COGNIZANT (CTS) -2002 Placement Paper

COGNIZANT (CTS) -2002

There were
* 5 sections
* 8 questions each (40 q totally)
* 60 minutes
* 5 different sets of question papers
* 1 Mark each
* 0.25 negative marking

CTS_BLACK

Vocabulary, strings, dominoes, functions, coding (each section 8 ques)

CTS_BROWN
Word series, numerical series, functions, figures, verbal (each section 8 ques)

CTS_VIOLET
Functions, strings, bricks, jigsaw puzzle, cryptic clues (each section 8 ques)

CTS_RED
1) 8 functions 2) 4 cryptic clues, 4 anagrams
3) 4 Tetris figures, 4 bricks 4) 8 strings 5) 4 jigsaw puzzles 4 number series

BROWN_2002

There were different papers for different sessions.
The paper had 5 sections, 5 * 8 = 40 Que's. totally.

Section 1: Functions

---------------------------
Q: 1 - 8
Certain functions were given & based upon the rules & the choices had to be made based on recursion. This is time consuming, but u can do it. Try to do it at the end. Start from the last section.

L(x) is a function defined. functions can be defined as
L(x)=(a,b,ab) or (a,b,(a,b),(a,(b,b)),a,(b,b)).... two functions were given A(x) & B(x) like
if l(x)=(a,b,c) then A(x)=(a) & B(x)=(b,c)
i.e., A(x) contains the first element of the function only.
& B(x) contains the remaining, except the first element.
then the other two functions were defined as
C(x) = * if L(x) = ()
A(x) if L(x) = () & B(x) != () & C(B(x)) otherwise
D(x) = * if L(x) = ()
** if B(x) = ()
A(x), if L(x) != () & B(x) != ()
D(D(x)),otherwise ;

Now the Questions are,

1 : if L(x) = (a,b,(a,b)) then C(x) is ?
(a): a (b): b (c): c (d): none

2 : if L(x) = (a,b,(a,b)) then find D(x)
same options as above

3 : if L(x) = (a,b,(a,b),(b,(b))) find C(x)

4 : -----------~~~~~~~~---------- find D(x)

5 : if L(x) = (a,(a,b),(a,b,(a,(b))),b) then find c(x)

6 : -----------~~~~~~~~---------- find D(x)

7 : if L(x) = (a,b,(a,b)) then find C(D(x))

8 :
-----------~~~~~~~~---------- find D(C(x))

Section 2: Word series
------------------------------
Q’s: 9 - 16
This is one of the easiest sections. Try to do it at first.

If S is a string then p, q, r forms the sub strings of S. For eg, if S = aaababc & p = aa,
q = ab, r =bc . Then on applying p à q on S is that ababaabc. Only the first occurrence of S has to be substituted. If there is no sub string of p, q, r on s then it should not be
substituted.

If S = aabbcc, R = ab, Q = bc. Now we define an operator R  Q when operated on S, R is replaced by Q, provided Q is a subset of S, otherwise R will be unchanged. Given a set S =… when R Q, P= = 672; R, Q  P operated successively on S, what will be new S? There will be 4 =: if s = aaababc & p = aa, q = ab, r = bc then applying p à q, q à r & r à p will give,
(a): aaababc (b): abaabbc (c): abcbaac (d): none of the a,b,c

10: if s = aaababc & p = aa q = ab r = bc then applying q à r & r à p will give,

11: if s = abababc & p = aa q = ab r = bc then applying p à q, q à r & r à p will give,

12: if s = abababc & p = aa q = ab r =bc then applying q à r & r à p will give,

13: if s=aabc & p=aa q=ab r=ac then applying p->q(2) q->r(2) r->p will give,
(2) Means applying the same thing twice.
14: Similar type of problem.
15) if s = abbabc p = ab q = bb r = bc then to get s = abbabc which one should be applied.
(a): p->q,q->r,r->p

16) if s = abbabc p = ab q = bb r = bc then to get s = bbbcbabc which one should be applied.
Let us consider a set of strings such as S = aabcab. We now consider two more sets P and Q that also contain strings. An operation Pà Q is defined in such a manner that if P is a subset of S, then P is to be replaced by Q. In the following questions, you are given various sets of strings on which you have to perform certain operations as defined above. Choose the correct alternative as your answer.
(Below are some ques from old ques papers)

a) Let S = abcabc, P = bc, Q = bb and R = ba. Then P à Q, Q à R and R à P, changes S to ________? (A) ............ (B) abcabc (C) ............ (D) none of A, B, C

b) Let S = aabbcc, P = ab, Q = bc and R = cc. Then P à Q, Q à R and R à P, changes S to _________? (A) ababab (B) ............ (C) ............ (D) none of A, B, C

c) Let S = bcacbc, P = ac, Q = ca and R = ba. Then P à Q, Q à R, P à R and changes S to ________? (A) ............ (B) ............ (C) bcbabc (D) none of A,B,C

d) Let S = caabcb, P = aa, Q = ca and R = bcb. Then P à Q, P à R, R àQ and changes S to ________? (A) ............ (B) ............ (C) ............ (D) none of A,B,C

Section 3: numerical series

------------------------------
Q’s: 17 - 24
This is little bit tough. proper guesses should be made.Find these problems in R.S.Aggarval's verbal & non verbal reasoning.
17: 2,20,80,100…
(a): 121, (b): 116 (c): (d):none
18: 10,16,2146,2218…

Section 4: figures

------------------------
19:
^ ^ ^
| -> <- | -> |
^ : ^ : ^ : ?
| -> <- | <- |

ans is :

^
| <-
^
| ->
Section 3: series (from other booklet): Transformations

17: 1 1 0 2 2 1 1 à 0 0 1 0 0 2 2
1 0 1 1 0 0 1 à 2 1 2 2 1 1 2
then 2 2 1 1 0 1 1 à ????
Ans: may be 0 0 2 2 1 2 2

18: 1 1 0 0 2 2 à 2 2 0 0 1 1
1 0 1 1 2 1 à 1 2 1 1 0 1

Section 5: Verbal
----------------------
Two words together forming compound words were given. The q's contained the second part of the compound word. The first word of the compound word had to be guessed. Then its meaning had to be matched with the choices.
(see old papers) like ...block head, main stream, star dust

Eg: OLD PAPERS

(1) -(head)- (a) purpose (b) man (c) obstacle

(2) (dust)- (a) container (b) celestial body (c) groom
(ans: c for star dust)

(3) (stream )-(a) mountain (b) straight (c)
(ans:a)

(4)
(crash)- (a) course (b) stock3

Red Set :

_________
i) Series Transformation:
-------------------------------
1) If 102101 à 210212 then 112112 à ?

2) If 102101 à 200111 then 112112 à ?

3) If 102101 à 101201 then 112112 à ?

Tips:
The 1st one all change 0->1, 1->2, 2->1 AND the 2nd on alternate do not change AND the 3rd it is just reverse of the original string.

ii) Target=127: Brick=24,17,13: Operation available = +, /, *, -
Again there r 4 choices. For ex choice b) 20,6,7

Q: 1) U have to make a Target =102; The Answer from the option is (6,17,2,1)

Q: 2) TARGET = 41; Five No’s were given, 25 22 16 5 1. U can use this no’s only once & can perform Operation +, *, -, /, (); Options were:
A) 25 22 16 5 B) 25 22 16 1 C) 25 22 5 1 D) 25 16 5 1)

4 SUCH QUESTI0NS ARE THERE.

iii) Cryptic Sentence - Form words
-------------------------------------------
A sentence is there .a cryptically clue is hidden in the sentence. Find out answer from the option.
1) A friend in Rome
a) aerodrome b) palindine c) palindrome d) condom
Ans: palindrome

2) Rowed them across
a) Crosswiz b) acropolis c) acroword d) crossword
Ans: crossword/crossover

3) Cuticle cutting the filly glass
a) Cubicle b) up hilly c) cut glass d) cutlass
Ans: cutlass

4)
Hat jumps upward in a water closet
a) Watch b) witch

Last section had meaningful words whose anagrams are nouns and have to choose the best adjective from the list to describe this noun:

Eg: shore (word given)
Choices are: a) roman b) Spanish c) Trojan d)....
Ans: c) Trojan

Shore is anagram (jumbled form of) 'horse' and Trojan -- horse is the best match
________________________________________________________________________

2)
This section had the funda of xOy where x and y represented strings of Gs. The test was to find the valid or invalid patterns with reference to the rules.

1)
L=list of objects

Eg) L = {a, b, c, d} where a,b,c,d are objects
P(L) was a function( don’t remember exactly).
M(L) was another function defined etc.
in the following questions P(x) etc were given to be found out.

Note: This may take considerable amount of time. So take intelligent guesses.