The document General Aptitude (Part - 2) Civil Engineering (CE) Notes | EduRev is a part of the Civil Engineering (CE) Course Topic wise GATE Past Year Papers for Civil Engineering.

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**Question 16: The CEO’s decision to quit was as shocking to Board as it was to __________.(a) myself(b) I(c) me(d) my [2019 : 1 Mark, Set-I]Answer:** (c)

- Board and me are receives of the action hence objective case of pronoun (me) is to be used.

**Question 17: The lecture was attended by quite _______ students, so the hall was not very_____. ****(a) few, quite ****(b) a few, quite ****(c) few, quiet ****(d) a few, quiet**** [2019 : 1 Mark, Set-I]****Answer:** (d)**Solution: **The lecture was attended by quite __a ____few__ students so the hall was not very __quiet__.

- quite a few indicates a fairly large number of units.
- quiet refer to making little or no noise.

**Question 18: They have come a long way in ________ trust among the users. ****(a) Creating ****(b) Created ****(c) Creation ****(d) Create**** [2019 : 1 Mark, Set-I]****Answer: **(a)**Solution: **They have came long way in __creating__ trust among the users.**Question 19: On a horizontal ground, the base of a straight ladder is 6 m away from the base of a vertical pole. The ladder makes an angle of 45° to the horizontal. If the ladder is resting at a point located at one-fifth of the height of the pole from the bottom, the height of the pole is_______ meters. ****(a) 15 ****(b) 25 ****(c) 35 ****(d) 30**** [2019 : 1 Mark, Set-I]****Answer:** (d)**Solution: **Here tan 45° = 1

Height of wall = 30 m**Question 20: ****If E = 10; J = 20 ; O = 30 ; and T = 40, what will be P + E + S + T?(a) 51 (b) 120 (c) 82 (d) 164 [2019 : 1 Mark, Set-I]**

P = 32, E = 10, S = 38, T = 40

P + E + S + T = 32 + 10 + 38 + 40 = 120

= 24 hours of correct clock

1 hour of IC = 96/97 hours of correct clock

= 99.958 hours of correct clock

= 99 hours + 0.95876 x 60 minutes of correct clock

= 99 hours + 57.525 min.

= 99 hours and approx 58 min.

So, correct time will be

2 PM, 11

= 12: 58 PM on 15th July

logQ = 10(z - x)

logR = 10(x - y)

logP + logQ + logR = 0

log (PQR) = log 1

PQR = 1

Area = 50 m

A + B + C= 13 ,..(i)

C + D + E= 13 ...(ii)

E + F + G = 13 ...(iii)

G + H + K= 13 ...(iv)

Adding [(i) + (ii) + (iii) + (iv)]

A + B + C +D + E + F+ G + H + K+ (C + E + G) = 13 x 4 = 52 ...(v)

Also A, B, C, D, E, F, G, H& K represents natural numbers from (1 to 9)

There sum will be given by

Substituting (iv) C + E + G - 7 ,..(vi)

Only possibly for sum 7 will be (1, 2, 4)

Now, C + E cannot be (1 and 2)

As eq. (ii) is C + D + E = 13

Now, D will become equal to 10 (which is not possible because digits 1 to 9 given)

If C = H from eq. (vi) C + E + G = 7

Now, E + G = 3

(Not possible in eq. (iii) E + F + G = 10,

F - 10 which is not possible)

So from eq. (vii) only possibility remains is E = H.

For duration of 7 units

4900 = k (7)^{2} ⇒ k = 100

Loss = kd^{2}

For duration of 4 units

= k(4)^{2} ⇒ 16k = 1600**Question 26: What of the following function(s) in an accurate description of the graph for the range(s) indicated?****(a) y = 2x + 4 for -3 < x < ~ 1 ****(b) y = |x - 1| for -1 < x < 2 **

(i) y = 2x + 4 is true in -3

On putting x = -3, y = -2 and x = -2,

y = 0 and x = -1, y = 2

(ii) y = |x - 1| is also true (x = -1, y = 2), (x = 0, y = 1) and (x = 1, y = 0)

(iv) y = 1 in (2 < x < 3) always true

(i), (ii) and (iv) are true.

C = kW

⇒ C= k(10)

⇒ k = 16

C1 = k(4)

C

Now total cost = 52k = 52 x 16 = 832

Any of A, B, C, D, E, F, G cannot be 5.

x =108/22.5

What is the sum of the first 50 terms?

Ail like terms will cancel out and we will be left with

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