101-500 Valid Exam Questions | Free 101-500 Dumps

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LPIC-1 Exam 101, Part 1 of 2, version 5.0 is a certification exam for Linux system administrators. It is designed to test the candidate's knowledge and understanding of the basics of Linux administration, including installation and configuration of Linux systems, command line usage, file management, and basic networking. 101-500 Exam is the first part of the LPIC-1 certification program, which is recognized globally as the standard for Linux certification.

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Lpi LPIC-1 Exam 101, Part 1 of 2, version 5.0 Sample Questions (Q99-Q104):

NEW QUESTION # 99
When installing XWindows the monitor wasn't recognised but now a data sheet with its specification is available. Which xorg.conf section must be edited to specify the correct values for several variables?

A. Device
B. Border
C. Monitor
D. Modeline
E. Screen

Answer: C

NEW QUESTION # 100
Which of the following statements is correct for a command line ending with a & character?

A. Thecommand is run in background of the current shell.
B. The result of the command defines if the next command will be run.
C. The command's output is redirected to/dev/null.
D. The command is run as a direct child of theinitprocess.

Answer: A

NEW QUESTION # 101
Which umask value will result in the default access permissions of 600 (rw-------) for files and 700 (rwx------) for directories? (Specify only the numerical umask value.)

Answer:

Explanation:
0077, 077
Explanation
The umask value that will result in the default access permissions of 600 (rw-------) for files and 700 (rwx------) for directories is 077. This is because the umask value is an octal number that represents the permissions that are not given to the files or directories. To calculate the umask value, we need to subtract the desired permissions from the maximum permissions, which are 666 for files and 777 for directories. For example, if we want the files to have the permissions 600 (rw-------), which means read and write for the owner and no permissions for the group and others, we need to subtract 600 from 666, which gives us 066.
Similarly, if we want the directories to have the permissions 700 (rwx------), which means read, write, and execute for the owner and no permissions for the group and others, we need to subtract 700 from 777, which gives us 077.The umask value is the minimum of these two values, which is 077. The first digit of the umask value is for the special permissions, such as setuid, setgid, and sticky bit, and it is usually set to 0. Therefore, the final umask value is 077. For more information on how to use the umask command, you can refer to the following articles:
* What Is umask in Linux, and How Do You Use It? - How-To Geek
* Umask command in Linux with examples - GeeksforGeeks
* What is UMASK and how to set UMASK in Linux/Unix?
* How to Set and Update the Default Umask Value - phoenixNAP

NEW QUESTION # 102
Which of the following commands display the IDs of all processes owned by root? (Choose two.) pgrep -c root

A. pgrep -u root
B. pgrep -f root
C.
D. pgrep -c 0
E. pgrep -U 0

Answer: B,D

NEW QUESTION # 103
Which of the following commands updates the already installed RPM package rpmname?

A. rpm -u rpmname
B. rpm --force rpmname
C. rpm -U rpmname
D. rpm --update rpmname
E. rpm -q rpmname

Answer: C

NEW QUESTION # 104
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