• The test pattern for TCS enrollment 2021 is given below.

At one time you clear the TCS online test, then students will be called for personal interview. Post the assessment, interview date and location will be communicated to the candidates personally.

• TCS Ninja Verbal ability questions will be based on Sentence Completion.

• Iteration, recursion, procedural vs OOp.
• C Language:
  • Call by value/reference, Basic and Derived Data types, Storage class, Scope and Visibility, Basics of Pointers, Basic header files, Library functions, Branching and Looping, Command line arguments, User-defined functions.

• Basic Search and Sort Algorithms (Tree traversal, Dynamic Programming, etc)

• Array, Stack, Queue, List (Tree, Hash table, etc)

• TCS Ninja Coding Question with Solution -> Click Here

• Only One question, 20 minutes.
• Choice of C / C++ / Java / Perl / Python 2.7.
• Provided an IDE to debug.
• For Java, the class name should be named Maze.
• Input to the program either through STDIN / Command line arguments, as per the instructions.
• Program should write the output to STDOUT.
• Public and private test cases based evaluation.

• While printing the output no leading or trailing spaces should be printed.
• Other than the required output, no other text should be printed.
• If the output is a number, no leading sign must be printed unless it is a negative number.
• No scientific notation (3.9265E + 2).
• All floating point numbers must contain that many decimal places as mentioned in the question.

• The test will progress as per specified in the above table. But Once you move from one group to another, you cannot go back to the previous group. For example, once you have moved to Quantitative Aptitude, you can’t go back to English and so on.
• The timer shown on the test screen will show the time resting to complete each group. Each group has independent timers and when the timer start the first question present on the screen.
• The test period for each group is fixed. For example: If you complete a group in limited time, the left over time will not be carried to the remaining groups.
• At the end of every group Click SUBMIT to move to the next group.

George 9 days work = 3/5
George 1 days work = 3/5 * 9
= 1/15.
George and Paul 4 days work = 2/5
George and Paul 1 days work = 2/5 * 4
= 1/10.
Paul 1 days work = George and Paul 1 days work – George 1 days work
= 1/10 – 1/15
= 1/30

So Paul takes 30 days to complete work.

• • A : C
    • Efficiency of A and C = 5 : 3
    • The Ratio of Work done is 3 : 5 = 3x : 5x
    • Given:
    • Difference between A’s and C’s work done is 6 days so,
    • 5x – 3x = 6
    • x = 3
    • Number of days taken by A = 3 * 3
    • =

    9 Days

    • Number of days taken by C = 5 * 3
    • =

    15 Days

  • B : C
    • Ratio of work done of B and C = 2:3
    • As we have calculated C’s total work done in 15 days so,
    • 3x = 15 days
    • x = 5
    • Therefore, Number of days taken by B = 2 * 5
    • =

    10 Days

    • Now we have total work done by a ,b and c in 9, 10 and 15 days respectively
    • Work done by B and C in initial 2 days = 2 * [ 1/10 + 1/15 ]
    • = 1/3
    • Thus, Remaining work = 1 – 1/3
    • =

    2/3

A do total work in 9 days so, 2/3 work is done in (9 * 2/3) = 6 days

Time Taken by Y = (2 * 100) / 3
= 66.66 unit.

Time taken by Z = (3 * 66.66) / 4
= 50 unit.

Ratio of time of X, Y and Z = 100 : 66.66 : 50

Fraction of work Y do = 66.66 / (100 + 66.66 + 50)
= 0.307
= 4/13.

A’s 2 day’s work = 1/20 x 2
= 1/10
(A + B + C) 1 day’s Work = (1/20 + 1/30 + 1/60)
= 6/60
= 1/10
Now,
Work done in 3 days = (1/10) + (1/10)
= 1/5
Now 1/5 work is done in 3 days.
Now whole work will be done in 3 * 5 = 15 days.

Babla work = 1/10
Ashu work = 1/15
Total work = 1/10 + 1/15
= 5/30
= 6 days.

Total amount to be paid for Babla = 6/10 * 5000
= 3000

Money should always be divided in the inversely proportion way.
So Babla will get money = 15 / (15 + 10) × 5000
= 3000

The money can pay George’s wage for 15 days
1 Day = 1/15 of the money

The money can pay Mark’s wage 10 for days
1 Day = 1/10 of the money

Find the Amount of money needed if both of them worked together:
1 Day = 1/15 + 1/10
= 1/6 of the money

Number of days the money can last:
1/6 of the money = 1 day
6/6 of the money = 1 x 6
= 6 days

A can do the total work in 8 days, and B can do it in 6 days.

Let the total work be 24 units.

Now,
A = 24/8 = 3
B = 24/6 = 4
A + B + C = 24/3 = 8

So Capacity of C = 1 unit.
In two days C will do 2 units which is 2/24^th part of the total work. So 1/12^th part.

As Jake , Paul and Hari can dig the well in 8 days,
Jack one day work = 1/16
Paul one day work = 1/24
Hari one day work = 1/x

Given,
(Jake, Paul and hari) 1 day work = 1/8

So, 1/16 + 1/24 + 1/x = 1/8
5/48 + 1/x = 1/8
1/x = 1/8 – 5/48
1/x = (6 – 5) / 48
= 1/48
1/x = 1/48

Therefore, Paul can finished work in 48 days.

No. of days taken by Thomas = 7 days.
So work done by thomas in a day = 1/7

No. of days taken by Raj = 9 days.
Work done by other in a day = 1/9

Work done by both in a day = 1/t

1/7 + 1/9 = 1/t
16/63 = 1/t.
16t = 63.
Days required by both (t) = 63/16.

Total no. of days taken to paint the house is (approx.) = 4 days .

• Let a man can do 1 unit of work in 1 hour.

• Total units of work = 4 * 8 * 5 = 160 units.

• Now work = 2 * 160 = 320 units

• Now 2 men work for 20 days. Let in x hours they have to work per day.

• Now total work = 2 * x * 20 = 40x

• 40x = 320

• So x = 320/40

• • =

  8 hours

• When t = -90 then (sin t, cos t) = (-1,0)

• When t=0 then (sin t, cos t) = (0,1)

Plot above two points in x-y plane.

Given that radius of sphere r = 3 cm
Given that thickness of the plate h = 1 cm

Given that a sphere is melt to produce a circular plate.
Since nothing is added or subtraced then volume of sphere and circular plate both will be equal.
Volume of sphere 
Volume of circular plate 
set both volumes equal to each other



36 = r²
6 = r
Hence the radius is 6 cm.

• In triangle PQR, by using heron’s formula

• S = P + Q + R / 2

• S = 6 + 8 + 12 / 2

• = 26 / 2 = 13

• S = 13

• Area of the triangle PQR = sqrt ( S (S – P) * (S – Q) * (S – R) )

• = sqrt ( 13 (13 – 6) * (13 – 8) * (13 – 12) )

• = sqrt ( 455 )

• Area of the triangle PQR =

21.33 cm²

Perimeter of rectangle = 18 cm
2 ( l + b ) = 18 cm
l + b = 18 / 2
= 9 cm
l + b = 9 cm

Diagonal of rectangle = √41 cm
We know that, Diagonal of rectangle = √ [ ( length )² + ( breadth )² ] √( l² + b²) = √41
l² + b² = ( √41 )²
l² + b² = 41
l² + b² + 2 lb – 2 lb = 41
( l + b )² – 2 lb = 41

Since , we have l + b = 9 , so now
( 9 )² – 2 l b = 41
81 – 2 lb = 41
2 lb = 81 – 41
2 lb = 40
lb = 40 / 2 = 20 cm ²
l × b = 20 cm²
Therefore, area of rectangular field = 20 cm²

Area of triangle formula = 1/2*(base)*(height)
Area of one triangle = 1/2 * (4) * (5)
= 10 cm².

Total area of rhombus = 4 * area of one triangle
= 4 * 10
= 40 cm²

Area of a rhombus is = 1/2 * (length of one diagonal) * (length of another diagonal)
= 1/2 * 8 * 10 cm²
= 40 cm²

Find the volume with diameter = 12 m

Radius = Diameter ÷ 2
Radius = 12 ÷ 2 = 6 m
Volume = π * r² * h
Volume = π * (6)² * 21
V₁ = 2,375 m³

Find the volume with diameter = 10 m

Radius = Diameter ÷ 2
Radius = 10 ÷ 2 = 5 m
Volume = π * r² * h
Volume = π * (5)² * 21
V₂ = 1650 m³

Find the volume of the metal used:

Volume = V₁ – V₂
= 2,375 – 1650
Volume = 725 m³

Area of path horizontal = 20 * 192.5
= 3850 dm²
Area of path vertical = 20 * 125
= 2500 dm²
Since center path is covered twice so total area,
= (3850 + 2500) – (20 * 20)
= 5950 dm²
Total cost = 1.32 * 59.5 m²
= 78.54 m²

Given that a square field is to be enclosed by four wires at height 1m, 2m, 3m and 4m.

Area of the square field = 31684 m²
Let’s find the perimeter of square

As we know

Area of square = x²
So,
x² = 31684 m²
Taking square root on both sides
√x² = √31684 m²
x = 178 m
So, the length of one side of square is x = 178 m

As we know the perimeter of the square is calculated as

P = 4x
P = 4 * 178
P = 712 m

Given that length of wire for each circuit is 5% more than the length of perimeter of square. So,

Length of wire = Length of perimeter  + 5% of Length of perimeter  
l = 712 + 0.05 * 712
l = 712 + 35.6
l = 747.6 m

As we need to enclose the square field by wires four times at different height; therefore,

Total Length of wire = 4l
Total Length of wire = 4 * 747.6
Total Length of wire = 2990.4 m

So, total length of wire is 2990.4 m

Outer length of the grassy plot = 112 m
Outer breadth of the plot = 78 m

Outer area of the plot = 112 x 78
= 8736 sq m
Width of the path = 2.5 m

Inner length of the plot = 112 – (2 x 2.5)
= 107 m
Inner breadth of the plot = 78 – (2 x 2.5)
= 73 m

Inner area of the plot = 107 x 73
= 7811 sq m

Area of the path = Outer area of the plot – inner area of the plot
= 8736 – 7811
= 925 m²