# TCS sample papers of 2013

**In this article I will let you know about the TCS and TCS sample papers of 2013**

## About TCS:

Tata Consultancy Services is a multi-national company of Information Technology which was founded by J R D Tata in the year of 1968.It visits most of the colleges for the campus placements for various designations.In this company there is a selection procedure which consists of three rounds which are as follows:

**1.Written Test**

**2.Technical Interview**

**3.HR Round**

Here is the section wise list of Sample paper

**1.Quantitative Analysis:**

1. If 3y + x > 2 and x + 2y≤3, What can be said about the value of y?

A. y = -1

B. y >-1

C. y <-1

D. y = 1

Answer: B

Multiply the second equation with -1 then it will become – x – 2y≥ – 3. Add the equations. You will get y > -1.

2. If the price of an item is decreased by 10% and then increased by 10%, the net effect on the price of the item is

A. A decrease of 99%

B. No change

C. A decrease of 1%

D. An increase of 1%

Answer: C

If a certain number is increased by x% then decreased by x% or vice versa, the net change is always decrease. This change is given by a simple formula −(x10)2=−(1010)2=−1%. Negitive sign indicates decrease.

3. If m is an odd integer and n an even integer, which of the following is definitely odd?

A. (2m+n)(m-n)

B. (m+n2)+(m−n2)

C. m2+mn+n2

D. m +n

Answer: C and D (Original Answer given as D)

You just remember the following odd ± odd = even; even ± even = even; even ± odd = odd

Also odd x odd = odd; even x even = even; even x odd = even.

4. What is the sum of all even integers between 99 and 301?

A. 40000

B. 20000

C. 40400

D. 20200

Answer: D

The first even number after 99 is 100 and last even number below 301 is 300. We have to find the sum of even numbers from 100 to 300. i.e., 100 + 102 + 104 + …………… 300.

Take 2 Common. 2 x ( 50 + 51 + ………..150)

There are total 101 terms in this series. So formula for the sum of n terms when first term and last term is known is n2(a+l)

So 50 + 51 + ………..150 = 1012(50+150)

So 2 x 1012(50+150) = 20200

5. There are 20 balls which are red, blue or green. If 7 balls are green and the sum of red balls and green balls is less than 13, at most how many red balls are there?

A. 4

B. 5

C. 6

D. 7

Answer: B

Given R + B + G = 17; G = 7; and R + G < 13. Substituting G = 7 in the last equation, We get R < 6. So maximum value of R = 6

6. If n is the sum of two consecutive odd integers and less than 100, what is greatest possibility of n?

A. 98

B. 94

C. 96

D. 99

Answer : C

We take two odd numbers as (2n + 1) and (2n – 1).

Their sum should be less than 100. So (2n + 1) + (2n – 1) < 100 ⇒ 4n < 100.

The largest 4 multiple which is less than 100 is 96

7. x2 < 1/100, and x < 0 what is the highest range in which x can lie?

A. -1/10 < x < 0

B. -1 < x < 0

C. -1/10 < x < 1/10

D. -1/10 < x

Answer: A

Remember:

(x – a)(x – b) < 0 then value of x lies in between a and b.

(x – a)(x – b) > 0 then value of x does not lie inbetween a and b. or ( −∞, a) and (b, −∞) if a < b

x2 < 1/100 ⇒

(x2−1/100)<0⇒(x2−(1/10)2)<0⇒(x−1/10)(x+1/10)<0

So x should lie inbetween – 1/10 and 1/10. But it was given that x is -ve. So x lies in -1/10 to 0

8. There are 4 boxes colored red, yellow, green and blue. If 2 boxes are selected, how many combinations are there for at least one green box or one red box to be selected?

A. 1

B . 6

C. 9

D. 5

Answer: 5

Total ways of selecting two boxes out of 4 is 4C2 = 6. Now, the number of ways of selecting two boxes where none of the green or red box included is only 1 way. (we select yellow and blue in only one way). If we substract this number from total ways we get 5 ways.

9. All faces of a cube with an eight – meter edge are painted red. If the cube is cut into smaller cubes with a two – meter edge, how many of the two meter cubes have paint on exactly one face?

A. 24

B. 36

C. 60

D. 48

Answer : A

If there are n cubes lie on an edge, then total number of cubes with one side painting is given by 6×(n−2)2. Here side of the bigger cube is 8, and small cube is 2. So there are 4 cubes lie on an edge. Hence answer = 24

10. Two cyclists begin training on an oval racecourse at the same time. The professional cyclist completes each lap in 4 minutes; the novice takes 6 minutes to complete each lap. How many minutes after the start will both cyclists pass at exactly the same spot where they began to cycle?

A. 10

B. 8

C. 14

D. 12

Answer: D

The faster cyclyst comes to the starting point for every 4 min so his times are 4, 8, 12, ……… The slower cyclist comes to the starting point for every 6 min so his times are 6, 12, 18, ……… So both comes at the end of the 12th min.

**2.Logical Reasoning:**

**square of the sum of its digits**. How many such numbers are there?

(1)1

(2)2

(3)3

(4)4

Ans: Option 1

Take N = 10a+b.

Given that, 10a+b+18 = (a+b)2

for a = 1 to 9, the L.H.S. will be, 28+b, 38+b, 48+b,…..,108+b.

As LHS is perfect square for the values of b = 1 to 9, only 28+b, 48+b, 58+b, 78+b can be equal to 36, 49, 64, 81 for b = 8, 1, 6, 3 respectively. But only 78+b = 81 for b = 3 So only one such number is possible. I.e, 632. A two digit number is 18 less than the

**sum of the squares of its digits.**How many such numbers are there?

(1)1

(2)2

(3)3

(4)4

Ans: Option 2

Only 47 and 67 satisfy the condition

3. For real number x, int(x) denotes integer part of x.int(x) is the largest integer less than or equal to x.int(1,2)=1,int(-2,4)=-3. Find the value of int(1/2)+int(1/2+ 100)+int(1/2+2/100)+….+int(1/2+99/100)

Sol: int (1/2) = 0

int (1/2 + 100 ) = 100

into (1/2 + 2/100) = 0

……

int ( 1/2 + 50/100 ) = 1

int (1/2 + 51 /100) = 1

…….

int (1/2 + 99/100) = 1

So 100 + 1 + 1 + …..50 times = 150

4. Given a square of length 2m. Its corners are cut such that to represent a regular octagon. Find the length of side of octagon

Sol:

Let x is the side of the octagon and x + 2y is the side of the square.

In the given octagon, y2+y2=x2⇒2y2=x2⇒y=x2√

But x2√+x+x2√=2

⇒2√x+x=2

⇒x=22√+1=22√+1×2√−12√−1=2(2√−1)

5. Find the number of ways a batsman can score a double century only in terms of 4’s & 6’s?

Assume the batsman scored x 4’s and y 6’s.

4x + 6y = 200 ⇒2x+3y=100 ⇒x=100−3y2=50−32y

As x is an integer, y should be a multiple of 2.

If

y = 0, x = 50

y = 2, x = 47

y = 4, x = 44

…

y = 32, x = 2

So total ways are (32-0)/2 + 1 = 17 ( if 0 6’s are possible) otherwise 16

6. 5000 voted in an election between two candidates.14% of the votes were invalid.The winner won by a margin approximately closer to 15%.Find the number of votes secured by the person

Invalid Votes = 14 % (5000) = 700

Valid Votes = 5000 – 700 = 4300

Assume the looser got x votes. Then the winner must have got x + 15% (x)

But x + x + 15% (x) = 4300

Solving x = 2000

So Looser got 2000 and winner got 2300

7. There are 100 wine glasses. I offered my servant to 3 paise for every broken glass to be delivered safely and forfeit 9 paisa for every glass broken at the end of day. He recieved Rs.2.40 .how many glass did he break.

a. 20 b. 73 c. 5 d. 8

If a glass has been broken, he has to loose 3 paisa + 9 paise = 12 paise

Assume K glasses got broken

100 x 3 – 12 x K = 240 ⇒K=5

8. A is 20 percent more efficient than B. If the two person can complete a piece of work in 60 days.in how many days. A working alone can complete the work

a. 80 b. 90 c. 100 d. 110

As A is 20% more efficient than B, If B’s per day work is 100 units then A’s 120.

Both persons together completes (100 + 120) units = 220 units a day.

They took 60 days to complete the work. So total work = 60 x 220

If A alone set to complete the work, he takes = 60×220120=110 days

9. A property was originally on a 99 years lease and two thirds of the time passed is equal to the four fifth of the time to come.how many years are there to go.

a. 45 b. 50 c. 60 d. 55

Assume x years have passed and y years to go

Given 23x=45y ⇒x=32×45y=65y

But x + y = 99

So 65y+y=99

Solving we get y = 45 years

10. In how many different ways can the letters of the word “LEADING” be arranged in such a way that the vowels always come together.

a. 360

b. 720

c. 480

d. 5040

Given letters are A, E, I, D, L, N, G

Of which AEI are vowels. Let us combine them into a single letter x. Now total letters are x, D, L, N, G

These letter are arranged in 5! ways. But 3 vowels can arrange themselves in 3! ways. So total ways 5! x 3! = 720

**3.Verbal:**

SYNONYMS:

1. PLAINTIVE

2. MIRAGE

3. MISANTHROPE

4. PROLIXITY

5. AVID

6. MUNDANE

7. PARLEY

8. GARNER

9. DECOROUS

10. DANGLE

ANTONYMS:

1. ASPERITY

2. RETROSPECTION

3. OSCILLATE

4. PLACATE

5. SAGE

6. CASTIGATE

7. CRITICAL

8. INDIGENOUS

9. TAPER

10. HIGH-HANDED

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