Infosys Placement Papers PDF & Previous Papers Questions 2020

Infosys Placement Papers with Answers PDF 2020

Infosys Placement Papers for the 2020 year can be found out on this page. We have collected all the Infosys Previous Year Papers for you to practice on this page. Analyzing Infosys Off-Campus Drive and Written Test Papers, the level of difficulty for Infosys is high.

PrepInsta provides you all Infosys Placement Papers with Answers PDF 2020 along with Infosys Previous Year Questions with solutions. You will clear the Infosys Written Test Papers and Infosys Online Test Papers round here.

Name Of The Section	Number Of Questions
Logical	15 Ques
Quants	10 Ques
Verbal	40 Ques
Total	65 Ques

IInfosys Placement Papers with Answers PDF 2020

Difficulty Level : High

Sections

Time

95 min

Adaptive/Non-Adaptive

Infosys is Non-Adaptive

Negative Markikng

Yes

Applicants: 3,00,000

Job Opening: 15,000

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Infosys Test Section of Exam 2020

There are the following sections in the test. Infosys Sample papers with solutions also you can find here.

Quants
Logical Reasoning
English Verbal

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

Topic Name	Time	Questions
Logical	35 mins	15 Ques
Quants	25 mins	10 Ques
Verbal	35 mins	40 Ques
Total	95 mins	65 Ques

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Infosys Quants

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Tips for this sections

Number of Questions – 10
Time Limit – 25 mins

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Infosys Logical Reasoning

Fact about this sections

Number of Questions – 15
Time Limit – 35 mins
Difficulty - Moderate

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Infosys Verbal English

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Number of Questions – 40
Time Limit – 35 mins
Difficulty - Easy-Medium

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Infosys Latest Preparation Test Paper 2020

There are the following sections in the test.

Quants
Logical Reasoning
English Verbal

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

Topic Name	Time	Questions
Logical	35	15
Quants	25	10
Verbal	35	40
Total	95 mins	65 ques

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Infosys Crytarithmetic Questions 2020

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 a numerically perfect solutions or the correct answers.

We have Desinged this Cryptarithmetic Page for begineers to help them understand the Basics of Cryptarithmetic.

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Additional Facts for Infosys

Name of Exam	Infosys
CTC	Rs. 3.5 LPA
Sections	3
Mode Conducted	Online
No. of Interviews	2
Type of test	Non - Adaptive

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Infosys Previous Papers 2020 – Quants

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More info about the sections

Number of Questions – 10
Time Limit – 25 mins

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Question #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

22.22%

1,50,000

36.51%

1,40,000

32.14%

1,20,000

9.13%

PrepInsta Explanation

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.

Question #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

22.44%

3×15P3

27.17%

15C3

41.34%

15C3/3

9.06%

PrepInsta Explanation

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.

Question #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

33.49%

16!/2^8

33.49%

(16!)^2/2^8

13.21%

None of These

19.81%

PrepInsta Explanation

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!

Question #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

21.09%

Rs 5

23.13%

Rs 20

29.93%

Rs 7

25.85%

PrepInsta Explanation

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.

Question #5

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

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

19.53%

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

29.69%

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

36.72%

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

14.06%

PrepInsta Explanation

Question #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

22.22%

48 , 12

38.89%

32 , 14

22.22%

44 , 11

16.67%

PrepInsta Explanation

Explanation:

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

Question #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!

13.04%

11! / (3! * 8!)

34.78%

3! * 8!

34.78%

11! / 3!

17.39%

PrepInsta Explanation

In 11 letters we have 3 vowels and 11 consonants .

Question #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

13.64%

Sugar, jam, apple

36.36%

Jam, rock, apple

31.82%

Apple, sugar, rock

18.18%

PrepInsta Explanation

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

Question #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?

37.5%

120

25%

3600

18.75%

4800

18.75%

PrepInsta Explanation

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

= 40

Question #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.

11.11%

33.33%

22.22%

33.33%

PrepInsta Explanation

Explanation : Correct option : 4

Question #11

Find LCM of 12, 15, 20, 27.

500

5.26%

540

52.63%

600

15.79%

None

26.32%

PrepInsta Explanation

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

Question #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

61.9%

16/3

28.57%

18/4

4.76%

None

4.76%

PrepInsta Explanation

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

Question #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.

29.41%

35.29%

5.88%

PrepInsta Explanation

Explanation : Correct answer is 3.

Question #14

Find the opposite of Literal.

Dishonest

57.14%

Truth

21.43%

Honest

17.86%

Care

3.57%

PrepInsta Explanation

Explanation : Opposite of Literal is Dishonest.

Question #15

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

234

15%

345

356

15%

400

65%

PrepInsta Explanation

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)

Question #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)

47.06%

√(dc)

17.65%

√(ac)

23.53%

√(c)

11.76%

PrepInsta Explanation

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)

Question #17

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

9.52%

14.29%

9.52%

66.67%

PrepInsta Explanation

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.

Question #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?

7.14%

64.29%

14.29%

PrepInsta Explanation

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

Question #19

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

58.33%

8.33%

25%

PrepInsta Explanation

Solution: 4+4+4=12 .

ONLY GIVES 2 AT UNIT PLACE

74+74+944=1092

O+L+E=7=N

Question #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

50%

0.1564

14.29%

0.213

14.29%

0.0987

21.43%

PrepInsta Explanation

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

= 256/20825

= 0.0128

Question #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.

55.56%

Only first statement is sufficient .

16.67%

Only second statement is sufficient.

11.11%

Neither both are sufficient.

16.67%

PrepInsta Explanation

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.

Question #22

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

222

46.67%

223

33.33%

345

6.67%

322

13.33%

PrepInsta Explanation

Explanation :

Solution: 3^3 + 3= 30

4^3 + 4= 68

5^3 + 5= 130

6^3 + 6= 222

Question #23

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

5, 10, 17, 26, 37, 50, ….. ?

123

244

10%

442

145

80%

PrepInsta Explanation

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

Question #24

Find LCM of 12, 15, 20, 27.

500

5.56%

540

72.22%

600

11.11%

none

11.11%

PrepInsta Explanation

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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Infosys Written Test Papers for Online Test – Logical Reasoning

More info about the sections

Number of Questions – 15
Time Limit – 25 mins
Difficulty - Hard

Infosys Previous Papers with Solutions PDF

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Infosys Latest Curriculum 2020

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Infosys Logical Reasoning

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Infosys Latest Preparation Test Paper 2020

Infosys Crytarithmetic Questions 2020

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Additional Facts for Infosys

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