TISL PLACEMENT PAPERS -------------- Placement Paper 10
Answer / guest
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TATA -IBM
-----------
LETTER SERIES
answers
- -------
1)e e f g g h i i --- j
2)a z a y b z b y c --- a e x y z z
3)d e f d e f g h i --- d e f g h g
4)c d e x y z f g h x y z i j k l m i
5)de f de g d e --- d j z h i h
6)a b c z a b c y a b c --- a b x y z x
7)d g b h i b j k b ---- b e l m n l
8)t s p t s q t s --- r s t v w r
9)q a r b s a t a r b ---- a b e r s s
10)b c c d e e f g -- e f g h i g
11)e f h i k l --- m n o p q n
12)a b cc d e ff g -- efghi h
13)a m n b op c --- depqr q
14)ttt sss qqq p --- opqrs p
15)ddffhhjj--- ijklm l
16)mnmnklopopkl ----- kopqr q
17)cddeeefff--- efghi f
18)gfde---- bcfgh c
19)dfhjl ----- jklmn n
20)abcijdefij---- ghijk g
21)efgefghefghi ---- e
22)bcbdedffghi ---- fghjk h
23)aababccdc --- cdefg d
24)aibcidef --- efghi i
25)cehl --- opqrs q
26)abdehimn --- pqrst s
27)becfdge --- efghi h
28)agbhc ---- dfghi i
29)adhko --- pqrst r
30)efghjklno --- pqrst q
31)aeibf --- cdgij j
32)aedhg-- hijkl k
33)zdwgt --- hijkl j
34)zeiyijxj--- ijklm k
35)dqreuvg--- vwxyz y
36)ksjtiuh--- vwxyz v
37)rsjtuhvw--- cdefg f
38)ieajfbk ---- cdefg g
39)hebijej ---- ghijk g
40)njfmiel --- dhijm h
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1)
main()
{
char *p1="name";
char *p2;
p2=(char*)malloc(20);
while(*p2++ = *p1++);
printf("%s\n",p2);
}
Ans:empty string.
2)
main()
{
int x=20,y=35;
x=y++ + x++;
y= ++y + ++x;
printf("%d%d\n",x,y);
}
Ans : 57 ,94
3)
main()
{
int x=5;
printf("%d%d%d\n",x,x<<2,x>>2);
}
ans: 5,20,16
4)
#define swap(a,b) a=a+b;b=a-b;a=a-b;
main()
{
int x=5;y=10;
swap1(x,y);
printf("%d %d\n",x,y);
swap2(x,y);
printf("%d %d\n",x,y);
}
int swap2(int a, int b)
{
int temp;
temp=a;
b=a;
a=temp;
return;
}
like that
ans: 10 ,5
10 ,5
5)
main()
{
char *ptr = " Ramco Systems"
(*ptr)++;
printf("%s\n",ptr);
ptr++;
printf("%s\n",ptr);
}
Ans: Samco Systems
amco systems
6)
main()
{
char s1[]="Ramco";
char s2[]= "systems";
printf("%s",s1);
}
Ans: compilation error giving that it can not
modified Lvalue.
7)
main()
{
char *p1;
char *p2;
p1=(char *)malloc(25);
p2=(char *)malloc(25);
strcpy(p1,"Ramco");
strcpy(p2,"systems");
strcat(p1,p2);
printf("%s",p1);
}
Ans: Ramcosystems
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Interview Questions
8)
The following variable is available in file1.c
static int average float;
Ans: all the functions in the file1.c can access
the variable.
9)
Ans : [2] . extern int x;
check the answer.
10) Another problem with
#define TRUE 0
somecode.
while(TRUE)
{
somecode
}
This will not go into the loop as TRUE is defined
as 0.
Ans : None of the above i.e (D).
11)
Ans :
[4]. A question in structures where the
members are dd,mm,yy.
mm:dd:yy
09:07:97
15) structure kswap
Ramco India
Ramco Systems Corporation
Ramco .... Limited.
After swaping the result will be
First two will be swaped
Ramco Systems Corporation.
Ramco India
Ramco .... Limited.
16) int x;
main()
{ int x=10;
x++;
change-value(x);
x++;
modify-value();
printf("First output:%d\n",x);
x++;
change-value(x);
printf("secpnd output:%d\n",x);
modify-value();
printf("Third output:%d\n",x);
}
modify-value()
{
return(x+=10);
}
change-value()
{
return(x+=1);
}
Ans: 12 , 1 , 1
17) main()
{
int x=10;y=15;
x = x++;
y = ++y;
printf("%d %d\n",x,y);
}
Ans: 11, 16
18)
main()
{
int a=0;
if(a==0) printf("Ramco Systems\n");
printf("Ramco Systems\n");
}
Ans: only one time "Ramco Systems" will be printed.
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SNAPDEAL QUESTIONS 1. A train is going at a speed of 60kmph towards Delhi and returned back at a speed of 30kmph. What is its average speed? ANS : (2*30*60)/(30+60) = 40kmph 2. How many different 4 letter words can be framed that have at least one vowel? ANS : 264 - 214 (total no of 4 digits words – no of words with no vowels) 3. Write an algorithm to find out a number from an array of numbers where only one number occurs once and rest all occurs twice. ANS : XOR all the numbers ,you will get the number with single occurrences . 4. Which among the following have the product of the distance between opposite sides of a regular polygon and it side equals one fourth of the area. A. hexagon B. octagon C. n=16 D. n=18 ANS : n=16.(area of regular polygon = apothem*perimeter/2 Apothem = distance between opposite sides/2 Area = (opp_side_dist * n * a )/4 Product of opp_side_dist and side of reg. polygon = opp_side_dist * a For n= 16 the ration becomes 1:4 5. Which of the following cannot be a relation between two variables? ANS = 4th diagram. 6.what will be the output of this program Void print (int n) { If (n>0) { printf(“hello”); print(n-1); } printf(“world”); } ANS : N times hello followed by N+1 times world. 7. Which among the following cannot be used for future prediction? ANS : 4th Diagram. 8. There are 25 horses. We have to find out the fastest 3 horses In one race maximum 5 horses can run. How many such races are required in minimum to get the result. ANS : 7 races (A. first run all horses = 5 races, eliminate 4th 5th of all races. B. Run horses who came 1st in those 5 races = 1 race , the horse coming first is the fastest Run horses a. 2nd and 3rd with the fastest horse (in first time race A) b. 2nd and 3rd coming horse in B. c. The horse who came 2nd with the horse(who came 2nd in race B) in race A You will have the fastest 3 horses.) 9. In a game of rolling dice you are given 2 dice and you have to roll them. Whatever is the outcome the player will win that many dollars. What should the game owner charge each player (optimum) so that he doesn’t have to bear any loss? ANS : $7 10. We have a function REV(“string”,m,n).This function is capable of reversing the caharacters in the string from mth location to nth location. e.g. REV(“abcd”,2,3)  the output will be acbd We need to swap a string from a position,e.g. SWAP(“abcdefg”,4)  output needs to be efgabcd. How can the REV function used do this. ANS : L = string length,N= position given in SWAP function. SWAP(“abcdefg”,4) = REV(REV(REV(“abcdefg”,N+1,L),1,N),1,L).
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