GATE 2017 CS (Set 1) question paper PDF and answer key

The official GATE 2017 Computer Science & Information Technology (Set 1) paper, organised by IIT Roorkee: 65 questions for 100 marks in 3 hours. Download the official question paper and answer key as PDFs. Below is how the marks were split by topic, the complete official answer key and 5 questions solved step by step.

Take this paper as a timed test Try the free demo

Question paper (PDF)Official answer key (PDF)

Source: the official paper and answer key published by IIT Roorkee for GATE 2017. Spotted a mistake? Email team@lemyte.com.

General Aptitude: 10 · 15 marks
Computer Science & Information Technology: 55 · 85 marks
MCQ / MSQ / NAT: 45 / 0 / 20
Marks to all: None

Where the marks were

GATE 2017 CS (Set 1) topic-wise marks

Computer Science & Information Technology questions only; General Aptitude adds 15 marks on top. The top three topics carried 33 of the 85 subject marks.

Topic	Questions	Marks
Computer Organization and Architecture	8	12
Programming and Data Structures	7	11
Theory of Computation	6	10
Databases	5	8
Computer Networks	5	8
Discrete Mathematics	5	7
Algorithms	4	6
Compiler Design	4	6
Operating System	4	6
Linear Algebra	3	5
Digital Logic	2	3
Calculus	1	2
Probability and Statistics	1	1

Free solved questions

5 solved questions from GATE 2017 CS (Set 1)

Question text and figures as in the official paper, the answer from the official key, and a worked solution. The other 60 solutions are in your report after you take the paper.

Q6 · General Aptitude · MCQ · 2 marks

The expression

\frac{x+y-|x-y|}{2}

is equal to:

(A)
the maximum of $x$ and $y$
(B)
the minimum of $x$ and $y$
(C)
1
(D)
none of the above

Answer (official key): B

Solution

If $x \ge y$ , then:

|x-y|=x-y.

So,

\frac{x+y-|x-y|}{2} = \frac{x+y-(x-y)}{2} = \frac{2y}{2} = y.

Here, $y$ is the minimum.

If $y \ge x$ , then:

|x-y|=y-x.

So,

\frac{x+y-|x-y|}{2} = \frac{x+y-(y-x)}{2} = \frac{2x}{2} = x.

Here, $x$ is the minimum.

Therefore, the expression is equal to the minimum of $x$ and $y$ .

The correct answer is Option B.

Q7 · General Aptitude · MCQ · 2 marks

Six people are seated around a circular table. There are at least two men and two women. There are at least three right-handed persons. Every woman has a left-handed person to her immediate right. None of the women are right-handed. The number of women at the table is:

(A)
2
(B)
3
(C)
4
(D)
Cannot be determined

Answer (official key): A

Solution

None of the women are right-handed, so all women are left-handed.

There are at least three right-handed persons among six people. Hence, there can be at most three left-handed persons.

So the number of women is at most 3.

Now suppose there are 3 women. Since every woman has a left-handed person immediately to her right, and all left-handed persons would then be women, each woman must have another woman immediately to her right. Around a circular table, this would force all six people to be women, contradicting the condition that there are at least two men.

Therefore, the number of women cannot be 3. Since there are at least two women, the number of women is exactly:

2.

Therefore, the correct answer is Option A.

Q18 · Digital Logic · Numerical · 1 mark

Consider the Karnaugh map given below, where x represents “don't care” and blank represents 0.

Karnaugh map for Q21

Assume for all inputs $(a,b,c,d)$ , the respective complements $(\overline{a},\overline{b},\overline{c},\overline{d})$ are also available. The above logic is implemented using 2-input NOR gates only.

The minimum number of gates required is __________.

Answer (official key): 1

Solution

Using the don't-care cells, the Karnaugh map can be grouped as one 4-cell group.

The simplified function is:

F=\overline{a}c.

Since complements of inputs are also available, this can be implemented using a single 2-input NOR gate as:

F = \overline{a}c = \overline{a+\overline{c}}.

So only one 2-input NOR gate is required.

Therefore, the answer is 1.

Q61 · Programming and Data Structures · Numerical · 2 marks

The output of executing the following C program is __________.

#include <stdio.h>

int total(int v) {
    static int count = 0;
    while (v) {
        count += v & 1;
        v >>= 1;
    }
    return count;
}

void main() {
    static int x = 0;
    int i = 5;
    for (; i > 0; i--) {
        x = x + total(i);
    }
    printf("%d\n", x);
}

Answer (official key): 23

Solution

The function total(v) counts the number of set bits in v, but count is declared as static, so it retains its value across calls.

Compute each call:

Call	Binary	Set bits added	Returned cumulative count
`total(5)`	`101`	2	2
`total(4)`	`100`	1	3
`total(3)`	`011`	2	5
`total(2)`	`010`	1	6
`total(1)`	`001`	1	7

Now x accumulates these returned values:

x=2+3+5+6+7=23.

Therefore, the output is 23.

Q62 · Computer Organization and Architecture · Numerical · 2 marks

A cache memory unit with capacity of $N$ words and block size of $B$ words is to be designed.

If it is designed as a direct-mapped cache, the length of the TAG field is 10 bits.

If the cache unit is now designed as a 16-way set-associative cache, the length of the TAG field is __________ bits.

Answer (official key): 14

Solution

For the same cache capacity and block size, changing from direct-mapped to 16-way set-associative reduces the number of index bits by:

\log_2 16 = 4.

Those 4 bits move from the index field into the tag field.

Given direct-mapped tag length:

10\text{ bits}.

Therefore, 16-way set-associative tag length is:

10+4=14\text{ bits}.

Therefore, the answer is 14.

The other 60 questions are solved in your report when you take GATE 2017 CS (Set 1) as a 3-hour test.

Take GATE 2017 CS (Set 1) as a test

Official answer key

GATE 2017 CS (Set 1) answer key

All 65 answers from the official key. Numerical answers are ranges; “or” means the key accepts either answer.

Download the official GATE 2017 CS (Set 1) answer key (PDF)

Q	Topic	Type	Marks	Answer
1	Verbal Aptitude	MCQ	1	D
2	Quantitative Aptitude	MCQ	1	D
3	Analytical Aptitude	MCQ	1	C
4	Verbal Aptitude	MCQ	1	C
5	Quantitative Aptitude	MCQ	1	C
6	Quantitative Aptitude	MCQ	2	B
7	Analytical Aptitude	MCQ	2	A
8	Verbal Aptitude	MCQ	2	B
9	Quantitative Aptitude	MCQ	2	D
10	Spatial Aptitude	MCQ	2	C
11	Databases	MCQ	1	A
12	Theory of Computation	MCQ	1	B
13	Computer Organization and Architecture	MCQ	1	D
14	Algorithms	MCQ	1	B
15	Algorithms	MCQ	1	C
16	Probability and Statistics	Numerical	1	0
17	Compiler Design	MCQ	1	B
18	Digital Logic	Numerical	1	1
19	Databases	Numerical	1	2.6
20	Discrete Mathematics	Numerical	1	18
21	Discrete Mathematics	MCQ	1	D
22	Computer Networks	MCQ	1	B
23	Operating System	MCQ	1	D
24	Theory of Computation	Numerical	1	4
25	Computer Organization and Architecture	MCQ	1	D
26	Computer Organization and Architecture	MCQ	1	C
27	Linear Algebra	MCQ	1	C
28	Programming and Data Structures	MCQ	1	B
29	Programming and Data Structures	MCQ	1	D
30	Discrete Mathematics	MCQ	1	B
31	Operating System	Numerical	1	3
32	Compiler Design	MCQ	1	C
33	Programming and Data Structures	MCQ	1	B
34	Computer Organization and Architecture	Numerical	1	0.05
35	Computer Networks	MCQ	1	D
36	Theory of Computation	MCQ	2	B
37	Calculus	MCQ	2	C
38	Computer Networks	MCQ	2	C
39	Linear Algebra	MCQ	2	A
40	Programming and Data Structures	MCQ	2	C
41	Databases	MCQ	2	D
42	Operating System	MCQ	2	D
43	Discrete Mathematics	MCQ	2	D
44	Theory of Computation	MCQ	2	D
45	Databases	MCQ	2	A
46	Programming and Data Structures	MCQ	2	A
47	Digital Logic	MCQ	2	B
48	Operating System	MCQ	2	B
49	Theory of Computation	MCQ	2	A
50	Linear Algebra	MCQ	2	B
51	Theory of Computation	MCQ	2	A
52	Algorithms	MCQ	2	A
53	Databases	Numerical	2	4
54	Compiler Design	Numerical	2	1024
55	Computer Networks	Numerical	2	11
56	Programming and Data Structures	Numerical	2	3
57	Computer Organization and Architecture	Numerical	2	1.49 to 1.52
58	Computer Organization and Architecture	Numerical	2	-16
59	Computer Networks	Numerical	2	86.5 to 89.5
60	Discrete Mathematics	Numerical	2	271
61	Programming and Data Structures	Numerical	2	23
62	Computer Organization and Architecture	Numerical	2	14
63	Algorithms	Numerical	2	5
64	Compiler Design	Numerical	2	2
65	Computer Organization and Architecture	Numerical	2	76

Questions about GATE 2017 CS (Set 1)

How many questions are in the GATE 2017 CS (Set 1) paper?

65 questions for 100 marks: 10 General Aptitude questions worth 15 marks and 55 Computer Science & Information Technology questions worth 85 marks. By type, there were 45 MCQs, 20 numerical answer (NAT) questions. The paper lasted 3 hours.

Is there negative marking in GATE 2017 CS (Set 1)?

Yes, for MCQs only. A wrong MCQ answer costs one-third of its marks (−⅓ for a 1-mark question, −⅔ for a 2-mark question). MSQ and numerical (NAT) questions have no negative marking, and an MSQ earns marks only when every correct option is chosen.

Which topics carried the most marks in GATE 2017 CS (Set 1)?

Outside General Aptitude, the biggest topics were Computer Organization and Architecture (12 marks), Programming and Data Structures (11 marks), Theory of Computation (10 marks). The full topic-wise split is in the table on this page.

Were any GATE 2017 CS (Set 1) questions awarded marks to all?

No. Every question was graded with the answer in the official key.

Where does this GATE 2017 CS (Set 1) answer key come from?

From the official answer key published by IIT Roorkee, which organised GATE 2017. Range answers for numerical questions and questions with more than one accepted answer are kept exactly as the key gives them.

GATE 2017 CS (Set 1) question paper PDF and answer key

GATE 2017 CS (Set 1) topic-wise marks

5 solved questions from GATE 2017 CS (Set 1)

GATE 2017 CS (Set 1) answer key

Questions about GATE 2017 CS (Set 1)

How many questions are in the GATE 2017 CS (Set 1) paper?

Is there negative marking in GATE 2017 CS (Set 1)?

Which topics carried the most marks in GATE 2017 CS (Set 1)?

Were any GATE 2017 CS (Set 1) questions awarded marks to all?

Where does this GATE 2017 CS (Set 1) answer key come from?

More GATE papers