Here is the 1st part of Robert Bosch Placement Papers. This is the latest placement paper of 2014 which is fully solved. We have provided the solution for each and every question. So practice this test before attending any Robert Bosch written test or exam. This sample Robert Bosch placement paper will definitely help you at the time of written exam.
1. Function used to display the character on the console is
A. fput() B.put() C. getch() D. none of these
Answer: B
2. Number of nodes in a perfect Binary tree is
A.2 * L - 1 B.2* L C.2* L +1 D.none of these
Answer: A
Explanation:
Priority queue uses two queues 1. For storing the data. 2. Other for storing the priorities.
3. Heterogeneous Linked List is implemented using:-
A.int pointer B.float pointer C.void pointer D.none of these
Answer: C
4. Number of queues needed to implement the priority queue:-
A.two B.one C.three D.none of these
Answer: A
Explanation:
Priority queue uses two queues 1. For storing the data. 2. Other for storing the priorities.
5. A variable used to store the address of another variable is termed as:-
A.variable B.function C.registers D.pointer
Answer: D
6. void main()
{
int i = 15;
int const &j = i;
cout << i << j;
i = 9;
cout << i << j<
}
A.15 garbage value 9 garbage value B. 15 15 9 9
C.15 garbage value 15 garbage value D.None of these
Answer: B
7. Virtual Memory consists of main memory and ______.
A.cache B.ROM C.RAM D.Swap files
Answer: D
8. Which class addressing system has less number of hosts ?
A.Class A B.Class B C.Class C D.Class D
Answer: D
9. Which of the following is a network layer protocol ?
A.TELNET B.FTP C. smtp D.none
Answer: A
10. #define A 10+10
main ()
{
int a;
a = A * A;
printf ("%d", a);
}
A.100 B.200 C.120 D.400
Answer: C
11. main()
{
int I;
char s[ ] = "man";
for(i=0;s[i];i++)
printf("%c %c %c %c",i[s],s[i],*(s+i),*(i+s));
}
A.mmm aaaa nnnn B.mmmm aaaa nnnn
C.aaaa mmmm nnnn D.none of these
Answer: B
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].
12. void main ()
{
int const *p = 10;
printf ("%d", ++ (*p));
}
A.10 B.12 C.compiler error D.Linker error
Answer: C
Explanation:
Cannot modify a constant value.
13. main ()
{
extern int i = 10;
printf ("%d", i);
}
A.10 B.9 C.compiler error D.none
Answer: C
14. main()
{
float f = 1.1;
double d = 1.1;
if ( f == d)
printf("True");
else
printf("False");
}
A.True B.False C.compiler error D.None of these
Answer: B
Explanation:
Due to precision Considerations, floating point numbers cannot used in if condition for comparison.
15. What is the name of the Command used to partition the disk ?
A.fdisk B.chkdisk C.sfdisk D.none of these
Answer: A
Explanation:
When you run the fdisk and format commands, the Master Boot Record (MBR) and file allocation tables are created.The MBR and file allocation tables store the necessary disk geometry that allows hard disk to accept, store, and retrieve data.
16. Equation for effective access time in paging(Memory management) ?
A.((1-p)*m+(p*s)) B.((1-p)*s+(p*m)) C.(1-p)*m D.none of these
Answer: A
Explanation:
Effective access time=((1-p)*m+(p*s)) where (1-p) is called hit ratio, p is called page fault ratio, 'm' is called main memory access time, 's' is called page fault service time.
17. The header file that is included to use manipulators is
A. iomanip.h B. manip.h C.math.h D.string.h
Answer: A
18. Which data structure is easier to implement the trees ?
A.Queues B.stacks C.linked lists D.Hashing table
Answer: C
19. Find post order traversal from the given tree ? tree (Image)
A. U4 U10 U3 U8 U2 U9 U11 U7 U6 U5 U1
B. U1 U2 U3 U4 U10 U8 U5 U9 U6 U11 U7
C. U4 U10 U3 U2 U8 U9 U11 U7 U6 U5 U1
D. none of these
Answer: A
20. Find pre order traversal from the given tree ? tree (Image)
A. U4 U10 U3 U8 U2 U9 U11 U7 U6 U5 U1
B. U1 U2 U3 U4 U10 U8 U5 U9 U6 U11 U7
C. U1 U2 U3 U4 U10 U8 U5 U6 U11 U7 u9
D.None of these
Answer: B
21. Main ()
{
int n=0;
while (n>32767)
printf ("%d", n);
n++
}
A.Prints value from 0 to 32767 B.Infinite loop
C.compiler error D.none
Answer: B
Explanation:
Indefinite loop here the loop will continue again & again i.e. after 32767 the go to -32768
22. If (fun ())
{
x++;
}
A.fun () returns 0 B.fun () return 1
C.fun () return -1 D.return a value other than 0
Answer: D
23. Name a x.25 addressing standard ?
A. X.121 B. X.251 C. 1EEE8.1 D.none of these
Answer: A
24. Short, int and long integers have how many bytes ?
A.2, 2, 4 B.Machine dependant. C.2, 4, 8 D.None of these
Answer: B
25. Main ()
{
int a = 5, b = 6;
int i = 0;
i = a > b ? a:b;
printf ("%d", i);
}
A.0 B.1 C.6 D.5
Answer: C
26. Int fun (char c)
{
int i;
static int y;
}
A.c, i are stored in stack and y stored in global memory space
B.c stored in stack and i, y are stored in global memory space
C. c is stored in text segment, y in global memory space and i in stack
D.none of these
Answer: A
27. How would you insert pre-written code into a current program ?
A.#read B.#get C.#include D.#pre
Answer: C
28. Structure may contain:-
A.Any other structure B.Any other structure expect themselves
C.Any other structure except themselves and pointed to themselves
D.None of the above
Answer: A
29. Windowing technique is a kind of
A.Flow control technique B.Error control technique
C.Both a & b D.None of these
Answer: A
30. Windowing technique is a kind of
A.Flow control technique B.Error control technique
C.Both a & b D.None of these
Answer: A
Robert Bosch Placement 2014:
1. Function used to display the character on the console is
A. fput() B.put() C. getch() D. none of these
Answer: B
2. Number of nodes in a perfect Binary tree is
A.2 * L - 1 B.2* L C.2* L +1 D.none of these
Answer: A
Explanation:
Priority queue uses two queues 1. For storing the data. 2. Other for storing the priorities.
3. Heterogeneous Linked List is implemented using:-
A.int pointer B.float pointer C.void pointer D.none of these
Answer: C
4. Number of queues needed to implement the priority queue:-
A.two B.one C.three D.none of these
Answer: A
Explanation:
Priority queue uses two queues 1. For storing the data. 2. Other for storing the priorities.
5. A variable used to store the address of another variable is termed as:-
A.variable B.function C.registers D.pointer
Answer: D
6. void main()
{
int i = 15;
int const &j = i;
cout << i << j;
i = 9;
cout << i << j<
}
A.15 garbage value 9 garbage value B. 15 15 9 9
C.15 garbage value 15 garbage value D.None of these
Answer: B
7. Virtual Memory consists of main memory and ______.
A.cache B.ROM C.RAM D.Swap files
Answer: D
8. Which class addressing system has less number of hosts ?
A.Class A B.Class B C.Class C D.Class D
Answer: D
9. Which of the following is a network layer protocol ?
A.TELNET B.FTP C. smtp D.none
Answer: A
10. #define A 10+10
main ()
{
int a;
a = A * A;
printf ("%d", a);
}
A.100 B.200 C.120 D.400
Answer: C
11. main()
{
int I;
char s[ ] = "man";
for(i=0;s[i];i++)
printf("%c %c %c %c",i[s],s[i],*(s+i),*(i+s));
}
A.mmm aaaa nnnn B.mmmm aaaa nnnn
C.aaaa mmmm nnnn D.none of these
Answer: B
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].
12. void main ()
{
int const *p = 10;
printf ("%d", ++ (*p));
}
A.10 B.12 C.compiler error D.Linker error
Answer: C
Explanation:
Cannot modify a constant value.
13. main ()
{
extern int i = 10;
printf ("%d", i);
}
A.10 B.9 C.compiler error D.none
Answer: C
14. main()
{
float f = 1.1;
double d = 1.1;
if ( f == d)
printf("True");
else
printf("False");
}
A.True B.False C.compiler error D.None of these
Answer: B
Explanation:
Due to precision Considerations, floating point numbers cannot used in if condition for comparison.
15. What is the name of the Command used to partition the disk ?
A.fdisk B.chkdisk C.sfdisk D.none of these
Answer: A
Explanation:
When you run the fdisk and format commands, the Master Boot Record (MBR) and file allocation tables are created.The MBR and file allocation tables store the necessary disk geometry that allows hard disk to accept, store, and retrieve data.
16. Equation for effective access time in paging(Memory management) ?
A.((1-p)*m+(p*s)) B.((1-p)*s+(p*m)) C.(1-p)*m D.none of these
Answer: A
Explanation:
Effective access time=((1-p)*m+(p*s)) where (1-p) is called hit ratio, p is called page fault ratio, 'm' is called main memory access time, 's' is called page fault service time.
17. The header file that is included to use manipulators is
A. iomanip.h B. manip.h C.math.h D.string.h
Answer: A
18. Which data structure is easier to implement the trees ?
A.Queues B.stacks C.linked lists D.Hashing table
Answer: C
19. Find post order traversal from the given tree ? tree (Image)
A. U4 U10 U3 U8 U2 U9 U11 U7 U6 U5 U1
B. U1 U2 U3 U4 U10 U8 U5 U9 U6 U11 U7
C. U4 U10 U3 U2 U8 U9 U11 U7 U6 U5 U1
D. none of these
Answer: A
20. Find pre order traversal from the given tree ? tree (Image)
A. U4 U10 U3 U8 U2 U9 U11 U7 U6 U5 U1
B. U1 U2 U3 U4 U10 U8 U5 U9 U6 U11 U7
C. U1 U2 U3 U4 U10 U8 U5 U6 U11 U7 u9
D.None of these
Answer: B
21. Main ()
{
int n=0;
while (n>32767)
printf ("%d", n);
n++
}
A.Prints value from 0 to 32767 B.Infinite loop
C.compiler error D.none
Answer: B
Explanation:
Indefinite loop here the loop will continue again & again i.e. after 32767 the go to -32768
22. If (fun ())
{
x++;
}
A.fun () returns 0 B.fun () return 1
C.fun () return -1 D.return a value other than 0
Answer: D
23. Name a x.25 addressing standard ?
A. X.121 B. X.251 C. 1EEE8.1 D.none of these
Answer: A
24. Short, int and long integers have how many bytes ?
A.2, 2, 4 B.Machine dependant. C.2, 4, 8 D.None of these
Answer: B
25. Main ()
{
int a = 5, b = 6;
int i = 0;
i = a > b ? a:b;
printf ("%d", i);
}
A.0 B.1 C.6 D.5
Answer: C
26. Int fun (char c)
{
int i;
static int y;
}
A.c, i are stored in stack and y stored in global memory space
B.c stored in stack and i, y are stored in global memory space
C. c is stored in text segment, y in global memory space and i in stack
D.none of these
Answer: A
27. How would you insert pre-written code into a current program ?
A.#read B.#get C.#include D.#pre
Answer: C
28. Structure may contain:-
A.Any other structure B.Any other structure expect themselves
C.Any other structure except themselves and pointed to themselves
D.None of the above
Answer: A
29. Windowing technique is a kind of
A.Flow control technique B.Error control technique
C.Both a & b D.None of these
Answer: A
30. Windowing technique is a kind of
A.Flow control technique B.Error control technique
C.Both a & b D.None of these
Answer: A
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