Calculating a file mode with umask 027 | LinuC Level 1 101 Exam original practice question

1.02.1Basic

Set umask to 027.

A normal program creates a regular file with the usual requested mode 666.

What mode is normally created?

Hint 1: Focus on the requirement

Start with the file creation base mode and remove the bits masked by umask.

Hint 2: Key terms

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

Base mode 666: The usual requested mode for a newly created regular file before masking.
umask: A process setting that removes permission bits from newly created objects.
umask 027: A mask that removes group write and all ordinary permissions for other users.
Mode 640: The resulting owner read/write, group read mode in this calculation.

Detailed explanation

Correct640

Correct. 0666 with umask 0027 produces 0640.

Incorrect650

Incorrect. 0650 does not follow the bitwise masking result.

Incorrect666

Incorrect. 666 is the requested mode before umask is applied.

Incorrect027

Incorrect. The umask value is not the resulting file mode.

An example you can run in a temporary verification environment.

tmp=$(mktemp -d); (umask 027; : >"$tmp/file"); stat -c '%a' "$tmp/file"; rm -rf "$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

umask removes permissions; it cannot add execute or other permissions that the creating program did not request.

Calculate files and directories from their different base modes: commonly 666 versus 777.

Question IDKL-101-378

EnvironmentPOSIX shell / GNU stat

Last technical review2026-08-13

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