GATE 2015 [Set-1]
Question 1 |
Didn’t you buy _________ when you went shopping?
any paper | |
much paper | |
no paper | |
a few paper |
Question 2 |
Which of the following combinations is incorrect?
Acquiescence – Submission | |
Wheedle – Roundabout | |
Flippancy – Lightness | |
Profligate – Extravagant |
Wheedle = Use endearments (or) flattery to persuade some to do something
Flippancy = Lack of respect (or) seriousness
Profligate = Recklessly extravagant
Question 3 |
Given set A = {2, 3, 4, 5} and Set B = {11, 12, 13, 14, 15}, two numbers are randomly selected, one from each set. What is probability that the sum of the two numbers equals 16?
0.20 | |
0.25 | |
0.30 | |
0.33 |
Total outcomes = 4C1 × 5C1 = 20
Probability = Favourable outcomes/ Total outcomes = 4/20 = 0.20
Question 4 |
Which of the following options is the closest in meaning to the sentence below?
She enjoyed herself immensely at the party.
She had a terrible time at the party. | |
She had a horrible time at the party. | |
She had a terrific time at the party. | |
She had a terrifying time at the party. |
→ Horrible and terrible means fearful.
→ Terrific = Wonderful
→ So, C is the Answer.
Question 5 |
Based on the given statements, select the most appropriate option to solve the given question.
If two floors in a certain building are 9 feet apart, how many steps are there in a set of stairs that extends from the first floor to the second floor of the building?
Statements:-
(I) Each step is 3/4 foot high.
(II) Each step is 1 foot wide.
Statement I alone is sufficient, but statement II alone is not sufficient. | |
Statement II alone is sufficient, but statement I alone is not sufficient.
| |
Both statements together are sufficient, but neither statement alone is sufficient. | |
Statement I and II together are not sufficient. |
Statement I:
Each step is 3/4 foot high.
No. of steps = 9/(3/4) = 12 steps
Statement I alone is sufficient.
Statement II:
Each step is 1 foot wide.
Statement II alone is not sufficient.
Question 6 |
The pie chart below has the breakup of the number of students from different departments in an engineering college for the year 2012. The proportion of male to female students in each department is 5:4. There are 40 males in Electrical Engineering. What is the difference between numbers of female students in the Civil department and the female students in the Mechanical department?
32 | |
33 | |
34 | |
35 |
There are 40 males in Electrical Engineering.
⇒ Number of females in Electrical Engineering = 4/5 × 40 = 32
Total number of students in Electrical Engineering = 40+32 = 72
This constitutes 20% of the strength of college.
Number of students in Civil department = 30/100 × 72 = 108
Number of female students in Civil department = 4/(4+5) × 108 = 48
Number of students in Mechanical department = 10/20 × 72 = 36
Number of female students in Mechanical department = 4/(4+5) × 36 = 16
Required Difference = 48 - 16 = 32
Question 7 |
Select the alternative meaning of the underlined part of the sentence.
The chain snatchers took to their heels when the police party arrived.
took shelter in a thick jungle | |
open indiscriminate fire | |
took to flight | |
unconditionally surrendered |
Question 8 |
The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c respectively. Of these subjects, the student has 75% chance of passing in at least one, a 50% chance of passing in at least two and a 40% chance of passing in exactly two. Following relations are drawn in m, p, c:
-
(I) p + m + c = 27/20
(II) p + m + c = 13/20
(III) (p) × (m) × (c) = 1/10
Only relation I is true | |
Only relation II is true | |
Relations II and III are true. | |
Relations I and III are true. |
= 1 - (1 - m) (1 - p) (1 - c) = 0.75 ----(i)
50% of students have a chance pass in atleast two
(1 - m)pc + (1 - p)mc + (1 - c) mp + mpc = 0.5 ----(ii)
40% of students have a chance of passing exactly two
(1 - m)pc + (1 - p)mc + (1 - c)mp = 0.4 ----(iii)
From equation (ii) and (iii) we can get
mpc = 0.1
⇒ m*p*c = 1/10
Statement (III) is correct.
→ Simplify eq (i), we get
⇒ p+c+m - (mp+mc+pc) + mpc = 0.75
⇒ p+c+m - (mp+mc+pc) = 0.65 ----(iv) → Simplify equation (iii), we get
⇒ pc + mc + mp - 3mpc = 0.4
From (iv) and (v)
p + c + m - 0.7 = 0.65
⇒ p + c + m = 1.35 = 27/20
Statement (I) is correct.
Question 9 |
The given statement is followed by some courses of action. Assuming the statement to be true, decide the correct option.
Statement:
There has been a significant drop in the water level in the lakes supplying water to the city.
Course of action:
-
(I) The water supply authority should impose a partial cut in supply to tackle the situation.
(II) The government should appeal to all the residents through mass media for minimal use of water.
(III)The government should ban the water supply in lower areas.
Statements I and II follow | |
Statements I and III follow | |
Statements II and III follow | |
All statements follow |
Banning of water in lower areas is not the solution of the problem.
Question 10 |
The number of students in a class who have answered correctly, wrongly, or not attempted each question in an exam, are listed in the table below. The marks for each question are also listed. There is no negative or partial marking.
What is the average of the marks obtained by the class in the examination?
2.290 | |
2.970 | |
6.795 | |
8.795 |
Total marks obtained = 21 × 2 + 15 × 3 + 11 × 1 + 23 × 2 + 31 × 5 = 299
Average marks = 299/44 = 6.795
Question 11 |
Which one of the following is True at any valid state in shift-reduce parsing?
Viable prefixes appear only at the bottom of the stack and not inside | |
Viable prefixes appear only at the top of the stack and not inside | |
The stack contains only a set of viable prefixes | |
The stack never contains viable prefixes |
A viable prefixes is prefix of the handle and so can never extend to the right of handle, i.e., top of stack.
So set of viable prefixes is in stack.
Question 12 |
Match the following:
(P) Condition coverage (i) Black-box testing (Q) Equivalence class partitioning (ii) System testing (R) Volume testing (iii) White-box testing (S) Alpha testing (iv) Performance testing
P-ii, Q-iii, R-i, S-iv | |
P-iii, Q-iv, R-ii, S-i | |
P-iii, Q-i, R-iv, S-ii | |
P-iii, Q-i, R-ii, S-iv |
Equivalence class partitioning ⇒ is a software testing technique that divides the input data of a software unit into partitions of equivalent data from which test cases can be derived, which is nothing but black box testing. Hence B-1.
Volume testing ⇒ Performance testing - C-4.
Alpha testing ⇒ System Testing D-2.
Question 13 |
For computers based on three-address instruction formats, each address field can be used to specify which of the following:
-
(S1) A memory operand
(S2) A processor register
(S3) An implied accumulator register
Either S1 or S2 | |
Either S2 or S3 | |
Only S2 and S3 | |
All of S1, S2 and S3 |
So as the question asks what can be specified using the address fields, implied accumulator register can't be represented in address field.
So, S3 is wrong.
Hence Option A is the correct answer.