General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE) PDF Download

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)
Solution: CEO’s decision to quit was as shocking to board as it was to me.

  • 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 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
General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)
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]

Answer: (b)
Solution: According to given coding
P = 32, E = 10, S = 38, T = 40
P + E + S + T = 32 + 10 + 38 + 40 = 120

Question 21: A faulty wall clock is known to gain 15 minutes every 24 hours. It is synchronized to the correct time at 9 AM on 11th July, What will be the correct time to the nearest minute when the clock shows 2 PM on 15th July of the same year? 
(a) 12:45 PM 
(b) 12:58 PM 
(c) 1:00 PM 
(d) 2:00 PM 
     [2018 : 2 Marks, Set-II]
Answer: (b)
Solution: 9 Am of 11 July of 2 PM on 15th July = 101 hours
General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE) hours of incorrect clock = 24 hours of correct clock
General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)
= 24 hours of correct clock
1 hour of IC = 96/97 hours of correct clock
General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)
= 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, 11th july + (99 hours and 58 minutes)
= 12: 58 PM on 15th July

Question 22: Given that General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE) = 10 for x ≠ y ≠ z, what is the value of the product PQR? 
(a) 0 
(b) 1 
(c) xyz 
(d) 10xyz
    [2018 : 2 Marks, Set-II]
Answer: (b)
Solution: logP = 10(y - z)
logQ = 10(z - x)
logR = 10(x - y)
logP + logQ + logR = 0
log (PQR) = log 1
PQR = 1

Question 23: The annual average rainfall in a tropical city is 1000 mm. On a particular rainy day (24-hour period), the cumulative rainfall experienced by the city is shown in the graph. Over the 24-hour period. 50% of the rainfall falling on a rooftop, which had an obstruction-free area of 50 m2, was harvested into a tank. What is the total volume of water collected in the tank in liters?
General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)
(a) 25,000 
(b) 18,750 
(c) 7,500 
(d) 3,125 
    [2018 : 2 Marks, Set-II]
Answer (c)
Solution: Cumulative rainfall = 300 mm
General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)
Area = 50 m2
General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)

Question 24: Each of the letters in the figure below represents a unique integer from 1 to 9. The letters are positioned in the figure such that each of (A+B+C), (C+D+E), (E+F+G) and (G+H+K) is equal to 13. Which integer does E represent?
General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)
(a) 1
(b) 4
(c) 6
(d) 7
    [2018 : 2 Marks, Set-II]
Answer: (b)
Solution: 
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 General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)
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)
General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)
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.

Question 25: In manufacturing industries, loss is usually taken to be proportional to the square of the deviation from a target. If the loss is Rs. 4900 for a deviation of 7 units, what would be the loss in Rupees for a deviation of 4 units from the target? 
(a) 400 
(b) 1200 
(c) 1600 
(d) 2800
    [2018 : 2 Marks, Set-H]
Answer: (c)
Solution: Loss = kd2 

For duration of 7 units
4900 = k (7)2 ⇒ k = 100
Loss = kd2   

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?
General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)
(a) y = 2x + 4 for -3 < x < ~ 1 
(b) y = |x - 1| for -1 < x <
(c) y = | |x | - 1| for -1 < x <
(d) y = 1 for 2 < x < 3
(a) (i), (ii) and (iii) only 
(b) (i), (ii) and (iv) only 
(c) (i) and (iv) only 
(d) (ii) and (iv) only
     [2018 : 2 Marks, Set-I]
Anwser: (b)
Solution: Put value and verify
(i) y = 2x + 4 is true in -3 < x < - 1
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.

Question 27: The price of a wire made of a superalloy material is proportional to the square of its length. The price of 10 m length of the wire is Rs. 1600. What would be the total price (in Rs.) of two wires of lengths 4 m and 6 m? 
(a) 768 
(b) 832 
(c) 1440 
(d) 1600 
    [2018 : 2 Marks, Set-I]
Answer: (b)
Solution: C ∝ W2
C = kW2 
⇒ C= k(10)2 = 100k = 1600
⇒ k = 16
C1 = k(4)2 = 16k
C2 = k(6)2 = 36k
Now total cost = 52k = 52 x 16 = 832

Question 28: Each of the letters arranged as below represents a unique from 1 to 9. The letters are positioned in the figure such that (A x B x C), (B x G x E) and (D x E x F) are equal. Which integer among the following choices cannot be represented by the letters A, B, C, D, E, F or G?
General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)
(a) 4 
(b) 5 
(c) 6 
(d) 9
Answer: (b)
Solution: 
General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)
Any of A, B, C, D, E, F, G cannot be 5.

Question 29: A fruit seller sold a basket of fruits at 12.5% loss. Had he sold it for Rs. 108 more, he would have made a 10% gain. What is the loss in Rupees incurred by the fruit seller? 
(a) 48 
(b) 52 
(c) 60 
(d) 108
     [2018 : 2 Marks, Set-I]
Answer: 12.5%x + 10%x = 108
x  =108/22.5
General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)

Question 30: Consider a sequence of number a1, a2, a3......, an where General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE) for each integer n > 0.
What is the sum of the first 50 terms?

General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)
    [2018 : 2 Marks, Set-I]
Answer: (c)
Solution: Sum of series will be
General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)
Ail like terms will cancel out and we will be left with
General Aptitude- 2 | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)

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FAQs on General Aptitude- 2 - Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)

1. What is the General Aptitude-2 exam?
Ans. The General Aptitude-2 exam is a standardized test that assesses a person's general aptitude skills, including numerical reasoning, verbal reasoning, and abstract reasoning. It is commonly used by employers and educational institutions to evaluate a candidate's problem-solving abilities, critical thinking skills, and overall cognitive aptitude.
2. How can I prepare for the General Aptitude-2 exam?
Ans. To prepare for the General Aptitude-2 exam, it is recommended to practice a variety of aptitude tests and familiarize yourself with the different question types. You can find sample questions and practice tests online or in books specifically designed for aptitude test preparation. Additionally, improving your mathematical, verbal, and logical reasoning skills through regular practice and study can greatly enhance your performance in the exam.
3. What are some tips for improving numerical reasoning skills for the General Aptitude-2 exam?
Ans. To improve your numerical reasoning skills for the General Aptitude-2 exam, it is helpful to practice mental arithmetic regularly. This can involve solving math problems without the use of a calculator, which helps in building speed and accuracy. Additionally, familiarize yourself with common mathematical concepts and formulas, such as percentages, ratios, and basic algebra. Practice solving numerical problems under time constraints to simulate exam conditions.
4. How can I enhance my verbal reasoning skills for the General Aptitude-2 exam?
Ans. Enhancing your verbal reasoning skills for the General Aptitude-2 exam can be done through regular reading and vocabulary building exercises. Read a variety of texts, including newspapers, magazines, and books, to expose yourself to different writing styles and improve your comprehension skills. Expand your vocabulary by learning new words and their meanings. Practicing verbal reasoning questions, such as identifying relationships between words or completing analogies, can also help improve your performance in this section.
5. What strategies can I use for improving abstract reasoning abilities for the General Aptitude-2 exam?
Ans. Improving abstract reasoning abilities for the General Aptitude-2 exam can be done by practicing pattern recognition and logical deduction exercises. Look for patterns and relationships in shapes, numbers, or symbols and try to identify the underlying rules governing them. Solve puzzles and riddles that require logical thinking and deduction. Additionally, familiarize yourself with different types of logical reasoning questions, such as syllogisms or series completion, and practice solving them to enhance your abstract reasoning skills.
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