Question Paper GATE 2016
The GATE 2016 AR was a bit unpredictable. Questions were slightly off the patterns this year. 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’. Cutoff marks 38.90
Part 1 Aptitude
Q1. An apple costs Rs. 10. An onion costs Rs. 8. (1mark)
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 10 s and 15 s, 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.
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.”Cluching is the best suited word. (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
Q5. M has a son Q and a daughter R. He has no other children. E is the mother of P and daughterinlaw of M. How is P related to M? (1 mark)
(A) P is the soninlaw of M. (B) P is the grandchild of M.
(C) P is the daughterin 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 yintercept
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 xintercept. So, now we have to find ‘c’ the yintercept.
for, y=0, lnx = 0.1 (given in the question)
Putting the value,
0 = 0.02×0.1 + 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 = 0.002×1.6 + 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, 2x3y=5 in the first quadrant. (2 marks)
(A) 14.95 (B) 15.25 (C) 15.70 (D) 20.35
Answer: (C) 15.25
Let orange line be, 3x+2y=14
and blue line be, 2x3y=5
Solving both equation, we get, (x=4 , y=1) the intersection points of the lines.
To get the required area in the first quadrant, we have to deduct the area of the triangle formed by the orange line, blue line & the xaxis from the area bounded by the orange line in the first quadrant.
Area of the triangle = 1/2 x base x height = 1/2 x B x A = 1/2 x 13/6 x 1 = 13/12 ………….(P)
Area bounded by orange line and X & Y axis = (1/2) x (xintercept) x (yintercept) = 1/2×14/2×7 = 98/6 …………….(Q)
So, required area = (Q) – (P) = 98/6 – 13/12 = 183/12 = 15.25
Q9. A flat is shared by four first year undergraduate students. They agreed to allow the oldest of them to enjoy some extra space in the flat. Manu is two months older than Sravan, who is three months younger than Trideep. Pavan is one month older than Sravan. Who should occupy the extra space in the flat. (2 marks)
(A) Manu (B) Sarvan (C) Trideep (D) Pavan
Answer: (C) Trideep
Q10. The velocity V of a vehicle along a straight line is measured in m/s and plotted as shown with respect to time in seconds. At the end of the 7 seconds, how much will the odometer reading increase by (in m)? (2 marks)
(A) 0 (B) 3 (C) 4 (D) 5
The odometer will read 5. (The area shaded under the graph). Odometer is an instrument for measuring the distance travelled by a wheeled vehicle. So, you have to just count the no. of squares made by triangles of the graph.
Part 2 – Architecture
Q1. The unit of Thermal Conductivity of a material is (1 mark)
(A) W/(m K) (B) W/(m2 K)
(C) m K/W (D) m2 K/W
Answer: (A ) W/(m K) (Watt per meter per Degree Kelvin)
Possible Intent: The examiner may need to know you the difference between Thermal Conductivity or Thermal Conductance.
kvalue – Thermal Conductivity
Thermal conductivity is the time rate of steady state heat flow through a unit area of a homogeneous material induced by a unit temperature gradient in a direction perpendicular to that unit area, W/m⋅K.
k = q(L/T)
Where,
L – Thickness of the specimen (m)
T – Temperature (K)
q – Heat flow rate (W/m2)
Cvalue – Thermal Conductance
Thermal conductance is the time rate of steady state heat flow through a unit area of a material or construction induced by a unit temperature difference between the body surfaces, in W/m2⋅K. Cvalue, hence, is the reciprocal of the Rvalue and can be expressed as Equation .
Consequently, the value of the thermal conductance can be calculated by dividing the thermal conductivity with the thickness of the specimen.
Q2. In a flat slab, the load is transferred through:
(A) Beam Action (B) Membrane Action (C) Plate Action (D) Vector Action
Answer: (C)
Generally, the term flat slab means a reinforced concrete slab with or without drops, supported generally without beams, by columns with or without flared column heads.
In flat slabs, the loads on the slab are directly transferred by plate action to the columns, whereas in conventional slabs with beams, the loads are first transferred from slabs to the beams and then from beams to the column.
Q3. Statutory setback of a building depends on (1 mark)
(A) Building height (B) Property boundary
(C) Width of access road (D) Ground coverage
Answer: (A)
Q4. As per Census of India 2011, Nagar Panchayat refers to (1 mark)
(A) Rural area surrounded by a growing city
(B) Urban area within a panchayat
(C) Rural area adjacent to a nagar
(D) Urban area with a statutory local government
Answer: (D)
Q5. ‘AMRUT’, the new scheme launched by Government of India, stands for (1 mark)
(A) Atal Mission for Rejuvenation and Urban Transformation
(B) Atal Mission for Renewal of Urban Transportation
(C) Atal Mission for Redevelopment of Urban Transportation
(D) Atal Mission for Renewal and Urban Transformation
Answer (A)
Notes: The purpose of Atal Mission for Rejuvenation and Urban Transformation (AMRUT) is to
(i) Ensure that every household has access to a tap with assured supply of water and a sewerage connection;
(ii) Increase the amenity value of cities by developing greenery and well maintained open spaces (parks); and
(iii) Reduce pollution by switching to public transport or constructing facilities for nonmotorized transport (e.g.
walking and cycling).
According to the initiative taken by the Government under Atal Mission for Rejuvenation and Urban Transformation
(AMRUT), the government is aiming to improve the existing basic infrastructure services like extending clean drinking
water supply, improving sewerage networks, laying of storm water drains, improving public transport services and
creating green public spaces like parks etc, with special focus on creating healthy open spaces for children.
The scheme Housing for All by 2022 and Atal Mission for Rejuvenation and Urban Transformation (AMRUT) were
launched on the same day June 2015.
Under this scheme the government is targeting 500 cities, these 500 cities are selected on the basis of population i.e. one
lakh and above, while the other criteria of selection will apply for certain locations like tourist popularity, cities located at
the stems of main rivers, certain popular hill towns and some select islands.
The initiative was announced by PM Narendra Modi on 25 June 2015, and is said to be aiming to transform 500 cities and
towns into efficient urban living spaces, with special focus on a healthy and green environment for children. it was also
reported that Cabinet has approved Rs 50,000 crore for this mission which is to be spent over the next five years.
– Here is the purpose under AMRUT
– Ensures basic infrastructure and sanitation.
– Focus on water supply and sewage.
– Promoting public transport.
– Enabling walking, cycling, greenery and open spaces.
– Improving Urban governance through reform.
– Capacity Building.
Q6. Spatial connectedness in GIS refers to (1 mark)
(A) Tomography (B) Topography (C) Topiary (D) Topology
Answer: (D)
Notes: Tomography imaging is the process where image slices of the object of interest are taken. These slices can then be reconstructed to produce a full image representing the original object. Computer Tomography (CT) is a widely used imaging service which has been in used for medical purposes.
Topography is the shape of Earth’s surface and its physical features, such as mountains, valleys, canyons, and other land forms.
Topiary is the art of ornamental gardening. Landscape designers grow outdoor topiary by trimming bushes and trees the way a sculptor works stone.
Remote Sensing is a general term for geodata collection at a distance from the subject area.
Photogrammetry is the science of making measurements from photographs, especially for recovering the exact positions of surface points.Moreover, it may be used to recover the motion pathways of designated reference points located on any moving object, on its components and in the immediately adjacent environment.
Q7. Lowemissivity coating on a glazing unit (1 mark)
(A) Increases the SHGC (B) Increases the VLT
(C) Reduces the SHGC (D) Increases the VLT and SHGC
Low emissivity coating on a curtain system leads to:
Answer: (C)
Visible Light Transmission (VLT)
The percentage of visible light that is transmitted through the glazing assembly. This is the essential characteristic for daylighting calculations. A perfectly clear window would have a VLT of 100 percent. Most practical assemblies for architectural use are between 35 and 80 percent.
Solar Heat Gain Coefficient (SHGC)
The percentage of total solar radiant energy that is transmitted through the assembly. This is the essential characteristic for solar gain calculations. For ordinary windows without special coatings, the SHGC and the VLT are the same and sometimes called the shading coefficient (SC). However, with modern coated windows, the SHGC is almost always lower than the VLT. Such window systems are generically referred to as lowemissivity or lowE and are used in most commercial construction.
Q8. The components measuring Human Development Index (HDI) of a nation are (1 mark)
(A) Life expectancy, Education and Per Capita Income
(B) Life expectancy, Economy and Housing
(C) Health, Development and Per Capita Income
(D) Housing, Health and Hygiene
Answer: (A) Life expectancy, Education & Gross national income per capita.
Q9. The concept of ‘Dry Garden’ is associated with
(A) Japanese Garden (B) Chinese Garden (C) Mughal Garden (D) Egyptian Garden
Answer: (A) Japanese garden
Elements
– A stone lantern representing four natural elements: earth, water, fire and wind
– Statues of male and female lions, placed at the entrance of the garden in order to protect the garden from
intruders, representing the two opposite forces: yin and yang (fire and water, male and female).
– Water basin known as a deer chaser, which keep deer away by making a special sound when filled up
– The koi fish swimming in ponds, which has a decorative meaning
– Typical Japanese bridge, called a moonbridge, whose purpose is to reflect artistic feelings.
Q10. Semi circular niche in a Mosque in the direction of Mecca is known as
(A) Sahn (B) Minbar (C) Qibla (D) Mihrab
Answer: (D) Mihrab
Q11. ‘Mushroom Column’ is a typical feature of (1 mark)
(A) Pantheon, Rome (B) Parthenon, Athens
(C) Chrysler Building, New York (D) Johnson Wax Building, Racine
Answer (D) Johnson Wax Building – Architect: Frank Lloyd Wright
Q12. Shodhan House, Ahmedabad was designed by (1 mark)
(A) Anant Raje (B) Le Corbusier
(C) Louis I. Kahn (D) B. V. Doshi
Answer: Le Corbusier
Q13. ‘Dharahara’ refers to (1 mark)
(A) Concept of Vastu Shastra
(B) Elevation feature of a Hindu Temple
(C) Heritage tower at Kathmandu
(D) Construction method of Buddhist Architecture
Answer: (C) Heritage tower at Kathmandu
Q14. As per CPWD Guidelines and Space Standards for Barrier Free Built Environment, the minimum
turning radius for a wheelchair in mm is (1 mark)
(A) 900 (B) 1200 (C) 1500 (D) 1800
Answer: (A)
Q15. In CMYK colour model, ‘K’ represents the colour (1 mark)
(A) White (B) Black (C) Blue (D) Green
Answer: Black
Q16. Super plasticizer is added in a concrete mix to (1 mark)
(A) Reduce the watercement ratio for a given workability
(B) Increase the watercement ratio for a given workability
(C) Reduce the flow
(D) Reduce the content of coarse aggregate
Answer: (A)
Q17. ‘Volume Zero’ is a film based on the architectural works of
(A) Peter Eisenman (B) Charles Correa
(C) Daniel Libeskind
Answer: (B) Charles Correa
Q18. The concept of Transit Oriented Development (TOD) refers to (1 mark)
(A) Transit oriented planning in institutional campuses
(B) High density development along a mass transit corridor
(C) Orienting residential blocks towards the transit in a gated community
(D) Spatial design for persons with disabilities along a mass transit corridor
Answer: (B)Transit–oriented development, or TOD, is an approach to development that focuses land uses around a transit station or within a transit corridor
Q19. Map is
(A) Isographic projection (B)Orthographic projection
(C) Axonometric Projection (D) Oblique projection
Answer: Orthographic projection
Q20. ‘Segregation at source’ is a concept associated with (1 mark)
(A) Solid waste management (B) Disaster management
(C) Traffic management (D) Water management
Answer:(A)
Q21. In a residential neighbourhood, the net area of residential plots is 50 percent of the total area. If the population is 8000, the ratio of net density to gross density of the neighbourhood is ________ (1 mark)
Answer: 2
Solution: Let, area of the plot be x.
Net density/ Gross density = (8000/0.5x)/(8000/x) = 2
Q22. As per the Ancient Monuments and Archaeological Sites and Remains Act, 2010, the extent of ‘prohibited area’ beyond the ‘demarcated protected area’ in meters is (1 mark)
(A) 50 (B) 100 (C) 200 (D) 500
Answer: 100m
Q23. ‘Indore Slum Networking Programme’ has been planned by (1 mark)
(A) B. V. Doshi (B) Dulal Mukherjee
(C) Anil Laul (D) Himanshu Parikh
Answer: (D) Himanshu Parikh
Q24. A hemispherical earth mound of 3 meter diameter is proposed to be constructed in a children’s park. If the proportion of soil and sand for the construction is 3:2, the estimated volume of soil in cubic meters is ________ (1 mark)
Answer: 42.43 CuM
Solution: Volume of sphere is 4/3(πr^{3})
Volume of hemisphere = ½ x4/3 (22/7×3/2×3/2×3/2) = 99/14
So, volume of mud = 3/5x(99/14) = 4.24 CuM
Q25. One litre of acrylic paint can cover 16 sqm of wall area for the first coat and 24 sqm for the second coat. The walls of a lecture hall measuring 12m × 8m × 4m (L × B × H) need to be painted with two coats of this paint. The hall has total glazed fenestration area of 12 sqm. The number of 4 litre
paint containers required will be __________ (2 marks)
Answer: Answer provided by the official GATE is 4
Q26. ‘Summit Curve’ is a term associated with the design of (1 mark)
(A) Roads and flyovers (B) Escalators and elevators
(C) Tensile fabric structures (D) Geodesic domes
Answer: (A) Roads and flyovers
Q27. The estimated number of bricks (unit size: 250 mm × 125 mm × 75 mm) for laying one course of a 250 mm thick brick wall using rattrap bond for a running length of 3.9 meter will be ___________ (2 marks)
Answer: 36
Q28. In 2001, the population and work force participation rate of a town were 30,000 and 30 percent respectively. The work force participation rate in the year 2011 increased to 34 percent. If the decadal population growth rate was 6 percent, the increase in the number of working people in the town in 2011 was __________ (2 marks)
Answer: 1812
Solution: Working population in 2001 = 9000
In 2011, total population = 30,000 + 6% of 30,000 = 31,800
Now, 34% of 31,800 is the work force = 10812
So, increase in work force = 10812 – 9000 = 1812 Ans.
Q29. In a 20 storey building with 3m floor to floor height, a passenger lift is hoisted by a steel rope.
Weight of the lift car is 750 kg and ultimate load the steel rope can carry is 39,000 kg. Assuming a
factor of safety of 20 for the steel rope and an average passenger weight of 75 kg, the passenger
capacity of the lift is ____________
Answer: 16
Solution: Usable strength of steel rope = 39000/20 = 1950 kg
Stength of steel rope considering liftcar weight = 1950 – 750 = 1200kg
Capacity of lift = 1200/75 = 16 person
Q30. A room is mechanically ventilated through four airconditioning ducts. The opening area of each duct is 0.35 sqm. The air velocity in the duct is 0.5 m/s. The temperature difference between the ambient air and supply air is 10 °C. Volumetric specific heat of air is 1250 J/m3 °C. Assuming one Ton of refrigeration (TR) equals 3.5 kW, the cooling load of the room in TR will be_____________ (2 marks)
Answer: 2.5
Solution: Total area of the duct = 4 x 0.35 = 1.4 sqm
Total air volume in the room collected in 1 hr (3600 sec) = 1.4 x 0.5 x 3600 = 2520 cum
Total energy required = ms∆t = 2520 x 1250 x 10 = 31500000 Joule
Load in TR = (31500000/3.5KW)/3600 = (31500000/3500)/3600 = 2.5
Q31. Match Group I with Group II
GROUP I  GROUP II  
1.  P  Nadyavarta
(Square or Rectangle) 

2.  Q  Swastika  
3.  R  Padmaka


4.  S  Chaturmuka  
5.  T  Karmuka 
Answer: 1P, 2Q, 3R, 4S
Q32. Match Group I with Group II
GROUP I  GROUP II  
1.  Villa Müller, Prague  P  Adolf Loos 
2.  Farnsworth, Illions  Q  Mies van der Rohe 
3.  Schröder House, Utrecht  R  Gerrit Rietveld 
4.  Dancing House, Prague  S  Frank Gehry 
Answer: 1P, 2Q, 3R, 4S
Villa Müller, Prague Known as an innovative landmark of early modernist architecture, the
Villa Müller embodies Loos’ ideas of economy and functionality. The spatial design, known as
Raumplan, is evident in the multilevel parts of individual rooms, indicating their function and symbolic importance. Raumplan is exhibited in the interior as well as the exterior.
Villa Müller, Prague by Adolf Loos
Q33. Find the reverberation time of room 4m x 3m x 3m (LBH) with the help of following data.
Wall  Ceiling  Floor  
Percentage area  30  70  40  60  100 
Absorption coefficient  0.4  0.1  0.6  0.1  0.1 
Answer: 0.37 (0.44 to 0.46 answer by official GATE)
Solution: t = 0.16x [(V/A)] = 0.16 [(4x3x3)/15.44] = 0.37 Sec
Calculation of ‘A’
Wall area = perimeter of wall x height = [2(4+3)] x [4] = 56 sqm = P
Ceiling area = 4×3 = 12 sqm = Q
Floor area = 4×3 = 12 sqm R
So, A = P[(0.4 x 30%) + (0.1 x 70%)] + Q[(0.6 x 40%) + (0.1 x 60%)] + R (0.1 x 100%)
= P[(0.4 x 0.3) + (0.1 x 0.7)] + Q[(0.6 x 0.4) + (0.1 x 0.6)] + R (0.1 x 1)
= 56[(0.4 x 0.3) + (0.1 x 0.7)] + 12[(0.6 x 0.4) + (0.1 x 0.6)] + 12 (0.1 x 1)
= 56(0.12 + 0.07) + 12(0.24 + 0.06) + 12 x 0.1
= 56 x 0.19 + 12 x 0.3 + 1.2
= 10.64 + 3.6 + 1.2
= 15.44
Q34. Match Group I with Group II
GROUP I  GROUP II  
1.  MohenjoDaro  P  Grid Iron Street Pattern 
2.  Babylon  Q  Hanging Garden 
3.  Kahun  R  Enclose court 
4.  Athens  S  Agora 
T  Forum 
Answer: 1P, 2Q, 3R, 4S
Q35. A fourstorey building with equal areas in each floor is required to be designed on a plot with FAR of 2.0. If the FAR is increased to 2.2, the percentage increase in ground coverage utilizing full FAR in both cases will be ___________ (2 marks)
Answer: 10
Solution: Let site area be 100 unit. FAR = 2 means built up area = 2×100 = 200 unit. As each floor has equal area, so each floor area = 200/4 = 50 unit. So ground coverage = 50 sqm.
In the second case, site area = 100 unit. Built up area = 220 as FAR=2.2
Ground coverage = 220/4 = 55
So ground coverage is increased from 50 to 55. Therefore, change in % = 10%
Q36. Match Group I with Group II
GROUP I  GROUP II  
1.  Palazzetto dello Sport, Rome  P  PL Nervi 
2.  Olympic Arena, Tokyo  Q  Kenzo Tange 
3.  Bird Nest, Beijing  R  Herzog & de Meuron 
4.  Olympic Stadia, Munich  S  Frie Otto 
T  Rozer Tailbr 
Answer: 1P, 2Q, 3R, 4S
Q37. The difference in invert levels between two pits separated by a distance of 30 meter is one meter. An intermediate pit is required to be constructed at a distance of 18 meter from the pit at higher level. Maintaining the same slope, the difference in invert levels of the new pit and the pit at lower level in mm will be _________ (2 marks)
Answer: 400
Solution: 30/1= 12/x => x = 12/30 = 0.4 m =400mm
Q38. Match Group I with Group II
GROUP I  GROUP II  
1.  Coagulation  P  Alum 
2.  Filtration  Q  Sand 
3.  Disinfection  R  Chlorine 
4.  Softening  S  Sodium Zeolite 
T  Ammonium Hydroxide 
Answer: 1P, 2Q, 3R, 4S
Q39. Match Group I with Group II
GROUP I  GROUP II  
1.  Acquisition  P  Ownership 
2.  Planning permission  Q  Land use 
3.  Building plan sanction  R  Construction Land cover 
4.  Mutation  S  Land 
T  Land cover 
Answer: 1P, 2Q, 3R, 4S
Q40. As per URDPFI guidelines, the number of Equivalent Car Space (ECS) required to accommodate ten cars, sixteen scooters and forty bicycles will be______________ (2 marks)
Answer: 21.2 (18 by official GATE answer)
Solution: ECS = 1 x10 + 0.2 x 16 + 0.2 x 40 = 21.2
Q41. Match Group I with Group II
GROUP I  GROUP II  
1.  Parliamentary  P  Hinge 
2.  Aldrop  Q  Bolt 
3.  Mortise  R  Lock 
4.  Panic  S  Bar 
Answer: 1P, 2Q, 3R, 4S
Q42. Hoop & Meridional force is found in (2 marks)
(A) Dome (B) Truss (C) Folded Plate (D) Space Frame
Answer: Dome
Q43. A solar photovoltaic system is proposed to be installed at the roof top of a hostel. The cost of installation and the annual maintenance are INR 2,40,000 and INR 6000 respectively. It is expected to generate 600 kWh of electricity per month. Assume unit price of electricity as INR 5. Ignoring the discount rate, the payback period of the investment in years is____________ (2 marks)
Answer: 8
Solution: Let the payback time be ‘P’ year.
So, 240000 + 6000P = 600x5x12xP
⇒ 240000 +6000P = 36000P
⇒ 240000 = 30000P
⇒ P = 8 Years
Q44. In a housing project, the number of LIG, MIG and HIG units are in the ratio of 1:1:2. The ratio of areas of the units is 3:5:8. Assume unit cost of construction is same for all the three types. For a no profit no loss situation, if 10% discount is offered to LIG and MIG units on sale price, extra charge in percentage payable per HIG unit will be ____________ (2 marks)
Answer: 5
Solution: Let,
No. of LIG = 1
No. of MIG = 1
No. of HIG = 2
Total area of LIG = 1×3 = 3 unit
Total area of MIG = 1×5 = 5 unit
Total area of HIG = 2×8 = 16 unit
Also assume,
Cost of construction for all LIG = 3unit
Cost of construction for all MIG = 5unit
Cost of construction for all HIG = 16 unit
As 10% is on LIG, so 10% 3 unit = 0.3 would be burden on HIG
As 10% is on MIG, so 10% 5 unit = 0.5 would be burden on HIG
Total 0.3 + 0.5 = 0.8 would be burden on HIG cost which totals 16 unit
So, % change in cost = (0.8/16)x100% = 5%
Q45. Match Group I with Group II
GROUP I  GROUP II  
1.  CASBEE  P  Japan 
2.  Greenmark  Q  Singapore 
3.  GRIHA  R  India 
4.  Estidama  S  UAE 
T  China 
Answer: 1P, 2Q, 3R, 4S
Q46. A pump is installed in an apartment building to lift water from ground level to the roof top water tank with the capacity of 10,000 litres. Total head of lift is 18 m and pumping time is 30 minutes to fill the tank completely. Assuming acceleration due to gravity (g) as 10 m/sec2 and efficiency of the pump as 80 percent, the power requirement of the pump in kW will be __________ (2 marks)
Answer: 1.25
Solution: Power = Energy/ Time = Work Done/ Time = Mgh/30min = 10000x10x18/1800 sec = 1000
As the pump is woking at 80% efficiency, power = 10/8 (1000) =1250 Watt =1.25 Watt
Q47. Match Group I with Group II
GROUP I  GROUP II  
1.  Space, Time & Architecture  P  Sigfried Giedion 
2.  The social logic of space  Q  Bill Hillier 
3.  Timeless way of building  R  Christopher Alexander 
4.  Form, space & order  S  Frank DK Ching 
T  Rob Krier 
Answer: 1P, 2Q, 3R, 4S
Q48. Match Group I with Group II
GROUP I  GROUP II  
1.  Pyranometer  P  Solar radiation 
2.  Heliodon  Q  Shadow Analysis 
3.  Totalstation  R  Land survey 
4.  Anenometer  S  Wind velocity 
T  Seismic intensity 
Answer: 1P, 2Q, 3R, 4S
Q49. A lamp source of 3200 candela is mounted on a wall at a height of 2 meter from the workplane. It subtends an
angle of incidence of 60o with the center of the work plane. The illumination at the centre of the work plane in Lux is __________ (2 marks)
Answer: 100 (by GATE official)
Solution: E(lux)= [ candela /distance square ]* Cosθ = [3200/4]*Cos60 = [800]*0.5 = 400
Basics: The “brightness” of light can mean different things: for example, the amount of light coming from a light source is luminous flux (lumens), the amount of light falling on a surface is illuminance (lux), and the amount of light reflected off a surface is luminance (cd/m2).
Q50. A CPM network of a construction project is given in the figure below. The activity durations are mentioned in weeks. The project completion time in weeks will be _____________ (2 marks)
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: 1P, 2Q, 3R, 4S
Q52. A steel Ibeam section is subjected to a bending moment of 96 kNm. 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 = 96x1000x0.1/24000×10^8 = 40 x 10^6 Pa = 40MPa
Bending stress = (My)/I
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: 1P, 2Q, 3R, 4S
Q54. Choose the best option
1  Nisargruna  P  Solid waste management 
2  VortexDEWAT  Q  Waste water treatment 
3  Swale  R  Ground Water Discharge 
4  BIPV  S  Renewable 
T  Desalination 
Answer: 1P, 2Q, 3R, 4S
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/mm2 respectively. The axial load carrying capacity of the column in kN is _____________ (2 marks)
Answer: 261 (585 – 595 by GATE official)
Solution:
Pu =0.4 fck Ac + 0.67 fy Asc
Pu = 0.4 x8 x 0.0625(11/100) + 0.67 x 150 x 0.0625×1/100 = 0.198+ 0.063 = 0.261 = 261kN
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
hi, the question related number of ecs as per URDPFI , please currect the solution as per URDPFI 18 is correct, check it on page no.289 of vol.01
S.No. Vehicle Type ECS
1 Car/taxi 1.00
2 Two Wheeler 0.25
3 Auto Rickshaw 0.50
4 Bicycle 0.10
5 Trucks/Buses* 2.5
6 Emergency Vehicles* 2.5
7 Rickshaw* 0.8
Hi Anand, Thank for valuable information. We have noted it.
QUES 49.
THE FORMULA = I/ D SQAURE. COS THETA
H= D.COS THETA
SO THE FORMULA IS= (I/HSQUARE). COS CUBE THETA
THAT IS WHY THE ANSWER IS 100.
in Q.33 calculation of reverberation time problem, total Area of wall is given as Perimeter X Height where height from the given question is 3m. But the solution uses 4m as height.
Please clarify.
Ranj, thanks for finding out an error. Wall of height is 3 meters. So, the wall area should be 42 sq.m.
Thank you admin. thereforethe solution changes,
t= 0.16 x [(V/A)] = 0.16x[(4x3x3x)/12.78] = 0.45 Sec
__________________________________________________
Calculation of ‘A’
Wall area (P) = Perimeter of wall x Height
= [2(4+3)]x3 = 14×3 = 42 Sqm
Ceiling (Q) = 4×3 = 12 Sqm
Floor Area (R) = 4×3 = 12 Sqm
So,
A = P [(0.4×30%) + (0.1 x 70%)] +
Q [(0.6 x 40%) + (0.1 x 60%)] +
R [0.1x 100%]
A = (42x 0.19) + ( 12x 0.3) + 1.2
= 7.98 + 3.6 + 1.2
A = 12.78
—————————————–
So the answer would be 0.45 Sec