DXC Technology Placement Papers 2020

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  • DXC  Technology curriculum pattern with Most repeated questions of similar pattern of DXC Technology and
  • DXC Technology sample question papers that were asked in DXC Technologies Previous year paper.

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DXC Placement Papers 2020

Difficulty Level

Sections

4

Time

60mins

Adaptive/Non Adaptive

No

Negative Marking

No

Applicant: 150000

Job opening: 10000/ year

Curriculum

DXC technology Previous year placement paper

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To know about DXC syllabus

DXC Technologies Placement Selection Process

DXC written test paper and latest curriculum are available that will help you in DXC online exam 2020

  • Quants
    • Number of Questions - 15 Questions
    • Time - 15 mins
  • English
    • Number of Questions - 15 Questions
    • Time - 15 mins
  • Logical
    • Number of Questions - 15 Questions
    • Time - 15mins
  • Technical/Computer Programming
    • Number of Questions - 15 Question
    • Time - 15mins

DXC Technology has four written sections :

  • Written Rounds(Quantitative +Verbal+Logical Reasoning+Computer Programming)
  • Technical Round
  • HR Round

All Branches – English Usage + Analytical Reasoning + Numerical Reasoning

  • Ques – 15 each
  • Total Time – 45 mins + 25 mins(WET)

Written English Test (WET) – Essay type question. Need to Write an article in 25 mins.

Branch Specific – English Usage + Analytical Reasoning + Numerical Reasoning + Domain Module (20 min)

  • Ques – 15 each/20 for Domain Module
  • Total time – 45 mins+ 25 mins(WET) +20 mins (Domain Module)

DXC Technology Placement Paper with Latest Solutions

DXC Technology has four written sections :

  • Written Rounds(Quantitative +Verbal+Logical Reasoning+Computer Programming)
  • Technical Round
  • HR Round

All Branches – English Usage + Analytical Reasoning + Numerical Reasoning

  • Ques – 15 each
  • Total Time – 45 mins + 25 mins(WET)

Written English Test (WET) – Essay type question. Need to Write an article in 25 mins.

Branch Specific – English Usage + Analytical Reasoning + Numerical Reasoning + Domain Module (20 min)

  • Ques – 15 each/20 for Domain Module
  • Total time – 45 mins+ 25 mins(WET) +20 mins (Domain Module)

DXC Technology Placement Paper with Latest Solutions

DXC Syllabus No. Of Questions Time Difficulty Cut Off
Quants 15 15 Medium 80 percentile
Logical 15 15 Hard 80 percentile
English 15 15 Medium 80 percentile
Computer Programming 15 15 Hard 75 percentile
Deloitte Logical Test 2020

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To know more about Logical

DXC Technology QUESTIONS WITH ANSWERS LOGICAL REASONING

Difficulty Level: Easy - high
Importance: Moderate
Suggested time to solve: 55secs - 1min 20secs

dxc previous year paper with solutions

Click here to visit Syllabus Page

To know more about Verbal

DXC Technology MOST REPEATED QUESTION PAPERS VERBAL ENGLISH

Difficulty Level: Easy-Medium
Importance: High
Suggested time to solve: 40secs - 1min 30secs

DXC Previous year Paper with Solutions

Click here to visit Syllabus Page

To know more about Technical

DXC Technology Latest Questions Paper 2019 COMPUTER PROGRAMMING

  • C/C++/Java Concepts
  • Computer Fundamentals
  • Operating Systems
  • Data Structures
  • DBMS

Difficulty Level: Medium
Importance: Easy - High
Suggested time to solve: 40secs - 60secs

Paid materials and Online Classes

Trending

Preparation Section

DXC Quants Question

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DXC Quants Dashboard

Difficulty Level: Med-High
Importance: Medium
Suggested time to solve: 1min each

Question #1

Article A was sold at 20% profit and Article B was sold at 60% profit. The selling price of Article B was 150% more than that of Article A. Cost price of Article B is what percent more than that of Article A?

A

87.5%

87.5%

40%
B

86.5%

86.5%

36%
C

23.5%

23.5%

20%
D

None

None

4%
  • PrepInsta Explanation

Solution: (SP(A))/(CP(A)) = 120/100____________ (1)

(SP(B))/(CP(B)) = 160/100____________ (2)

(SP(B))/(sp(A)) = 250/100____________

Given (1)/ (2) =(CP(B))/(CP(A)) x100/250 = 120/100 x 100/160 =(CP(B))/(CP(A)) = 15/8 = (CP(B)-CP(A))/(CP(A)) x100 =(15-8)/8 x 100 =87.5%

Question #2

How many pairs of integers (x, y) exist such that the product of x, y and HCF (x, y) = 1080?

A

8

8

29.17%
B

7

7

25%
C

9

9

22.92%
D

12

12

22.92%
  • PrepInsta Explanation

We need to find ordered pairs (x, y) such that xy * HCF(x, y) = 1080.
Let x = ha and y = hb where h = HCF(x, y) => HCF(a, b) = 1.
So h3(ab) = 1080 = (23)(33)(5).
We need to write 1080 as a product of a perfect cube and another number.

Four cases:
1. h = 1, ab = 1080 and b are co-prime. We gave 4 pairs of 8 ordered pairs (1, 1080), (8, 135), (27, 40) and (5, 216). (Essentially we are finding co-prime a,b such that a*b = 1080).

2. h = 2, We need to find number of ways of writing (33) * (5) as a product of two co-prime numbers. This can be done in two ways - 1 and (33) * (5) , (33) and (5)
number of pairs = 2, number of ordered pairs = 4

3. h = 3, number of pairs = 2, number of ordered pairs = 4

4. h = 6, number of pairs = 1, number of ordered pairs = 2

Hence total pairs of (x, y) = 9, total number of ordered pairs = 18.
The pairs are (1, 1080), (8, 135), (27, 40), (5, 216), (2, 270), (10, 54), (3, 120), (24, 15) and (6, 30).

Question #3

A prime number p greater than 100 leaves a remainder q on division by 28. How many values can q take?

A

8

8

17.65%
B

12

12

38.24%
C

9

9

23.53%
D

15

15

20.59%
  • PrepInsta Explanation

q can be 1.
If q =2, number would be of the form 28n + 2 which is a multiple of 2.

Similarly, when q = 4, number would be of the form 28n + 4 which is again a multiple of 2. Any number of the form 28n + an even number will be a multiple of 2.

When q = 7, number would be of the form 28n + 7 which is a multiple of 7.

So, the only remainders possible are remainders that share no factors with 28. Or numbers that are co-prime to 28.

There is a formula for this and a shorter way of finding the number of numbers co-prime to a given natural number. A more detailed discussion on this is provided here and here.

1, 3, 5, 9, 11, 13, 15, 17, 19, 23, 25 and 27. q can take 12 different values

Question #4

The difference between the compound interest and the simple interest on a certain sum at 12% p.a. for two years is Rs.90. What will be the value of the amount at the end of 3 years?

A

9000

9000

24.14%
B

6250

6250

34.48%
C

8530.80

8530.80

20.69%
D

8780.80

8780.80

20.69%
  • PrepInsta Explanation

Solution:

The difference in the simple interest and compound interest for two years is on account of the interest paid on the first year\'s interest, when interest is reckoned using compound interest, interest being compounded annually.
Hence 12% of simple interest = 90 => simple interest = 90/0. =750.

As the simple interest for a year = 750 @ 12% p.a., the principal = 750/0.2 = Rs.6250.

If the principal is 6250, then the amount outstanding at the end of 3 years = 6250 + 3(simple interest on 6250) + 3 (interest on simple interest) + 1 (interest on interest on interest) = 6250 + 3(750) + 3(90) + 1(10.80) = 8780.80.

Question #5

Two persons A and B start moving at each other from point P and Q respectively which are 1400 Km apart. Speed of A is 50 Km/hr and that of B is 20 Km/hr. How far is A from Q when he meets B for the 22nd time?

A

1000 km

1000 km

16%
B

400 km

400 km

68%
C

800 km

800 km

4%
D

1400 km

1400 km

12%
  • PrepInsta Explanation

Total distance travelled by both of them for 22nd meeting = 1400 + (21 * 2 * 1400) = 43 * 1400
Distance travelled by each will be in proportion of their speed :
Therefore, distance travelled by A = 50/(50 + 20) * 43 * 1400 = 43000 (Note - Always do complicated calculations at last because things cancel out generally)
Now, for every odd multiple of 1400, A will be at Q and for every even multiple of 1400 A will be at P. So, at 42000 Km (1400 x 30, even multiple) A will beat P. So at their 22 meeting, A will be 1000 Km from P, therefore, 400 Km from Q.

Question #6

A and B take part in a duel. A can strike with an accuracy of 0.6. B can strike with an accuracy of 0.8. A has the first shot, post which they strike alternately. What is the probability that A wins the duel?

A

7/10

7/10

14.29%
B

15/23

15/23

38.1%
C

2/3

2/3

28.57%
D

11/17

11/17

19.05%
  • PrepInsta Explanation

A can win in the following scenarios:
A strikes in first shot.
A misses in the first shot, B misses in the second, A strikes in the third.
A misses in the 1st shot, B misses in the 2nd, A misses the 3rd, B misses the 4th and A strikes the 5th.
And so on…
P(A in 1st shot) = 0.6
P(A in 3rd shot) = 0.4 * 0.2 * 0.6 {A misses, then B misses and then A strikes}
P(A in 5st shot) = 0.4 * 0.2 * 0.4 * 0.2 * 0.6 {A misses, then B misses and then A misses, then B misses, then A strikes}
Overall probability = Sum of all these

= 0.6 + 0.4 * 0.2 * 0.6 + 0.4 * 0.2 * 0.4 * 0.2 * 0.6 + …
Which is nothing but
0.6 + (0.4 * 0.2) * 0.6 + (0.4 * 0.2)2 * 0.6 + (0.4 * 0.2)3 * 0.6…

This is an infinite geometric progression with first term 0.6 and common ratio 0.4 * 0.2
Required Probability = a1−r = 0.61−0.08 = 0.60.92 = 3046 = 1523

Question #7

If log2X + log4X = log0.25√6 and x > 0, then x is

A

6^(-1/6)

6^(-1/6)

23.81%
B

6^(1/6)

6^(1/6)

38.1%
C

3^(-1/3)

3^(-1/3)

28.57%
D

6^(1/3)

6^(1/3)

9.52%
  • PrepInsta Explanation

log2x + log4x = log0.25√6
We can rewrite the equation as:
log2x + log2xlog24 = log0.25√6
log2x * 32 = log0.25√6
=> log2x * 3 = 2log0.25√6
=> log2x3 = -log46
=> log2x3 = −log26log24
=> log2x3 = −log266

=> 2log2x3 = -log26
2log2x3 + log26 = 0
log26X6 = 0
6x6 = 1
x6 = 16
x = 16−−√6

Question #8

Traders A and B buy two goods for Rs. 1000 and Rs. 2000 respectively. Trader A marks his goods up by x%, while trader B marks his goods up by 2x% and offers a discount of x%. If both make the same non-zero profit, find x

A

25%

25%

31.58%
B

12.5%

12.5%

36.84%
C

37.5%

37.5%

21.05%
D

40%

40%

10.53%
  • PrepInsta Explanation

SP of trader A = 1000 (1 + x).
Profit of trader A = 1000 (1 + x) – 1000.
MP of trader B = 2000 (1 + 2x).
SP of trader B = 2000 (1 + 2x) (1 – x).

Profit of trader B = 2000(1 + 2x) (1 – x) – 2000.
Both make the same profit => 1000(1 + x) – 1000 = 2000(1 + 2x) (1 – x) – 2000
1000x = 2000 – 4000x2 + 4000x – 2000x – 2000
4000xx2 -1000x = 0
1000x (4x – 1) = 0
=> x = 25%

Question #9

Let the nth term of AP be defined as tn, and sum up to 'n' terms be defined as Sn. If |t8| = |t16| and t3 is not equal to t7, what is S23?

A

23(t16 - t8)

23(t16 - t8)

22.73%
B

zero

zero

45.45%
C

23t11

23t11

22.73%
D

Cannot be determined

Cannot be determined

9.09%
  • PrepInsta Explanation

t8|=|t16|. This can happen under two scenarios t8 = t16 or t8 = – t16.
If t8 = t16, the common difference would be 0 suggesting that t3 would be equal to t7.
However, we know t3 is not equal to t7, so the common difference cannot be zero.
This tells us that t8 = – t16 Or, t8 + t16 = 0.
If t8 + t16 = 0, then t12 = 0. t12 = t8 + 4d, and t16 – 4d So, t12 = t8+t162
For any two terms in an AP, the mean is the term right in between them.
So, t16 is the arithmetic mean of t8 and t16.
So, t12 = 0.
Now, S23 = 23 * t12. We know that average of n terms in an A.P. is the middle term.
This implies that sum of n terms in an A.P., is n times the middle term. So, S23 = 0.

Question #10

Two circles are placed in an equilateral triangle as shown in the figure. What is the ratio of the area of the smaller circle to that of the equilateral triangle

Geometry - triangles: CAT Geometry Triangles Area

A

π:36√3

π:36√3

12%
B

π:18√3

π:18√3

40%
C

π:27√3

π:27√3

44%
D

π:42√3

π:42√3

4%
  • PrepInsta Explanation

In-radius of equilateral triangle of side a = a2√3
Diameter of larger circle = a2√3

Let us say common tangent PQ touches the two circle at R, center of smaller circle is I.
Now, PQ is parallel to BC. AR is perpendicular to PQ. Triangle PQR is also an equilateral triangle and AORID is a straight line. (Try to establish each of these observations. Just to maintain the rigour.)

AD = √32

a
RD = a√3
AR = √32a - a√3 = 3a−2a2√3 = a2√3
AR = 13 AD
Radius of smaller circle = 13 radius of larger circle
Radius of smaller circle = 13 * a2√3 = a6√3
Area of smaller circle = πr2
π (a6√3)2 = πa2108
Area of triangle = √34a2
Ratio = πa2108 : √34a2 = π:27√3

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DXC Logical Question

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DXC Logical Dashboard

Question #1

Directions for Ques 1 to 2:

Six techies- A,B,C,D,E,F attended a Tech-conference in Mumbai. Each of them was working in differnt companies Microsoft, Yahoo, Google, FB, Cisco, TCS in any order and each of them attended the conference on a different day among- Monday, Tuesday, Wednesday, Thrusday, Friday, Saturday not necessarily in the same order. The following information is also given:

  1. A attended the con ferenceon Monday and the person who works in Microsoft  attended the conference on Saturday.
  2. B and D attended the conference on the consecutive days.
  3. Exactly one person attended the conference between E and C and C was working in Google.
  4. F was working in Yahoo.
  5. C attended the conference on the next day of FB employee.
  6. The person who was working in Cisco attended the conference on Friday.

Which of the following person was working in TCS?

 

A

B

B

22.22%
B

C

C

44.44%
C

D

D

11.11%
D

E

E

22.22%
  • PrepInsta Explanation

Monday        A

Tuesday        E/C

Wednesday   F         Yahoo

Thrusday      C/E

Friday           B/D      Cisco

Saturday       D/B     Microsoft

From Statement 5:

Day        Person     Company

Monday      A                   FB

Tuesday      C               Google

Wednesday F               Yahoo

Thrusday     E                TCS

Friday         B/D            Cisco

Saturday     D/B        Microsoft

Question #2

Which of the following person attended the conference on Friday?

A

E

E

12.5%
B

D

D

12.5%
C

B

B

37.5%
D

Cannot be determined

Cannot be determined

37.5%
  • PrepInsta Explanation

Monday        A

Tuesday        E/C

Wednesday   F         Yahoo

Thrusday      C/E

Friday           B/D      Cisco

Saturday       D/B     Microsoft

From Statement 5:

Day        Person     Company

Monday      A                   FB

Tuesday      C               Google

Wednesday F               Yahoo

Thrusday     E                TCS

Friday         B/D            Cisco

Saturday     D/B        Microsoft

Question #3

Direction for Ques 3 to 6:

A quiz has three rounds of two questions each. In the first round, each correct answer carries 20 points;each incorrect answers carries a penalty of 10. If both the questions are answered correctly, a bonus of 10 points is awarded. In the second round, each correct and incorrect answer carries the same number of points as in the first round. However, an additional penalty of 10 points is awarded if both the questions were answered incorrectly. In the third round,  each correct answer is awarded with 40 marks and an incorrect one carries a penalty of 20 points.

In how many ways can a score of 40 be achieved in the quiz?

A

4

4

28.57%
B

2

2

42.86%
C

1

1

14.29%
D

3

3

14.29%
  • PrepInsta Explanation

Solution:

40 can be scored by:

  1. 50-30+20=40
  2. 10+10+20=40

Hence 2 is the correct answer.

Question #4

What is the probability of scoring 100 in the quiz?

A

2/9

2/9

28.57%
B

4/27

4/27

42.86%
C

2/27

2/27

14.29%
D

1/9

1/9

14.29%
  • PrepInsta Explanation

There are 3x3x3=27 possible scores. Out of these 3 there are only 2 ways of scoring 100. Hence 2/27 is the answer.

Question #5

If two answers are incorrect, the minimum possible score is:

A

50

50

42.86%
B

70

70

28.57%
C

60

60

14.29%
D

80

80

14.29%
  • PrepInsta Explanation

The maximum penalty for 2 incorrect answers would occur when they happen in the last round. Hence the score would be 50+50-40=60

Question #6

If only two answers were correct, what is the probability that the score would be 20?

A

1/3

1/3

50%
B

1/6

1/6

33.33%
C

2/3

2/3

16.67%
  • PrepInsta Explanation

There are only two ways of scoring 20 that is 50+10-40 or 10+50-40. In both the cases there are three right answers. Hence the required probability is zero.

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dxc previous year paper with solutions

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English Verbal Dashboard

Question #1

Choose the word which close to opposite of the word:

BLASPHEMOUS

A

irreligious

irreligious

40%
B

inferior

inferior

15%
C

reverent

reverent

25%
D

blarney

blarney

20%
  • PrepInsta Explanation

Currently there is no PrepInsta Explanation.

Question #2

INSIPID : PIQUANT : :

A

tasty : bland

tasty : bland

37.5%
B

relish : spice

relish : spice

18.75%
C

sweet : dessert

sweet : dessert

31.25%
D

flat : spicy

flat : spicy

12.5%
  • PrepInsta Explanation

Solution: Relation of opposites

Question #3

Fill in the blanks:

......... and ecstasy are two sides of a puzzle called life.

A

pain

pain

41.18%
B

penury

penury

11.76%
C

agony

agony

11.76%
D

anguish

anguish

5.88%
E

elation

elation

29.41%
  • PrepInsta Explanation

Currently there is no PrepInsta Explanation.

Question #4

The first line [A] and the last line [f] of each question are fixed. Arrange the other four lines in a logical sequence:

A. But Obama, while acknowledging that Detroit's auto majors would see the move as an added burden on an ailing industry in the short term, made clear that the way forward wa to embrace green technology.

B. Such that political commitment to fight climate change seems lacking in India. Attempts to set fuel emission standards have been stymied by inter-agency squabbling on exactly how much norms should be set.

C. It was only with the recent intervention of the Prime Minister's Office that a consensus was reached.

D. Fuel economy regulations will go some way in reducing India's import bill since we import 78 per cent of our oil.

E. In doing so, he has sent a clear signal that his administration is preparedto act on climate change even in the face of opposition from interest groups such as the car industry.

F. The government should follow Obama's example and act quickly to fight climate change and cut oil imports.

A

DEBC

DEBC

8.33%
B

EBDC

EBDC

41.67%
C

EBCD

EBCD

33.33%
D

DBCE

DBCE

8.33%
E

ECDB

ECDB

8.33%
  • PrepInsta Explanation

Currently there is no PrepInsta Explanation.

Question #5

Choose the correct sequence:

(I)I'm no different.

(II)When I was diagnosed with breast cancer, some years ago, Ihad surgery several times,  and also chemotherapy and radiotherapy.

(III)Today now that I have recovered fully, I thank them all to have bee there for me and have saved my life. If it were not because of them I would have not been writing this.

(IV)The sheer number of individual professionals I encountered was huge- maybe 100, over the years.

(V) When you have undergone intensive NHS treatment, it's routine to feel greatful to the NHS staff who saved your life.

A

V I II IV III

V I II IV III

20%
B

I II III IV V

I II III IV V

10%
C

II I III V IV

II I III V IV

40%
D

None of these

None of these

30%
  • PrepInsta Explanation

Currently there is no PrepInsta Explanation.

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dxc placement paper with solutions

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Computer Programming Dashboard

DXC Technology Latest Questions Paper 2019 COMPUTER PROGRAMMING

  • C/C++/Java Concepts
  • Computer Fundamentals
  • Operating Systems
  • Data Structures
  • DBMS

Difficulty Level: Medium
Importance: Easy - High
Suggested time to solve: 40secs - 60secs

 

Q- Given an array arr[] of N non-negative integers representing height of blocks at index i as Ai where the width of each block is 1. Compute how much water can be trapped in between blocks after raining.
Structure is like below:
|  |
|_|
We can trap 2 units of water in the middle gap.

Bars of Input {3,0,0,2,0,4}

Total Trapped Water= 3+3+1+3= 10

Input:
The first line of input contains an integer T denoting the number of test cases. The description of T test cases follows. Each test case contains an integer N denoting the size of the array, followed by N space separated numbers to be stored in array.

Output:
Output the total unit of water trapped in between the blocks.

Constraints:
1 <= T <= 100
3 <= N <= 107
0 <= Ai <= 108

Example:
Input:
2
4
7 4 0 9
3
6 9 9

Output:
10
0

Explanation:
Testcase 1: Water trapped by block of height 4 is 3 units, block of height 0 is 7 units. So, total unit of water trapped is 10 units.

Solution :

int maxi(int arr[],int ind,int n){
int max=arr[ind];
for(int i=ind+1;i<n;i++){
if(arr[i]>max) max=arr[i];
}
return max;
}
int main() {
//code
int t; cin>>t;
while(t--){
int n; cin>>n; int arr[n];
for(int i=0;i<n;i++) cin>>arr[i];
if(n<2){
cout<<0<<endl;
continue;
}
int lmax=arr[0],rmax=maxi(arr,1,n),sum=0;
for(int i=1;i<n;i++){
if(arr[i]>lmax) lmax=arr[i];
if(arr[i]==rmax and i<n-1)
rmax=maxi(arr,i+1,n);
int h=min(lmax,rmax);
sum+=max(h-arr[i],0);
}
cout<<sum<<endl;
}
return 0;
}

Question #1

Convert the following Infix expression to Postfix form using a stack
x + y * z + (p * q + r) * s, Follow usual precedence rule and assume that the expression is legal.

A

xyz*+pq*r+s*+

xyz*+pq*r+s*+

52%
B

xyz*+pq*r+s+*

xyz*+pq*r+s+*

24%
C

xyz+*pq*r+s*+

xyz+*pq*r+s*+

16%
D

xyzp+**qr+s*+

xyzp+**qr+s*+

8%
  • PrepInsta Explanation

Explanation: The Infix Expression is x + y * z + (p * q + r) * s.
(x y z ) + (p * q + r) * s. ‘+’, ‘*’ are present in stack.
(x y z * + p q * r) * s. ‘+’ is present in stack.
x y z * + p q * r + s * +. Thus Postfix Expression is x y z * + p q * r + s * +.

Question #2

Making the push operation costly, select the code snippet which implements the same.(let q1 and q2 be two queues)

A

public void push(int x)
{
if(empty())
{
q1.offer(x);
}
else{
if(q1.size()>0)
{
q2.offer(x);
int size = q1.size();
while(size>0)
{
q2.offer(q1.poll());
size--;
}
}
else if(q2.size()>0)
{
q1.offer(x);
int size = q2.size();
while(size>0)
{
q1.offer(q2.poll());
size--;
}
}
}
}

public void push(int x)
{
if(empty())
{
q1.offer(x);
}
else{
if(q1.size()>0)
{
q2.offer(x);
int size = q1.size();
while(size>0)
{
q2.offer(q1.poll());
size--;
}
}
else if(q2.size()>0)
{
q1.offer(x);
int size = q2.size();
while(size>0)
{
q1.offer(q2.poll());
size--;
}
}
}
}

44.44%
B

public void push(int x)
{
if(empty())
{
q1.offer(x);
}
else
{
if(q1.size()>0)
{
q1.offer(x);
int size = q1.size();
while(size>0)
{
q2.offer(q1.poll());
size--;
}
}
else if(q2.size()>0)
{
q2.offer(x);
int size = q2.size();
while(size>0)
{
q1.offer(q2.poll());
size--;
}
}
}
}

public void push(int x)
{
if(empty())
{
q1.offer(x);
}
else
{
if(q1.size()>0)
{
q1.offer(x);
int size = q1.size();
while(size>0)
{
q2.offer(q1.poll());
size--;
}
}
else if(q2.size()>0)
{
q2.offer(x);
int size = q2.size();
while(size>0)
{
q1.offer(q2.poll());
size--;
}
}
}
}

22.22%
C

public void push(int x)
{
if(empty())
{
q1.offer(x);
}
else
{
if(q1.size()>0)
{
q2.offer(x);
int size = q1.size();
while(size>0)
{
q1.offer(q2.poll());
size--;
}
}
else if(q2.size()>0)
{
q1.offer(x);
int size = q2.size();
while(size>0)
{
q2.offer(q1.poll());
size--;
}
}
}
}

public void push(int x)
{
if(empty())
{
q1.offer(x);
}
else
{
if(q1.size()>0)
{
q2.offer(x);
int size = q1.size();
while(size>0)
{
q1.offer(q2.poll());
size--;
}
}
else if(q2.size()>0)
{
q1.offer(x);
int size = q2.size();
while(size>0)
{
q2.offer(q1.poll());
size--;
}
}
}
}

16.67%
D

public void push(int x)
{
if(empty())
{
q1.offer(x);
}
else
{
if(q1.size()>0)
{
q2.offer(x);
int size = q1.size();
while(size>0)
{
q2.offer(q2.poll());
size--;
}
}
else if(q2.size()>0)
{
q1.offer(x);
int size = q2.size();
while(size>0)
{
q2.offer(q1.poll());
size--;
}
}
}
}

public void push(int x)
{
if(empty())
{
q1.offer(x);
}
else
{
if(q1.size()>0)
{
q2.offer(x);
int size = q1.size();
while(size>0)
{
q2.offer(q2.poll());
size--;
}
}
else if(q2.size()>0)
{
q1.offer(x);
int size = q2.size();
while(size>0)
{
q2.offer(q1.poll());
size--;
}
}
}
}

16.67%
  • PrepInsta Explanation

Explanation: Stack follows LIFO principle, hence a new item added must be the first one to exit, but queue follows FIFO principle, so when a new item is entered into the queue, it will be at the rear end of the queue. If the queue is initially empty, then just add the new element, otherwise add the new element to the second queue and dequeue all the elements from the second queue and enqueue it to the first one, in this way, the new element added will be always in front of the queue. Since two queues are needed to realize this push operation, it is considered to be costlier.

Question #3

Which of the following is not an application of priority queue?

A

Huffman codes

Huffman codes

19.44%
B

Interrupt handling in the operating system

Interrupt handling in the operating system

30.56%
C

Undo operation in text editors

Undo operation in text editors

30.56%
D

Bayesian spam filter

Bayesian spam filter

19.44%
  • PrepInsta Explanation

Explanation: Undo operation is achieved using a stack.

Question #4

What is a memory efficient double linked list?

A

Each node has only one pointer to traverse the list back and forth

Each node has only one pointer to traverse the list back and forth

34.48%
B

The list has breakpoints for faster traversal

The list has breakpoints for faster traversal

13.79%
C

An auxiliary singly linked list acts as a helper list to traverse through the doubly linked list

An auxiliary singly linked list acts as a helper list to traverse through the doubly linked list

20.69%
D

A doubly linked list that uses bitwise AND operator for storing addresses

A doubly linked list that uses bitwise AND operator for storing addresses

31.03%
  • PrepInsta Explanation

Explanation: Memory efficient doubly linked list has only one pointer to traverse the list back and forth. The implementation is based on pointer difference. It uses bitwise XOR operator to store the front and rear pointer addresses. Instead of storing actual memory address, every node store the XOR address of previous and next nodes.

Question #5

Which of the below diagram is following AVL tree property?
i.data-structure-questions-answers-avl-tree-q3
ii.data-structure-questions-answers-avl-tree-q3a

A

only i

only i

21.88%
B

only i and ii

only i and ii

28.13%
C

only ii

only ii

34.38%
D

i is not a binary search tree

i is not a binary search tree

15.63%
  • PrepInsta Explanation

Explanation: The property of AVL tree is it is height balanced tree with difference of atmost 1 between left and right subtrees. All AVL trees are binary search tree.

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Interview Preparation

dxc technology questions answers 2020

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DXC Technologies Placement Selection Process

DXC Technology has four written sections :

  • Written Rounds
    • Quantitative
    • Verbal
    • Logical Reasoning
    • Computer Programming
  • Technical Round
  • HR Round

Click here to visit to know more

About Interview Experience

Student Details

  • Name: Sam
  • College:  Amity college
  • Type of Interview: On-Campus

Student Interview

Recently DXC Technology, one of the renowned IT companies visited our college for an on-campus interview. The eligibility criteria for this drive was:

Eligibility criterion:

  1. BE/B.Tech (All Streams)
  2. MCA/MSc (Comp Science)
  3. Minimum 65% in BE/B.Tech
  4. Minimum 60% in 10th and 12th

Since I fulfilled this criterion, so I was eligible to participate in the interview. The selection process had three rounds.

  1. Group Discussion
  2. Technical Interview
  3. HR Interview

Round 1- Written Test

For the first round I referred to some of the previous year papers for practice. The pattern of the test was as follows:

  1. Quantitative:  This section included questions on HCF and LCM, Divisibility, probability speed, and distance, Algebra, equations, progression, profit and loss, ratio and percentage, averages, geometry, probability, Clocks and calendars, Geometry, Coordinate geometry, etc.
  2. Logical reasoning: This section included questions on topics: Coding, Directional Sense, seating arrangement, visual reasoning, blood relations, flowcharts, statement, and conclusion, Data Sufficiency, etc.
  3. Computer programming: This section had questions on Data Types, Iteration, Recursion, Decision, Trees, Graphs, Queues, Hash Tables, Heaps, Searching and Sorting, Polymorphism, Encapsulation, Complexity Theory, Searching and sorting, etc.
  4. Logical: This section had questions on Contextual vocabulary, Synonyms, Antonyms, Jumbled up sentences, Prepositions, and Conjunctions.

Tenses and articles etc.

Round 2-Group Discussion

Next was the Group discussion round. All the students who cleared the test successfully were qualified for the Group discussion round. For this round, all the participants were divided into groups, and each group was given a topic. The topics were related to everyday activities and events prevalent around society. My group got the topic “How is IT industry influencing the economy?”.

In this round, we were being judged for our general knowledge, confidence, convincing ability, and communication skills. This round also went well. The list was out within an hour, and I was selected for the next round as well.

Round 3- Technical Interview

This was more of a resume round where I was asked all the technical questions on the previous projects that I had done in my final term internship and the one that was mentioned in my resume. The interviewer also asked me the below-mentioned questions:

  1. Explain SDLC?
  2. List the different storage class specifiers in C?
  3. What is the necessity & advantages of OOPS?
  4. Write the programming code to swap two numbers without using the third variable?
  5. What is networking?
  6. What was your role in your major project?

Mainly, he was checking the details written on my resume. I answered all the questions correctly, and fortunately passed this round as well.

Round 4- HR Interview or Face to Face Interview Round

It was the most comfortable and most relaxed round of all the above. It was more of a cake walk for me. Here the interviewer asked me some personal questions like,

  1. What are your future plans?
  2. What are your strengths professionally?
  3. Tell us about an accomplishment you are most proud of
  4. Do you want to pursue higher education?

Moreover, the interviewer was testing my fundamental knowledge and communication skills. I answered all the raised questions in a polite, clear and confident way. This helped me to clear this round as well.

A Week later I got an email from the company that I got selected for the Job. Hence my interview ended on a happy note.

DXC Technology recent placement paper

Click here t visit DXC interview experience

Interview Experience

DXC Technology latest interview experience is available here.We have collected an interview experience of various students.DXC Technology Experience of every candidate has given here and you can get all the idea of the interview questions asked in the DXC Technology.

We have set of the interview list. The list is given below :

Analytics

Total Number of Questions - 60

Quants - 15

Logical - 15

Verbal - 15

Computer Programming - 15

Overall Time - 60 min

Quants Time - 15 min

Logical Time - 15 min

Verbal Time - 15 min

Computer Programming Time - 15 min

75%

% of people clearing the written round

60%

% of people clearing the interview

20%

% of people joining the company

Quants

Cut off

Difficulty

Importance

Logical

Cut off

Difficulty

Importance

Verbal

Cut off

Difficulty

Importance

Computer Programming

Cut off

Difficulty

Importance

DXC PLACEMENT PAPERS SECTION NUMBER OF QUESTIONS SECTIONAL TIME DIFFICULTY IMPORTANCE
Aptitude 15 15 Medium Medium
Logical Reasoning 15 15 High Medium
Verbal English 15 15 Medium High
Computer Programming 15 15 Medium Very High

DXC Technology Placement Paper Based FAQ”s

What is the selection process for DXC Technologies ?

The selection process of the company consists of four rounds. These rounds are as follows:

  • Written Exam
  • Group Discussion
  • Technical Interview
  • HR Interview

Does DXC Technology hire through oncampus or offcampus ?

Yes, it is hire through offcampus generally . But sometimes it is also hiring through on campus.

From where you have to study for preparation of DXC online written pattern ?

You can get all the preparation material of DXC from PrepInsta . It is provided the best and important topics of each and every section precisely .

What skills are required in DXC Technologies for computer programming ?

Skills like NET ,C, JAVA, C++ are required to get eligible in DXC Technologies.

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