Question Bank - GATE ARCHITECTURE...Out of 20 years Question Bank, 10 years (2016-2007) are dealt with all questions and next 10 years (2006-1997), only numerical questions are provided

Introduction

The best way to prepare for an exam like GATE is through comprehensive study of previous year

question papers. Solving the previous year’s GATE questions help aspirants to understand the exam

pattern, knowing the level of questions and predict the pattern. At the same time you may be aware that

just knowing the answers of previous year question paper is just not enough.

For example if the question is: The teahouse is a feature of which type of landscape architecture? And you

learnt that the answer ‘Japanese Garden’. It is best to support the answer with addition notes & figures

about different types of gardens i.e. French, English, and Chinese etc. One reason for providing such

notes is that it is rarely possible that in the next few years, the same question will be repeated. But it is

quite possible that if a question is asked form related topic, you should answer it if you have gone through

addition studies or notes.

Providing answer with essential notes & explanation is the main features of this Question Bank. It’s been

tried to cover the maximum part of the syllabus through providing supportive notes.

For further reading on particular topics, we have also provided QR codes & short links. Just scan or type

the links to reach the web resources.

All illustrations are color printed. Paper published by National Center for Biotechnology Information, US

suggests that there is positive effects of color illustration on cognitive process.

Out of 20 years Question Bank, 10 years (2016-2007) are dealt with all questions and next 10 years

(2006-1997), only numerical questions are provided with solutions. All it makes it the complete question

bank.

This book should provide an edge to your study. Hopeful that it will make you confident and feel easy on

question pattern. Best wishes for your preparation.

Contents

Question Paper GATE 2016 (Page 11016 – 502016)

Question Paper GATE 2015 (Page 12015 - 732015)









Numericals GATE 2006 (Page N1 – N17)








Numericals GATE 1998 (Page N56–N58)

Numericals GATE 1997 (Page N59 - N60)

References (x1)

GATE 2016

2016-2

Question Paper GATE 2016

The GATE 2016 AR was a bit unpredictable. Questions were slightly off the patterns this year in comparison

to 2015. Numericals were twisted. Questions were also asked from current affairs. For example, in Earthquake

2015 in Nepal, many heritage buildings collapsed such as ‗Dharahara‘. When Charles Correa died in 2015 in

Mumbai, a movie was made in the memory of called ‗Volume Zero‘.

Cut-off was 38.90.

Part 1- Aptitude

Q1. An apple costs Rs. 10. An onion costs Rs. 8. (1 mark)

Select the most suitable sentence with respect to grammar and usage.

(A) The price of an apple is greater than an onion.

(B) The price of an apple is more than onion.

(C) The price of an apple is greater than that of an onion.

(D) Apples are more costlier than onions.

Answer: (C)

Q2. It takes 10s and 15s, respectively, for two trains travelling at different constant speeds to completely

pass a telegraph post. The length of the first train is 120 m and that of the second train is 150 m. The

magnitude of the difference in the speeds of the two trains (in m/s) is ____________. (1 mark)

(A) 2.0 (B) 10.0 (C) 12.0 (D) 22.0

Answer: (A)

Speed of the first train = length/time = 120/10 = 12m/s

Speed of the second train = length/time = 150/15 = 10m/s

2.0 m/s is the difference in the train speed.

Q3. The number that least fits this set: (324, 441, 97 and 64) is ________. (1 mark)

(A) 324 (B) 441 (C) 97 (D) 64

Answer: (C).

97 is the odd number out. All other numbers are squares.

324 = 18*18

441 = 21*21

64 = 8*8

Q4. The Buddha said, “Holding on to anger is like grasping a hot coal with the intent of throwing it at

someone else; you are the one who gets burnt.” (1 mark)

Select the word below which is closest in meaning to the word underlined above.

(A) burning (B) igniting (C) clutching (D) flinging

Answer: (C) clutching

GATE 2016

2016-3

Q5. M has a son Q and a daughter R. He has no other children. E is the mother of P and daughter-in-law

of M. How is P related to M? (1 mark)

(A) P is the son-in-law of M. (B) P is the grandchild of M.

(C) P is the daughter-in law of M. (D) P is the grandfather of M.

Answer: (B)

Q6. A straight line is fit to a data set (ln x, y). This line intercepts the abscissa at ln x = 0.1 and has a

slope of −0.02. What is the value of y at x = 5 from the fit? (2 marks)

(A) −0.030 (B) −0.014 (C) 0.014 (D) 0.030

Answer: (A)

The equation of a line is

y = mx + c where m is the slope & c is the y-intercept

Now, In this question x is replaced with lnx

So, the equation of line becomes,

y = mlnx + c

or, y = -0.02lnx + c

We have given with abscissa which is essentially x-intercept. So, now we have to find 'c' the y-intercept.

for, y = 0, lnx = 0.1 (given in the question)

Putting the value,

0 = -0.02x0.1 + c

or, c = 0.002

So, the equation of line becomes,

y = -0.02lnx + 0.002

putting x = 5 (asked in the question)

y = -0.002ln5 + 0.002 = -0.002x1.6 + 0.002 = -0.03

(ln5 = 1.6)

Q7. The overwhelming number of people infected with rabies in India has been flagged by the World

Health Organization as a source of concern. It is estimated that inoculating 70% of pets and stray dogs

against rabies can lead to a significant reduction in the number of people infected with rabies. (2 marks)

Which of the following can be logically inferred from the above sentences?

(A) The number of people in India infected with rabies is high.

(B) The number of people in other parts of the world who are infected with rabies is low.

(C) Rabies can be eradicated in India by vaccinating 70% of stray dogs.

(D) Stray dogs are the main source of rabies worldwide.

Answer: (A)

Q8. Find the area bounded by the lines 3x+2y=14, 2x-3y=5 in the first quadrant. (2 marks)

(A) 14.95 (B) 15.25 (C) 15.70 (D) 20.35

Answer: (C) 15.25

GATE 2016

2016-43

The longest duration time is 18 weeks during which all activity paths would be completed. So, the completion

time for the project would be 18 weeks.

Answer: 18

Q51. Match Group I with Group II

GROUP I GROUP II

1. Diagrid P Hearst Tower, New York

2. Outrigger truss Q Taipai 101

3. Suspended floor R HSBC, Hong Kong

4. Cable stayed S Millennium Dome, London

T Sears Tower

Answer: 1-P, 2-Q, 3-R, 4-S

Hearst Tower, New York is the world headquarters of the Hearst Corporation, housing the numerous

publications and communications companies of the media conglomerate under one roof, including, among

others, Cosmopolitan, Esquire, Marie Claire, Harper's Bazaar, Good Housekeeping, and Seventeen.

The six-story base was designed by the architect Joseph Urban and the tower designed by the architect Norman

Foster. The uncommon triangular framing pattern (also known as a diagrid) required about 20% less than a

conventional steel frame. The diagrid (also called diagonal grid) is a framework of diagonally intersecting

metal, concrete or wooden beams that is used in the construction of buildings and roofs.

Figure: Hearst Tower, New York.

GATE 2016

2016-44

Taipei 101

Taipei 101, formerly known as the Taipei World Financial Centre, is a 106-storey 508m-high mixed use

building located in Taiwan and designed by Taiwanese architecture firm C.Y. Lee & Partners. The building

also received an award from Guinness World Records for using the world's fastest elevators.

The tower's architectural design is inspired by Chinese ethos. The design of the eight segments of the tower is

based on the Chinese lucky number eight. The whole structure resembles a bamboo growing up high, which

symbolises upward progress and everlasting strength in Chinese culture.

The tower features slanted exterior curtain walls with the facade system comprised of double glazed blue-green

glass and aluminium panels.

Taipei 101 uses a passive tuned mass wind damper system for protecting the high-rise building against

vibration due to wind. It involves a 5.5m diameter sphere shaped mass block weighing 660t suspended from

level 92 to level 87. It is considered as the world's largest tuned mass damper (TMD) of its kind.

Figure: Taipei 101.

GATE 2016

2016-45

Figure: Taipei 101. Structural System. Outrigger truss

Figure: Taipei 101.TMD (Tuned Mass

Damper)

Figure: Building height..

GATE 2016

2016-46

HSBC, Hong Kong

With its prefabricated steel parts, aluminium cladding, plug-in service modules and reconfigurable spaces,

HSBC‘s office building is still one of the most modern-looking buildings around, even though it opened in

1985. The building is more akin to the structure of a bridge, with floors suspended from the distinctive chevron-

shaped trusses supported by eight sets of masts that run the height of the building. The structure is not hidden

and is a feature of the façade. Services are accommodated on the periphery, allowing clear floor spans.

Figure: HSBC, Hong Kong & its simplified structural system.

Figure: Structural system simplified.

GATE 2016

2016-47

It doesn‘t have a regular floor plan; floors get smaller the higher up they are in the building, and the distance

they are set back from the façade to create small atria varies between floors. The lower floors rising from the

public plaza on the ground level are divided into two by the banking hall atrium that runs from the front to the

back of the bank. Further variation is introduced with double-height spaces on the floors that meet the bottom

of each truss. Express lifts serve only certain floors; further access to individual levels is via escalators.

Millennium Dome

The Millennium Dome was constructed to be the home of a very large exhibition that was to celebrate the

coming of the third millennium. The Millennium Dome was designed by Richard Rogers. Rogers is known

around the United Kingdom for his very functional, semi-modernist design flair. He has worked on the Lloyd's

Building, and the Court of Human Rights building that is in Strasbourg.

Figure: Millennium Dome, London.

.

Figure: Inside view, Millennium Dome, London.

GATE 2016

2016-48

Design and Construction

The building is constructed out of tensioned fabric over a skeleton of steel. Of the many large domes worldwide

which share this construction scheme, the Millennium Dome is one of the largest. Symbolism is key to the

design of the dome and there are many symbolic pieces of the structure. It has 12 supports that jut out from the

cloth ceiling that represent the months of the year and the hours on a clock face. This is an attempt to pay

homage to the role of Greenwich Mean Time since the Prime Meridian passes just to the west of the building.

The circular dome also has a diameter of 365 metres to represent the days of the year. The centre of the dome is

a full 52 metres tall to represent the 52 weeks in each year.

The top of the structure is comprised of a thick glass/fibre fabric that has been coated with PTFE. It is a very

durable material that is commonly used in similarly styled buildings. Due to the fact that the roof is held

together using 12 struts, the building is not technically a "dome" by architectural standards. The roof was

designed to actually be lighter than the air inside of the building. This helps its structural integrity significantly.

Current Use

After the failure of the Millennium Exhibition and the resulting scandal over the cost of the Millennium Dome,

the dome was sold and has been rebranded as O2 Arena. It is a major exhibition and events venue in London

that often hosts concerts by famous bands.

Sears Tower, New York (Now

called Willis Tower)

As designed by architect Bruce

Skidmore Owings and Merrill

(SOM), the structure was a "bundled

tube" system of nine squares with

sides of 75 feet (for an overall 225 x

225 ft), sheathed in a curtain wall of

dark tinted glass.

Figure: Sears Tower, Bundled Structure.

Figure: Sears Tower, The 103rd floor of the tower is the location of the famous skydeck..

GATE 2016

2016-49

Q52. A steel I-beam section is subjected to a bending moment of 96 kN-m. The moment of inertia of the

beam section is 24,000 cm4. The bending stress at 100 mm above the neutral axis of the beam in MPa will

be ____________ (2 marks).

Answer: 40

Solution: Bending stress = 96x1000x0.1/24000x10-8

= 400 x 106 Pa = 400MPa

M = the internal bending moment about the section‘s neutral axis

y = the perpendicular distance from the neutral axis to a point on the section

I = the moment of inertia of the section area about the neutral axis

Q53. Match Group I with Group II

GROUP I GROUP II

1. Radiance P Lighting Analysis

2. Odeon Q Acoustic design

3. Rayman R Outdoor thermal emission

4. Primavera S Construction management

T Air flow

Answer: 1-P, 2-Q, 3-R, 4-S

Figure: Sears Tower.

GATE 2016

2016-50

Q54. Choose the best option

1 Nisargruna P Solid waste management

2 Vortex-DEWAT Q Waste water treatment

3 Swale R Ground Water Discharge

4 BIPV S Renewable

T Desalination

Answer: 1-P, 2-Q, 3-R, 4-S

Q55. A 250 mm × 250 mm RCC column is reinforced with one percent steel. The permissible

compressive stress of concrete and steel are 8 N/mm2 and 150 N/mm

2 respectively. The axial load

carrying capacity of the column in kN is _____________ (2 marks).

Answer: 826 (585 - 595 by GATE official)

Solution:

Pu =0.4 fck Ac + 0.67 fy Asc

Pu = 0.4 x (8N/mm2) x 62500 mm

2 (1-1/100) + 0.67 x 150N/mm

2 x 62500 mm

2 (1/100)

= 62500 [(0.4 x 8 x 0.99) + (0.67 x 150 x 0.1)]

= 62500 [3.168 + 10.050]

= 62500x13.218 = 826125N = 826 kN

where

Pu= factored axial load on the member,

fck= characteristic compressive strength of the concrete,

Ac= area of concrete,

fy= characteristic strength of the compression reinforcement, and

Asc= area of longitudinal reinforcement for columns.

END OF THE QUESTION PAPER 2016

GATE 2015

2015-22

29. The saturation level of a colour represents

(A) distribution

(B) brilliance

(C) darkness

(D) warmth

Notes:

Color saturation refers to how vivid and intense a color is. For example, a display with poor color saturation will

look washed out or faded. When a color's saturation level is reduced to 0, it becomes a shade of gray.

Answer : B

30. Invert level of a pipe at a given cross section refers to the

(A) highest point of the internal surface

(B) lowest point of the internal surface

(C) highest point of the external surface

(D) lowest pot of the external surface

Notes:

Invert level of a pipe is the level taken from the bottom of the

inside pipe as shown below.

The level at the crown of the pipe is the Invert level plus the internal diameter of the pipe plus the pipe wall

thickness. It may necessary to use in calculations when measurements are taken from the crown of a pipe.

Answer : B

31. The command DVIEW in AutoCAD permits to view

(A) a selected portion of the drawings in detail

(B) the entire screen on the monitor

(C) a perspective of the drawing

(D) a damaged part of the drawing

Answer : C

32. Match the Land use categories of Group-I with their respective Colour codes in Group_II as per

practice in India

Group-I Group-II

P Residential 1 Red

Q Commercial 2 Grey

R Industrial 3 Blue

S Public/Semi-public 4 Violet

5 Yellow

Figure: Invert level of pipe.

GATE 2015

2015-23

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

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

(C) P -1, Q- 2, R – 4, S- 5

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

Answer : A

33. A rectangular beam section of size 300 mm (Width) X 500 mm (depth) is loaded with a shear force of 600

kN. The maximum shear stress on the section in N/mm2 is_____________

Notes:

Maximum shear stress in a beam section = 3/2(F/A) .

F = shear Force & A = cross-sectional Area

So, Maximum shear

= 3/2[(600x1000N)/(300mmx500mm)]

= 6N/mm2

Figure: Derivation of Shear Force.

GATE 2015

2015-24

Answer : 6

34. In a 50 meter section of a waste water pipe. if the gradient is 1 in 80, then the fall in millimeter is

Notes:

1:80 => for 80m , fall is 1m

So, for 1m, fall will be 1/80m

So, for 50m, fall will be 1/80x50 = 0.625m = 625mm

Answer : 625

35. A 15 meter long and 3 meter wide driveway needs to be paved with 300 mm X 300 mm square Tiles. If

each packet contains 30 numbers of tiles, then the number of packets to be procured to pave the whole area

is__________________

Notes:

Area covered by each tile is 300mmx300mm = 0.3mx0.3m = 0.09sqm

As each packet contains 30 numbers of tiles, area covered by each packet = 0.09x30 = 2.7sqm

Total area o the roadways = 15mx3m = 45sqm

So, no. of required packets= 45sqm/2.7sqm = 16.67 = 17 packets

Answer : 16.5 to 17.0

36. Match the Monuments in Group-I with their Features in Group-II

Group-I Group-II

P Panch Mahal Fathepur, Sikri 1 Painted Stone Figures

Q Meenakshi Temple, Madurai 2 Intricate Red Sand Stone Carvings

R Jor-Bangla Temple, Bishnupur 3 Granite Statues

S Sun Temple, Konark 4 Khondalite Stone Work

5. Terracotta Carvings

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

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

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

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

Notes:

GATE 2015

2015-25

Panch Mahal is a five-story palace in Fatehpur Sikri,

Uttar Pradesh, India.

The Panch Mahal, also known as "Badgir" meaning

wind catcher tower, was commissioned by Akbar the

Great. This structure stands close to the Zenana

quarters (Harem) which supports the supposition that

it used for entertainment and relaxation. This is an

extraordinary structure employing the design

elements of a Buddhist Temple; entirely columnar,

consisting of four stories of decreasing size arranged

asymmetrically upon the ground floor, which

contains 84 columns. These columns, that originally

had jaali (screens) between them, support the whole

structure. Once these screens provided purdah (cover)

to queens and princess on the top terraces enjoying the cool breezes and watching splendid views of Sikri

fortifications and the town nestling at the foot of the ridge.

The pavilion gives a majestic view of the fort that lies on its left. The pool in front of the Panch Mahal is called the

Anoop Talao. It would have been filled with water, save for the bridge, and would have been the setting for musical

concerts and other entertainment. The ground floor has 84 columns, the first story has 56 columns and the second

and third stories have 20 and 12 columns respectively. The topmost story has 4 columns supporting a chhattri.

There are 176 columns in all and each is elegantly carved with no two alike.

Meenakshi Sundareswarar temple (twin temples, Dravidian architecture) is one of the biggest temples in India.

The original temple built by Kulasekara Pandyan was in ruins. The plan for the current temple structure was laid by

Viswanath Naik and was completed by Tirumalai Nayakar. The Aadi, Chittirai and the Maasi, and Veli streets

surround the temple. Both temples are adorned with exquisite carvings & sculptures and gold plated vimanams.

There are 12 massive gopurams in the temple, the four tallest gopurams at the outer walls (The tallest is the

southern gopuram, measuring 49 metres). There are four entrances. The main entrance is to the Meenakshi Amman

Figure: Panch Mahal.

Figure: Meenakshi Temple, Madurai.

GATE 2015

2015-54

Sensible cooling can be aided by evaporative cooling. To reduce the underground temperature, the ground can be

shaded using vegetation and can be wetted by sprinkling water. This water seeps through and dampens the tunnel

walls. Consequently, air from the tunnel is evaporatively cooled as it passes through the tunnel. Another variation

possible is to use buried pipes instead in place of tunnel.

Radiant cooling systems typically use chilled water running in pipes in thermal contact with the surface. The

circulating water only needs to be 2-4°C below the desired indoor air temperature.Heat is removed by the water

flowing in the hydronic circuit once the heat from different sources in the space is absorbed by the actively cooled

surface – ceiling, floor or walls.

Answer : C

46. Match the Vibrator Types in Group-I with their related Areas of Application in Group-II

Group-I

P Needle Vibrator

Q Shutter Vibrator

R Surface Vibrator

S Table Vibrator

Group-II

1 Concrete Pavement

2 Pre-cast Concrete Unit

3 Beam-Column Junction

4 Retaining Wall

5 Slip Forming

Figure: Radiant cooling

GATE 2015

2015-55

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

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

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

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

Notes:

TYPES OF CONCRETE VIBRATORS FOR COMPACTION

Since concrete contains particles of varying sizes, the most satisfactory compaction would perhaps be obtained by

using vibrators with different speeds of vibration. Polyfrequency vibrators used for compacting concrete of stiff

consistency are being developed. The vibrators for compacting concrete are manufactured with frequencies of

vibration from 2800 to 15000 rpm. The various types of vibrators used are described below:

Immersion or Needle Vibrators:

Immersion or needle concrete vibrators

This is perhaps the most commonly used vibrator. It essentially

consists of a steel tube (with one end closed and rounded)

having an eccentric vibrating element inside it. This steel tube

called poker is connected to an electric motor or a diesel

engine through a flexible tube. They are available in size

varying from 40 to 100 mm diameter. The diameter of the

poker is decided from the consideration of the spacing between

the reinforcing bars in the form-work.

The frequency of vibration varies upto 15000 rpm. However a

range between 3000 to 6000 rpm is suggested as a desirable

minimum with an acceleration of 4g to 10g.

The normal radius of action of an immersion vibrator is 0.50 to

1.0m. However, it would be preferable to immerse the vibrator

into concrete at intervals of not more than 600mm or 8 to 10 times the diameter of the poker.

The period of vibration required may be of the order of 30 seconds to 2 minute. The concrete should be placed in

layers not more than 600mm high.

External or Shutter Vibrators

These vibrators are clamped rigidly to the form work at the pre-determined points so that the form and concrete are

vibrated. They consume more power for a given compaction effect than internal vibrators.

These vibrators can compact upto 450mm from the face but have to be moved from one place to another as

concrete progresses. These vibrators operate at a frequency of 3000 to 9000 rpm at an acceleration of 4g.

Figure: Immersion or Needle Vibrators

GATE 2015

2015-56

The external vibrators are more often used for pre-casting of thin in-situ sections of such shape and thickness as can

not be compacted by internal vibrators.

Surface concrete vibrator

These are placed directly on the concrete mass. These best

suited for compaction of shallow elements and should not

be used when the depth of concrete to be vibrated is more

than 250 mm.

Very dry mixes can be most effectively compacted with

surface vibrators. The surface vibrators commonly used are

pan vibrators and vibrating screeds. The main application

of this type of vibrator is in the compaction of small slabs,

not exceeding 150 mm in thickness, and patching and repair

work of pavement slabs. The operating frequency is about

4000 rpm at an acceleration of 4g to 9g.

Vibrating Table

The vibrating table consists of a rigidly built steel platform

mounted on flexible springs and is driven by an electric motor. The

normal frequency of vibration is 4000 rpm at an acceleration of 4g to

7g.

The vibrating tables are very efficient in compacting stiff and harsh

concrete mixes required for manufacture of precast elements in the

factories and test specimens in laboratories.

Answer: B

Figure: External or Shutter Vibrators

Figure: Surface concrete vibrator

Figure: Vibrating Table

GATE 2015

2015-57

47. Match the type of Temporary Structure in Group-I with their corresponding Functions in Group-II

Group-I

P Scaffolding

Q Formwork

R Shoring

S Underpinning

Group-II

1 To support unsafe structure

2 To support platforms for workmen and materials at raised height during construction

3 Removal of water from pits

4 Mould for RCC Structure

5 Strengthening the existing foundation

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

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

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

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

Notes:

Scaffolding is a temporary platform constructed for reaching heights above arms' reach for the purpose of building

construction, maintenance, or repair. It is usually made of lumber and steel and can range from simple to complex

in design, depending on its use and purpose.

Figure: Scaffolding (metal & wood, both can be.)

GATE 2015

2015-58

Formwork is another term for shuttering. Formwork is a structure, usually temporary, used to contain poured

concrete and to mould it to the required dimensions and support until it is able to support itself. It consists primarily

of the face contact material and the bearers that directly support the face contact material.

Shoring is a general term used in construction to describe the process of supporting a structure in order to prevent

collapse so that construction can proceed. It can be done with the help of shores or props.

They can be used under the following circumstances:

1. When walls bulge out

2. When walls crack due to unequal settlement of foundation and repairs are to be carried out to the cracked

wall.

3. When an adjacent structure needs pulling down.

4. When openings are to be newly made or enlarged in a wall.

Figure: Formwork

Figure: Shoring

GATE 2015

2015-66

Robert Owen, a British social reformer and socialist, pioneer in the cooperative movement. In 1800, Owen moved

to New Lanark, Scotland, where he bought mills.There he reconstructed the community into a model industrial

town with good housing and sanitation, nonprofit stores, schools, and excellent working conditions. Mill profits

increased. The New Lanark experiment became famous in England and abroad, and Owen's ideas spread. He

instigated the reform that resulted in the passage of the Factory Act of 1819—a watered down version of his

proposals, but still a landmark in social reform. He also proposed the formation of self-sufficient cooperative

agricultural-industrial communities. By 1817 he had formulated the goal of the eight-hour day and coined the

slogan: "Eight hours labour, Eight hours recreation, Eight hours rest".

Arcosanti is the a prototype for architect Paolo Soleri's vision of an arcology. The idea of an arcology is the

combination of architecture and ecology, a concept first conceived by Paolo Soleri in the 1950s.

Arcosanti is an experimental town in Arizona. He began construction in 1970, to demonstrate how urban conditions

could be improved while minimizing the destructive impact on the earth. He taught and influenced generations of

architects and urban designers who studied and worked with him there to build the town.

Answer: C

53. The housing stock of a town has total number of 9090 dwelling units. Present population of the town is

45,450. Assuming an average household size of 4.5, the housing shortage in percentage is________

Notes:

Present Population = 45,450

Household size = 4.5

So, required no of dwelling units = 45,450/4.5 =10100

Current no of dwelling units = 9090

Shortage = 10100 – 9090 = 10

Answer : 10

54. A hall is 15 m long and 12 m wide. If the sum of areas of the floor and ceiling is equal to the sum of the

area of its four walls, then the volume of the hall in cubic meter is________

Figure: Arcosanti by Paolo Soleri's vision of an arcology.

GATE 2015

2015-67

Notes:

Area of floor =15 x 12 = 180

Area of ceiling = 15 x 12 = 180

Let the height of the wall be ‗H‘ meter

According to question,

Area of Floor + Area of ceiling = Area of four walls

So, 180 + 180 = wall perimeter x height of wall

So, 360 = [2(15+12)] x H

So, 360 = 54H

So, H = 20/3 meter

Therefore, volume of room = Length x Breadth x Height = 15 x12 x20/3 = 1200 cubic meter

Answer : 1200

55. The actual roof area of a building is 3,60,000 sqm, which on a site plan measures 25 sq cm. The scale of

the site plan is 1:____

Notes:

Let the scale of the plan be 1:X

So, (1/X) 2 = (25 sq cm/ 360000 sqm)

= 25/360000 x10000 sq cm……………………..(1 sqm = 100cm x 100 cm = 10000 sq cm)

= (1/12000x12000) = (1/12000)2

So, 1/X = 1/12000

Answer : 12000

56. If the annual net come from a commercial property is Rs 22,000/- and the interest rate is 8%, then the

capitalized value in rupees of the property in perpetuity is________________

Notes:

Let the value of the property is X

According to question, 8% of X is 22000

(X) x 8/100 = 22000

8X/100 = 22000

X = 275000 Answer

Answer : 275000

57. A five storey building is constructed on a 100 m x 50 m plot having coverage of 60% (option 1).

Alternatively, a four storied building is constructed on the same plot with a 50% ground coverage (option2).

The ratio of FARs between options1 and 2 is ______________________

Notes:

FAR = Total built up area/ Plot area

GATE 2015

2015-68

Total built up area in option 1 = (coverage area) X (no. of floors) = (100x50x0.6) X (5) = 15000

Total built up area in option 2 = (coverage area) X (no. of floors) = (100x50x0.5) X (4) = 10000

Plot area is common for both options = 100x50 = 5000 sqm

So, FAR 1/FAR 2 = [15000/5000]/[10000/5000] = 1.5

Answer : 1.5

58. If a roof is treated with a layer of thermal insulation material, the internal heat gain is reduced by 60%.

The U-value of the roof (without thermal insulation) is 3 Wm2/degree centigrade. Assuming a constant

temperature difference between indoor and outdoor, the U-value of the thermal insulation layer in

Wm2/degree centigrade is_____

Notes:

We will start from the beginning with concept.

U-value = 1/ R-value

Following is the basic equation of heat flow,

…………………………………………………………(1)

Where,

is rate of heat flow

k = Thermal conductivity coefficient

A= surface area of the wall

L= wall thickness

Tf-Ti = temperature difference

R-value is inversely related to the thermal conductivity constant and is also related to thickness, thus:

So, the equation (1) becomes ,

GATE 2015

2015-69

Let,

R-value of thermal insulation material = R1

R-value of the roof = R2 = 1/U-value = 1/3 …………(given)

R-value of combined set = R1+2


∆Q/∆t = A (Tf-Ti)/R2……………………………………(heat transfer rate when there is no insulation)

0.4 x (∆Q/∆t) = A (Tf-Ti)/R1+2…………………………..(heat transfer rate with insulation)

1/0.4 = R1+2/ R2

1/0.4 = R1+2/ (1/3)

1/ R1+2 = 1.2

Now as per addition of thermal resistance,

R1+2 = R1 + R2

1/1.2 = R1 + 1/3

R1 = 2 Answer

More about U-values:

U-values measure how effective a material is

an insulator. The lower the U-value is, the

better the material is as a heat insulator. For

example, here are some typical U-values for

building materials:

a cavity wall has a U-value of 1.6

W/m²

a solid brick wall has a U-value of

2.0 W/m²

a double glazed window has a U-

value of 2.8 W/m².

The cavity wall is the best insulator and the

double glazed window is the worst insulator.

Note that you do not need to remember any

U-values for the exam.

Most household heat is lost through the windows and roof as shown in the figure.

Relationship between K-value, R-value & U-value:

K-value (Thermal Conductivity)

Thermal conductivity (also known as Lambda) is the rate at which heat passes through a material, measured in

watts per square metre of surface area for a temperature gradient of one kelvin for every metre thickness.This is

expressed as W/mK. Thermal conductivity is not affected by the thickness of the product.

The lower the conductivity, the more thermally efficient a material is.

Figure: Infrared mapping. Most household heat is lost through

the windows and roof as shown in the figure.

GATE 2015

2015-70

Example:

PIR Board: Lambda = 0.022 W/mK

Glass Fibre Roll: Lambda = 0.044 W/mK

R-Value (Thermal Resistance)

Thermal resistance is the ability of a material to prevent the passage of heat. It‘s the thickness of the material (in

metres) divided by its conductivity. This is expressed as m2K/W.

If the material consists of several elements, the overall resistance is the total of the resistances of each element. The

higher the R-value, the more efficient the insulation.

Example:

PIR Board: 0.022 W/mK and 100mm thick; R-value = 0.1 metres ÷ 0.022 = 4.54 m2K/W

Glass Fibre Roll: 0.044 W/mk and 100mm thick; R-value = 0.1 metres ÷ 0.044 = 2.27 m2K/W

N.B. Surfaces and cavities also provide thermal resistance which must be taken into account when calculating U-

values. There are standard figures for the resistances of surfaces and cavities.

U-Value (Thermal Transmittance)

Thermal transmittance, commonly known as the U-value, is a measure of the rate of heat loss of a building

component. The U-value is the sum of the combined thermal resistances of all the elements in a construction,

including surfaces, air spaces, and the effects of any thermal bridges, air gaps and fixings.

The U-value is expressed in watts per square metre, per degree kelvin, or W/m2K.

Calculating U-Values:

Start by calculating the thermal resistances of each element (R-values).

The R-value is the thickness of the product in metres ÷ Lambda (thermal conductivity).

Add the R-values of all materials used in the application (including any air gaps) and calculate the reciprocal. The

reciprocal = 1 ÷ total of all R-values

Example:

PIR Board 0.022 W/mK100mm thick + Glass Fibre Roll 0.044 W/mK100mm thick

Total combined R-value = 4.54 + 2.27 = 6.81 m2K/W

U-value = 1 ÷ 6.81 = 0.147 W/m2K

Answer : 2

59. A simply supported beam having effective span of 5 meter is carrying a centrally concentrated load of

16kN. The maximum bending moment in the beam in kN-m is_________

Notes:

Maximum bending moment will at center = PL/4 = 20

Answer

GATE 2015

2015-71

Answer : 20

60. A landscaped garden with irregular profile and minor undulation, measuring35,000 sqm has a total

surface area covered with 20% brick paving, 15% cement concrete paving, and rest with grass. The peak

intensity of rainfall in that region is 70 mm/hr. Tile coefficient of runoff for brick paving, cement concrete

paving and grass is 0.8, 0.9 and 0.5 respectively. The estimated quantity of runoff in cubic meter/hr for the

entire garden area is______________

Notes:

Brick paving area = 20% of 35000 = 7000 sqm

Concrete paving area = 15 %of 35000 = 5250 sqm

Grass cover area = 65% of 35000 = 22750 sqm

Water runoff by brick paving = 0.8 x 7000 sqm area x 0.07 m of rain fall = 392.0 cum

Water runoff by concrete paving = 0.9 x 5250 sqm area x 0.07 m of rain fall = 330.75 cum

Water runoff by grass cover = 0.5 x 22750 sqm area x 0.07 m of rain fall = 796.25 cum

Answer : 1510 to 1530

61. The number of standard cement bags required to prepare 1400 kg of concrete in the ratio of 1:2:4 (mixed

by weight batching)

Notes:

1:2:4 → 1part cement : 2 part sand : 4 part aggregates

So, To make 7 kg of concrete, amount of cement required = 1 kg

So, To make 1 kg of concrete, amount of cement required = 1/7 kg

So, To make 1400 kg of concrete, amount of cement required = (1/7)x 1400 = 200 kg

1 bag of cement contains 50 kg of cements

So, no of bags required = 200kg/50 kg = 4 bags

Answer : 4

62. A class room measuring 10 m (L) x 8 m (B) x 2.7 m (H) require an Illumination level of 500 lux on the

desk level using 40 W fluorescent lamps with rated output of 5000 lumens each. Assuming utilization factor

of 0.5 and maintenance factor of 0.8, the number of lamps required is_____

GATE 2015

2015-72

Answer : 20

Notes:

500 lux = 500 lumen/meter square

Total lumen required = (500 lumen/meter square) x (Area of the room) = 500 x10x8

Lumen output of one lamp = 5000 lumen (given)

Actually the lumen output of the lamp will be less than 5000 lumen because of utilization factor and maintenance

factor.

So, net lumen output = 5000 x 0.5 x 0.8

Required no. of lamps = Total lumen required / lumen of on lamp = 500x10x8 /5000x0.5x0.8 = 20 Ans.

Luminous flux:

Luminous flux is the rate of energy radiation in the form of light waves. The unit is lumen.

Lumen:

Lumen is the unit of luminous flux. It represents the flux emitted in unit solid angle of one steradian by a point

source having a uniform intensity of one candela. Thus a uniform point source of one candle power emits 4π

lumens.

Q. Illumination level required for precision work is around

(A) 50 lm/m2

(B) 100 lm/m2

(C) 200 lm/m2

(D) 500 lm/m2.

Q. The illumination level in houses is in the range

(A) 10-20 lumen/m2

(B) 30 - 50 lumen/m2

(C) 40-75 lumen/m2

(D) 100-140 lumen/m2.

Q. One lumen per square meter is the same as

(A) One lux

(B) One candela

(C) One foot candle

(D) One lumen meter.

Q. Candela is-the unit for

(A) Light flux

(B) Luminous intensity

(C) Brightness

(D) Luminous efficiency.

GATE 2015

2015-73

63. Area of tense steel per meter width of a reinforced concrete slab is 335 sq mm. If 8 mm rods are used as

reinforcement, then centre to centre spacing of the reinforcement in mm is

Notes:

Total area of steel is 335 sq mm. (which is spread in 1m of width)

Area of 8 mm rod = ∏r2

= 3.14 x 4mmx 4mm = 50.24 sq mm

So, total no. of rods spread in 1m of width = 335/50.24 = 6.67

So, 7 rods are spread in 1 m of width.

So, distance between two rods will be 1m/6.67 = 150 mm

Answer : 145 to 155

64. The population of a town as per Census 2011 was 22,730 and the population as per census 2001 was

15,770. Considering arithmetic projection of growth, the projected population in 2016 will be

Notes:

Arithmetic projection of growth → Arithmetic progression, example 2, 4, 6, 8, 10 …

Geometric projection of growth → Geometric progression, example 2, 4, 8, 16, 32 …

In 10 years, growth in population = 22,730 – 15,770 = 6960

So, in 1 year, population growth would be = 6960/10 = 696

So, in 5 years, population growth would be = 696x5 = 3480 (Arithmetic projection)

So, after 5 years that is in 2016, the population would be = 22730 + 3480 = 26210

Answer : 26178 to 26210

65. Two concrete mixers of capacity 200 liters each are used in a construction site to produce 20 cubic meter

of concrete. Ingredient charging, mixing and discharge times are 3 minutes, 7 minutes and 1 minutes

respectably. Assuming a tune loss of 5 minutes per hour of operation, the total time in hours for the mixers

to produce the required amount of concrete will be_________.

Notes:

Capacity of the concrete mixers = 2x 200 liters = 2x (200/1000) cubic meter = 0.4 cubic meter

According to question, time taken to produce 0.4 cubic meter of concrete = 3+7+1 = 11 minutes

Time taken to produce 20 cubic meter of concrete = (20/0.4) x 11 minutes = 550 minutes.

As tune loss of 5 min is in an hour

So for 550 minutes, (550/55) x 5min = 50 minutes.

So total time required = 550 + 50 = 600 minutes = 10 hours.

Answer : 10

END OF THE QUESTION PAPER

GATE 2013

2013-7

Q.13 If threshold of hearing has a sound level of zero decibels and the sound level in a broadcasting

studio is 100 times the threshold of hearing, its value in decibels would be

(A) 0 (B) 10 (C) 20 (D) 100

Notes: dB=10log(100I/I) =20

Answer(C)

The decibel (abbreviated dB) is the unit used to measure the intensity of a sound. The decibel scale is a little

odd because the human ear is incredibly sensitive. Your ears can hear everything from your fingertip brushing

lightly over your skin to a loud jet engine. In terms of power, the sound of the jet engine is about

1,000,000,000,000 times more powerful than the smallest audible sound. That's a big difference!

On the decibel scale, the smallest audible sound (near total silence) is 0 dB. A sound 10 times more powerful is

10 dB. A sound 100 times more powerful than near total silence is 20 dB. A sound 1,000 times more powerful

than near total silence is 30 dB. Here are some common sounds and their decibel ratings:

Near total silence - 0 dB

A whisper - 15 dB

Normal conversation - 60 dB

A lawnmower - 90 dB

A car horn - 110 dB

A rock concert or a jet engine - 120 dB

A gunshot or firecracker - 140 dB

You know from your own experience that distance affects the intensity of sound -- if you are far away, the

power is greatly diminished. All of the ratings above are taken while standing near the sound.

Any sound above 85 dB can cause hearing loss, and the loss is related both to the power of the sound as well as

the length of exposure. You know that you are listening to an 85-dB sound if you have to raise your voice to be

heard by somebody else. Eight hours of 90-dB sound can cause damage to your ears; any exposure to 140-dB

sound causes immediate damage (and causes actual pain).

Q.14 The width to height ratio of the front facade of Parthenon (without the pediment) is

(A) 9:4 (B) 4:9 (C) 1:1.618 (D) 1.618:1

Figure: Golden ratio

GATE 2013

2013-8

The Fibonacci Sequence

The 1:1.618 ratio is said to be related to the Fibonacci Sequence, a series of numbers discovered by

Leonardo of Pisa (known as Fibonacci) who was a 12th-13th Century Italian mathematician.

In the Fibonacci Sequence, each number is made by adding the previous two together.

ie

1

1

2 (1+1)

3 (1+2)

5 (2+3)

8 (3+5)

13 (5+8)

21 (8+13)

34 (13+21)

etc

From the number '3' onwards, all the numbers in the sequence - if divided by the previous number - give a

result which is approximately1.618

eg

3 divided by 2 = 1.5



13 divided by 8 =1.625

21 divided by 13 = 1.615

34 divided by 21 = 1.619

When we draw a square within a Golden Ratio rectangle, the rectangular shape left over is also of Golden Ratio

proportions. This remaining rectangular shape is approximately 1.618 times smaller than the previous Golden

Ratio rectangle it sits within. This can be repeated endlessly.

Answer(A)

Figure: Thumb rule to Golden Ratio

Scan for more on

golden ratio

GATE 2013

2013-41

Architect : Norman Foster, Foster + Partners

Major cantilever and structural clusters form the technological

aesthetic!

Answer (A)

Q.37 Match the Five Year Plans listed under Group I with their corresponding feature from Group II

Group I

P. First Five Year Plan

Q. Fourth Five Year Plan

R. Seventh Five Year Plan

S. Tenth Five Year Plan

Group II

1. Formation of HUDCO

2. Establishment of TCPO

3. Introduction of JNNURM

4. Announcement of National Housing Policy

5. Passing of Urban Land Ceiling and Regulation Act

(A) P-5,Q-2,R-4,S-3 (B) P-2,Q-1,R-4,S-3

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

Answer (B)

Q.38 Match the landscape designers listed under Group I with their appropriate contribution

from Group II

Group I

P. Lancelot ‘Capability’ Brown

Q. Andre Le Notre

R. Joseph Paxton

S. Frederick Law Olmstead

Group II

1. The Well-tempered Garden

2. Kew Garden

3. Versailles Garden

4. Crystal Palace

5. Central Park

(A) P-3,Q-1, R-4, S-2 (B) P-5, Q-3, R-4, S-2

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

GATE 2013

2013-42

Answer (D)

Figure: Lancelot ‘Capability’ Brown changed the face

of eighteenth century England, designing country

estates and mansions, moving hills and making flowing

lakes and serpentine rivers, a magical world of green

Figure : Royal Botanic Gardens, Kew by Lancelot ‘Capability’ Brown

GATE 2013

2013-43

Figure: The king of gardeners and the King’s own

gardener, Le Nôtre made the French formal garden

famous throughout Europe. He designed the finest

gardens of the 17th century and made Versailles into his

absolute masterpiece. His talents earned him a colossal

fortune and an international reputation.

Figure:The gardens of Versailles as they look today

GATE 2013

2013-44

Frederick Law Olmsted

Unsurpassed in the field of landscape architecture,

Frederick Law Olmsted, Sr., defined and named the

profession and designed many of America's most beloved

19th-century parks and landscapes, including New York's

Central Park, Brooklyn's Prospect Park, the Biltmore Estate,

and the U.S. Capitol grounds. His commitment to public

works stemmed in part from his abolitionist stance: by

creating elegant and equitable public spaces for all, he

hoped to show the inherent beauty possible in a free society.

He also hoped to refute the long-held sentiment that the

only source of culture was through noblesse oblige, instead

insisting that his parks be publicly funded. During his

remarkable 40-year career, beginning in the mid-1850s,

Olmsted and his partners and employees created the first

park systems, urban greenways, and suburban residential

communities in this country. He and his colleague, Charles

Eliot, were pivotal figures in the movement to create scenic

reservations to preserve natural treasures such as Yosemite, Yellowstone, and Niagara Falls. Also, site planner

for the Great White City of the 1893 World‟s Columbian Exposition, planner of Boston‟s “Emerald Necklace”

of green space.

Figure:The Crystal Palace was a glass and cast iron structure built in London,

England, for the Great Exhibition of 1851. The building was designed by Sir

Joseph Paxton, an architect and gardener, and revealed breakthroughs in

architecture, construction and design.

Figure: Joseph Paxton was an English

gardener and architect who is chiefly

remembered for designing one of most

famous buildings of Victoria's reign

GATE 2013

2013-45

Central Park, New York

Figure: A timeless gardening classic by Christopher Lloyd, one of Britain's most highly respected plantsmen,

updated for the 21st century. With a new foreword by Anna Pavord.

GATE 2012

2012-19

Answer(B) Antarala is a small antechamber or foyer between

the garbhagriha (shrine) and the mandapa, more typical of north Indian temples.

Amalaka: a stone disk,

usually with ridges on the

rim, that sits atop the temple's

main tower. According to

one interpretation, the

amalaka represents a lotus,

and thus the symbolic seat for the deity below. Another interpretation is that it symbolizes the sun, and is thus the

gateway to the heavenly world. The amalaka itself is crowned with a kalasha (finial), from which a temple banner

is often hung.

Entrance Porch (Ardhamandapa): The entrance porch formed a transitional area between the outside world and

the mandapa or hall. Most temple buildings have some sort of transitional space between the central shrine

(garbhagrha) and the outside world, but only the largest, most developed temples will have all of these elements.

Hall (Mandapa): A hall in the temple, forming a transitional space between the ardhamandapa and

mahamandapa. In smaller or less architecturally developed temples, this was usually omitted.

Great Hall (Mahamandapa): The temple's main entrance-hall, separated from the central shrine (garbhagrha), by

a short vestibule named the antarala. Just about every temple has some sort of entrance-hall between the central

shrine (garbhagrha) and the outside world, but only the largest and most developed temples have all of the

transitional members. At Khajuraho, a mahamandapa is often distinguished by transepts (bumped-out portions

perpendicular to the temple's main axis).

Vestibule (Antarala): a transitional space between a temple's main hall and the inner sanctum (garbhagrha) where

the image of the temple's primary deity would be housed. The antarala was found only in the largest temples, and

in many smaller ones was omitted entirely. This architectural element marks the liminal space between the exterior

world and the divine world, and at Khajuraho the exterior panels on these elements are the primary sites for large

panels with sexually explicit scenes (particularly on the Vishvanath and Kandariya Mahadev temples). This

http://bit.ly/29ETZaF

More on Temple

Architecture.

Figure: The Khajuraho temples are a pinnacle of the North Indian Nagara

architectural style.

http://en.wikipedia.org/wiki/Garbhagriha

http://en.wikipedia.org/wiki/Mandapa

http://bit.ly/29ETZaF

http://bit.ly/29esOi0

GATE AR Numericals 2006-1997

N5

Room Index (K) value

A Room Index value is a number that describes the ratios of the rooms length, width and height. These

dimensions directly effect the performance of a specific luminaire, so determining your Room Index value is

very important when considering luminaires for your room. The Room Index (K) value is determined by this

formula;

K = [L x W] / [[L + W] x Hm]

where:

K = Room Index Value

L = Room Length in Metres

W = Room Width in Metres

Hm = Distance between Mounting Height of luminaire and the Working Plane in Metres

NOTE #1: As a simple rule of thumb, a space with a K value greater than 3 is a very efficient space to light,

and a space with a K value less than three is considerably less efficient to light.

Q5. Two 15 cm cubes of concrete are cast in a Building Materials Testing Laboratory, one of grade M15

and the other M20. The specified characteristic compressive strength of the stronger block at 28 days

will be

(A) 15 N/mm2

(B) 20 N/mm2

(C) 1.5 N/mm2

(D) 0.88 N/mm2

Answer: (A) 15 N/mm2

Notes:

Letter M represents mix (concrete mix) and the number followed after M represents the compressive strength to

be attained in N/sq.mm at 28 days age when a standard cube of 150x150x150 mm is subjected to standard

compressive test. Both combinedly termed as Grade of Concrete.

Thus, a concrete of M30 grade should attain a compressive strength of 30 N/sq.mm at 28 days age.

Units/SI

N= Newton, for easy understanding 1 N= 100g considering gravity

mm= millimeter

Figure: Concrete Test Cubes


N6

Q6. For a two way RCC slab, the Length to Width (L/W) ratio should be

(A) 0.5 <_ L/W <_ 2.0

(B) 2.0 <_ L/W <_ 3.0

(C) 2.5 <_ L/W <_ 4.0

(D) 0.0 <_ L/W <_ 0.5

Answer: (A) 0.5 <_ L/W <_ 2.0

Notes:

One-way slabs carrying predominantly uniform load are designed on the assumption that they consist of a

series of rectangular beams 1 m wide spanning between supporting beams or walls.

Above figure explains the share of loads on beams in two perpendicular directions depends upon the aspect

ratio ly /lx of the slab, lx being the shorter span. For large values of ly, the triangular area is much less than the

trapezoidal area. Hence, the share of loads on beams along shorter span will gradually reduce with increasing

ratio of ly /lx. In such cases, it may be said that the loads are primarily taken by beams along longer span. The

deflection profiles of the slab along both directions are also shown in the Figure. The deflection profile is found

to be constant along the longer span except near the edges for the slab panel. These slabs are designated as one-

way slabs as they span in one direction (shorter one) only for a large part of the slab when ly /lx > 2.

Two-way slabs which are subjected mostly to uniformly distributed loads resist them primarily by bending

about both the axis shown below and in this aspect ratio should be less than 2 ,it is designed by method of

coefficients method given is IS 456-2000..


N7

Slab section in one way and two way slab:

Q7. The distance between two points on a map of scale 1:40,000 is 3.6 cm. The distance between the same

two points in an aerial photograph is 6 cm. The scale of the aerial photograph is

(A) 1: 12000

(B) 1: 2400

(C) 1: 240000

(D) 1: 24000

Answer: (D) 1: 24000

Solution:

Let, The distance between two points on the ground = x

And the scale of the aerial photograph = y



N8

3.6/x = 1/ 40,000................................................(a)

=> x = 3.6*40,000

6/x = 1/y ............................................................(b)

Solving equation (a) & (b)

y = x/6 = 3.6*40,000/6 = 24,000 Answer

Q8. A camera is used to shoot an aerial photograph from a flight. The scale of the photograph is 1:40000.

If the flying height above mean sea level is 7500 m, and the mean ground level is 1500 m, then the focal

length of the camera lens is

(A) 70 mm

(B) 150 mm

(C) 200 mm

(D) 50 mm

Answer: (B) 150 mm

Solution:

Properties of triangle:

1:40000 = f:6000

f = 6000/40000 = 6/40m = 0.15 m =

150mm

Q 9. In a multipurpose hall, 30 tube lights of 40 W each are switched on for 5 hours. The electric meter

will record a power consumption of

(A) 6 units

(B) 12 units

(C) 30 units

(D) 60 units

Answer: (A) 6 units

Solution:

1 unit = 1000Watt-hout ( 1000W cooler can operate for 1 hour)

Total wattage of tube light = 30 x 40 W = 1200 Watt

So total Watt-Hour = 1200 Watt x 5 Hours = 6000 Watt-Hour = 6kWH = 6 units. Answer.

Common Data Questions

A town with an area of 340 hectares has 15,000 households. The number of occupied dwelling units is 12,400

of which 15% are in dilapidated condition. 5% of the households are below poverty line and unable to afford

any type of dwelling. (DU = Dwelling Unit)

Q10. The present density of the town (in DU/acre) is approximately

(A) 15

(B) 18

(C) 37

(D) 44


N9

Answer: (A) 15

Solution:

100m x 100m = 1 hectare

1 acre = 4046.86 sq.m.

So, present Density = Dwelling units/Area = 12400 units/340 hectares = 12400 units/340x100x100 square

meter = 12400 dwelling units/ (3400000/4046.86 Acres) = 14.76 DU/acre = 15 DU/Acre Answer.

Q11. The housing need of the town is approximately

(A) 1860DU

(B) 2600 DU

(C) 3320 DU

(D) 4460 DU

Answer: (D) 4460 DU

Q12. The housing demand of the town is approximately

(A) 1240 DU

(B) 1860 DU

(C) 2470 DU

(D) 3710 DU

Answer: (D) 3710 DU

Notes:

Although there is no set definition of housing „need‟ and „demand‟ they can broadly be described as

follows. Housing „demand‟ is a market driven concept and relates to the type and number of houses that

households will choose to occupy based on preference and ability to pay. Housing „need‟ is an indicator of

existing deficit: the number of households that do not have access to accommodation that meets certain

normative standards. This measure mainly refers to the level of need for more or improved social housing.

The term „housing requirement‟ is sometimes used to combine the two measures to generate an overall picture

of the housing market.

„need‟ refers to “shortfalls from certain normative standards of adequate accommodation” and „demand‟ refers

to “the quantity and quality of housing which households will choose to occupy given their preferences and

ability to pay (at given prices)”.The term „housing requirement‟ is sometimes used to combine the two

concepts of need and demand when referring to the overall housing market as opposed to social

housing on its own–where „need‟ is the primary consideration.

Common Data Questions :

The population of an urban area for the period 1971- 2001 is given in the folowing table:

Year Population in

‘000s

Increase in population in

‘000s

% increase in

Population

1971 80

1981 95 15 18.75

1991 115 20 21.05

2001 140 25 21.74


N10

Q13. The projected population (in thousands) for 2021 using Arithmetic Increase method is

(A) 160

(B) 180

(C) 220

(D) 240

Answer: (B) 180

Solution:

As increase in population is 15, 20, 25 (in „000), we will take average of three i.e. (15+20+25)/3 = 20.

So, population in 2011 will be 140+20 = 160

And population in 2021 will be 160+20 = 180 Answer.

Q14. The projected population ( in thousands) for 2021 using Geometric Increase method is

(A) 168.66

(B) 180

(C) 203.18

(D) 253

Answer: (C) 203.18

Solution:

% increase in population given is 18.75, 21.05 and 21.74. We will take average. (18.75+21.05+21.74)/3 = 20.5

So, population in 2011 will be 140x1.205 = 168.6

And population in 2021 will be 140x1.205x1.205 = 203.18 Answer

More about:

If a sequence of values follows a pattern of adding a fixed amount from one term to the next, it is referred to

as an arithmetic sequence. The number added to each term is constant (always the same).

The fixed amount is called the common difference, d.

To find any term of an arithmetic sequence:

where a1 is the first term of the sequence,

d is the common difference, n is the number of the term to find.

To find the sum of a certain number of terms of an arithmetic sequence:

where Sn is the sum of n terms (n

th partial sum),

a1 is the first term, an is the nth term.

Examples:

Question Answer

1. Find the common difference for this arithmetic

sequence

5, 9, 13, 17 ...

1. The common difference, d, can be found by

subtracting the first term from the second term, which

in this problem yields 4. Checking shows that 4 is the

difference between all of the entries.

2. Find the common difference for the arithmetic 2. The formula indicates that 6 is the value being


N11

sequence whose formula is

an = 6n + 3

added (with increasing multiples) as the terms

increase. A listing of the terms will also show what is

happening in the sequence (start with n = 1).

9, 15, 21, 27, 33, ...

The list shows the common difference to be 6.

3. Find the 10th term of the sequence

3, 5, 7, 9, ...

3. n = 10; a1 = 3, d = 2

The tenth term is 21.

Linked Answer Questions:

Statement for Linked Answer Question:

While testing the strength of a steel beam, it is found that

Poisson‟s ratio for steel = 0.3 and Young‟s modulus of Elasticity = 2.1 x 106

Q15. If the lateral strain of the bean is 1 unit, the longitudinal strain will be

(A) 7 x 106 units

(B) 0.3 units

(C) 2.1 x 106 units

(D) 3.33 units

Answer: (D) 3.33 units

Solution:

Poisson‟s ratio= lateral strain / longitudinal strain = 0.3 (given)

0.3 = 1 / longitudinal strain

=> longitudinal strain = 3.33 Answer

Q16. The longitudinal stress for the same steel will be approximately

(A) 7 x 106 units

(B) 0.3 units

(C) 2.1 x 106 units

(D) 3.33 units


N42

(D) 3 days

Q5. In a seminar room of area 200 sq.m., 4 m. Height and total absorbing power of 120 m2 sabines, what

is the reverberation time?

(A) 0.24 secs.

(B) 1.06 secs.

(C) 1.52 secs

(D) 4.16 secs

Answer: (B) 1.06 secs.

Solution,

Volume of the room = 200sq.m.*4m = 800cu.m.

reverberation time = 0.16*(V/A) = 0.16*(800/120) = 1.06 Seconds Answer

Q6. A town has a basic employment of 25,000 workers. If the basic: non-basic ratio is 1:2.5 and the

workers dependency ratio is 1:4, what is the population size of the town?

(A) 2,50,000

(B) 4,37,500

(C) 3,50,000

(D) 3,12,500

Answer: (B) 4,37,500

Solution:

Basic employment population = 25000

Non-basic employment popuation = 2.5*25000 = 62500

Total employment population = 25000 + 62500 = 87500

Dependent population = 4*87500 = 350000

So, town population is 350000+87500 = 437500 Answer

Q7. What is the rate of ventilation due to wind action if the free area of the window is 1 sq.m., and the

wind speed is 1 m/hr. Assume the wind to be perpendicular to the window.

(A) 1.0 cu.m./hour

(B) 0.6 cu.m./hour

(C) 0.3 cu.m./hour

(D) 0.1 cu.m./hour

Answer: (A) 1.0 cu.m./hour

Solution:

Volume of air intake by window in 1 hour = Area of window x wind speed = 1sq.m. * 1m/hr = 1 cu.m/hr

Q8. The diagram below shows the relative distribution of different types of housing within a total

residential area of 150 hectares. If the net density of the plotted housing area is 350 ppha., how many

people will be accommodated there?


N43

(A) 20,000-25,000

(B) 25,000-30,000

(C) 30,000-35,000

(D) 35,000-40,000

Answer: (C) 30,000-35,000

Solution:

Percentage division of plots:

Builders‟ group housing = 25%

Cooperative Society Housing = 12.5%

Plotted Housing = 62.5%

So, Area under plotted housing = 0.625*150 ha = 93.75 ha

So, population under plotted housing = 93.75ha*350ppha = 32812 Answer.

Q9. A residential plot of 20 metre frontage and 25 metre depth is governed by the development

regulations of maximum FAR of 200 and maximum plot coverage of 50%. Up to what maximum height

can the plot be built?

(A) 2 floors

(B) 3 floors

(C) 4 floors

(D) 10 floors

Answer: (C) 4 floors

Solution:

Area of the plot = 20m x 25m = 500 sq.m.

Maximum builtup area = FAR*Plot Area = 2*500 = 1000

Coverage area of the building at ground floor = 0.5*Plot Area = 0.5*500 = 250 sq.m.

So, no. of floors = Maximum built-up area / ground floor coverage = 1000/250 = 4 Answer.

Q10. A rectangular room (internal dimension 5 m x 3 m) is made of 250 mm walls. Calculate the volume

of concrete needed for 25 mm Damp Proof Course.

(A) 0.425 cu.m.

(B) 4.25 cu.m.

(C) 0.4 cu.m.

(D) 4.0 cu.m.


N51

Solution:

We just can‟t add 92 dBA and 88 dBA because they are Log derivations. But we can add their intensity of

sound.

Let intensity of sound of source 1 be I1

And intensity of sound of source 2be I2

So,

92 = 10log(i1/Io) => I1/Io = 109.2

.................................................(A)

88 = 10log(I2/Io) => I2/Io = 108.8

.................................................(B)

Adding (A) & (B),

(I1/Io) + (I2/Io) = 109.2

+ 108.8

Taking log both side,

Log[(I1/Io) + (I2/Io)] = log[109.2

+ 108.8

]

Log[(I1+I2)/Io] = log[2215850537]

Multiplying by 10 both side,

10*Log[(I1+I2)/Io] =10* log[2215850537]

10*Log[(I1+I2)/Io] =10* 9.34

10*Log[(I1+I2)/Io] = 93.4 Answer

Q4. The velocity head of water supply line is measured in terms of

(A) m/sec

(B) m/sec2

(C) m

(D) m2/sec

Answer: (C) m

Notes:

Bernoulli’s theorem is one of the most used equations in fluid engineering. The theorem expresses

conservation of energy in a flow system by relating velocity, pressure and elevation.

The term ½(v2/g) commonly is referred to as the velocity head and has units of height (metre) of the flowing

fluid. It can be thought of as the amount of potential energy required to accelerate a fluid to its actual flowing

velocity. It also is the amount of head generated when fluid velocity drops to zero.

Q5. A loan of Rs. 8,00,000 has been granted by a financial institution to an individual for the

construction of his house. The loan has to be repaid by way of annuity at the rate of 13.5 % interest per

annum in 15 equal instalments from the year in which the loan is taken by the house owner. Calculate

the yearly instalment for repaying the loan.

Answer: Rs. 1,09,605

Solution:

The general formula for the future value:

S = P (1+i)n

Where:

S = the future value after n years

P = principle amount

i = rate of interest

n = number of years


N52

Future value of an annuity is given by the following formula

Sn = R { ( [1+i] n

– 1 ) / i }

Where:

Sn = future value of an annuity which has duration of n years

R = constant periodic flow [annual instalment]

To calculate the yearly instalment, the future value of annuity has to be equated to the general formula

for the future value

R { ( [1+i] n

– 1 ) / i } = P (1+i)n

But for this given problem, we just need to calculate „Sn‟ with given R = Rs. 800000, n = 15 years and i =

13.5%.

So,

Sn = R { ( [1+i] n

– 1 ) / i }

= R { ( [1+13.5 / 100] 15

– 1 ) / 13.5 / 100 } = 8,00,000 (1+13.5 / 100)15

= 1,09,605.16

The yearly instalment is Rs. 1,09,605

More about annuity:

What Are Annuities?

Annuities are essentially a series of fixed payments required from you or paid to you at a specified frequency

over the course of a fixed time period.

Since the present and future value calculations for ordinary annuities and annuities due are slightly different,

we will first discuss the present and future value calculation for ordinary annuities.

Calculating the Future Value of an Ordinary Annuity

If you know how much you can invest per period for a certain time period, the future value of an ordinary

annuity formula is useful for finding out how much you would have in the future by investing at your given

interest rate. If you are making payments on a loan, the future value is useful in determining the total cost of the

loan.

Let's now run through Example 1. Consider the following annuity cash flow schedule:

To calculate the future value of the annuity, we have to calculate the future value of each cash flow. Let's

assume that you are receiving Rs.1,000 every year for the next five years, and you invested each payment at

5%. The following diagram shows how much you would have at the end of the five-year period:


N53

Since we have to add the future value of each payment, you may have noticed that if you have an ordinary

annuity with many cash flows, it would take a long time to calculate all the future values and then add them

together. Fortunately, mathematics provides a formula that serves as a shortcut for finding the accumulated

value of all cash flows received from an ordinary annuity:

where C = Cash flow per period

i = interest rate

n = number of payments

Using the above formula for Example 1 above, this is the result:

= 1000*[5.53]=

Rs. 5525.63

Calculating the Present Value of an Ordinary Annuity

If you would like to determine today's value of a future payment series, you need to use the formula that

calculates the present value of an ordinary annuity. This is the formula you would use as part of a bond pricing

calculation. The PV of an ordinary annuity calculates the present value of the coupon payments that you will

receive in the future.

For Example 2, we'll use the same annuity cash flow schedule as we did in Example 1. To obtain the total

discounted value, we need to take the present value of each future payment and, as we did in Example 1, add

the cash flows together.

Again, calculating and adding all these values will take a considerable amount of time, especially if we expect

many future payments. As such, we can use a mathematical shortcut for PV of an ordinary annuity.

where C = Cash flow per period

i = interest rate

n = number of payments

The formula provides us with the PV in a few easy steps. Here is the calculation of the annuity represented in

the diagram for Example 2:

= Rs. 1000*[4.33]= Rs. 4329.48


N54

http://bit.ly/29uMsei

Q6. An activity in a CPM network has a duration of 4 days. The free float for the activity is 10 days and

the total float is also 10 days. Find the maximum delay that can be allowed for the activity from the

occurrence of the preceding event.

Q.7 Workout the net and the gross population densities for a neighbourhood, given the following:

a. Ground coverage = 30%

b. FAR = 1.5

c. Plinth area per residential block = 372 sqm

d. Number of dwelling units per floor in each block = 2

e. Average family size = 4.5

f. Area under residential plots = 62 %

g. Area under access roads = 2%

h. Area under others including major roads = 36%

Solution:

Plot area = 372sq.m./30% = 372/0.3 = 1240 sq.m.

Total builtup area = 1.5*1240 = 1860

No. of floors = 1860 / 372 = 5

No. of dwelling units = 2*5 = 10

Population = 4.5*10 = 45

Net residential area = 62% of 1240 sq.m. = 768.8 sq.m.

So, Net Density = 45/768.8 = 0.0585 people/sq.m. = 585 people/hectare

And. Gross Density = 45 / 1240 = 0.0363 people/sq.m. = 363 people / hectare

Q8. A primary road within a city has to bend along a horizontal curve having a radius of 150m. What

should be the design speed of the road at that section if the maximum super-elevation of 0.07 is not to be

exceeded and the safe limit of transverse coefficient of friction is 0.15?

Solution:

R = V2 / 127 * (e + f)

where

http://bit.ly/29uMsei


N56

1998

Q1. Cement (in cu.m) required for preparing 10 cu.m.of cement concrete in the proportion of 1:2:4 is

(A) 0.80

(B) 1.00

(C) 1.20

(D) 1.40

Answer: (D) 1.40

Solution:

1:2:4 = cement : sand : aggregate

Volume of cement = 10*(1/7) = 1.42 Answer.

Because dry mix volume decrease when we add the water so we take more volume so increase the volume

1 C.M. to 1.52 C.M.(1+0.52)

Now for quantity make total of proportion

1+1+2 = 4 total volume if material.

For Cement

1 * (1.52 / 4 ) = 0.38 meter cube

For sand

1 * (1.52 / 4 ) = 0.38 meter cube

For Aggregate

2 * (1.52 / 4 ) = 0.76 meter cube

So you get the quantity of all materials for 1 meter cube.

Cement: sand : aggregate

M 5 : 1:4:8

M10 : 1:3:6

M15 : 1:2:4

M20 : 1:1.5:3

M25 : 1:1:2

Q2. Permissible stress in bending compression (N/mm2) for M20 grade concrete is

(A) 3.0

(B) 5.0

(C) 7.0

(D) 8.5

Answer: (C) 7.0

Notes:

Click/scan here to

practice more:

Concrete Mix

http://bit.ly/29lcAam

http://bit.ly/29lcAam


N58

Solution:

For Arithmetic rate:

tn = S1 + (n-1) d

Sn = 3000000 + (3-1)*500000 = 40 lakhs

For Geometric rate:

tn=arn−1

tn = 3000000*(35/30)3 = 4763888

Q7. An aircraft flying at an altitude of 5000m above mean sea level takes aerial photographs of a terrain

having an average elevation of 1000m above mean sea level.:

A. Find the scale of photograph if focal length of camera is 20 cm.

B. Find the area covered in ground by each photo format of 23 cm x 23 cm.

Let scale of the photograph = x

X = 0.2m / 4000m = 1 / 20000 = 1:20000

Area covered in ground by each photo format of 23 cm x 23 cm

= (20000*23 cm x 20000*23 cm)

= (200*23m x 200*23m)

=(4600m x 4600m)

= 21160000 sq.m.

= (21160000 sq.m.) / (100*100) ha

= 2116 ha Answer

Scan for more on AP,

GP & HP

or visit:

http://bit.ly/29n02Nx

http://bit.ly/29n02Nx

************************* END OF PREVIEW *************************

visit www.gatearchitecture.com for more information.

Documents

Question Bank - GATE ARCHITECTURE...Out of 20 years Question Bank, 10 years (2016-2007) are dealt with all questions and next 10 years (2006-1997), only numerical questions are provided