Calculating a directory mode with umask 002 | LinuC Level 1 101 Exam original practice question

1.02.1Basic

Set umask to 002 in a collaborative shell and create a directory with mkdir.

What mode is normally created?

Hint 1: Focus on the requirement

Use the directory base mode and remove only the bits named by the mask.

Hint 2: Key terms

Review the short definitions before comparing the ownership and permission choices.

Base mode 777: The usual requested mode for a newly created directory before masking.
umask 002: A mask that removes other-user write permission while retaining group write.
Mode 775: The resulting rwx owner/group and r-x other directory mode.
Directory execute: The permission needed to traverse directory paths.

Detailed explanation

Incorrect755

Incorrect. 755 would also remove group write, which umask 002 does not mask.

Correct775

Correct. 0777 with umask 0002 produces 0775.

Incorrect664

Incorrect. 664 is a regular-file-style mode.

Incorrect002

Incorrect. The umask itself is not the resulting directory mode.

An example you can run in a temporary verification environment.

tmp=$(mktemp -d); (umask 002; mkdir "$tmp/team"); stat -c '%a' "$tmp/team"; rm -rf "$tmp"

Environment: POSIX shell / GNU stat
Use a disposable file or directory when checking chmod calculations.
Exact permission display can vary slightly with filesystem and distribution defaults.

Foundation review

Directory creation normally starts from 777, so its execute bits remain unless the mask removes them.

Do not apply the regular-file base mode when calculating a new directory's permissions.

Question IDKL-101-379

EnvironmentPOSIX shell / GNU stat

Last technical review2026-08-13

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