Infosys Previous Papers with Solutions PDF

Infosys Placement Papers with Solutions 2021

Infosys Previous Year Placement Papers for 2021 graduates can be found out on this page. The pattern for Infosys Placement Papers 2021 is completely changed now. PrepInsta provides you all the updated information regarding the change in pattern of Infosys Placement Papers 2021.

Infosys Previous Year Paper 2021

Name of the Section	No. of questions
Reasoning Ability	15 Question
Mathematical Ability	10 Question
Verbal Ability	20 Question
Pseudo Code	5 Question
Puzzle Solving	4 Question

Note: Registered Candidates will appear for the Infosys online Test on the confirmed date and time using the link and One-Time Password shared with them by Infosys Talent Acquisition Team.

Section

Time Duration

100 minute

Adaptive/ Non- Adaptive

Adaptive

Negative Marking

Detailed Course Syllabus

(Updated) Infosys Placement Preparation for 2021

Infosys has completely changed its pattern this year. Get detailed information about Infosys Exam Pattern 2021.

Quants
Logical Reasoning
English Verbal
Pseudo Code
Puzzle Solving

Note: Questions from “Cryptarithmetic” may be asked in Logical Section. In total, around 1 to 2 questions are asked from Cryptarithmetic in Infosys Placement Exam.

Infosys Placement Paper for Freshers

Section	No. of questions	Total Time
Reasoning Ability	15 Questions	25 mins
Mathematical Ability	10 Questions	35 mins
Verbal Ability	20 Questions	20 mins
PseudoCode	5 Questions	10 mins
Puzzle Solving	4 Questions	10 mins
Total	54 Questions	100 mins

Infosys Previous Year Quants Questions

Below mentioned the Updated Infosys Quants Topics for Infosys Placement Papers 2021. Make sure that you also like our Facebook page here we generally post the latest questions here.

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Number of Questions – 10
Time Limit – 35 mins

Infosys Aptitude Questions

Infosys Logical Reasoning Questions and Answers

Here below you will get detailed information about Infosys Logical Reasoning Curriculum.

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Number of Questions – 15
Time Limit – 25 mins

Infosys Logical Questions

Infosys Cryptarithmetic Questions 2021-2022

Definition of Cryptarithmetic : An arithmetic Question in which verbal letters have been replaced with numerals and which is answered by finding all possible pairings of digits with letters that produce numerically perfect solutions or the correct answers.

We have Designed this Cryptarithmetic Page for beginners to help them understand the Basics of Cryptarithmetic.

Prepare with Previous Year Cryptarithmetic Question

Infosys Cryptarithmetic Questions

Infosys Verbal English Questions and Answers

Infosys recent curriculum Verbal English Topics and questions and answers are given below:-

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Number of Questions – 20 questions
Time Limit – 20 mins

Infosys Verbal Questions

Infosys PseudoCode Question and Answer 2021 Topics

The Pseudocode round will have cover questions from the following topics:

Basic of C
C++
Data Structures

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Number of Questions- 5 Question
Total Time- 10 minutes

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Infosys Puzzle Solving Section

This is the newly introduced section of Infosys Placement Papers 2021. You will be provided a few Mathematical Puzzle questions to solve.

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Number of Questions- 4 Questions
Total Time- 10 Question

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1. The number of votes not cast for the PNC Party increased by 25% in the National General Election over those not cast for it in the previous Assembly Polls, and the PNC Party lost by a majority twice as large as that by which it had won the Assembly Polls. If a total 2,60,000 people voted each time, how many voted for the PNC Party in the previous Assembly Polls?

1,10,000

21.23%

1,50,000

30.56%

1,40,000

34.4%

1,20,000

13.82%

Total Votes = 2,60,000

Let x voters voted against the party in the Assembly Poll.

Then votes in favour = 260000 – x

Therefore, majority of votes by which party won in previous poll = 260000– x – x = 260000 – 2x

Next year votes against the PNC party increase by 25%

So, votes against the party in general election = 1.25 x

And votes polled in favour of the party = total votes – votes against = 260000 – 1.25x

Therefore, majority of votes by which party lost in general election

= 1.25x – (260000 – 1.25x) = 2.5x – 260000

It is given that, PNC Party lost by a majority twice as large as that by which it had won the Assembly Polls, Therefore

2.5x – 260000 = 2(260000 – 2 x)

⇒ 2.5x – 260000 = 2 260000 – 4x

⇒ 6.5x = 3260000⇒x == 1,20,000

Therefore, votes polled by the voters for the party in Assembly Polls for previous year

= (2,60,000 – x) = (2,60,000 – 1,20,000) = 1,40,000.

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2. A train is going from Mumbai to Pune and make 5 stops on the way. Three persons enter the train after it has started from Mumbai with 3 different tickets. How many different sets of tickets may they have had?

15P3

20.73%

3×15P3

23.46%

15C3

44.31%

15C3/3

11.49%

Since three persons enter the train after it has started from Mumbai, so they could have entered at:

First station (from where they could have bought tickets for the 2nd, 3rd, 4th or 5th station or for Pune i.e. 5 tickets).

Second station (from where they could have bought tickets for 3rd, 4th or 5th stations or for Pune i.e. 4 tickets)

3rd station (from where they could have bought tickets for the 4th or 5th stations or for Pune ie 3 tickets).

4th station (from where they could have bought tickets for the 5th station or for Pune i e. 2 tickets).

5th station (from where they could have bought a ticket for Pune i.e. 1 ticket)

Thus, we can see that there are 5+4+3+2+1=15 tickets available out of which 3 tickets are to be selected This can be done in 15C3 ways.

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3. Spider has one sock and one shoe for each of its 8 legs. In how many different ways can the spider put on its socks and shoes, assuming that in each leg the sock has to be put on before the shoe?

(8!)^2/2

33.33%

16!/2^8

35.59%

(16!)^2/2^8

13.79%

None of These

17.29%

Option # 2

Let the spider's legs be numbered 1. 2, 3, …, 8.

Let a_k, and b_k, denote the Sock and the shoe respectively for the k^thleg where 1 ≤ k ≤ 8.

So a total of sixteen objects are there (a₁b₁a₂b₂a₃b₃…. a₈b₈) which are to be arranged such that a_kcomes before b_k

16 objects can be arranged in 16! ways, Out of these the arrangements in which a_kcomes before b_k=12×16!

Similarly out of the remaining, the arrangement in which a₂comes before b₂

=12×12×16! and so on

Finally arrangements in which a_kcomes before b_k=(12)8×16!

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4. One man can do as much work in one day as a woman can do in 2 days. A child does one-third the work in a day as a woman. If an estate owner hires 39 pairs of hands—men, women and children in the ratio 6 : 5 : 2 and pays them in all Rs. 1, 113 at the end of the day’ s work, what must the daily wages of a child be, if the wages are proportional to the amount of work done ?

Rs 14

21.12%

Rs 5

19.9%

Rs 20

24.8%

Rs 7

34.18%

Ratio of number of men, women and children = 6 : 5 : 2 and the total number of men, women and children = 39.

Therefore the number of men = 18, women = 15 and children = 6.

Ratio of work done by men:women: children = 6 : 3 : 1

∴ Ratio of work done by 18 men, 15 women and 6 children

= (18 × 6) : (15 × 3) : (6 × 1) = 108 : 45 : 6

Hence Rs.1113 would be divided in this ratio.

That makes Rs756 for men, Rs 315 for women and Rs42 for children. Hence 6 children earn Rs42 in a day. So the daily

wage of a child is equal to 42/6 = Rs7

Hence, option d.

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5. Let S denote the infinite sum 2 + 5x + 9x² + 14x³ + 20x⁴ + …, where | x | < 1, then S equals

{(2-x)/(1-x)^3}

30.89%

{(2-x)/(1+x)^3}

24.38%

{(2+x)/(1-x)^3}

28.99%

{(2+x)/(1+x)^3}

15.74%

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6. In a hockey tournament, there are 12 teams, divided into 2 groups. Each group has 6 teams. Teams of each group will play once against each other. From each group, 4 top teams will qualify for the second round. In the second round, each team will play against every other team in its group once. 4 top teams of the second round will qualify for the semifinal round, where they will play against each other once. The top 2 teams will proceed to the final. The final will be played in the form of the best of three matches. The minimum and a maximum number of matches in this tournament will respectively be:

47 , 15

31.67%

48 , 12

39.39%

32 , 14

16.36%

44 , 11

12.58%

Explanation:

12 teams divided into 2 groups and using permutations and Combination we get 47 maximum and 15 minimum tournaments.

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7. Raj was given a task to arrange the letters of the word ‘SWITZERLAND”. In how many ways can he arrange the letters of the word “SWITZERLAND” so that the positions of the vowels are used by vowels only and positions of the consonants are used by the consonants only?

11! / 8!

9.97%

11! / (3! * 8!)

39.09%

3! * 8!

36.92%

11! / 3!

14.02%

In 11 letters we have 3 vowels and 11 consonants .

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8. If apple is coded as jam, the jam is coded as chalk, chalk is coded as rock, rock is coded as sugar and sugar is coded as apple, which one of the following sets does contain only those things that are edible?

Sugar, jam, rock

11.43%

Sugar, jam, apple

32.96%

Jam, rock, apple

20.22%

Apple, sugar, rock

35.39%

Explanation : The correct answer is : Apple, sugar, rock

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9. If a man is given 5 numerical digits and 4 alphabets, how many alphanumeric words can be formed using exactly 2 digits and 3 alphabets?

46.57%

120

23.39%

3600

10.41%

4800

19.64%

Explanation: 5 $C_{ 2}* 4 C^{ 3}$

= 40

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10. Which statement has to be correct from the following?

1)Tourists are able to clearly see the City of New York on top of the Empire State Building.

2)From the Empire state building, tourists are able to clearly see the city of new york.

3) Tourists are able to see the city of new york clearly from the top of the empire state building.

4) From the top of the Empire state building, the city of New York was clearly seen by the tourists.

9.36%

15.53%

41.55%

33.56%

Explanation : Correct option : 4

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11. Find LCM of 12, 15, 20, 27.

500

3.83%

540

77.5%

600

6.36%

None

12.31%

Explanation :

12 = 2×2×3 = 2^2 ×3^1

15 = 3× 5

20 = 2×2×5 = 2^2×5

27 = 3×3×3 = 3^3

HCF = 1

LCM = 2^2 × 3^3 × 5 = 540

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12. A does a work in 10 days, B in 6 days. Find the number of days they can complete the work together?

15/4

74.04%

16/3

12.31%

18/4

4.52%

None

9.13%

Explanation : Number of days ther are work together = 1/10 + 1/6 = 4/15 = 15/4 days.

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13. Which statement is grammatically correct?

1) Broadly speaking, the curriculum includes all experiences which the student may have within the environment of the school.

2) To speak broadly, curriculum includes all experiences which the student may have within the environment of the school.

3) Broadly speaking, the curriculum means all the experiences of the student in the school environment.

4) All the experiences of the student in the school environment are curriculum to speak broadly.

35.6%

18.32%

31.34%

14.75%

Explanation : Correct answer is 3.

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14. Find the opposite of Literal.

Dishonest

72%

Truth

10.45%

Honest

11.31%

Care

6.24%

Explanation : Opposite of Literal is Dishonest.

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15. In a farm, 50 hens lay 200 eggs in 2 days, how many days that 20 hens to lay 400 eggs?

234

15.25%

345

12.83%

356

10.73%

400

61.2%

Solution: Apply unitary method,

50 hens --- 200 eggs ---- 2 days

50 hens --- 100 eggs ---- 1 day

1 hen --- 2 eggs ---- 1 day

20 hens --- 40 eggs ---- 1 day

So 20 hens can lay 40 eggs in 1 day

then 20 hens can lay 400 eggs in 10 days (40*10=400)

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16. The population of a town 3 years ago was ‘ b ‘ and the population of the town 3 years from now will be ‘ c’. What is the current population of the town?

√(bc)

74.19%

√(dc)

7.56%

√(ac)

10.23%

√(c)

8.02%

Let the population increases at the rate of r per year and present population is x, then

b[1+(r/100)]^3 = x

[1+(r/100)]^3 = x/b

Also

x[1+(r/100)]^3=c

x*x/b= c or x^2= bc

Hence

x= √(bc)

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17. 4 men can repair a road in 7 hours. How many men are required to repair the road in 2 hours?

11.91%

9.45%

7.48%

71.16%

Explanation :

No. of Men x Time taken / Work = Constant

So, here we apply

M1 x T1/ W1 = M2 x T2 / W2

Given that, M1 = 4 men, T1 = 7 hours; T2 = 2 hours, we have to find M2 =?

Note that here, W1 = W2 = 1 road, ie. equal workload.

Clearly, substituting in the above equation we get,

4 * 7 = M2 * 2

M2= 14 men.

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18. Ram is a five-year elder to his youngest sibling Shreya. Shreya is two years younger than her brother Ritesh. Ritesh is 13 years old and is Ram’s brother. How old will ram be in two years from now?

7.24%

71.16%

15.86%

5.74%

Solution:

Let shreya's age be x, then Ram's age = x+5

Ritesh's age= x+2

given,

x+2=13

therefore, x=11

hence,

Ram's age= 16

Ram's age after 2 yrs=16+2=18 yrs

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19. NO+NO+TOO=LATE where E=2 value of O+L+E

66.4%

12.91%

12.09%

8.6%

Solution: 4+4+4=12 .

ONLY GIVES 2 AT UNIT PLACE

74+74+944=1092

O+L+E=7=N

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20. Out of 52 cards, 4 cards are drawn at random. What is the probability of getting 1 king, 1 queen, 1 jack?

0.0128

44.49%

0.1564

20.66%

0.213

19.56%

0.0987

15.29%

Solution: 4c1*4c1*4c1*52c1/52c4

= 256/20825

= 0.0128

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21. Statement: The Total money earned by Salesman.

Statement:

They sold 3 cars at Rs. 3475.
He gained 20% commission for each car.

Both statements 1 and 2 together are sufficient.

67.27%

Only first statement is sufficient .

14.44%

Only second statement is sufficient.

7.68%

Neither both are sufficient.

10.61%

Explanation:

Solution:

According to question, Selling Price of 3 cars = Rs. 3475

Commission earned = 20%

So, 1158 * 20% = 1158*20100=Rs. 231.60

Total earning= 231.60 * 3= Rs. 695

Both statements 1 and 2 together are sufficient to answer the question asked but neither statement alone is sufficient.

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22. Find the next number in series 9, 30, 68, 130, ?

222

49.09%

223

29.34%

345

13.01%

322

8.56%

Explanation :

Solution: 3^3 + 3= 30

4^3 + 4= 68

5^3 + 5= 130

6^3 + 6= 222

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23. What will be the 11th term in the given series:

5, 10, 17, 26, 37, 50, ….. ?

123

17.99%

244

12.98%

442

4.47%

145

64.56%

Solution: 5-10 = 5

17- 10 = 7

26 – 17 = 9

37 – 26 = 11

We can see that an A.P. is formed of odd numbers .

So the series will be 5, 10, 17, 26, 37, 50, 65, 82, 101, 122, 145…

The 11th term is 145

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24. Find LCM of 12, 15, 20, 27.

500

4.75%

540

82.2%

600

4.55%

none

8.49%

Explanation :

Sol: 12 = 2×2×3 = 2^2 ×3^1

15 = 3× 5

20 = 2×2×5 = 2^2×5

27 = 3×3×3 = 3^3

HCF = 1

LCM = 2^2 × 3^3 × 5 = 540

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