mathematical series problem

4 series .
>(1) 2,11,22,121 ,? ans 242 (2)102,211,1020,1210,?7
another type
>string S=abc now substrings P=ab Q=ad such that
>if p->q, then S=adc (according to the
firstoccurance)based on these type
>(a) S=abcabc ;P->Q Q->R R->P what would be
theresulting S ,given P=ab, Q=ba R=bc
>(b) S=aaabbb; P->Q Q->P R->P (according to the
firstoccurrence)P=….Q=……R=……
>two questions on some figures given (see
compitionmaster )
------------------------------------------------

Questions 17-20:
------------------------
In this section, four series of numbers are provided.
You have to choose
the next term of each given series from the supplied
alternatives.

17. There is a series of numbers as 102,210,1020,1121,... .
The next term is
(A) ............ (B) ............ (C) ............
(D) none of A,B,C
18. There is a series of numbers as 110,122,1021,1110,... .
The next term is
(A) ............ (B) ............ (C) ............
(D) none of A,B,C
19. There is a series of numbers as 2,11,22,121,... . The
next term is
(A) ............ (B) ............ (C) ............
(D) none of A,B,C
20. There is a series of numbers as . The
next term is
(A) ............ (B) ............ (C) ............
(D) none of A,B,C
--------------------------------------
old papers:
I
1. 2D, 3C, 4B, 5A. Ans: 6Z
2. C5, B0, A5, 90. Ans:
3. 25, 19, 31, 1f, 28. Ans:
4. 34, 52, 2b, 43, 3c. Ans:
II.
1. 102, 111, 120, 122. Ans:
2. 110, 212, 1021, 1200. Ans:
3. 102, 211, 1020, 1122. Ans:
4. 2, 11, 22, 121. Ans:
------------------------------------------


Series Transformation
1) If 102101->210212 then 112112->?
a)
b)
c)
d)

2) if 102101-> 200111 then 112112->?
Again there r 4 choices.

3) If 102101->101201 then 112112->?
Again there r 4 choices.

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

Section 2 : Word series
Q's : 9 - 16

This is one of the easiest section. Try to do it at first.
if S is a string then p,q,r form the substrings of S.
for eg, if S=aaababc & p=aa q=ab r=bc
then on applying p->q on S is that ababaabc
only the first occurance of S has to be substituted.
if there is no substring 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->q, q->r & r->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->r & r->p will
give,
11: if s=abababc & p= aa q=ab r=bc then applying p->q, q->r & r->p will
give,
12: if s=abababc & p= aa q=ab r=bc then applying q->r & r->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: similiar type of prob.
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 which also contain strings. An operation
P->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.

(the below are some ques from old ques papers)

21. Let S=abcabc, P=bc, Q=bb and R=ba. Then P->Q, Q->R, R-
>P changes S to
(A) ............ (B) abcabc (C) ............
(D) none of A,B,C
22. Let S=aabbcc, P=ab, Q=bc and R=cc. Then P->Q, Q->R, R-
>P changes S to
(A) ababab (B) ............ (C) ............
(D) none of A,B,C
23. Let S=bcacbc, P=ac, Q=ca and R=ba. Then P->Q, Q->R, P-
>R changes S to
(A) ............ (B) ............ (C) bcbabc
(D) none of A,B,C
24. Let S=caabcb, P=aa, Q=ca and R=bcb. Then P->Q, P->R, R-
>Q 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 probs 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, ??

like these other series were given.

section 3 : series (from other booklet)
transformations

17: 1 1 0 2 2 1 1 ---> 0 0 1 0 0 2 2
1 0 1 1 0 0 1 ---> 2 1 2 2 1 1 2
then
2 2 1 1 0 1 1 ---> ????
ans may be 0 0 2 2 1 2 2

18: 1 1 0 0 2 2 ---> 2 2 0 0 1 1
1 0 1 1 2 1 ---> 1 2 1 1 0 1
-----------------------------------------------
Series.

1. Interchange of letters in a word and the adjacent letters are also to
be changed. given letters series like [also few condotions]
AAABBB=
ABABAB=
LET QUESTION IS ABBAAB
If we apply 25 on this it means we have to interchange the letters
at positions 2 and 5, and we have to change the adjacent letters 2 and 5
from A to B and B to A.
That is q's A B B A A B
after Step 1 i.e interchange 2 and 5.

now change adjacent elements of 2 and 5...finally answer becomes
Ans: B A A B B A

//Hint: As per question papers 5 questions above like but numbers
change.

REMAINING 3 QUESTIONS:
6. To get AAABBD from BBBAAA what ot apply:-
a) 25 b) 34 c)25 & 34 d) none
------------------

0 comments:

Post a Comment