GATE CE 2017 Official Paper: Shift 2 (Previous Year Paper)

The sentence is in the perfect continuous tense which can be noticed because of the usage of "been". 

When you see "would have" in a sentence it means that the action didn't actually happen, because something else didn't happen first.

From the sentence we know that the "event" has already happened. 

Hence the correct answer is "had been" as it would convey the correct tense, which is the past perfect continuous.

Option 2 is correct.

Let us look at the meanings of all the words:

Through: having finished using or doing something.

Pretty: (of a person, especially a woman or a child) attractive in a delicate way without being truly beautiful.

Complete: finish making or doing.

Sloppy: careless and unsystematic; excessively casual.

Haphazard: lacking any obvious principle of organization.

Example sentence: I've got some work to do but I should be through in an hour if you can wait.

Option 2 is correct.

Faces available on the cards are: 2, 3, red, and blue.

Proposition: If a card has an even value on one side, then its opposite face is red.

We need to check the card with face 2 whether its opposite is red or not. The opposite side will be RED, if the given proposition is TRUE.

No need to check the card with face 3 as it is not even number.

No need to check the card with face red as it can have either even or odd number as its opposite number. (The converse of the given proposition may or may not be true).

We need to check the card with face blue whether its opposite number is even or odd. Its opposite number will be odd, if the given proposition is TRUE.

$81 \times {\left( {\frac{{16}}{{25}}} \right)^{x + 2}} \div {\left( {\frac{3}{5}} \right)^{2x + 4}} = 144$

$\Rightarrow 81 \times \frac{{{{\left( {\frac{{16}}{{25}}} \right)}^{x + 2}}}}{{{{\left( {\frac{3}{5}} \right)}^{2x + 4}}}} = 144$

$\Rightarrow 81 \times \frac{{{{\left( {\frac{4}{5}} \right)}^{2x + 4}}}}{{{{\left( {\frac{3}{5}} \right)}^{2x + 4}}}} = 144$

$\begin{array}{l} \Rightarrow 81 \times {\left( {\frac{4}{3}} \right)^{2x + 4}} = 144\ \Rightarrow {81 \times\left( {\frac{4}{3}} \right)^{2x}} \times {\left( {\frac{4}{3}} \right)^4} = 144 \end{array}$

$\begin{array}{l} \Rightarrow {\left( {\frac{{16}}{9}} \right)^x} = \frac{{144}}{{256}}\ \Rightarrow {\left( {\frac{{16}}{9}} \right)^x} = \frac{9}{{16}} = {\left( {\frac{{16}}{9}} \right)^{ - 1}} \end{array}$

⇒ x = -1

Two dice are thrown simultaneously.

Total number of possibilities = 6² = 36

The possible perfect squares = 1, 4, 9, 16, 25, 36

Possible outcomes to get the product as 1 = (1, 1)

Possible outcomes to get the product as 4 = (1, 4), (4, 1), (2, 2)

Possible outcomes to get the product as 9 = (3, 3)

Possible outcomes to get the product as 16 = (4, 4)

Possible outcomes to get the product as 25 = (5, 5)

Possible outcomes to get the product as 36 = (6, 6)

Total number of possibilities = 8

Probability = 8/36 = 2/9

Explanation:

Option 1 can be safely concluded from the given passage.

Option 2 is incorrect because it says 'Customers were served in an arbitrary order which is not given in the passage.

Option 3 is incorrect because it says 'the system is likely to operate on a last-come-first-served basis' which is not given in the passage.

Option 4 is incorrect because it says 'Entering the center early ensured that one would have shorter service times' which is not given in the passage.

Hence option 1 is correct.

A map shows the elevations of Darjeeling (D), Gangtok (G), Kalimpong (K), Pelling (P), and Siliguri (S).

Kalimpong is at a lower elevation than Gangtok.

⇒ K < G       ----(1)

Pelling is at a lower elevation than Gangtok.

⇒ P < G       ----(2)

Pelling is at a higher elevation than Siliguri.

⇒ S < P        ----(3)

Darjeeling is at a higher elevation than Gangtok.

⇒ G < D       ----(4)

From inequalities (1) and (4), we get

⇒ K < G < D

From inequalities (2) and (3), we get

⇒ S < P < G

i. Pelling is at a higher elevation than Kalimpong (K < P): There is no clear relation between K and P, so it may or may not TRUE.

ii. Kalimpong is at a lower elevation than Darjeeling (K < D): It is TRUE as K < G < D.

iii. Kalimpong is at a higher elevation than Siliguri (S < K): There is no clear relation between S and K, so it may or may not TRUE.

iv. Siliguri is at a lower elevation than Gangtok (S < G): It is TRUE as S < P < G.

P, Q, R, S, T and U are seated around a circular table.

R is seated two places to the right of Q.

P is seated three places to the left of R.

S is seated opposite U.

According to the above information, there are two possibilities:

Case 1: The arrangement is as shown below.

After switching the seats of P and U, the arrangement becomes

P is immediately right of Q.

Case 2:

The arrangement is as shown below.

After switching the seats of P and U, the arrangement becomes

T is immediately to the left of P.

So, T is immediately to the left of P or P is immediately to the right of Q.

Let Budhan cycles at a speed of x km/hr and walks at a speed of y km/hr.

Budhan covers a distance of 19 km in 2 hours by cycling one fourth of the time and walking the rest.

I.e. Budhan cycled for half an hour and had walk for remaining time.

⇒ 0.5x + 1.5y = 19       ----(1)

The next day he cycles (at the same speed as before) for half the time and walks the rest (at the same speed as before) and covers 26 km in 2 hours.

⇒ x + y = 26      ----(2)

By solving the equations (1) and (2), x = 20 km/hr and y = 6 km/hr

So, the speed at which Budhan walks is 6 km/hr.

At 28^th minute and at 39^th minute elevator directly moves form second floor to fifth floor.

So, statement (i) is not correct.

Elevator stays at ground floor for 21 minutes whereas it stays for 19 minutes at fourth floor.

So, statement (ii) is not correct.

3x₁ + 2x₂ = c₁

3x₁ + x₂ = c₂

\(\left[ {\begin{array}{*{20}{c}} 3&2\ 4&1 \end{array}} \right] = \left[ {\rm{A}} \right]\)

[A] – λ[I] = 0

\(\left[ {\begin{array}{*{20}{c}} {3 - {\rm{\lambda }}}&4\ 2&{1 - {\rm{\lambda }}} \end{array}} \right] = 0\)

3 + λ² – 4λ – 8 = 0

λ² – 4λ – 5 = 0

w = f (x, y)

Partial differentiation with respect to x and y gives:

$\frac{{{\rm{dw}}}}{{{\rm{dt}}}} = \frac{{\partial {\rm{w}}}}{{\partial {\rm{x}}}}\frac{{{\rm{dx}}}}{{{\rm{dt}}}} + \frac{{\partial {\rm{w}}}}{{\partial {\rm{y}}}}\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$

If scope of construction work is well-defined with all its drawings, specification quantities and estimates, then lump sum contract is used.

Concept:

The dry density (γ_d) of soil is given by,

\({{\rm{γ }}{\rm{d}}} = \frac{{\left( {1 - {{\rm{η }}{\rm{a}}}} \right){{\rm{G}}{\rm{}}}{{\rm{γ }}{\rm{w}}}}}{{1 + {\rm{e}}}}\)

η_a - Percentage of air voids

G - Specific gravity of soil solids

e - Voids ratio

For zero air voids ⇒ volume of air is zero, hence η_a = 0, a_c = 0                       

Also, S + a_c = 1

Hence, S = 1      

So above equation becomes, \({{\rm{γ }}{\rm{d}}} = \frac{{\left( {1 - 0} \right){{\rm{G}}{\rm{}}}{{\rm{γ }}{\rm{w}}}}}{{1 + \frac{{{\rm{w}}{{\rm{G}}{\rm{}}}}}{{\rm{S}}}}}\)

⇒ \({{\rm{γ }}{\rm{d}}} = \frac{{{{\rm{G}}{\rm{}}}{{\rm{γ }}{\rm{w}}}}}{{1 + {\rm{w}}{{\rm{G}}{\rm{}}}}}\)

Explanation

Skempton’s Pore Pressure coefficient are A and B,

where,

A → depends upon the degree of saturation and over consolidation ratio

B → depends upon degree of saturation of soil

Value of B lies between 0 and 1. For completely saturated soil, B = 1 and for completely dry soil, B = 0.

Value of A may be as large as 2 to 3 for very loose saturated fine sand and it can be less than zero for over consolidated clay.

Pressure diagram for partially submerged soil with surcharge is shown below:

But no option is matching.

If the water table is at ground table then Option b is the answer.

Explanation:

The specific speed is of a fluid machine is a type number representative of its performance. To compare the performance of different pumps (or turbines), it is necessary to have some terms that will be common to all the pumps (or turbines), the term ordinarily used for this purpose is the specific speed. This can be used to predict one pump (or turbine) behavior based on tests of similar, but different sized pumps (or turbines).

${N_{{s_T}}}$ (specific speed for turbines) = $N{P^{\frac{1}{2}}}/{H^{\frac{5}{4}}}$

${N_{{s_P}}}$ (specific speed for the pump) = $N{Q^{\frac{1}{2}}}/{H^{\frac{3}{4}}}$

For a turbine, ${N_{{s_T}}}$ is the speed of a member of the same homologous series as the actual turbine, so reduced in size as to generate unit power under a unit head of the fluid. Similarly, for a pump, ${N_{{s_P}}}$ is the speed of a hypothetical pump with reduced size but representing a homologous series so that it delivers unit flow rate at a unit head. The specific speed (${N_s}$) is, therefore, not a dimensionless quantity.

As per Noise pollution (Regulation and control) Rules 2000 of India.

$\begin{array}{l} {\rm{EL}}{{\rm{R}}_1} = \frac{{{\rm{\Delta T}}}}{{{\rm{\Delta h}}}} = \frac{{15.5 - 15.0}}{{60 - 10}} = 0.01^\circ \frac{{\rm{c}}}{{\rm{m}}} = ;10^\circ \frac{{\rm{c}}}{{{\rm{km}}}};\ {\rm{EL}}{{\rm{R}}_2} = \frac{{15.0 - 14.3}}{{130 - 60}} = 0.01^\circ \frac{{\rm{C}}}{{\rm{m}}} = ;10^\circ \frac{{\rm{c}}}{{{\rm{km}}}} \end{array}$

Adiabatic lapse rate (ALR) is 9.8°C/km for dry adiabatic and 6.c/km for wet adiabatic condition.

∵ ELR > ALR

So, it is super adiabatic and the atmosphere is said to be highly unstable.

Heterotrophs cannot produce organic compounds from inorganic sources and therefore rely on consuming other organisms in the food chain. In activated sludge process, in the presence of food mass and oxygen heterotrophic bacteria converts food mass into biomass.

Concept:

Equilibrium cant $= \frac{{{\rm{G}}{{\rm{V}}^2}}}{{127{\rm{R}}}}$

V - Design speed (kmph)

R - Radius of curvature (m)

For B.G track G = 1.676

$\begin{array}{l} {\rm{Cant}} = \frac{{1.676{\rm{;}}{{\rm{V}}^2}}}{{127{\rm{;R}}}}{\rm{m}}\ {\rm{e}} = \frac{{1.315{\rm{;}}{{\rm{V}}^2}}}{{\rm{R}}}{\rm{cm}} \end{array}$

The essence of the intersection control is to resolve these conflicts at the intersection for the safe and efficient movement of both vehicular traffic and pedestrians.

The safety within a roundabout and the efficiency of a roundabout can be increased respectively by decreasing the entry radius and increasing the exit radius.

Note:

There are two methods of intersection controls 

1. time sharing and

2. space sharing.

The type of intersection control that has to be adopted depends on the traffic volume, road geometry, cost involved, importance of the road etc.

Resection method of orientation is employed when the plane table occupies a position not yet plotted on the drawing sheet.

Concept:

Total degree of freedom when axial and shear deformations in all members are allowed. 

D_k = 3 + 3 + 3 + 3 + 1 + 1 = 14

When axial and shear deformations are negligible,

The degree of freedom equals to the number of members would be deducted i.e. 6

∴ Kinematic indeterminacy for the frame if axial and shear deformation is considered negligible, (D_k) = 14 - 6 = 8.

Hence, reduction in degree of kinematical indeterminacy = 14 – 8 = 6

Mistake Point:

Kindly do read the Question again as it is asked about the reduction in the degree of kinematical indeterminacy i.e 14 – 8 = 6

Concept:

The correct expression as per IS 456: 2000 to find Target mean strength is:

fcm = fck + (k × σ)

fcm = Target mean strength

fck = Characteristic strength

σ = Standard deviation

 For not more than 50% of results are likely to fall k = 0.

Since characteristic strength has been defined as the value below which not more than 50% of results are likely to fall.

Hence,

Target mean strength = f_t = fck + (0× σ) = Characteristic strength (f_ck) = 25 N/mm²

Shear strain in an element is positive when the angle between two positive faces (or two negative faces) is reduced. The strain is negative when the angle between two positive (or two negative) faces increase.

Angle between 1 & 4 is increased by 0.0005 rad.

∴ γ_xy = – 0.0005 = 0.001 K

∴ K = – 0.5

In plate load test bearing capacity of clay does not depend upon size of footing.

q_up (for plate) = q_uf (for clay) = 180 kPa

Target is having inclination of 45° with x-axis

$\begin{array}{l} \Rightarrow \frac{{dy}}{{dx}} = \tan 45^\circ \ \Rightarrow \frac{{d\left( {xln;x} \right)}}{{dx}} = 1\ \Rightarrow \ln x + \frac{x}{x} = 1 \end{array}$

∴ At x = 1, y = 1 × ln1 = 0

Explanation:

\(I = \mathop \smallint \limits_0^1 \frac{{{{\left( {{{\sin }^{ - 1}}x} \right)}^2}}}{{\sqrt {1 - {x^2}} }}dx\)

Put sin^-1 x = t

\(\begin{array}{l} \frac{{dx}}{{\sqrt {1 - {x^2}} }} = dt\ = \mathop \smallint \limits_0^{\frac{\pi }{2}} {t^2}dt\ = \left. {\frac{{{t^3}}}{3}} \right]_0^{\frac{\pi }{2}}\ = \frac{{{\pi ^3}}}{{24}} \end{array}\)

\(\begin{array}{l} A = \left[ {\begin{array}{{20}{c}} 1&5\ 6&2 \end{array}} \right];\left[ B \right] = \left[ {\begin{array}{{20}{c}} 3&7\ 8&4 \end{array}} \right]\ A{B^T} = \left[ {\begin{array}{{20}{c}} 1&5\ 6&2 \end{array}} \right]\left[ {\begin{array}{{20}{c}} 3&8\ 7&4 \end{array}} \right]\ = \left[ {\begin{array}{{20}{c}} {3 + 35}&{8 + 20}\ {18 + 14}&{48 + 8} \end{array}} \right]\ = \left[ {\begin{array}{{20}{c}} {38}&{28}\ {32}&{56} \end{array}} \right] \end{array}\)

Explanation:

y” – 4y’ + 3y = 2t – 3t²

f(D) = D² – 4D + 3

$\begin{array}{l} P.I = \frac{I}{{f\left( D \right)}}\left( {2t - 3{t^2}} \right)\ = \frac{1}{{{D^2} - 4D + 3}}\left( {2t - 3{t^2}} \right)\ = \left[ {\frac{1}{{1 - D}} - \frac{1}{{3\left( {1 - \frac{D}{3}} \right)}}} \right]\left( {t - \frac{{3{t^2}}}{2}} \right)\ \left( {1 + D + {D^2} + \ldots } \right)\left( {t - \frac{{3{t^2}}}{2}} \right) - \frac{1}{3}\left( {t - \frac{D}{3} + \frac{{{D^2}}}{9} + \ldots } \right) \times \left( {t - \frac{{32{t^2}}}{2}} \right)\ = \left( {t - \frac{{3{t^2}}}{2} + 1 - 3t - 3} \right) - \frac{1}{3}\left( {1 - \frac{{3{t^2}}}{2} + \frac{1}{3} - 8 - \frac{1}{3}} \right) \end{array}$

= – 2 – 2t – t²

1. Le chatelier Test: This test is used to measure the soundness of OPC due to lime. Lime & Magnesia are two primary compounds responsible for soundness of cement.

2. Vee Bee Test: It is one of the methods of measuring the workability of concrete.

3. Blaine Air Permeability: It is used to measure fineness of cement.

4. The Vicat Apparatus: It is used to measure setting time and consistency of concrete.

$\begin{array}{l} {{\rm{\delta }}_1} = \frac{5}{{384}}\frac{{w{L^4}}}{{EI}}\ = \frac{5}{{384}} \times \frac{{6 \times {4^4}}}{{10000;m}} = 20;mm \end{array}$

$\begin{array}{l} {{\rm{\delta }}_{\rm{L}}} = \frac{{P{L^3}}}{{48EI}}\ = \frac{{120 \times {2^3}}}{{48 \times 10000;m}} \end{array}$

= 20 mm

∴ δ₁ = δ₂

Explanation:

For P:

Σf_v = 0

⇒ V = 80 N

∴ Shear force at support = S_P = 80 N.

Moment at support = Pl ⇒ 80 × 8 = 640 Nm.

∴ M_P = 640 Nm.

For Q:

Σf_v = 0

⇒ V = wx ⇒ 20 × 8 = 160 N

∴ Shear force at support = S_Q = 160 N.

Moment at support = M_Q ⇒ $\frac{wx^2}{2}$ 

$\therefore \frac{20;\times;8^2}{2}⇒640;Nm$

For R:

Σf_v = 0

⇒ V = 0 = S_R

Moment at support = M_R = 640 Nm

Hence,

S_P < S_Q > S_R & M_P = M_Q = M_R

Explanation:

As per IS 1200 (part 3), 1976, Cl No- 4, brick work is measured in the following way:

No deduction is made for following.

• Opening each up to 0.1 m².
• Ends of beams rafters up to 0.05 m² in section.

When the ground is not uniform levels shall be taken before the start, after site clearance and after the completion of the work and the quantity of excavation in cutting computed from these levels.

Concept:

End Bearing Resistance = Effective Stress at Base × N_q

End Bearing Resistance = qNq

Q_P1 = qN_q 

Q_P2 = qN_q 

Find $\frac{{{Q_{{P_1}}}}}{{{Q_P}_2}}$

Calculation:

Q_P1= 18 × 20 × N_q  

Q_P1= 18 × (20N_q ) = 360 N_q

Q_P2 = qN_q 

= (19 – 10) × 20 × N_q 

= 9 × (20 N_q) = 180 N_q

(Assuming γ_w = 10 kN/m³)

$\frac{{{Q_{{P_1}}}}}{{{Q_P}_2}} = 2$

Hence, by 100%.

When σ₁ = σ₃

q = 0, $p = \frac{{2{\sigma _1}}}{2} = {\sigma _1}$

So stress point lies on p-axis of p-q plot

If σ₁ > σ₃

$\begin{array}{l} q = \frac{{{\sigma _1} - {\sigma _3}}}{2};so;q > 0\ p = \frac{{{\sigma _1} + {\sigma _3}}}{2};so;p > 0 \end{array}$

So, stress point p & q are positive.

P: d = 60t

Q: d = 60t²

Space Headway (S) = 60t² – 60t

For space headway to be max,

$\frac{{ds}}{{dt}} = 0$

120t – 60 = 0

⇒ 60(2t – 1) = 0

⇒ t = ½ hr

Maximum space Headway

= 60 × ½ - 60 × (½)²

= 60 × ½ - 60 × ¼

= 30 – 15 = 15 km

Case-I: Undrained condition

${\rm{F}}.{\rm{O}}.{\rm{S}} = \frac{{{\rm{resisting;shear;strength}}}}{{{\rm{Acting;shear;stress}}}} = ;\frac{{80}}{{50}} = 1.6$

Case-II: Drained condition

$\begin{array}{l} {\rm{F}}.{\rm{O}}.{\rm{S}} = \frac{{{\rm{\bar \sigma }}\tan \phi ' + c'}}{{{\rm{Acting;shear;stress}}}}\ = \frac{{\left[ {2 \times 18 + 4\left( {20 - 9.81} \right)} \right] \times \tan 18^\circ + 20}}{{50}} = 0.9 \end{array}$