B Search

Q:

I. A. B, C. D, E and F are six members of a family.

II. One couple has parents and their children in the family.

III. A is the son of C and E is the daughter of A.

IV. D is the daughter of F who is the mother of E.

Which of the following pairs is the parents of the couple ?

A) AB	B) BC
C) AF	D) CF

Answer & Explanation Answer: B) BC

Explanation:

E is the daughter of A and F is the mother of E So, A is the father of E and hence the husband of F. Now,

D is the daughter of F So. D and E are the daughters of A and F Also. A is the son of C.

Now, only B remains Thus. B and C are the parents of A.

Clearly. B and C are the parents of the couple

Report Error

View Answer Report Error Discuss

Filed Under: Blood Relations

Q:

What is relative path and absolute path?

Answer

Absolute path-- Exact path from root directory.

Relative path-- Relative to the current path.

Workspace

Report Error

View answer Workspace Report Error Discuss

Subject: Operating Systems

Q:

If a box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains no defective bulbs.

A) 5/12	B) 7/12
C) 3/14	D) 1/12

Answer & Explanation Answer: D) 1/12

Explanation:

Total number of elementary events = $10 C_{5}$

Number of ways of selecting no defective bulbs i.e., 5 non-defective bulbs out of 7 is $7 C_{5}$ .

So,required probability = $7 C_{5}$ / $10 C_{5}$ = 1/12.

Report Error

View Answer Report Error Discuss

Filed Under: Probability

Q:

A box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains exactly one defective bulb.

A) 5/12	B) 7/12
C) 3/14	D) 1/12

Answer & Explanation Answer: A) 5/12

Explanation:

Total number of elementary events = $10 C_{5}$

Number of ways of selecting exactly one defective bulb out of 3 and 4 non-defective out of 7 is $3 C_{1} * 7 C_{4}$

So,required probability = $3 C_{1} * 7 C_{4}$ / $10 C_{5}$ = 5/12.

Report Error

View Answer Report Error Discuss

Filed Under: Probability

Q:

A is the brother of B. B is the brother of C. D is the father of A. Based on these three statements, which of the following statements cannot be definitely true ?

A) B is the brother of A.	B) B is the son of D.
C) A is the brother of C.	D) C is the brother of A.

Answer & Explanation Answer: D) C is the brother of A.

Explanation:

A is the brother of B and B is the brother of C. So, C may be the brother or sister of A.

Report Error

View Answer Report Error Discuss

Filed Under: Blood Relations

Q:

E is the son of A. D is the son of B. E is married to C. C is B's daughter. How is D related to E ?

A) Brother	B) Uncle
C) Father-in-Law	D) Brother-in-law

Answer & Explanation Answer: D) Brother-in-law

Explanation:

C is D's daughter and D is B's son. So, D is the brother of C. E is a male married to C.

So, E is the husband of C, whose brother is D. So, D is the brother-in-law of E.

Report Error

View Answer Report Error Discuss

Filed Under: Blood Relations

Q:

A bag contains 50 tickets numbered 1,2,3,4......50 of which five are drawn at random and arranged in ascending order of magnitude.Find the probability that third drawn ticket is equal to 30.

A) 551/15134	B) 1/2
C) 552/15379	D) 1/9

Answer & Explanation Answer: A) 551/15134

Explanation:

Total number of elementary events = $50 C_{5}$
Given,third ticket =30

=> first and second should come from tickets numbered 1 to 29 = $29 C_{2}$ ways and remaining two in $20 C_{2}$ ways.

Therfore,favourable number of events = $29 C_{2} * 20 C_{2}$

Hence,required probability = $29 C_{2} * 20 C_{2}$ $/ 50 C_{5}$ =551 / 15134

Report Error

View Answer Report Error Discuss

Filed Under: Probability

Q:

What are the different job scheduling in operating systems?

Answer

Scheduling is the activity of the deciding when process will receive the resources they request.

FCS ---> FCSFS stands for First Come First Served. In FCFS the job that has been waiting the longest is served next.

Round Robin Scheduling--->Round Robin scheduling is a scheduling method where each process gets a small quantity of time to run and then it is preempted and the next process gets to run. This is called time-sharing and gives the effect of all the processes running at the same time

Shortest Job First ---> The Shortest job First scheduling algorithm is a nonpreemptive scheduling algorithm that chooses the job that will execute the shortest amount of time.

Priority Scheduling--->Priority scheduling is a scheduling method where at all times the highest priority process is assigned the resource.

Workspace

Report Error

View answer Workspace Report Error Discuss

Subject: Operating Systems

Searching for "B"

Quick Links