, large tools, etc. It has a hoist in a fixed machinery house or on a trolley that runs horizontally along rails, usually fitted on a single beam (mono-girder) or two beams (twin-girder).

Q.7. A Rock-cut style of architecture is represented by (A) Shyama Rama Temple, Bishnupur

(C) Kandariya Mahadeva Temple, Khajuraho

(B) Kailasa Temple, Ellora

(D) Sanchi Stupa, Sanchi

Ans. B



Name of temple

Shyama Rama Temple

Kailasa Temple, Ellora

Kandariya Mahadeva temple, Khajuraho

Sanchi Stupa, Sanchi



It was built by King Raghunath Singha. It stands on a low square plinth and consists of an ambulatory pathway with a porch opened by three arches on the four sides of the temple. The central shikhara is octagonal, while the rest four are square. The walls are richly decorated with terracotta carvings

It is one of the largest rock-cut ancient Hindu temples located in Ellora, Maharashtra, India. Its construction is generally attributed to the 8th century Rashtrakutas. The temple architecture shows traces of Pallava and Chalukya styles.

It was built by the Chandela rulers. The temple architecture is an assemblage of porches and towers which terminates in a shikhara or spire, a feature which was common from the 10th century onwards in the temples of Central India.

The Great Stupa at Sanchi is one of the oldest stone structures in India. It was originally commissioned by the emperor Ashoka in the 3rd century BC and was built during the Mauryan period. It consists of a central chamber where the relics of Lord Buddha are placed. Four ornamental gateways facing four directions and a balustrade surrounding the Stupa were later added in the first century BC.

(A) Smart City Mission

(C) Swachh Bharat Mission

(B) Digital India Mission

(D) Atal Innovation Mission

Ans. A Explanation: Refer Q.7 GATE-2017 for more information.

Q.9. In mass transportation, LRTS stands for (A) Light Rail Transit System

(C) Light Rail Transportation System

(B) Linear Rail Transit System

(D) Linear Rail Transportation System

Ans. A Explanation: Light Rail Transit System medium capacity mode of mass rapid transport which straddles between the heavy capacity Metro rail and the low capacity bus services. It is a form of rail transit that utilizes equipment and infrastructure that is typically less massive than that used for heavy rail modes i.e. commuter/regional, and metro rail/subway. LRT may be at grade, partially grade separated or completely elevated. It is a low cost, low axle load, eco-friendly, electrically propelled system with 7


no local pollution and low noise and vibrations. The earliest form of LRT is the horse-drawn carriage. LRT differs from the Metro rail in that the train length is short, segregated right of way is not essential, may have road level crossings, coaches can go around sharp bends and no signaling and train control is essential.

Q.10. The structural grid type shown in the adjacent figure is a (A) Tartan Grid

(C) Rectangular Grid

(B) Square Grid

(D) Irregular Grid

Ans. A

Tartan Grid

Square grid

Rectangular Grid

It is a design of straight lines of It is a design of straight lines of It is a design of straight lines of similar varying widths and distances, similar widths and distances, widths and distances in two ways, crossing at right angles. crossing at right angles. crossing at right angles.

Q.11. Assuming other variables remaining constant, the Tropical Summer Index (A) Increases with increase in air velocity

(C) Decreases with increase in globe temperature

(B) Decreases with increase in wet-bulb temperature

(D) Increases with increase in vapour pressure

Ans. D Explanation: "Tropical Summer Index" (TSI) is defined as the air/globe temperature of the still air at 50% Relative Humidity (RH) which produces the same overall thermal sensation as the environment under investigation. This index takes into account all four environmental variables (air temperatur e, globe temperature, humidity, air velocity) in proportion to their influence on the thermal sensation. Determining the thermal comfort conditions in les thermal sensation. A simple and approximate equation for the rapid determination of TSI values for any combination of environmental variables is as follows: TSI = 1/3 tw + 3/4 tg - 2 V ½ Where: TSI = Tropical Summer Index, root of air velocity (m/s)½.


Wet-bulb Temperature (°C),


= Globe temperature (°C), V½ = Square

(A) Heritage Rejuvenation Implementation Development Aayog Yojana (B) Heritage Review Implementation Development Augmentation Yojana (C) Heritage City Development and Augmentation Yojana (D) Heritage City Improvement and Development Aawas Yojana Ans. C



Explanation: HRIDAY (Heritage City Development and Augmentation Yojana) Programme: The HRIDAY scheme was launched on 21st January, 2015, with a focus on holistic development of heritage cities. The scheme aims to by encouraging aesthetically appealing, accessible, informative & secured environment. The Scheme has duration of 4 years (Completing in November 2018). The Scheme is being implemented in 12 identified Cities namely, Ajmer, Amaravati, Amritsar, Badami, Dwarka, Gaya, Kanchipuram, Mathura, Puri, Varanasi, Velankanni and Warangal. The Scheme supports development of core heritage infrastructure projects which shall include revitalization of urban infrastructure for areas around heritage assets identified / approved by the Ministry of Culture, Government of India and State Governments. These initiatives shall include development of water supply, sanitation, drainage, waste management, approach roads, footpaths, street lights, tourist conveniences, electricity wi ring, landscaping and such citizen services.

Q.13. As per the Urban and Regional Development Plan Formulation and Implementation (URDPFI) guidelines, the (A) 1-10 years

(C) 20 -30 years

(B) 11-15 years

(D) 35 -45 years

Ans. C Explanation: Refer Q.44 GATE-2004 for more information.

Q.14. The Hall of Nations, New Delhi, was designed by (A) Charles Correa

(C) Joseph Allen Stein

(B) Raj Rewal

(D) A. P. Kanvinde

Ans. B Explanation: Refer Q.30 GATE-2011 for more information.

Q.15. As per the National Building Code of India 2016, the minimum turning radius (in meters) required for fire tender movement is (A) 8.0

(C) 9.0

(B) 8.5

(D) 9.5

Ans. C

Q.16. Sidi Bashir Mosque wi (A) Ajmer

(C) Ahmedabad

(B) Allahabad

(D) Amritsar

Ans. C Explanation: The Sidi Bashir Mosque is located in the city of Ahmedabad, Gujarat. Due to its unique construction, the mosque is also called Jhulta Minar or Shaking Minarets. There are two minarets in the mosque, each of which is three stories tall with carved balconies. A gentle shaking of either minaret results in the other minaret vibrating after a few seconds, though the connecting passage between them remains free of vibration. The actual cause of this is unknown as of yet.



(A) Road Intersection.

(C) Open Kitchen

(B) Fenestration

(D) Auditorium

Ans. A Explanation: Refer Q. 17 GATE-2016 for more information.

(A) Land renewal

(C) Land reclamation

(B) Land rejuvenation

(D) Land readjustment

Ans. D Explanation: Refer Q.70 GATE-2005 for more information.

Q.19. Bamboo is a type of (A) Shrub

(C) Evergreen Tree

(B) Timber

(D) Perennial Grass

Ans. D Explanation: As per Indian Forest Act, 1927, bamboo was defined as a tree under the Act, its inter-state movement requires permit when in transit in other states. Consequent to the amendment, felling or transportation of bamboos growing in non-forest areas will not require any permits. The Indian Forest (Amendment) Bill, 2017 was introduced in Lok Sabha by Mr. Harsh Vardhan, Minister of Environment, Forest, and Climate Change on December 18, 2017. Under the Act, the definition of tree includes palms, bamboos, stumps, brush-wood, and canes. The Bill amends this definition of tree to remove the word bamboos. The Bill permits felling and transit of bamboo grown in non-forest areas and categorized bamboo as . However, bamboo grown on forest lands would continue to be classified as a tree.

(A) Economic well-being

(C) Education

(B) Physical infrastructure

(D) Life expectancy

Ans. Marks to All for option A or B or C.

Q.21. The unit of measurement of Damp Proof Course (DPC) in building construction is (A) kg

(C) sq.m.

(B) cu.m.

(D) m.

Ans. C Explanation: Refer Q.51 A GATE-2014 for more information.

Q.22. Which of the following is NOT a Building Information Modeling software tool (A) Adobe Illustrator

(C) Autodesk Revit

(B) Bentley MicroStation

(D) Graphisoft ARCHICAD

Ans. A Explanation: Adobe Illustrator is a graphic designing software tool. 10


Q.23. The concentric circles in a solar chart represent (A) Azimuth angle (B) Altitude angle (C) Horizontal shadow angle (D) Vertical shadow angle Ans. B Explanation: Azimuth Lines diagram.

Azimuth angles run around the edge of the

Altitude Lines Altitude angles are represented as concentric circular dotted lines that run from the center of the diagram out. Date Lines Date lines start on the eastern side of the graph and run to the western side and represent the path of the sun on one particular day of the year. In Ecotect, the first day of January to June is shown as solid lines, while July to December is shown as dotted lines. Hour Lines/ Analemma Hour lines are shown as figure-eight-type lines that intersect the date lines and represent the position of the sun at a specific hour of the day. The intersection points between date and hour lines give the position of the sun.

Q.2 in the room is _________ Sabin (up to one decimal place). Ans. 5.3 to 5.7 Explanation: Here, Volume (V)= 27 m3 . Reverberation Time (R.T.) = 0.8 sec. Absorption area (m2sabine) A=? R.T.= 0.16 V/A A = (0.16 x 27) / 0.8 = 5.4 Sabin Q.25. A 25 storeyed building has 5 lifts. The resulting waiting time is 35 secs number of lifts required for reducing waiting time to 25 secs, without increasing the lift speed, is _________ Ans. 7 Explanation: Resulting waiting time (T) = Return travel time (RT) / No. of Lifts. Here, Expected Resulting waiting time (T) = 25 Sec. Return travel time (RT) = 175 Sec. Thus, No. of Lifts required = RT / T = 175 /25 = 7 Q. 26

Q. 55 carry two marks each.

Q.26. Match the planning documents in Group-I with their respective government schemes in Group-II Group I P. Integrated Cluster Action Plan Q. Service Level Improvement Plan R. Housing for All Plan of Action S. City Livelihood Center Development Plan

Group II 1. NULM 2. Make in India 3. RuRBAN Mission 4. PMAY 5. AMRUT

(A) P-4, Q-1, R-5, S-2

(C) P-5, Q-1, R-4, S-3

(B) P-3, Q-5, R-4, S-2

(D) P-3, Q-5, R-4, S-1

Ans. D 11


Government Schemes NULM Make in India

Full Form of the Scheme

Planning Document of the Scheme / Target of the Scheme

National Urban Livelihood Mission Make in India

City Livelihood Centre Development Plan

RuRBAN Mission

Rural urban Mission

Transform India into a global design and manufacturing hub. Integrated Cluster Action Plan


Pradhan Mantri Aawas Yojana

Housing for All Plan of Action


Atal Mission for Rejuvenation and Urban Transformation

Service Level Improvement Plan

Q.27. Associate the fire safety requirements for high rise buildings in Group-I with corresponding standards of the National Building Code of India 2016 in Group-II Group I P. Minimum Refuge area Q. Maximum Travel distance R. Maximum Occupant load S. Minimum Stair case width

Group II 1. 12.5 sqm/person 2. 2.0 m 3. 0.3 sqm/person 4. 12.0 ton 5. 30.0 m

(A) P-4, Q-1, R-5, S-2

(C) P-3, Q-5, R-1, S-2

(B) P-3, Q-5, R-4, S-1

(D) P-4, Q-5, R-1, S-3

Ans. C

Q.28. Match the photometric quantities in Group-I with their respective units in Group-II Group I P. Illuminance Q. Luminous Intensity R. Luminance S. Luminous Efficacy

Group II 1. Candela 2. Candela/sqm 3. Lumens/sqm 4. Lumens/watt 5. Lumens

(A) P-3, Q-2, R-5, S-4

(C) P-5, Q-1, R-2, S-3

(B) P-5, Q-4, R-2, S-1

(D) P-3, Q-1, R-2, S-4

Ans. D

Q.29. Associate the symbols in Group I with their meanings in Group II Group I


Group II

1. Hearing impaired




2. Emergency lamp


3. Electrical and Electronic waste disposal


4. Biohazard

5. Speech impaired (A) P-5, Q-3, R-1, S-2

(C) P-1, Q-3, R-4, S-5

(B) P-1, Q-5, R-3, S-4

(D) P-5, Q-3, R-4, S-2

Ans. D

Q.30. Match the elements in Group-I with the building components in Group-II P. King post Q. Grade Beam R. Metal decking S. Jamb

Group I

Group II 1. Curtain Glazing 2. Door 3. Plinth 4. Intermediate Floor 5. Truss

(A) P-5, Q-3, R-4, S-1

(C) P-2, Q-4, R-5, S-3

(B) P-2, Q-4, R-3, S-1

(D) P-5, Q-3, R-4, S-2

Ans. D Element


King Post



Grade Beam

Metal Decking


Q.31. Match the iconic architectural examples in Group-I with their predominant structural systems in Group-II. Group I P. S. Maria del Fiore Cathedral, Florence Q. Notre Dame Cathedral, Paris R. Olympic Arena, Tokyo S. Bahai Temple, Delhi

Group II 1. Shell 2. Suspended roof 3. Space frame 4. Double-walled dome 5. Flying buttress

(A) P-1, Q-3, R-5, S-4

(C) P-4, Q-5, R-2, S-1

(B) P-4, Q-1, R-2, S-3

(D) P-5, Q-4, R-3, S-2

Ans. C Monument

Predominant Structural Systems used in the Monument

S. Maria del Fiore Cathedral, Florence

Double walled dome




Notre Dame Cathedral, Paris

Flying Buttress

Olympic Arena, Tokyo

Suspended Roof

Bahai Temple, Delhi.

Shell Structure

Q.32. Match the building materials in Group-I with their distinctive properties in Group-II. Group I P. Cement Q. Steel R. Wood S. Glass

Group II 1. Charring 2. Brittle 3. Evaporation 4. Tensile strength 5. Setting Time

(A) P-5, Q-3, R-2, S-1

(C) P-1, Q-4, R-5, S-2

(B) P-5, Q-4, R-1, S-2

(D) P-4, Q-3, R-1, S-5

Ans. B

Q.33. Match the built forms in Group-I with their descriptions in Group-II Group I P. Agora Q. Ziggurat R. Mastaba S. Synagogue

Group II 1. Custodial precincts 2. Place of Jewish worship 3. Built in diminishing stages of masonry with buttressed wall 4. Market place or public square 5. Tomb made of mud bricks

(A) P-1, Q-4, R-3, S-2

(C) P-4, Q-3, R-5, S-2

(B) P-4, Q-3, R-1, S-5

(D) P-3, Q-1, R-5, S-2

Ans. C 15





Market place or public square (Greek Period)



Built in diminishing stages of masonry with buttressed wall (Assyrian Period)


Tomb made of mud bricks (Egyptian Period)


Place of Jewish worship

Q.34. Match the building configuration characteristics in Group-I with their seismic consequences in Group-II. Group I P. Re-entrant corner Q. Floating column R. Irregular storey stiffness S. Gap between adjacent buildings

Group II 1. Soft storey 2. Stress concentration at corner 3. Load path discontinuity 4. Vertical asymmetry 5. Pounding

(A) P-3, Q-1, R-2, S-4

(C) P-4, Q-3, R-1, S-5

(B) P-2, Q-3, R-1, S-5

(D) P-3, Q-5, R-2, S-1

Ans. B



Building configuration

Re-entrant corner.

Floating Column

Irregular storey stiffness


Seismic consequence Movement of the wings of an L-shaped building during an earthquake results in high shear stresses combined with a stress concentration at the re-entrant corner. It is aggravated by torsional effects which develop since the center of mass and the center of rigidity cannot coincide in this form. A 15% plan setback is required in both directions in L-shaped buildings to be considered a re-entrant corner. A column is supposed to be a vertical member starting from foundation level and transferring the load to the ground. The term floating column is also a vertical element which ends (due to architectural design/ site situation) at its lower level (termination Level) rests on a beam which is a horizontal member.

In earthquake resistant design, the soft story and the weak story irregularities are reciprocal to a significant difference between the stiffness and the resistance of one of the floors of a building and the rest of them. Both configurations are known in architectural terms as: the open floor.

Because of insufficient separations, structural pounding can occur between adjacent buildings during strong ground motions. Many times, due to small gap between adjacent buildings, primary reason of the collision is collision with the neighboring buildings.

Q.35. Match the landscaping terms in Group-I with their descriptions in Group-II. Group I P. Xeriscaping Q. Drip line R. Swale S. Turf block paver

Group II 1. Wide vegetated drain 2. Tree rings 3. Outermost circumference of a tree canopy 4. Solution to topsoil erosion and water permeability 5. A little or no irrigation

(A) P-5, Q-3, R-1, S-4

(C) P-2, Q-3, R-1, S-5

(B) P-3, Q-5, R-1, S-4

(D) P-5, Q-2, R-4, S-1 17


Ans. A

Landscape Term


Drip line


Turf block paver

Description Xeriscaping means conservation of water through creative landscaping. It reduces or eliminates the need for supplemental water from irrigation. Plants whose natural requirements are appropriate to the local climate are emphasized, and care is taken to avoid losing water to evaporation and run-off.

Tree drip line is the area defined by the outermost circumference of a tree canopy where water drips from and onto the ground.

A swale is a shallow trench dug along the land's contour, with a berm on the downhill side. All points along a contour line are exactly the same height above sea level. Therefore, a trench along the contour slows the water and spreads it across the contour line.

It is a permanent solution to soil erosion problems. It allows vegetation to thrive and Ideal for pedestrian and overflow traffic areas requiring solid traction and/or erosion control. Turf paver block blends naturally with the allows for a breathable pavement, making it an environmentally friendly alternative to heat producing solid concrete or asphalt surfaces.

Q.36. Match the planning principles in Group-I with their descriptions in Group-II Group I P. Transit Oriented Development Q. Core Periphery Theory R. Bid Rent Theory S. Cluster Theory

Group II 1. Four stage model of regional development 2. Compact and walkable mixed-use development 3. Geographic concentration of inter-connected institutions 4. Change of land price with relative distance from the CBD 5. Interactive and participatory planning process

(A) P-2, Q-1, R-4, S-3

(C) P-4, Q-2, R-5, S-3

(B) P-2, Q-1, R-5, S-3

(D) P-2, Q-3, R-5, S-4

Ans. A



Q.37. Match the cities in Group-I with their planners in Group-II Group I P. Islamabad Q. Tel Aviv R. Bhubaneswar S. Brasilia

Group II 1. Patrick Geddess 2. C.A. Doxiadis 3. Lucio Costa 4. B. V. Doshi 5. O. Koenigsberger

(A) P-2, Q-4, R-1, S-3

(C) P-2, Q-1, R-5, S-3

(B) P-4, Q-1, R-5, S-2

(D) P-2, Q-3, R-4, S-5

Ans. C

Q.38. Match the Temples in Group-I with their Dynastic period in Group-II Group I P. Brihadeshvara Temple Q. Kailasanatha Temple R. Bhitragaon Temple S. Lad Khan Temple

Group II 1. Gupta 2. Chalukya 3. Lodhi 4. Chola 5. Pallava

(A) P-4, Q-5, R-1, S-2

(C) P-2, Q-5, R-1, S-3

(B) P-5, Q-1, R-2, S-3

(D) P-4, Q-1, R-2, S-5

Ans. A Temple

Brihadeshvara Temple (Thanjavur, Tamil Nadu)

Kailasanatha Temple (Kanchipuram, Tamil Nadu)

Bhitragaon Temple (Kanpur, Uttar Pradesh)

Dynasty Period


Chola Period

Pallava Period

Gupta Period



Lad Khan Temple (Aihole, Karnataka)

Chalukya Period

Q.39. Match the Buildings in Group-I with their Architects in Group-II Group I P. Guggenheim Museum, Bilbao Q. The Shard, London R. Commerz Bank, Frankfurt S. Heydar Aliyev Centre, Baku

Group II 1. Richard Rogers 2. Norman Foster 3. Frank Gehry 4. Renzo Piano 5. Zaha Hadid

(A) P-3, Q-4, R-2, S-5

(C) P-2, Q-4, R-1, S-5

(B) P-3, Q-4, R-1, S-2

(D) P-2, Q-5, R-4, S-3

Ans. A Name of the Building Architect of the Building

Guggenheim Museum, Bilbao

The Shard, London

Commerz Bank, Frankfurt

Heydar Aliyev Centre, Baku

Frank Gehry

Renzo Piano

Norman Foster

Zaha Hadid


Q.40. Match the following urban conservation themes in Group-I with their respective descriptions in Group-II Group I P. Restoration Q. Reconstitution R. Reconstruction S. Replication

Group II 1. Piece by piece re-assembly 2. Returning to previous stage 3. Physical addition 4. Re-creation of vanished elements 5. Reproduction of an exact copy

(A) P-2, Q-5, R-4, S-3

(C) P-3, Q-2, R-1, S-4

(B) P-2, Q-1, R-4, S-5

(D) P-3, Q-1, R-3, S-5

Ans. B

Q.41. A Single Phase Neutral (SPN) electrical circuit has a power consumption of 330 W. Considering a voltage of 110 V and power factor of 0.8, the electrical current drawn is__________ Amp (up to one decimal place). Ans. 3.7 to 3.8



Power (Wats) = Voltage (Volt) x Current (Ampere). (P = V x I x power factor. ) Here, 330 W = 110 V x Current (Ampere) x 0.8 (Power Factor) Thus, Current (I) = 3.75 Amp. Q.42. A building with 100 sq.m. roof area is connected to a 72 cum rainwater collection tank. If the rainfall is 60 mm per hour and the loss during water storage is 20%, then the time taken to fill the tank completely is _________ hours. Ans. 15 Roof Area of the building = 100 sq.m. Rainfall is 60 mm per hour. = 0.06 m per hour. Thus, water collection on the roof per hour = 100 x 0.06 = 6 m3 per hour. Thus, time to fill the 72 cum storage tank without loss = 72/6= 12 hours. But, there is loss of 20% during water storage. Thus, loss of water storage per hour = (20/100) x 6 = 1.2 m 3 Thus, water getting stored per hour = 6 1.2 = 4.8 m3 Thus, to fill the 72 cum capacity water storage tank the time required = 72 / 4.8 = 15 hours. Q.43. The planning norms for provision of schools in a given town is shown in the table below. Schools

Population norm

Land requirement per school

Elementary School

One per 2500 persons

0.4 hectare

Primary School

One per 5000 persons

1.0 hectare

Secondary School

One per 12500 persons

2.0 hectare

Total land area required for providing all types of schools for a population of 2,00,000 is ____________ hectares. Ans. 104 Explanation: Schools

Population norm

Land requirement No. of Schools for Area Required for Schools per school (x) population of 2,00,000 (xy) (y)


1 per 2500 persons

0.4 hectare



Primary School

1 per 5000 persons

1.0 hectare



Secondary School 1 per 12500 persons

2.0 hectare






Q.44. In a mixed-use development on a 2.0-hectare site with 2.0 FAR, the ratio of residential to commercial floor area is 3:2. The minimum parking (in ECS) needed per 100 sq.m. of residential and commercial floor area is 1.0 and 1.25 respectively. Considering full FAR utilization, the total parking requirement is ________ECS. Ans. 440 Explanation: Site Area = 2.0 hectare = 20,000 sq.m. Maximum FAR = 2.0. Thus, Maximum Built-up Area = 40,000 sq.m. It is given that, Ratio of residential: commercial floor area = 3:2 Thus, Area for residential as per 3:2 ratio= 24,000 sq.m. & Area for commercial as per 3:2 ratio= 16,000 sq.m. Minimum parking (in ECS) needed per 100 sq.m. of residential area = 1. Thus, ECS required for 24,000 sq.m. of residential area = 240 Minimum parking (in ECS) needed per 100 sq.m. of commercial area = 1.25. Thus, ECS required for 16,000 sq.m. of commercial area = 200 Thus, total ECS required for 40,000 sq.m. of maximum built-up area = 240 + 200 = 440



Q.45. A plotted housing scheme on a site of 12 hectare has 60% saleable area. The average unit cost of land development is INR 300 million per hectare. If the profit margin is 20%, then the selling price of land per hectare is _______ million INR. Ans. 600 Total area of site = 12 hectare. Saleable area = 0.6 x 12 = 7.2 hectare. Average unit cost of land development = 300 Million / hectare. To sale 60% of land, we need to develop entire 12-hectare land (Not only 7.2 hectare of saleable area) Thus, total cost of land development of 12 hectare = 12 x 300 = 3600 Million. A profit margin of 20% is expected over entire selling price. Thus, selling price with 20% profit margin = 120/100 x 3600 = 4320 Million. Thus, selling price per hectare (for only 7.2-hectare saleable area) = 4320 / 7.2 = 600 Million / Hectare.

Q.46. An isolated enclosure shown in the Figure has inlet P and outlet Q of 2 sq.m. each, on the opposite walls. The outdoor wind speed is 5 m/sec. If the coefficient of effectiveness is 0.6, the rate of natural ventilation in the enclosure due to wind action is____________ cum/hr. Ans. 21600 Explanation:

Volume of air getting extracted from the Inlet / Outlet = Area of Inlet / Outlet x Speed. (V = A x S) Here, The area of inlet and outlet = 2 sq.m. and wind speed = 5 m /sec. Thus, volume of air getting extracted from inlet P per second = 2 m2 x 5 m/sec = 10 m3 / sec. As we know, 1 hr. = 60 min = 3600 sec. Volume of air getting extracted inside from inlet P per hour = 10 m3 x 3600 / hr. = 36000 m3 /hr. If the co-efficient of effectiveness is 1 then, amount of air getting extracted from inlet = amount of air extracted from outlet. Here, co-efficient of effectiveness = 0.6 Thus, rate of natural ventilation per hour = 0.6 x amount of air coming inside from inlet P per hour = 0.6 x 36,000 = 21,600 m3 / hour. Q.47. A 5m Ă— 5m Ă— 3m room has four 230 mm thick external brick walls. Total wall fenestration is 10 sq.m. The temperature difference between indoor and outdoor is 2 degC. The air to air transmittance values for 230 mm thick brick wall and 200 mm thick aerated concrete block wall are 2.4 and 1.7 W/sq.m. degC respectively. If the brick walls are replaced with the aerated concrete block walls, then the change in conductive heat flow through the walls is ______W. Ans. 69.5 to 70.5 Explanation: Here, Area of 4 walls of the room = 4 (5x3) = 60 m 2, Height= 3m. Total area of wall fenestration (Openings) = 10 m 2 Thus, net external wall area= 50 m2 Thermal transmittance value of 230 mm thick brick wall (U-value of wall) = 2.4 W/sq.m. deg C Thermal transmittance value of 200 mm thick aerated concrete block wall (U-value of aerated concrete wall) = 1.7 W/sq.m. deg C Difference between outdoor and indoor temper




Conductive heat flow through brick wall= (U-

= (2.4 x 50 x 2) = 240 watts

Conductive heat flow through aerated concrete block wall= (U= (1.7 x 50 x 2) = 170 watts Thus, change in conductive heat flow through the wall due to change of materials = 240 -170 = 70 Watts.

-likely time

dura Ans. 7.8 to 7.9 Explanation: PERT value of time duration / Expected Time (te) = (to + 4tm + tp) 6 Where, to = Optimistic time, tm = Most likely time, tp = Pessimistic time. Thus, (te) = (to + 4tm + tp) = {4+ (4 x 8) + 11} = 7.83 days. 6


Q.49. In the Figure, the negative bending moment at point A of the cantilever is ____________ kNm. Ans. 224 to 226 end, B.M.max= -WL Here, due to point load B.M.max= (-WL) = (- 20 x 5) = -100 kN.m. B.M.max at fixed end= (-wL2/2) Here, due to UDLB.M.max= (--wL2/2) = (- 10 x 52) / 2 = -125 kN.m. Thus, total B.M. moment at point A (fixed end) = -100 + (-125) = -225 kN.m. Q.50. The water consumption of a high-rise apartment building with 60 dwelling units having an average household size of 5 persons is 135 lpcd. Assuming 80% of the total use is met with recycled water supply, the daily domestic demand for the building is __________ liters. Ans. 8100 Total dwelling units = 60. Average household size = 5. Thus, total population in building = 300. Thus, total water consumption demand of the building per day = 135 lpcd x 300 = 40,500 liters. It is given that, 80% of the total use is met with recycled water. Thus, (80 x 100) x 40,500 = 32,400 liters demand is met with recycled water. Thus, the net daily domestic demand of water of the building = 40,500

32,400 = 8100 liters.

Q.51. In India, for 1.0 cum of M-20 grade concrete, the number of cement bags required is ________ (up to two decimal places). Ans. 5 to 9. Composition of M20 grade of concrete is 1:1.5:3 (Cement: Sand: Aggregate) (Total 5.5) i.e. 5.5-unit volume of M20 grade of concrete contains 1-unit volume of cement. Thus, 1 cu.m. Volume of M20 grade of concrete contains 0.19 cu.m. volume of cement. Wet concrete hardens after some time. Generally, volume of dry Concrete = 1.54 to 1.57 times Volume of wet concrete Thus, for 1 cu.m. Volume of wet M20 grade of concrete contains, 0.19 x 1.5 = 0.285 cu.m. volume of c ement. 1 cu.m. of cement = 1440 kg. 23


Thus, 0.285 cu.m. of cement = 410 kg. The standard weight of each cement bag = 50 kg.

Q.52. The sound power level of an outdoor non-directional point source is 90 dB. Considering an atmospheric impedance of 400 rayls, the sound pressure level at 10 m distance from the source is ___________ dB Ans. 58 to 60 Explanation: The sound power or acoustic power: It is the sound energy constantly transferred per second from the sound source. A sound source has a given constant sound power that does not change if it is placed in a different room environment. Sound power: It is a theoretical value that is not measurable. It is calculated and expressed in watts and as sound power level LW in decibels. A sound source produces sound power and this generates a sound pressure fluctuation in the air. Sound power is the distance independent cause of this, whereas sound pressure is the distance-dependent effect. The relationship between Sound Power Level and Sound Level is expressed as,

Putting values of Lw= 90, r = 10 m. Lp= 59.00 dB (No role of atmospheric impedance)

Q.53. The live load and dead load in a three storeyed residential building, transferred through a single column, is 12 tons and 18 tons respectively. If the soil bearing capacity is 10 ton/sq.m. and the factor of safety is 1.5, the area of column footing is ____________ sq.m. (up to one decimal place). Ans. 4.0 to 5.0 Given, Live load = 12 ton. Dead load = 18 ton. Total load = 30 ton. Factor of Safety = 1.5. Thus, Factored total load = Factor of Safety x Total Load = 1.5 x 30 = 45 ton Soil bearing capacity is 10 ton/sq.m. Area of Column footing = (Factored total load / Soil Bearing Capacity) = 45 / 10 = 4.5 sq.m. Q.54. The indoor illumination requirement for a building is 350 Lux. If the daylight factor is 2.7 and the design sky illuminance is 9000 Lux, then the required supplementary artificial lighting is _________ Lux. Ans. 107 Explanation: The three components contributing to daylight factor are: (a) sky component (SC), (b) externally reflected component (ERC), and (c) internally reflected component (IRC). Thus, DF = SC + ERC + IRC Daylight factor is also defined as the ratio of indoor illumination to the corresponding outdoor illumination can be taken as constant. This constant ratio, expressed in percentage, is the daylight factor (DF), given by: DF = (E i / E o) x 100 where, E i = indoor illumination at the point of consideration & E


= outdoor illumination from unobstructed sky.

Here, DF = 2.7, E o = 9000 Lux. Assume the existing indoor illumination (E


Thus, E I = (DF x E o ) / 100 = (2.7 x 9000 ) /100 = 243 Lux. (existing indoor illumination). The indoor illumination required = 350 Lux. Thus, supplementary artificial lighting required = 350


243 = 107 Lux.


Q.55. Two design options of a business building on a 10.0-hectare site are being compared for built up area. Floor to floor height of Option A is 3.6 m and that of Option B is 4.5 m. If the maximum allowable building height is 45 m with same ground coverage for both options, the additional built up area achievable in Option A over Option B is ____ percent. Ans. 20 Explanation: Site Area

Option A

Option B

10 Hectare (10,000 sq.m.)

10 Hectare (10,000 sq.m.)

Floor to Floor Height

3.6 m

4.5 m

Maximum allowable height of the building

45 m

45 m

Assume ground coverage as (Same for Both as

5000 sq.m. (x)

5000 sq.m. (x)

Max. floors possible within height limit (45 m)

12 floors

10 floors

Maximum built-up area which can be consumed


60,000 sq.m. (xy)

Additional Built-up area in option A above option B

50,000 sq.m. (xy)

10,000 sq.m.

% of additional Built-up area in option A above

(10,000 /50,000 x 100) = 20





Q. 5 carry one mark each.

Q.1. A right-angled cone (with base radius 5 cm and height 12 cm) as shown in the figure below, is rolled on the ground keeping the point P fixed until the point Q (at the base of the cone, as shown touches the ground again. By what angle (in radians) about P does the cone travel?

(A) 5 /12

(C) 24 /5

(B) 5 /24

(D) 10 /13

Ans. D Explanation: Slant height of cone= 13 cm. Shaded area = 2 r = 10 (Circumference of Base of cone) i.e. Circumference covered by the cone in 1 rotation. Circumference of total path with radius 13 cm = 2 r = 2 x

x 13

= 26 Circumference of path) x 2 Thus, angle covered by the rotating cone= (10 /26 ) x 2 = 10 / 13


Q.2. As the two speakers became increasingly agitated, the debate became ________. (A) Lukewarm

(C) Forgiving

(B) Poetic

(D) Heated

Ans. D

Q.3. cars. R indicates that to his knowledge, S has at least one car. Only one of P, Q and R is right. The number of cars owned by S is. (A) 0

(C) 3

(B) 1

(D) Cannot be determined

Ans. A



Q.4. In a company with 100 employees, 45 earn Rs. 20,000 per month. 25 earn Rs. 30,000. 20 earn Rs. 40,000. 8 earn Rs. 60,000 and 2 earn Rs. 1,50,000. The median of the salaries is (A) Rs. 20,000.

(C) Rs. 32,300.

(B) Rs. 30,000.

(D) Rs. 40,000.

Ans. B Explanation: To find the median, the data needs to be arranged in ascending order. Here, the salary of 100 employees will be arranged in ascending order as, 20,000 (1st employee), 20,000 (2 nd



Rule to find out median of odd number of observations (n) is, median = (n+1) th term. 2 Rule to find out median of even number of observations (n) is, median is mean of = (n/2) th & {(n/2) +1}th term. Here, number of observations are 100. Thus median = mean of (100/2) th term+ {(100/2) +1}th term = mean of (50th term + 51th term) Thus, Median of salaries = (30,000 +30,000)/2 = 30,000.

Q.5. He was one of my best ______ and I felt his loss ________ (A) Friend, keenly

(C) Friend, keener.

(B) Friends, keen

(D) Friends, keenly.

Ans. D

Q. 6

Q. 10 carry two marks each.

Q.6. The growth of bacteria (lactobacillus) in milk leads to curd formation. A minimum bacterial population density of 0.8 (in suitable units) is needed to form curd. In the graph below, the population density of lactobacillus in 1 liter of milk is plotted as a function of time, at two different temperatures 25 0c and 37 0c. Consider the following statements based on the date shown in the graph: i. The growth in bacterial population stops earlier at 37 0C as compared to 25 0C. ii. The time taken for curd formation at 25 0C is twice the time taken at 370C.



Which one of the following options is correct? (A) Only i

(C) Both i and ii

(B) Only ii

(D) Neither i and ii

Ans. A

Q.7. caste-marks and their own choice of facial hair on parade, being again reprimanded by the commander-inStronger instance than this of European prejudice with relation to this country has never

According to this paragraph, which of the statements below is most accurate? (A)

-in-chief was moved by this demonstration of his prejudice.





(D) The commander-in-chief was exempt from the European prejudice that dictated how the sepoys were to dress. Ans. C

Q.8. Let S1 be the plane figure consisting of the points (x, y) given by the inequalities

I y +2 I

2 be the plane figure given by the inequalities xS2. The area of S is

(A) 26

(C) 32

(B) 28

(D) 34

Ans. C Explanation: For plane figure S1: Ix-1I

-2 < x-1 < 2

i.e. -1 < x < 3 (adding 1 on both sides) -3 < y+2 < 3 i.e. -5 < y < 1 (Subtracting 2 from both sides) For plane figure S2: x-


means if x=0 then y=2 & if y=0 then x=-2 Also,

Thus, in above equation of S 2, If x=3 then y=5. Here, area of union = Area of rectangle + Area of Triangle. = (6 X 4) + (1/2 X 4 X 4) = 32 28


I x - 1 I 1



Q.9. What is the sum of the missing digits in the subtraction problem below? 5____ - 48_89 1111 (A) 8

(C) 11

(B) 10

(D) Cannot be determined

Ans. D Explanation: The figures can be (50100, 48989) and (50000, 48889). Thus, sum of the missing digits can be either 8 or 10.

Q.10. Two very famous sportsmen Mark and Steve happened to be brothers and played for country K. Mark

Which one of the following can be inferred from this conversation? (A) Mark was known to play better than James.

(C) James and Steve were good friends.

(B) Steve was known to play better than Mark.

(D) James played better than Steve.

Ans. B


Q. 25 carry one mark each.

Q.1. In a multi-storied building, the type of plumbing system suitable for reusing the sullage for non-potable use is (A) Single stack system

(C) One pipe system

(B) Partially ventilated single stack system

(D) Two pipe system

Ans. D Explanation: In two pipe system, the soil water (containing human excreta) and waste water (Containing floor water of bathroom, kitchen, washing area, etc.) both are separated from each other in two different stacks. The soil water pipe needs to be connected to either main municipal sewer connection or need to be disposed of in septic tank near the site. The waste water however, can be treated, recycled on site by various techniques and can be used for non-potable uses like gardening, flushing, floor washing, etc.

Q.2. Which of the following processes is NOT adopted in solid waste management? (A) Incineration

(C) Flocculation

(B) Pyrolysis

(D) Sanitary landfill

Ans. C Explanation: Flocculation process often known as Coagulation- Flocculation process for chemical treatment of drinking water. This technique is applied prior to sedimentation and filtration (e.g. rapid sand filtration) to enhance the ability of a treatment process to remove particles. Coagulation is a process used to neutralize charges and form a gelatinous mass to trap (or bridge) particles thus forming a mass large enough to settle or be trapped in the filter by using alum as coagulant. Flocculation is gentle stirring or agitation to encourage the particles thus formed to agglomerate into masses large enough to settle or be filtered from solution.



Q.3. Tuscan and Composite orders are associated with (A) Greek Architecture

(C) Byzantine Architecture

(B) Islamic Architecture

(D) Roman Architecture

Ans. D Order




The column shaft height is 7 times its diameter and typically plain rather than fluted, with a simple Tuscan style capital and base.

The composite order is a mixed order, combining the volutes of the Ionic order capital with the acanthus leaves of the Corinthian order.


Q.4. In which of the following models does the private partner own the revenue as well as the risk associated with the project for limited period of time? (A) Build, Own, Operate (BOO)

(C) Design, Build, Finance, Operate (DBFO)

(B) Build, Own, Operate, Transfer (BOOT)

(D) Design, Bid, Build (DBB)

Ans. B Explanation: Build-Own-Operate-Transfer (BOOT) arrangement involves the transfer of responsibility for constructing, financing and operating a single facility to a private sector partner for a fixed period of time. At the end of that period, the responsibility reverts to the public party.

Q.5. The unit for measuring sound absorption in a room is (A) Sabin

(C) Decibel

(B) Phon

(D) Hertz

Ans. A

Q.6. The design element provided to ensure the safety of a vehicle traveling at a prescribed speed along the curved segment of a highway is (A) Shoulder

(C) Median

(B) Super-elevation

(D) Footpath

Ans. B Explanation: Super-elevation is the amount by which the outer edge of a curve on a road or railway is banked above the inner edge. It helps to offset centripetal forces developed as the vehicle goes around a curve. Along with friction on the road, super-elevation keeps a vehicle from going off the road. 30


Q.7. Select the right option representing strategic components arranged in ascending order of specified minimum area under smart city mission of Government of India. (A) Greenfield development- Redevelopment- Retrofitting. (B) Redevelopment- Greenfield development- Retrofitting (C) Retrofitting- Redevelopment- Greenfield development (D) Redevelopment- Retrofitting- Greenfield development. Ans. B Explanation: Smart City Mission of Govt. of India has three options for area-based proposal. The strategic components of Area-based development in the Smart Cities Mission are city improvement (retrofitting), city renewal (redevelopment) and city extension (Greenfield development) plus a Pan-city initiative in which Smart Solutions are applied covering larger parts of the city. The appropriate site for either of three types of development: retrofitting means addition of smart features in the existing core city (approx. 500 acres), redevelopment means proposal with entire demolition of existing built mass in core city (approx. 50 acres) or Greenfield development means entire fresh development at the fringe part of the city (approx. 250 acres).

Q.8. Which of the following processes is NOT used for corrosion resistance of cast iron? (A) Painting

(C) Quenching

(B) Epoxy coating

(D) Galvanizing

Ans. C Explanation: Quenching is the most commonly used process to harden steel by rapid cooling technique. In this process, cast steel and iron are heated to between 815 and 900 °C (1,500 to 1,650 °F), with careful attention paid to keeping temperatures throughout the workpiece uniform. This follows soaking of material in air and followed by rapid cooling. This can prevent the formation of all crystal structure, resulting in amorphous metal or "metallic glass".

Q.9. In Geographic Information System, DEM represents information on (A) Vegetation cover

(C) Water Table

(B) Soil Type

(D) Topography

Ans. D Explanation: DEM stands for Digital Elevation Model. DEMs are typically used to represent the bare-earth terrain, void of vegetation and manmade features.

Q.10. As per the CPWD handbook on Barrier Free and Accessibility 2014, Government of India, the minimum length of a straight ramp in meter to raise a wheelchair to the plinth level of 600 mm will be ______ Ans. 7.2 m As per CPWD handbook on Barrier Free and Accessibility 2014, maximum gradient of ramp should be 1:12. It means to raise a height of 1 m (1000 mm), it requires 12 m (12000 mm) length of ramp. Thus, to raise a wheelchair to height 600 mm, it requires 12 X 600 = 7200 mm = 7.2 m ramp Q.11. A pointed arch having two centers and radii greater than the span is known as (A) Lancet arch

(C) Roman arch

(B) Gothic arch

(D) Drop arch

Ans. A



Lancet Arch

Gothic Arch

Roman Arch

Drop Arch

Q.12. Slenderness ratio of a column is represented as: (A) Effective length/ Cross-sectional area

(C) Actual length / Cross- sectional area

(B) Effective length/Radius of gyration

(D) Actual length / Radius of gyration.

Ans. B

Q.13. ArchiCAD, Autodesk Revit, Digital Project Designer (CATIA) and Vector works Architect are examples of (A) Statistical Analysis Software

(C) BIM Software

(B) GIS Software

(D) Image Processing software

Ans. C Explanation: BIM stands for Building Information Modeling. It is an intelligent 3D model-based process that equips architecture, engineering, and construction professionals with the insight and tools to more efficiently plan, design, construct, and manage buildings and infrastructure.

Q.14. In one liter of paint, volume of solid pigment and volume of non-volatile binder are 400 cc and 600 cc respectively. The pigment volume concentration number of the paint is _____ Ans. 40 Explanation: cc stands for cubic centimeters, a unit of volume. So, 1000 cc means 1 liters. Here, pigment volume concentration = (400/1000) X 100 = 40

Q.15. Liquidated damage refers to the (A) Cost borne by the contractor to rectify defects within defect-liability period. (B) Compensation paid on breach of contract to the affected party by the other party (C) Money paid by the insurance company to the owner of insured property if is damaged. (D) Money earned by te owner from selling damaged property through auction. Ans. B (As per GATE-Official Key) Explanation: Refer UPSC-CPWD 2015, Q.71 for more information.

Q.16. Excellence in Design for Greater Efficiency (EDGE) programme DOES NOT focus on (A) Lower carbon emission

(C) Cost effectiveness

(B) Greater resource efficiency

(D) Labour safety.

Ans. D



Q.17. (A) Expansion joint in large span concrete members. (B) Interface between an already setting concrete and fresh batch of concrete (C) Structural crack arrested by embedding metal rods. (D) Joining of two similar metals in vacuum Ans. B Explanation: An advancing face of a concrete pour, which could not be covered by fresh concrete before expiry of initial setting time (due to an unscheduled stoppage or delay on account of breakdown in plant, inclement weather, low rate of placement or any other reason), is called a cold joint.

Q.18. The grade-separated interchange suitable for 3-legged road intersection is (A) Trumpet

(C) Diamond

(B) Full Clover leaf

(D) Partial Clover leaf

Ans. A Intersection




Trumpet interchange is a popular form of three leg interchange. If one of the legs of the interchange meets a highway at some angle but does not cross it, then the interchange is called trumpet interchange.

Full Clover leaf

It is a four-leg interchange and is used when two highways of high volume and speed intersect each other with considerable turning movements. The main advantage of cloverleaf intersection is that it provides complete separation of traffic. In addition, high speed at intersections can be achieved. However, the disadvantage is that large area of land is required. Therefore, these are provided mainly in rural areas.


Diamond interchange is a popular form of four-leg interchange found in the urban locations where major and minor roads crosses. The important feature of this interchange is that it can be designed even if the major road is relatively narrow.

Partial clover leaf

A partial cloverleaf interchange is a modification of a cloverleaf interchange. It may be used in lieu of a diamond when development or other physical conditions prohibit construction in a quadrant, or where heavy left turns are involved. A continuous flow design is required where two major facilities intersect.



Q.19. used for computing (A) Housing density

(C) Housing Price

(B) Housing Shortage

(D) Housing affordability

Ans. B Explanation: Housing Shortage is defined as insufficient housing / lack of housing to accommodate the existing population. The standard household size depends upon the existing infrastructure available in the area, location of the area, etc. If the household size exceeds this standard size, it shows sharing of single dwelling unit by one or more households.

Q.20. The Pritzker Architecture Prize for the year 2016 has been awarded to (A) Alejandro Aravena

(C) Stephen Breyer

(B) Frei Otto

(D) Yung Ho Chang

Ans. A

Year Pritzker Award Winners 2018 B. V. Doshi 2017 Rafael Aranda, Carme Pigem and Ramon Vilalta 2016 Alejandro Aravena 2015 Frei Otto 2014 Shigeru Ban 2013 Toyo Itto The complete list of winners is available at . The architectural work of all prize

Q.21. (A) Le Corbusier

(C) Gordon Cullen

(B) Louis Kahn

(D) Kevin Lynch

Ans. C (Refer GATE-2010, Q.17 for more information.)

Q.22. Which of the following tree has a columnar form? (A) Delonix Regia

(C) Polyalthia Longifolia

(B) Tamarindus Indica

(D) Callistemon lanceolatus

Ans. C TreeCommon Name


Tamarind / Imli


crimson or lemon bottlebrush

Botanical Name

Delonix Regia

Tamarindus Indica

Polyalthia Longifolia

Callistemon lanceolatus

Tree / Form of Branch



Q.23. Minimum points required for GRIHA certification is (A) 35

(C) 50

(B) 40

(D) 60

Ans. A (As per official GATE Answer Key- A, B and C all three answers are given as correct) Explanation: As per New Rating thresholds, below table shows the GRIHA rating with respect to points achieved in Building Design and Construction. New Rating Thresholds 25-40 41-55 56-70 71-85 86 or more

GRIHA Rating 1 Star 2 Star 3 Star 4 Star 5 Star

Q.24. The CARTOSAT 2C satellite recently launched by ISRO (A) is a geo-synchronous satellite.

(C) Was launched using a GSLV rocket

(B) is a part of IRNSS GPS satellite system

(D) Has high spatial resolution.

Ans. D Explanation: CARTOSAT Satellite was launched by ISRO in June 2016. The applications of the satellite include cartographic applications, urban and rural applications, coastal land use and regulation, utility management like road network monitoring, water distribution, creation of land use maps, precision study, change detection to bring out geographical and manmade features and various other Land Information System (LIS) and Geographical Information System (GIS). The satellite used for remote sensing purpose are Polar Synchronous and they are launched by Polar Satellite Launch Vehicle (PSLV). Thus, above question can also be answered by deleting A, B, C options.

Q.25. The principle of Eminent Domain is the power to (A) Restrict exercise of rights in land through zoning and environmental laws (B) Control land use (C) Retain land use (D) Acquire and take possession of property in order to promote public interest. Ans. D (Refer Q.47, GATE-2009 for more information)

Q. 26

Q. 55 carry two marks each.

Q.26. Match the classical urban planning theories in Group-I with their proponents in Group II Group I P. Concentric Zone Model Q. Sector Model R. Multiple Nuclei Model S. Factorial Ecology

Group II 1. Berry and Horton 2. Homer Hoyt 3. Ernest Burgess 4. Shevky and Bell 5. Harris and Ullman

(A)P-4, Q-1, R-3, S-5

(C) P-2, Q-4, R-5, S-1

(B) P-3, Q-2, R-3, S-5

(D) P-3, Q-2, R-5, S-1

Ans. D 35


Urban Planning Theory Proponent

Concentric Zone Model

Sector Model

Multiple Nuclei Model

Factorial Ecology

Ernest Burgess

Homer Hoyt

Harris and Ullman

Berry and Horton

Graphical Presentation of theory

Q.27. Match the development schemes by Government of India in Group I with their objectives in Group II Group I P. PMAY Q. AMRUT R. NRuM S. HRIDAY

Group II 1. Housing for All 2. Rural cluster development 3. Heritage city development 4. Urban mobility improvement 5. Urban rejuvenation

(A) P-1, Q-5, R-4, S-3

(C) P-3, Q-5, R-1, S-2

(B) P-1, Q-5, R-2, S-3

(D) P-4, Q-2, R-1, S-5


Long Form of Scheme Pradhan Mantri Aawas Yojana Atal Mission for Rejuvenation and urban transformation National Rurban Mission Heritage City Development and Augmentation Yojana

Objectives 1. Housing for All 2. Urban Rejuvenation 3. Rural cluster development 4. Heritage city development

Q.28. In year 2001, a district with 4000 manufacturing jobs had a 10% share of total manufacturing jobs within the state. In year 2011, the state recorded 15% drop in manufacturing jobs whereas, share of the district in total manufacturing jobs within the state increased to 15%. Additional manufacturing jobs created in the district between year 2001 and 2011 is ________ Ans. 1100 Year 2001 2011 Net Increase

Jobs in District 4000 5100 1100

% of jobs in District compared to state 10% 15%

Q.29. Match the instruments in Group I with the corresponding tests in Group II Group I P. Pycnometer R. Los Angeles Apparatus

Group II 1. Initial and final setting time 2. Abrasion test 3. Surface hardness test 4. Slump test 5. Apparent specific gravity

(A) P-5, Q-3, R-2, S-1

(B) P-5, Q-4, R-2, S-1 36

Total Jobs in State 40,000 34,000 (-15%)


(C) P-3, Q-2, R-1, S-5

(D) P-2, Q-3, R-4, S-1

Ans. A Instrument


Los Angeles Apparatus

Used for testing

Apparent specific gravity

Surface hardness test

To determine the Los Angeles abrasion value. It is also used to find the suitability of aggregates for use in road construction.


Q.30. A site has a unidirectional slope of 30 0 with the horizontal along its longer side. The projected dimensions of the site on the horizontal plane measures 30m X 40m. Using cut and fill method the site has to be levelled parallel to the horizontal plane. The minimum amount of earth to be excavated in cubic meter is ______ Ans. 3463 m3 Explanation: Draw Cross-Section of site as shown in figure below. Volume of existing site section = (Cross Section Area) X Length of shorter side = [1/2 X 40 X (tan30 X40)] X 30 = 13856 m3


This volume needs to be converted into cuboid of dimensions 30 m X 40 m X h (2) Equating 1 & 2, h= 11.546 m (BF) CF = BC-BF = (tan300 x 40)


= 11.548 As per 30-60-90 principle, m Thus, volume of area need to be cut and filled = ½ X 20 X 11.546 X 30 = 3463 m3


Initial and final setting time


Q.31. A circular plate inclined at an angle ø with horizontal plane generates an ellipse as top view with major axis and minor axis of 5cm and 2.5 cm respectively. The value of ø in degrees is ____ Ans. 60 Explanation: The above problem can be illustrated as shown in figure. Thus, in triangle, Cos ø = 5/10 = ½. Thus, ø = Cos-1 (1/2) = 600.



Q.32. Match the terminologies in Group I with their description in Group II Group I P. Pruning Q. Felling R. Hoeing S. Mulching

Group II 1. Cutting of trees 2. Removing broken branches from trees for better growth. 3. Maintaining moisture content in soil by a protective layer 4. Indiscriminate cutting of branches to reduce the size of a tree 5. Loosening the ground to remove weeds

(A) P-2, Q-1, R-5, S-3

(C) P-2, Q-1, R-3, S-4

(B) P-2, Q-1, R-4, S-3

(D) P-1, Q-2, R-3, S-1

Ans. A Pruning- Removing broken branches from trees for better growth

Feeling- Cutting of trees

Hoeing- Loosening the ground to remove weeds

Mulching- Maintaining moisture content in soil by a protective layer



Q.33. Calculate the volume of cement in cubic meter required for making 10 cubic meters of M20 grade Plain Cement Concrete Work, assuming the ratio of dry concrete mix to wet concrete mix as 1.52. Ans. 2.88 Explanation: Composition of M20 Grade Concrete (C: S: A) = 1: 1.5: 3. (Total 5.5) i.e. 5.5-unit volume of concrete contains 1-unit volume of cement. Thus, 1 cu.m. volume of concrete contains 0.19 cu.m. volume of cement. Thus, 10 cu.m. volume of concrete contains 1.9 cu.m. volume of cement. For wet concrete mix, it requires 1.52 times more volume as given by the ratio of dry concrete mix: wet concrete mix. Thus, for making 10 cu.m. of wet concrete mix, 1.52 x 1.9 = 2.88 cu.m. of cement. Q.34. The optimistic, most-likely and pessimistic time for developing a new product are 12 months, 15 months and 17 months respectively. Calculate the expected time in months. Ans. 14.84 months Expected Time (te) = (to + 4tm + tp) 6 Where, to = Optimistic time, tm = Most likely time, tp = Pessimistic time. Thus, (te) = (to + 4tm + tp)/6 = {12+ (4 x 15) + 17}/ 6 = 14.84 months

Q.35. A fluorescent light source consumes 40 W electric power and has luminous efficacy of 40 lm/W. Illumination in lux at a distance of 3 m from this light source is ______ Ans. 177.78 Explanation: To find out Illumination produced from light source. E = I / d2 Where, E= Illumination on surface (lux), I= Illumination intensity from source (cd) d= distance from light source Here, I = Luminous efficacy of lamp X Power consumption of lamp = 40 lm/W X 40 W = 1600 lm. E= 1600 / 32 = 177.78 lux.

Q.36. Fee of contractor for a project has the following provisions. Basic fee= 15% of actual cost of work incurred. Bonus = 20% of savings from estimated cost of work Penalty = 20% of cost overrun. If the estimated cost of the project is Rs. 60,000 and the actual cost is Rs. 70,000. Then the total fee of contractor in Rupees is _______ Ans. 8,500 Estimated Cost of Project Actual Cost of Project Cost Overrun Basic Fee (15% of Actual cost) Bonus Penalty (20% of overrun) Thus, Fees of Contractor (Basic Fee- Penalty)

Rs. 60,000 Rs. 70,000 Rs. 10,000 0.15 X 70,000 = 10,500 Rs. Nil (Project cost exceeds estimated cost) 0.20 X 10,000 = 2000 Rs. 10,500-2000 = 8,500 Rs.



Q.37. For a symmetrical two-dimensional truss as shown in the above figure, vertical force in kN acting on the member PQ is ______

Ans. 0 Q.38. Associate the structural systems in Group I with the buildings in Group II. Group I P. Folded Plates Q. Shell R. Tensegrity S. Pneumatic (A) P-3, Q-4, R-1, S-2

Group II 1. Kurilpa Bridge, Brisbane 2. Eden Project, Cornwall. 3. Riverside Museum, Glasgow. 4. MIT Auditorium, Boston. 5. 30, St. Mary Axe, London

(B) P-5, Q-4, R-3, S-1

(C) P-3, Q-2, R-1, S-5 (D) P-1, Q-3, R-4, S-2

Ans. A Structural System

Used in Building- Architect

Folded Plates

Riverside Museum, GlasgowZaha Hadid


MIT Auditorium, Boston- Eero Saarinen


Kurilpa Bridge, Brisbane- Cox Rayner Architects with Arup





Diagrid System

Eden Project, Cornwall- Nicholas Grimshaw

30, St. Mary Axe, London (Also known as Gherkin Tower)Norman Foaster & partners

Q.39. Match the distinguished housing projects in Group I with their architects in Group II Group I P. Nagakin Capsule Tower, Tokyo, Japan Q. Tara Apartment, New Delhi, India R. Habitat 67, Montreal, Canada S. Byker Wall, New Castle, England

Group II 1. Walter Gropius 2. Moshe Safdie 3. Ralph Erskine 4. Charles Correa. 5. Kisho Kurokawa

(A) P-5, Q-4, R-2, S-3

(C) P-5, Q-2, R-1, S-4

(B) P-1, Q-3, R-4, S-5

(D) P-5, Q-4, R-2, S-1

Ans. A Housing Project Name of Architect

Nakagin Capsule Tower, Tokyo, Japan Kisho Kurokawa

Tara Apartment, New Delhi, India Charles Correa


Housing Project

Byker Wall, New Castle, England

Name of Architect

Ralph Erskine


Habitat 67, Montreal, Canada Moshe Safdie


Q.40. Match the architectural movements in Group I with their proponents in Group II P. Deconstruction Q. Historicism R. Metabolism S. Art Nouveau

Group I

Group II 1. Joseph Paxton 2. Kenzo Tange 3. Walter Gropius 4. Victor Horta 5. Frank O. Gehry

(A) P-5, Q-1, R-2, S-4

(C) P-5, Q-2, R-3, S-3

(B) P-5, Q-4, R-2, S-3

(D) P-2, Q-4, R-1, S-5

Ans. A Architectural Movement



Frank O. Gehry


Joseph Paxton


Kenzo Tange

Art Nouveau

Victor Horta

Type of work



Q.41. A proposed housing will have HIG, MIG and LIG units on a site measuring 60,750 sq.m. The buildable area of each category of units with respect to the total buildable area will be 30%, 50% and 20% respectively. The maximum allowable FAR is 2.5, ground coverage 45% and height 15m. The maximum buildable area in sq.m. of HIG units, considering a floor height of 3m for all categories will be _____ Ans. 41,006.25 sq.m. Site Area = 60,750 Sq.m. Maximum FAR = 2.5 Maximum Built-up area consuming full FAR = 2.5 X 60,750 = 1,51,875 sq.m. Maximum ground coverage = 45% = 0.45 X 60750 = 27,337.5 Sq.m. HIG MIG Max. buildable area 30 % = (0.30 X 1,51,875) No need to calculate = 45,562.5 sq.m. Max. ground coverage 30 % of 27,337.5 sq.m. No need to calculate = 0.30 X 27,337.5 = 8,201.25 sq.m. Max. floor possible as per height restriction = 15/3= 5 floors. Thus, maximum buildable area of HIG units = 5 X 8,201.25 = 41,006.25 sq.m.

LIG No need to calculate No need to calculate

Q.42. One acre of agricultural land has been given on a lease till perpetuity at an annual rent of Rs. 10,000 to be paid at the end of each year. Net Present Value of the land parcel in Rupees assuming discount rate of 5% per annum is _____ Ans. 2,00,000 Explanation: Net Present Value

= Annual Rent (annual net income) x (100 / interest rate)

(Capitalized value in perpetuity) = 10,000 x (100/5) = 2,00,000

Q.43. Match the planning tasks in Group I with the tools of analysis in Group II Group I P. Population Projection Q. Regional Resource allocation R. Trip distribution S. Design of water distribution network (A) P-3, Q-1, R-4, S-2

Group II 1. Input-Output Analysis 2. Hardy Cross Method 3. Cohort Analysis 4. Gravity Model 5. Moving Observer Method (C) P-5, Q-1, R-3, S-4

(B) P-3, Q-5, R-4, S-1

(D) P-1, Q-3, R-5, S-2

Ans. A

Q.44. Match the parameters in Group I with their units in Group II Group I P. Traffic Flow Q. Traffic Density R. Right of Way S. Traffic Signal Cycle Length

Group II 1. Meter 2. Cycles/Second 3. Seconds 4. Vehicle/km 5. PCU/hr.

(A) P-5, Q-4, R-1, S-2

(C) P-5, Q-2, R-4, S-3

(B) P-5, Q-4, R-1, S-3

(D) P-4, Q-5, R-1, S-3

Ans. B



Parameters Traffic Flow Traffic Density Right of Way Traffic Signal Cycle Length

Description Traffic flow of a particular road shows the average passenger car units (PCU) that travels per hour by that road. Traffic density of a particular section of the street is measured in terms of maximum vehicles that can travel along the particular length of that street Right of way is the term used to show the horizontal extent of the road. At signalized intersection, the traffic signal cycle is measured in seconds.

Q.45. Match the international events in Group I with their directives in Group II Group I P. Earth Summit, Rio de Janeiro, 1992 Q. UN Framework Convention on Climate Change, New York, 1992. R. UN Sustainable Development Summit, New York, 2015 S. Habitat II, Istanbul, 1996

Group II 1. Kyoto Protocol 2. Agenda 21 3. Heritage Conservation 4. Agenda 2030 5. Housing for All

(A) P-1, Q-5, R-4, S-3

(C) P-2, Q-1, R-4, S-5

(B) P-1, Q-5, R-2, S-3

(D) P-2, Q-1, R-5, S-4

Ans. C

Q.46. A drainage basin of 180 hectares comprises 40% wooded area, 45% grassed area and 15% paved area. Runoff coefficients for wooded, grassed and paved areas are 0.01, 0.2 and 0.95 respectively. The composite runoff coefficient for the drainage basin is _____ Ans. 0.2365 Area of drainage basin= 180 hectares Distribution Wooded Area (40%) Grassed Area (45%) Paved Area (15%) Area in hectares 72 81 27 Runoff coefficients 0.01 0.2 0.95 Quantity of runoff taking as 0.72i 16.2i 25.65i constant (Q= C x i X A) Total runoff (Q1) 0.72i + 16.2i +25.65i = 42.57 i Total runoff on drainage basin C X i X 180. (C- Composite runoff coefficient of entire drainage basin) of 180 hectares. (Q2) As, Q1 = Q2. 42.57 i = C X 180 i Thus, C= 0.2365 Q.47. Value of bending moment in kN-m at point C for a beam as shown in the figure is

Ans. 28 kN-m. Explanation: Here, RB + RD = 20 +40 KN = 60 KN. Also, Bending Moment about point D = 0. 44


Thus, RB x 5

40 x 2

20 x 7.5 = 0.

RB = 46 KN. Thus, RD = 14 KN. Thus, bending moment in kN-m at point C = RD x 2 m = 14 X 2 = 28 kN-m.

Q.48. Water flows through constricted circular pipe whose diameter at the constricted end is half of the non constricted end. Velocity of water at the non-constricted end is 2 m/s. Velocity of water in m/s at the constricted end using the principle of continuity of flow is ______ Ans. 8 m/s Explanation: The volume of liquid Q passing through area A1 & A2 in unit time is, Q= A1. V1 & Q= A2. V2 Where, V1 & V2 is the velocity of the liquid at that point respectively. The quantities must be equal for steady flow. Thus, A1. V1 = A2. V2 Thus, V1 / V2 = A2 / A1 Assume, A1 = Area at constricted end, A2 = Area at non-constricted end V1 = Velocity at constricted end, V2 = Velocity at non-constricted end (2m/s) As per given condition of diameter, d1/d2 = 0.5. Thus, A1 / A2= (d1/d2 )2 = 0.25 A2. Thus, A2 / A1 = V1 / V2 = 4. Thus, V1 = 4 X 2 = 8 m/s.

Q.49. Match the planning techniques in Group I with their salient features in Group II Group I P. Land Pooling Q. Action Plan R. Land Sharing S. Transfer of Development Rights

Group II 1. Assigning specific task on a short time horizon 2. Assembly privately owned land parcels for development 3. Agreement for reallocation of land between occupiers and owners 4. Assigning specific task on a long-time horizon 5. Incentive based voluntary shifting of FAR of a plot to another plot

(A) P-1, Q-5, R-4, S-3

(C) P-2, Q-1, R-3, S-4

(B) P-2, Q-1, R-3, S-5

(D) P-4, Q-2, R-1, S-5

Ans. B

Q.50. Match the land use classes in Group I with the use zones in Group II Group I P. Transportation Q. Commercial R. Public and Semi-Public S. Recreational (A) P-4, Q-1, R-3, S-5

Group II 1. Sports Complex 2. Heritage and Conservation areas 3. Burial ground 4. BRT Corridor 5. Service Sector (C) P-4, Q-5, R-1, S-2

(B) P-5, Q-3, R-1, S-2

(D) P-4, Q-5, R-3, S-1

Ans. D Explanation: Refer Q.22, GATE-2015 for more information.

Q.51. In 2011, the population of a town was 5,00,000 and the number of housing units were 1,00,000. Calculate the additional number of dwelling units (DU) required by 2031 so that there is no housing shortage. The assumptions are i. 5% decadal increase in population. 45


ii. New Du to be completed by 2021 is 10,000. iii. Number of DU which will become non-habitable by 2031 is 5000. iv. Average household size is 4.5. Ans. 17,500 Year Population Decadal increase Net Increase in Population in 1st decade Population Decadal increase Net Increase in Population in 2nd decade Population Number of DU as per population (Considering average HH size 4.5) Existing DU Housing Shortage Number of DU as per population (Considering average HH size 4.5) New DU completed by 2021 Net Shortage upto 2021 Net Housing shortage after 2021

2011 5,00,000 -

5,00,000/4.5 1,11,111 1,00,000 11,111

2021 5% 25,000 5,25,000

2031 5% 26,500 5,51,250



1,16,666 10,000 1,111 (1,16,666-1,11,111) +1,111 = 6666

Number of DU as per population (Considering average HH size 4.5) Net Housing shortage upto 2031

5,51,25 1,22,500 (1,22,500-1,16,666) +6.666 = 12,500 5,000 12,500+5,000 17,500

DU which will become inhabitable by 2031 Net Shortage upto 2031 Additional number of DU required by 2031 Q.52. Match the equipment in Group I with their applicants in Group II Group I P. PIR Q. FCU R. OLED S. BIPV

Group II 1. Air Conditioning 2. Lighting 3. Power Generation 4. Motion detection 5. Daylight Sensing

(A) P-4, Q-5, R-2, S-1

(C) P-4, Q-1, R-2, S-3

(B) P-1, Q-4, R-5, S-3

(D) P-4, Q-2, R-5, S-1

Ans. C Equipment




A passive infrared sensor (PIR sensor) is an electronic sensor that measures infrared (IR) light radiating from objects in its field of view. They are most often used in PIR-based motion detectors.






A Fan Coil Unit (FCU) is a simple device consisting of a heating and/or cooling heat exchanger or 'coil' and fan. It is part of an HVAC system found in residential, commercial, and industrial buildings.

An organic light-emitting diode (OLED) is a lightemitting diode (LED) in which the emissive electroluminescent layer is a film of organic compound that emits light in response to an electric current. OLEDs are used to create digital displays in devices such as television screens, computer monitors.

Building Integrated Photovoltaics (BIPV) System. Building Integrated Photovoltaics (BIPV) is the integration of photovoltaics (PV) into the building envelope. The PV modules serve the dual function of building skin replacing conventional building envelope materials and power generator

Q.53. Associate the historic buildings in Group I with their predominant materials in Group II Group I P. Lingaraj Temple, Bhubaneshwar, India Q. Victoria Memorial, Kolkata, india R. Padmanabhapuram Palace, Thuckalay, India S.

Group II 1. Red Sandstone 2. Timber 3. Terracotta tiles 4. Sandstone and laterite 5. Marble

(A) P-1, Q-2, R-3, S-5

(C) P-2, Q-1, R-3, S-4

(B) P-1, Q-4, R-3, S-5

(D) P-4, Q-5, R-2, S-1

Ans. D Building

Predominant Material

Lingaraj Temple, Bhubaneshwar, India

Sandstone and laterite




Victoria Memorial, Kolkata, india


Padmanabhapuram Palace, Thuckalay, India



Red Sandstone

Q.54. As per National Building Code of India 2005, the maximum number of occupants per unit exit width of a doorway is 60, where unit exit width is 500 mm. The maximum permissible occupants in a theatre having four number of 2.2 m wide doors will be ______ Ans. 960 Unit exit width is 500 mm (0.5 m). In the problem, door width is 2.2 m. Thus, effective width of door as per unit standards need to be considered as 2.0 m. instead of actual width 2.2 m. Thus, maximum number of occupants per door of 2 m width = (2/0.5) x 60 = 240 Thus, maximum number of occupants in a theatre with 4 doors = 240 x 4 = 960 Q.55. A room measures 3 m (width) X 4m (length) X 3m (height). The outdoor temperature is 36 0 C. The volumetric specific heat of air is 1300 J/Cu.m. 0C. The ventilation heat flow rate in Watts required to attain an internal room temperature of 260 C with 3 air changes per hour is _____ Ans. 390 W Explanation: Total volume of room = 3 x 4 x 3 = 36 m 3 The amount of air changes in 3 turns per hour = 36 x 3 = 108 m 3 Ventilation heat flow rate in Watts = Total energy required

= 108 x 1300 x 10 = 1404000 Joule = 1404000 3600 = 390 W 48

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This book will provide the past trends of questions asked in the exams with their solutions with detailed explanations for better understand...