Practice Quiz for Sudo Placement
Question 1 |
How many 10 letter words can be formed from the letters in the word SUDO PLACEMENT, starting with S, ending with T and space after the first four characters(repetition is not allowed)?
1 ,814,400 | |
3, 628, 800 | |
108 | |
810 |
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Question 1 Explanation:
Here, repetition of digits are not allowed.
So, distinct letters present in SUDO PLACEMENT are 12.
Total possible combinations are :-
S _ _ _ _ _ _ _ _ T
10 9 8 7 6 5 4 3
First and last letters are fixed so rest places can be filled in the above shown ways.
Total combination possible is 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 = 10! / 2! = 1 ,814,400
Question 2 |
In a 100 meters sprint, Usain Bolt beats Tyson Gay by 1 meter. In another 200m sprint, Tyson Gay beats Asafa Powell by 4 meters. By how many meters Bolt will beat Powell in a long run of 1 KM?
28 meters | |
298 meters | |
32 meters | |
29.8 meters |
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Question 3 |
What is the probability of composite numbers in first 100 numbers(1 - 100)?
0.26 | |
0.62 | |
0.74 | |
0.76 |
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Question 4 |
5 male and 3 female workers can finish a work in 10 days, 8 male and 12 female workers can finish the same work in 5 days. How many female workers(only) are required to finish the same work in a day?
360 | |
720 | |
180 | |
90 |
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Question 5 |
What will be the output of following C program?
main() { char g[] = "geeksforgeeks"; printf("%s", g + g[6] - g[8]); }
geeks | |
rgeeks | |
geeksforgeeks | |
forgeeks |
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Question 5 Explanation:
char g[] = “geeksforgeeks”; // g now has the base address string “geeksforgeeks” // g[6] is ‘o’ and g[1] is ‘e’. // g[6] – g[1] = ASCII value of ‘o’ – ASCII value of ‘e’ = 8 // So the expression g + g[6] – g[8] becomes g + 8 which is // base address of string “geeks” printf(“%s”, g + g[6] – g[8]); // prints geeksHence, option (A) is correct
Question 6 |
Find minimum number of multiplications in matrix chain multiplication(MCM) of matrix given in sequence A2 x 3, B3 x 4, C4 x 3, D3 x 2.
66 | |
64 | |
62 | |
60 |
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Question 7 |
Find the inorder and postorder of the binary tree with the given preorder:
60, 40, 20, 10, 30, 33, 50, 44, 51, 90, 70, 65, 80, 110, 100, 95, 99, 120.
In order: 110, 100, 99, 90, 80, 70, 65, 60, 51, 50, 44, 40, 33, 30, 20, 10.
Postorder: 110, 120, 100, 95, 99, 70, 80, 65, 60, 40, 50, 51, 44, 20, 30, 33, 10 | |
Inorder: 10, 20, 30, 33, 40, 44, 50, 51, 60, 65, 70, 80, 90, 95, 99, 100, 110, 120
Postorder: 10, 33, 30, 20, 44, 51, 50, 40, 65, 80, 70, 99, 95, 100, 120, 110, 90, 60 | |
In order: 10, 33, 30, 20, 44, 51, 50, 40, 60, 65, 80, 70, 99, 95, 100, 120, 110,
Postorder: 10, 20, 30, 33, 40, 44, 50, 51, 60, 65, 70, 80, 90, 95, 99, 100, 110 | |
In order: 10, 33, 30, 20, 44, 51, 60, 65, 80, 70, 99, 95, 100, 120, 110,
Postorder: 110, 100, 99, 90, 80, 70, 65, 60, 51, 50, 44, 40, 33, 30, 20, 10. |
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Question 8 |
Here are the two concurrent process P1, P2 with respective codes:
P1 code:
while (true) // infinite condition { A :____; printf("%d", 1); printf("%d", 1); B:____; }P2 code:
while (true) // infinite condition { C:____; printf("%d", 0); printf("%d", 0); D:____; }What should be the binary semaphore operation on A,B,C,D respectively and what must be the intial values of semaphore M,N inorder to get the output 110011001100....? Where P is down and V is up operation respectively.
A = P(N), B = V(M), C = P(M), D = V(N); M = 0, N = 1; | |
A = P(N), B = V(M), C = P(M), D = P(N); M = N = 1; | |
A = P(N), B = V(N), C = P(M), D = V(M); M = 1, N = 0; | |
A = P(N), B = V(N), C = P(M), D = V(M); M = N = 1; |
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Question 9 |
Consider the relation R (ABCDE):
FD = { A → B, B → C, C → D, D → E}
Find out the highest normal form.
1 NF | |
2 NF | |
3 NF | |
BCNF |
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Question 9 Explanation:
Here candidate Key is A and B -> C , C -> D , D -> E all are. (Non prime attribute -> Non prime attribute.).
This type of FD must not be present in 3NF therefore highest normal form of this FDs are 2NF.
Option (B) is correct.
Question 10 |
When and where RARP is used intentionally or effectively?
At the time of network booting where no space to store IP address (or diskless network) for address resolution. | |
In broadcasting to get IP address of Network. | |
To get the access in private network whenever it is required. | |
None of the above. |
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There are 10 questions to complete.