TCS Placement Papers 2022

TCS Previous Papers with Solutions for Recruitment 2022

TCS Placement based Papers with Solutions are given below. If you're looking for TCS Placement Paper 2022 for your Recruitment Test? Don't worry you will find all the TCS Previous Year based Papers with Solutions on PrepInsta. Around 91% of the students who got selected in TCS had prepared from PrepInsta, so you're at the best place to learn.

TCS Online Test has totally of 6 sections and a total of 92 questions and the time given is 180 mins. We suggest you study from our  TCS Model papers and TCS Sample papers given on our website and a good amount of questions will be of a similar pattern in the TCS Written Test 2022.

Name Of The Section Number Of Questions
Quants 26 Ques
Verbal 24 Ques
Reasoning 30 Ques
Programming Logic 10 Ques
Hands On (Coding 1) 1 Code
Hands On (Coding 2) 1 Code
Total 92
Other Important Pages:

Section

5

Time Duration

180 minutes

Adaptive/ Non- Adaptive

Non-Adaptive

Negative marking

No

Applicants

3,50,000+

Job Openings

21,000

Syllabus

TCS Written Test Overview

Very Important: TCS from 2021 has introduced FUV (Fill up Values) type of questions as well. Some questions in the test will be MCQ type which will have 4 options and one correct answer and some questions will be of FUV type of questions wherein you have to enter exact answer up to 2 correct decimal points.

TCS Placement Papers have been divided into following sections in the exam known as Round 1 and Round 2:

ROUND 1 includes:

  • Quants
    • Number of Questions - 26 Questions (40 mins)
  • English
    • Number of Questions - 24 Questions (30 mins)
  • Reasoning Ability

    • Number of Questions - 30 Questions (50 mins)

ROUND 2 includes:

  • Programming Logic

    • Number of Questions - 10 Questions (15 mins)
  • Coding Ability
    • Hands On 1 - 1 Question (15 mins)
    • Hands On 2 - 1 Question (30 mins)

Quants Curriculum 2021:-

  • Total Number of Questions - 26 Questions
  • Negative Marking - No
  • Time - 40 mins

You can expect to get MCQ types questions for this sections that will carry 1 mark each. TCS NQT has negative marking scheme only for MCQ questions.

Topics -

Make sure that you visit our Preparation section below and solve sample previous year questions, we have segregated questions topic-wise.

English Curriculum 2021

  • Total Number of Questions - 24 Questions
  • Negative Marking - No
  • Time - 30 mins
Topics -

There will be no negative marking in the exam. For practicing more sample papers visit our Verbal dashboard.

Coding Ability Curriculum 2020:-

  • Total Number of Questions - 2 Question
  • Negative Marking - No
  • Time

    • Coding Question 1 - 15 mins
    • Coding Question 2 - 30 mins

This is the most important section of the exam, without getting the code output in TCS you can't get into the next round, even if you have scored 100% in other sections.

There will be one code and you have to get the output of the same. The previous year sample questions for TCS can be found out on our TCS Coding Ability Dashboard.

The languages that you're allowed to code in are -

  • C
  • C++
  • Java
  • Perl
  • Python

TCS Coding Questions

Programming Logic Curriculum 2020:-

  • Total Number of Questions - 10 Questions
  • Negative Marking - No
  • Time - 15 mins

Topics -

Make sure that you visit our Preparation section below and solve the previous year sample questions, we have segregated questions topic wise.

Reasoning Ability Curriculum 2020

  • Total Number of Questions - 2 Question
  • Negative Marking - No
  • Time

Topics to prepare for this section:

This being a new section will require more concentration and practice. Visit our TCS Reasoning Ability dashboard to score full marks in it.

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1. In how many ways can we give change for rs 100 using 1 rupee and 2 rupee coins? For example for 5 rs we can give three ways(1,1,1,1) (1,1,1,2) )(,1,2,2)

52 ways

52 ways

24.58%

50 ways

50 ways

21.04%

51 ways

51 ways

44.02%

53 ways

53 ways

10.35%

Let X be the number of 1 rupee and Y be the number of 2 rupees coin

X+Y= 100

Take x = 100, Y=0

Likewise,Y=1, X=98

It will go till Y=50, X=0

Therefore there are 51 ways

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

Oops!

2. We are driving along a highway at a constant speed of 55 miles per hour (mph). You observe a car one-half mile behind you. The car is moving fast and zooms past you exactly one minute later. How fast is this car traveling (mph) if its speed is constant?

85

85

49.1%

70

70

24.98%

82

82

16.38%

80

80

9.54%

The car moves 1/2 mile in 1 minute.

Speed of the car = 1/2 mile for 1 min.

30 miles for 60 min(1 hr).

Therefore, the speed of the car is 30 mph.

Also, the speed of the observant car is 55 mph.

And both are moving in the same direction, we must add the speeds.

So, speed of the car = (55 + 30) mph

= 85 mph.

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

Oops!

3. What is the next number in the series 70,54,45,41?

35

35

21.63%

40

40

70.8%

44

44

5.13%

45

45

2.43%

We are given to select the next number in the following series :

70, 54, 45, 41, . . .

Let a(n) denote the n-th term of the given series.

Then, we observe that

a(1)=70,\\\\a(2)=54=70-16=a(1)-4^{2},\\\\a(3)=45=54-9=a(2)-3^{2},\\\\a(4)=41=45-4=a(3)-2^{2},

Repeating the above rule, we get

a(5)=a(4)-1^{2}=41-1=40.

Thus, the next term in the given series is 40.

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

Oops!

4. George can do some work in 8 hours. Paul can do the same work in 10 hours while Hari can do the same work in 12 hours. All the three of them start working at 9 AM. while George stops work at 11 am, the remaining two complete the work, approximately when will the work be finished?

11:30 AM

11:30 AM

11.02%

12 noon

12 noon

17.26%

1 pm

1 pm

58%

12.30 PM

12.30 PM

13.72%

George can do the work in 8 hours

⇒ 1 hour = 1/8 of the work

Paul can do the work in 10 hours

⇒1 hour = 1/10 of the work

Hari can do the work in 12 hours

⇒ 1 hour = 1/12 of the work

Find the amount of work that can be completed in an hour:

⇒ 1 hour = 1/8 + 1/10 + 1/12 = 37/120

Find the amount of work completed in 2 hours (9 am to 11 am):

⇒ 1 hour = 37/120

⇒2 hours = 37/120 x 2 = 37/60

Find the amount of work remain:

Work remain = 1 - 37/60 = 23/60

Find the amount of work Paul and Hari can do in an hour:

1 hour = 1/10 + 1/12 = 11/60

Find the number of hours needed for them to complete the rest of the work:

Number of hours needed = 23/60 ÷ 11/60 = 2.09 hours = 2 hours

Find the time the work is completed:

11am + 2 hours = 1 pm

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

Oops!

5. Shush and Komal can do a job together in 7 days. Shush is 7/4 times as efficient as Komal. The same job can be done by Shush alone in

49/4 Days

49/4 Days

25.65%

49/3  Days

49/3  Days

15.14%

11 Days

11 Days

47.55%

28/3 Days

28/3 Days

11.66%

Shush completes work in s days.

so in one day \frac{1}{s}th part of the day.

Similarly;

Komal can complete \frac{1}{k}th part of the day.

\frac{1}{s} + \frac{1}{k} = \frac{1}{7}

Shush is \frac{7}{4}times more efficient

therefore, \frac{1}{s} = \frac{7}{4}\times \frac{1}{k}

\frac{4}{7}\times \frac{1}{s} = \frac{1}{k}[\latex]

so, last equation becomes

\frac{1}{s} + \frac{4}{7}\times = \frac{1}{7}[\latex]

S = 11 Days

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

Oops!

6. a,b,c are positive numbers such that a+b+ab= 8, b+c+bc=15 and c+a+ca = 35. What is the value of a+b+c+abc?

34

34

13.9%

36

36

56.23%

35

35

17.63%

33

33

12.24%

Acc to the question, there are 3 equations given:

a+b+ab= 8-----(i)

b+c+bc=15------(ii)

c+a+ca = 35---------(iii)

Now adding 1 to both the sides of equ (i) we get,

a+b+ab+1= 8+1

or, (a+1)(b+1)=9.

Similarly we get, (b+1)(c+1)=16 and (a+1)(c+1)=36.

now starting from the second equation,

We know that 16 has many factors but we have to brute force for satisfying the equation.

We know, 16 = 2 * 8 = (1+1) * (7+1) now, comparing it with equ(ii) we get, b=1,c=7.

Now putting the value of c in any two of the remaining equation we get,

(a+1)*(c+1)=36

or, (a+1)*(7+1)=36

or,(a+1) = 36/8 = 4.5

or, a = 4.5 - 1 = 3.5.

Now, a+b+c+abc = 3.5 + 1 + 7 + 3.5*1*7 = 36.

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

Oops!

7. The 260th term of the series a,b,b,c,c,c,d,d,d,d,e,e,e,e,e,f,f,f,f,f,f,….is:

U

U

13.89%

V

V

23.12%

W

W

51.82%

T

T

11.17%

Given,

First-term = a

third term = b

sixth term = c

We see that Each alphabet repeats for the number of times its position in the English Alphabet,

This series will come to end at z repeating itself for 26 times.

Last term z will be \frac{26\times 27}{2}[\latex] = 350th term

We can say that ,

N th alphabet will be existing from Σ(n-1)+1 to ΣN

ΣN ≥ 260

Let n = 23

Σ(n-1) +1 = 254

Σn = 276

Therefore , 23rd Alphabet will exist from 254 to 276 terms .

23rd Alphabet is W

The 260th term of the series is W

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

Oops!

8. It is possible to pair up all the numbers from 1 to 70 so that the positive difference between the numbers in each pair is always the same. For example, one such pairing up is (1,2), (3,4), (5,6),….(69,70). Here the common difference is 1. What is the sum of all such common differences?

630

630

46.1%

640

640

26.14%

650

650

17.63%

660

660

10.13%

Here,
Two things have to be taken into account:
• Common difference
• No. of such pairs

For simplicity,
Consider range as 1 - 10
Here, when the common difference exceeds 5, numbers are not paired as per the condition.

Like if you take 6 as common difference, not all the pairs will have same value, because it is impossible.

•°• For range 1 - n, we have (n/2) common differences.

Now,
Coming to the given range 1 - 70
n = 70
=> (n/2) = 35

•°• No. of common differences that will satisfy the condition = 35

Now, Find sum of these differences.
1 + 2 + 3 + …… + 35
= (35 × 36) ÷ 2
= 630

•°• Required sum of all such common differences = 630

 

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

Oops!

9. A  regular polygon with 12  sides (dodecagon) is inscribed in a square of area 24 square units as shown in the figure where four of the vertices are mid points of the sides of the square . The area of the dodecagon in square units is.

19

19

14.45%

20

20

33.46%

17

17

14.67%

18

18

37.42%

Let r be half the side of the square.

Consider one of the twelve triangle segments made by a side of the dodecagon and the centre of the dodecagon ( = the centre of the square).

The angle at the centre is 360° / 12 = 30°.

The two sides of the triangle that meet at the centre each have length r, as this is the radius of the circumscribed circle.

The area of this small triangle is then (1/2) r² sin 30° = r² / 4.

The dodecagon is made up of twelve of these triangles, so

area(dodecagon) = 12 × r² / 4 = 3 r².

As r is half the side of the square,

area(square) = 4 r².

Therefore

area(dodecagon ) / area(square) = 3 / 4

=> area(dodecagon) = 3 × 24 / 4 = 3 × 6 = 18

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

Oops!

10. In how many ways can we give change for 100 using 1 rupee and 2 rupee coins? For example for 5 rs, we can give in three ways : (1,1,1,1,1) (1,1,1,2) (1,2,2)

51

51

79.71%

48

48

7.42%

50

50

9.2%

49

49

3.67%

Here we have to find out the number of ways we can distribute 100 rupees money to rupee 1 and rupee 2.

The simplest method is to find out the number of ways is to calculate the number of coins we use to break partitions. As an example, the ways are

1, 1, 1,..........................................,1 (100 times 1)

1, 1, 1, ......................................, 1 and 2 (98 times 1 and 1 time 2)

1, 1, 1, ....................................., 1, 2 , 2 ( 96 times 1 and 2 times 2)

.............................

proceeding in this way, our last combination for taking the coins be

2, 2, 2, ........................, 2 (50 times 2)

So, here, we can find the numbers of ways by counting the number of 2 we took and is equal to 50 + 1 = 51.

Answer: 51

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

Oops!

11. In the figure shown, a triangle is divided into nine stripes of equal height each parallel to the same side of the triangle. The shaped stripes have a total area of 135 square units. What is the area of the triangle in square units?

143 square units.

143 square units.

18.62%

243 square units.

243 square units.

34.24%

135 square units.

135 square units.

35.65%

240 square units.

240 square units.

11.49%

Triangle Image

A = Area of the full triangle

Stripe 1

is a triangle that is similar to the full triangle.
It has height =\frac{1}{9}× height of the larger triangle.
Area(1) = \frac{</span><span id="MathJax-Span-86" class="mstyle"><span id="MathJax-Span-87" class="mrow"><span id="MathJax-Span-88" class="mfrac"><span id="MathJax-Span-89" class="mn">1}{</span><span id="MathJax-Span-90" class="mn">9}</span></span></span></span><span id="MathJax-Span-91" class="mo">^{</span></span><span id="MathJax-Span-92" class="mn">2}× Area of larger triangle =\frac{</span><span id="MathJax-Span-97" class="mstyle"><span id="MathJax-Span-98" class="mrow"><span id="MathJax-Span-99" class="mfrac"><span id="MathJax-Span-100" class="mn">1}{</span><span id="MathJax-Span-101" class="mn">81}[\latex]</span></span></span></span><span id="MathJax-Span-102" class="mi">A</span></span></span></span></span></span></p> <p class="q-text qu-display--block">Stripes <span class="qtext_span qtext_math _1sNWenR7FDgH9GLRfri58K _1xzGCdFnEIP2F4OzuIN1m9"><span id="MathJax-Element-18-Frame" class="MathJax" role="presentation" data-mathml="<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn></math>"><span id="MathJax-Span-106" class="math"><span id="MathJax-Span-107" class="mrow"><span id="MathJax-Span-108" class="mn">1 </span></span></span></span></span><span class="q-box qu-userSelect--text">and <span class="qtext_span qtext_math _1sNWenR7FDgH9GLRfri58K _1xzGCdFnEIP2F4OzuIN1m9"><span id="MathJax-Element-19-Frame" class="MathJax" role="presentation" data-mathml="<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn></math>"><span id="MathJax-Span-109" class="math"><span id="MathJax-Span-110" class="mrow"><span id="MathJax-Span-111" class="mn">2</span></span></span></span></span></span><span class="q-box qu-userSelect--text"> form a triangle that is similar to a larger triangle.<br /> This triangle has height <span class="qtext_span qtext_math _1sNWenR7FDgH9GLRfri58K _1xzGCdFnEIP2F4OzuIN1m9"><span id="MathJax-Element-20-Frame" class="MathJax" role="presentation" data-mathml="<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mstyle displaystyle="false" scriptlevel="0"><mfrac><mn>2</mn><mn>9</mn></mfrac></mstyle><mo>×</mo></math>"><span id="MathJax-Span-112" class="math"><span id="MathJax-Span-113" class="mrow"><span id="MathJax-Span-114" class="mo">=[latex]\frac{</span><span id="MathJax-Span-115" class="mstyle"><span id="MathJax-Span-116" class="mrow"><span id="MathJax-Span-117" class="mfrac"><span id="MathJax-Span-118" class="mn">2}{</span><span id="MathJax-Span-119" class="mn">9}× height of the larger triangle.
Area(1 and 2)=\frac{</span><span id="MathJax-Span-130" class="msubsup"><span id="MathJax-Span-131" class="mrow"><span id="MathJax-Span-132" class="mo"></span><span id="MathJax-Span-133" class="mstyle"><span id="MathJax-Span-134" class="mrow"><span id="MathJax-Span-135" class="mfrac"><span id="MathJax-Span-136" class="mn">2}{</span><span id="MathJax-Span-137" class="mn">9</span></span></span></span><span id="MathJax-Span-138" class="mo">}^{</span></span><span id="MathJax-Span-139" class="mn">2}× Area of larger triangle =\frac{</span><span id="MathJax-Span-144" class="mstyle"><span id="MathJax-Span-145" class="mrow"><span id="MathJax-Span-146" class="mfrac"><span id="MathJax-Span-147" class="mn">4}{</span><span id="MathJax-Span-148" class="mn">81}A
Area(2)=\frac{</span><span id="MathJax-Span-156" class="mstyle"><span id="MathJax-Span-157" class="mrow"><span id="MathJax-Span-158" class="mfrac"><span id="MathJax-Span-159" class="mn">4}{</span><span id="MathJax-Span-160" class="mn">81}</span></span></span></span><span id="MathJax-Span-161" class="mi">A</span><span id="MathJax-Span-162" class="mo">−\frac{</span><span id="MathJax-Span-163" class="mstyle"><span id="MathJax-Span-164" class="mrow"><span id="MathJax-Span-165" class="mfrac"><span id="MathJax-Span-166" class="mn">1}{</span><span id="MathJax-Span-167" class="mn">81}</span></span></span></span><span id="MathJax-Span-168" class="mi">A</span><span id="MathJax-Span-169" class="mo">= \frac{</span><span id="MathJax-Span-170" class="mstyle"><span id="MathJax-Span-171" class="mrow"><span id="MathJax-Span-172" class="mfrac"><span id="MathJax-Span-173" class="mn">3}{</span><span id="MathJax-Span-174" class="mn">81}</span></span></span></span><span id="MathJax-Span-175" class="mi">A

Stripes 1,2, and 3 forms a triangle that is similar to a larger triangle.
This triangle has height =\frac{</span><span id="MathJax-Span-191" class="mstyle"><span id="MathJax-Span-192" class="mrow"><span id="MathJax-Span-193" class="mfrac"><span id="MathJax-Span-194" class="mn">3}{</span><span id="MathJax-Span-195" class="mn">9} × height of the larger triangle.
Area(1,2, and 3)=\frac{</span><span id="MathJax-Span-209" class="msubsup"><span id="MathJax-Span-210" class="mrow"><span id="MathJax-Span-211" class="mo"></span><span id="MathJax-Span-212" class="mstyle"><span id="MathJax-Span-213" class="mrow"><span id="MathJax-Span-214" class="mfrac"><span id="MathJax-Span-215" class="mn">3}{</span><span id="MathJax-Span-216" class="mn">9</span></span></span></span><span id="MathJax-Span-217" class="mo">}^{</span></span><span id="MathJax-Span-218" class="mn">2} × Area of larger triangle = \frac{</span><span id="MathJax-Span-223" class="mstyle"><span id="MathJax-Span-224" class="mrow"><span id="MathJax-Span-225" class="mfrac"><span id="MathJax-Span-226" class="mn">9}{</span><span id="MathJax-Span-227" class="mn">81}A
Area(3)=\frac{</span><span id="MathJax-Span-235" class="mstyle"><span id="MathJax-Span-236" class="mrow"><span id="MathJax-Span-237" class="mfrac"><span id="MathJax-Span-238" class="mn">9}{</span><span id="MathJax-Span-239" class="mn">81}</span></span></span></span><span id="MathJax-Span-240" class="mi">A</span><span id="MathJax-Span-241" class="mo">− \frac{</span><span id="MathJax-Span-242" class="mstyle"><span id="MathJax-Span-243" class="mrow"><span id="MathJax-Span-244" class="mfrac"><span id="MathJax-Span-245" class="mn">4}{</span><span id="MathJax-Span-246" class="mn">81}</span></span></span></span><span id="MathJax-Span-247" class="mi">A</span><span id="MathJax-Span-248" class="mo">=\frac{</span><span id="MathJax-Span-249" class="mstyle"><span id="MathJax-Span-250" class="mrow"><span id="MathJax-Span-251" class="mfrac"><span id="MathJax-Span-252" class="mn">5}{</span><span id="MathJax-Span-253" class="mn">81}</span></span></span></span><span id="MathJax-Span-254" class="mi">A</span></span></span></span></span></span>[\latex]</p> <p class="q-text qu-display--block">and so on…</p> <p class="q-text qu-display--block">Area<span class="qtext_span qtext_math _1sNWenR7FDgH9GLRfri58K _1xzGCdFnEIP2F4OzuIN1m9"><span id="MathJax-Element-34-Frame" class="MathJax" role="presentation" data-mathml="<math xmlns="http://www.w3.org/1998/Math/MathML"><mo stretchy="false">(</mo><mn>4</mn><mo stretchy="false">)</mo><mo>=</mo><mstyle displaystyle="false" scriptlevel="0"><mfrac><mn>7</mn><mn>81</mn></mfrac></mstyle><mi>A</mi></math>"><span id="MathJax-Span-258" class="math"><span id="MathJax-Span-259" class="mrow"><span id="MathJax-Span-260" class="mo">(</span><span id="MathJax-Span-261" class="mn">4</span><span id="MathJax-Span-262" class="mo">)</span><span id="MathJax-Span-263" class="mo">=[latex]\frac{</span><span id="MathJax-Span-264" class="mstyle"><span id="MathJax-Span-265" class="mrow"><span id="MathJax-Span-266" class="mfrac"><span id="MathJax-Span-267" class="mn">7}{</span><span id="MathJax-Span-268" class="mn">81}A

Area(5)=\frac{</span><span id="MathJax-Span-276" class="mstyle"><span id="MathJax-Span-277" class="mrow"><span id="MathJax-Span-278" class="mfrac"><span id="MathJax-Span-279" class="mn">9}{</span><span id="MathJax-Span-280" class="mn">81}A

Area(6)=\frac{</span><span id="MathJax-Span-288" class="mstyle"><span id="MathJax-Span-289" class="mrow"><span id="MathJax-Span-290" class="mfrac"><span id="MathJax-Span-291" class="mn">11}{</span><span id="MathJax-Span-292" class="mn">81|}A

Area(7)=\frac{</span><span id="MathJax-Span-300" class="mstyle"><span id="MathJax-Span-301" class="mrow"><span id="MathJax-Span-302" class="mfrac"><span id="MathJax-Span-303" class="mn">13}{</span><span id="MathJax-Span-304" class="mn">81}A

Area(8)=\frac{</span><span id="MathJax-Span-312" class="mstyle"><span id="MathJax-Span-313" class="mrow"><span id="MathJax-Span-314" class="mfrac"><span id="MathJax-Span-315" class="mn">15}{</span><span id="MathJax-Span-316" class="mn">81}[\latex]</span></span></span></span><span id="MathJax-Span-317" class="mi">A</span></span></span></span></span></p> <p class="q-text qu-display--block">Area<span class="qtext_span qtext_math _1sNWenR7FDgH9GLRfri58K _1xzGCdFnEIP2F4OzuIN1m9"><span id="MathJax-Element-39-Frame" class="MathJax" role="presentation" data-mathml="<math xmlns="http://www.w3.org/1998/Math/MathML"><mo stretchy="false">(</mo><mn>9</mn><mo stretchy="false">)</mo><mo>=</mo><mstyle displaystyle="false" scriptlevel="0"><mfrac><mn>17</mn><mn>81</mn></mfrac></mstyle><mi>A</mi></math>"><span id="MathJax-Span-318" class="math"><span id="MathJax-Span-319" class="mrow"><span id="MathJax-Span-320" class="mo">(</span><span id="MathJax-Span-321" class="mn">9</span><span id="MathJax-Span-322" class="mo">)</span><span id="MathJax-Span-323" class="mo">=[latex]\frac{</span><span id="MathJax-Span-324" class="mstyle"><span id="MathJax-Span-325" class="mrow"><span id="MathJax-Span-326" class="mfrac"><span id="MathJax-Span-327" class="mn">17}{</span><span id="MathJax-Span-328" class="mn">81}[\latex]</span></span></span></span><span id="MathJax-Span-329" class="mi">A</span></span></span></span></span></p> <p class="q-text qu-display--block">So depending on which stripes are shaded, you should be able to figure out the area in terms of <span class="qtext_span qtext_math _1sNWenR7FDgH9GLRfri58K _1xzGCdFnEIP2F4OzuIN1m9"><span id="MathJax-Element-40-Frame" class="MathJax" role="presentation" data-mathml="<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>A</mi></math>"><span id="MathJax-Span-330" class="math"><span id="MathJax-Span-331" class="mrow"><span id="MathJax-Span-332" class="mi">A</span></span></span></span></span><span class="q-box qu-userSelect--text">, set this <span class="qtext_span qtext_math _1sNWenR7FDgH9GLRfri58K _1xzGCdFnEIP2F4OzuIN1m9"><span id="MathJax-Element-41-Frame" class="MathJax" role="presentation" data-mathml="<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>135</mn></math>"><span id="MathJax-Span-333" class="math"><span id="MathJax-Span-334" class="mrow"><span id="MathJax-Span-335" class="mo">=</span><span id="MathJax-Span-336" class="mn">135</span></span></span></span></span></span><span class="q-box qu-userSelect--text">, then solve for <span class="qtext_span qtext_math _1sNWenR7FDgH9GLRfri58K _1xzGCdFnEIP2F4OzuIN1m9"><span id="MathJax-Element-42-Frame" class="MathJax" role="presentation" data-mathml="<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>A</mi></math>"><span id="MathJax-Span-337" class="math"><span id="MathJax-Span-338" class="mrow"><span id="MathJax-Span-339" class="mi">A</span></span></span></span></span></span><span class="q-box qu-userSelect--text">.</span></p> <p>So [latex]\frac{135}{A} = \frac{45}{81} gives an area of 243 square units.

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

Oops!

12. In the IT department of backrub, the administrator password is changed every month. The team of administrator spread across the globe, receive an 8 digits number via email. This number is to be prefixed with a single-digit number and suffixed with a single-digit number to get the actual password. The password is divisible by 11 and 8. If the team received 54218345 this month. Then which of the following pairs give valid prefix and suffix respectively?

8,7

8,7

16.67%

7,8

7,8

39.43%

6,8

6,8

19.05%

8,6

8,6

24.85%

By backtracking

Option a)

The password would be 8653513649

8653513649 ÷ 11 = 786683059

8653513649 ÷ 8 = 1081689206 Remainder 1

⇒ Number cannot be divisible by 8

Option b)

The password would be 7653513648

7653513648 ÷ 11 = 695773968

7653513648 ÷ 8 = 956689206

⇒ The number is both divisible by 8 and 11

Option c)

The password would be 3653513644

3653513644 ÷ 11 = 332137604

3653513644 ÷ 8 = 456689205 Remainder 4

⇒ The number cannot be divisible by 8

Option d)

The password would be 2653513647

2653513647 ÷ 11 = 241228513 Remainder 4

2653513647 ÷ 8 = 331689205 Remainder 7

⇒ The number cannot be divisible by 8 and 11

Answer: (b) 7, 8

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

Oops!

13. The people of the land of unsymmetric to ensure that there is very little symmetry in life. Even their unit of length (no one knows what it is) is called as unseen. It is their pride that the front wheel and the rear wheels of their bikes are of different sizes but are made so that they will not slip on the road. In a certain model of bike, the radius of the front wheel is 28 enzymes and that of the back wheels is 14 unsyms after driving a certain distance if a front wheel had revolution 54 times how many revolutions would the back wheel have undergone?

111

111

13.68%

110

110

21.39%

109

109

15.27%

108

108

49.65%

Find the circumference of the front wheel:

Circumference = 2πr

Circumference = 2π(28) = 56π unsyms

Find the total distance covered by the front wheel:

1 revolution = 56π unsyms

54 revolutions = 56π x 54 = 3024π unsyms

Find the circumference of the back wheel:

Circumference = 2πr

Circumference = 2π(14) = 28π unsyms

Find the number of revolutions made by the back wheel:

Number of revolutions = 3024π ÷ 28π = 108

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

Oops!

14. Rs. 3000 is distributed among A, B, and C such that A gets 2/3rd of what B and C together get and C gets ½ of what A and B together get. Find C’s share.

1000

1000

59.13%

1004

1004

18.93%

2000

2000

14.19%

1001

1001

7.75%

Step-by-step explanation:

first of all a+b+c=3000 --------(1)eqn

second: 2/3(b+c)=a; -----------(2)eqn

third: (a+b)/2=c; -----------------(3)eqn

so we got 3 equation and 3 variable

a+b+c=3000; (1)eqn

a+b-2c=0; (3)eqn

subtracting both we get c=1000

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

Oops!

15. Two finals are scheduled - The Wimbledon match and the World Cup Cricket at the same time. Anu wants to watch the Wimbledon finals and her brother Vinu wants to watch WCC final. They decide to roll a tetrahedral die twice. The tetrahedral Is numbered 1,2,3,4 on its four sides and all numbers are equally likely to appear. Anu rolls fist and then Vinu rolls. If the number on the first roll Is strictly greater than the number on the second roll. Anu wins and gets to watch Wimbledon. What Is the probability that Anu will get to watch Wimbledon?

\frac{3}{10}

\frac{3}{10}

15.77%

\frac{3}{8}

\frac{3}{8}

54.75%

\frac{1}{8}

\frac{1}{8}

16.82%

\frac{5}{8}

\frac{5}{8}

12.66%

The probability that Anu will get to watch the Wimbledon match is 3/8

Step-by-step explanation:

Probability of any number appearing = 1/4

Anu wins in the following condition

1) If 1 occus on the first roll then there is no chance of her winning because in the next roll the number appeaing will be either 1 or greater than 1

2) If 2 appears on 1st roll then Anu will win if 1 appears on the second roll

3) If 3 appears on the 1st roll then Anu will win if 1 or 2 appears on the second roll

4) If 4 appears on the first roll then Anu will win if 1, 2, 3 appears on the second roll

Therefore, the probability of Anu winning

\frac{1}{4}\times \frac{1}{4} + \frac{1}{4}\times \frac{2}{4} + \frac{1}{4}\times \frac{3}{4} = \frac{3}{8}

 

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

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Interview Experience

Read more TCS Interview Experiences here

  • Name: Abhishek Jain
  • College: AKTS University
  • Branch: Information and Technology
  • Type of interview: On Campus Interview

Round 1: Online Round

For TCS Previous Year Papers these are the following sections in TCS 2022 Placement Papers –

Do note that you need to do all the rounds well as all of them have sectional cut-offs in TCS Previous Year Placement Papers. Even if you can't score well in one round you won't be selected even though you had done the other rounds well.

Written Test

This is round contains 20 questions and most of them were Fill the Blanks and Sentence Completion Questions.

This section has quantitative questions and Reasoning questions. Please refer to previous year's TCS sample papers and prepare. Prepare for TCS specific quant topic, thanks for letting me know PrepInsta.

TCS C MCQ

This round is of 20 minutes. It is based on C language and please prepare all the basic concepts of C to clear this section.

Coding Test

This year we didn't have code in Command-Line Programming and we could code in any of the following languages - C, C++, Java, Python, etc.

Round 2: Technical Round

After TCS Written Round we had this face to face round where you will be tested what you have done in years of your engineering. I had revised my engineering subjects basics. The person interviewing you may ask you your strong subject so please tell the subject that you have prepare well (I told C and Java).

I was asked to write a simple program to find all the prime numbers less than a number. I wrote a separate function for prime numbers and used a for loop to check all the numbers less than N, to check if they are prime or not. If they were prime I displayed the number. Then he gave coin weight balance puzzle which I stumbled with but after some clues, I understood the problem, but I couldn’t give an accurate solution. The next series of questions were from C, which I managed to answer. Some of my friends got questions from OS and DBMS too in TCS Previous Year Sample Placement Papers. It depends on your luck, so do prepare subjects like Networks, DBMS, OS, JAVA, C, C++ for tcs off campus sample placement paper.

For Non-CS/IT people questions were on projects resume and branch based questions like circuit diagram of the project and basic questions, definitions.

TCS Question Paper Round 3: Managerial Round

I was asked about my projects here. I was asked about my interests. My views and ideas of technology were tested. My friends had a few questions from programming here. This round basically checks your attitude and behavior. Please try to be the person whom you are, if you pretend here, the recruiters are smart enough to know that. If you don’t know something, then tell you are not sure about it. Also ask any query regarding the company when you are given a chance, failing to do so will show your lack of interest for the company in TCS 2021 Placement Papers.

Round 4: HR Round

After TCS Question Paper Round this round will be a round where your knowledge for TCS will be tested. Do prepare your introduction and also about TCS very well. If you had participated in CodeVita or testimony, please mention that in your introduction. You should be able to convince the HR about why you want to join TCS. Again, try to be yourself here. If had done the other rounds well, then you are mostly in. Ask 1 or queries about the company or your role when you are given the chance. Before leaving, greet the person who was interviewing you and leave.

Proper aptitude and 2 week’s preparation is enough to get the job via TCS Previous Year Sample Placement Papers provided you are determined. I felt the first round to be toughest Solve as much as TCS Previous Year Sample Papers that will be very crucial So, prepare well, all the best! for both TCS off campus placement paper and On-Campus Paper.

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Importance70%

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Prepinsta has provided the most important topics and that can come in TCS with a lot of questions will be similar and good amount of questions will be of the same pattern.

No, there will be no negative marking in any section or topic for the exam.

This year coding will be an important section as TCS looks to higher lesser number of Fresher Graduates via TCS online test papers for Recruitment.

The Words Like “Placement Papers” and “Previous Year Papers” are used here for Google Search Purposes only. All these placement papers and previous year papers are freely available on the internet, we are only charging students for the PrepInsta’s Mock Test experiences and Analytics as well as preparation for the exam. Prepinsta does not guarantee any recurrence of the questions in the exam however we believe that the questions should be of the similar pattern in the exam

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