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Title: Aptitude 8 Post by: Tanya on May 11, 2007, 01:06:22 PM Aptitude 8 BROWN 2002 There were different papers for different sessions. The paper had 5 sections, 5 * 8 = 40 Q'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 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 Section 4 : figures 19: ^ ^ ^ | -> <- | -> | ^ : ^ :: ^ : ? | -> <- | <- | ans is : ^ | <- ^ | -> all probems are very easy.(see cts_oldcts13 file) some are mirror images, some r rotated clockwise/anti Section 5 : Verbal if u have a very good vocab. then this section is managable. two words together forming a compound words were given. the q's contained the second part of the compound word. the first word of the compuond word had to be guessed. then its meaning had to be matched with the choices. if the word is "body" then its meaning of its first part is.. its really tough to guess.. the words were however very simple some words which i can remember are, head, god, (see old papers) like block head main stream star dust Eg: OLD PAPERS (1) -(head)- (a) purpose (b) man (c)obstacle (d)(ans:c for blockhead) >(2) (dust)- (a) container(b)celestial body (c)groom(d)(ans: c for star dust) >(3) (stream )-(a) mountain (b) straight (c) (d) (ans:a) >(4) (crash)- (a) course (b) stock3 anagram >first find the anagram of the given word & then >choose the meaning of the anagram from the options. >1. latter ->rattle 2..spread 3.risque 4.dangled(ansjogged)… |