TCS  Placement Paper   Aptitude - Numerical   LBRCE, Mylavaram-09 Dec 2010

#### TCS  Placement Paper   Aptitude - Numerical   LBRCE, Mylavaram-09 Dec 2010

• Posted by  FreshersWorld
7 Jan, 2012

Hi, Friends

First of all thanks to Freshersworld.com  and all of the previous paper posters.

Selection process: (in my college)

1) Written test
( 179 out of 300 were selected)
2) Technical round (113  out of 179 were  selected)
3) HR interview ( 65 out of 113 were selected)

I was selected. The key to crack all rounds is confidence and do not bluff the HR. Say what you are, if they think you will fit into TCS they will select you otherwise find the company that is best fit for you

Written Test:

Written test consists of 35 questions 80 mins. It is a online test.

You have to prepare some of the previous papers in this site to know the how the Q are asking. one thing you should solve and discuss with your friends because many of the papers have a few mistakes and i have attmpted 34 out of 35. All are questions in previous papers. I have found so many questions for which already answers are given, but the questions now i am explaining is some what difficult.

Alok and Bhanu play the following min-max game. Given the expression
N =9+ X + Y - Z

Where X, Y and Z are variables representing single digits (0 to 9), Alok would like to maximize N while Bhanu
would like to minimize it. Towards this end, Alok chooses a single digit number and Bhanu substitutes this for a variable of her choice (X, Y or Z). Alok then chooses the next value and Bhanu, the variable to substitute the value. Finally Alok proposes the value for the remaining variable. Assuming both play to their optimal strategies, the value of N at the end of the game would be
a) 0 b) 27 c) 18 d) 20
Ans: 20
X+Y-Z  = 11, then add the 9 or 12 or 24 or 38 which is given and get N=  20 or 23 or 35 or 49
X-Y-Z  is    2
X*(Y-Z) is 18, then add the given constant to get N

Explanation:
It was so lengthy and i am so lazy i can't type it all here but answer is 100% correct and i will explain some strategy
let N=X+Y-Z

If first A gives 9 then B had two ways if he puts in Z then A as usualy give 9 and 9 (because X or Y) and N=9+9-9=9.
If he puts in X or Y then A has to give a no to maximize 9+Y-Z
here Y-Z =4 (i will explain later )  so N=9+4=13
here B has to put in Z or X, Y by which he should minimize N, if he puts in Z, N=9+9-9=9.if he puts in X or Y, N=9+4=13
Here B go in a way which will minimize N,so min(9,13) is 9 and puts in  Z then A as usualy give 9 and 9and N=9+9-9=9

If first A gives 8, then B had two ways if he puts in Z then A as usualy give 9 and 9 (because X or Y) and N=9+9-8=10.

If he puts in X or Y then A has to give a no to maximize 8+Y-Z
Here Y-Z =4 (i will explain later )  so N=8+4=12
Here B has to put in Z or X,Y by which he should minimize N,if he puts in Z ,N=9+9-8=10.if he puts in X or Y, N=8+4=12
Here B go in a way which will minimize N,so min(10,12) is 10 and puts in  Z then A as usualy give 9 and 9and N=9+9-8=10.

If first A gives 7, then B had two ways  if he puts in Z then A as usualy give 9 and 9 (because X or Y) and N=9+9-7=11.

If he puts in X or Y then A has to give a no to maximize 7+Y-Z
Here Y-Z =4 (i will explain later )  so N=7+4=11
Here B has to put in Z or X,Y by which he should minimize N,if he puts in Z ,N=9+9-7=11.if he puts in X or Y, N=7+4=11
here B go in a way which will minimize N,so min(11,11) is 11 and puts anywhere to get N=11

If first A gives6, then B had two ways
If he puts in Z then A as usualy give 9 and 9 (because X or Y) and N=9+9-6=12.
If he puts in X or Y then A has to give a no to maximize 6+Y-Z
Here Y-Z =4 (i will explain later )  so N=6+4=10
Here B has to put in Z or X,Y by which he should minimize N,if he puts in Z ,N=9+9-6=12.if he puts in X or Y, N=6+4=10

Here Bgoina way which will minimize N,so min(10,12) is 10 and puts in X,Y so N=6+4=10
If first A gives 5, then B had two ways
If he puts in Z then A as usualy give 9 and 9 (because X or Y) and N=9+9-5=13.
If he puts in X or Y then A has to give a no to maximize 6+Y-Z
Here Y-Z =4 (i will explain later ) so N=5+4=9
Here B has to put in Z or X,Y by which he should minimize N, if he puts in Z , N=9+9-5=13.if he puts in X or Y, N=5+4=9

Here Bgoina way which will minimize N, so min(9, 13) is 9and puts in X,Y so N=5+4=9

so B choose or
like this if       A gives    N=           B   can not minimize under          so A choose to give 7 and N=11
9               9,13                        9
8              10,12                       10
7              11,11                       11
6               12,10                      10
5               13,9                        9
4                14,8                      8
3               15, 7                      7
2                 16,6                      6
1                  17,5                       5
0                18,4                        4
So, A has to maximize and he will give 7 becuse max(N) =11 occur. that is so A choose to give 7 and N=11

now why X-Y=4 ? do as above you can get a table like below
so B choose or
if  A gives    X-Y=           B   can not minimize under          so A choose to give 5 or 4 and X-Y=4
9           0,9                     0
8            1, 8                    1
7            2,7                     2
6             3,6                    3
5             4,5                    4
4             5,4                   4
3             6,3                   3
2             7,2                   2
1               8,1                  1
0               9,0                  0

That is all, I hope u understand.
Every thing remains same and in TR and HR some routine questions but i gave different answeres. That makes all different.

My mail id is  siva0ram@gmail.com   (passionate to help)

Thanks for your patience and All the best!

See you in TCS.

2009-2016 downloadmela.com. All rights reserved.