For an outside dimension specified as 1.0625 +.0025/-.0003 inches, what is the total tolerance width?

Unlock all questions

This demo includes only 20 questions. Upgrade to access hundreds of questions, flashcards, exam simulations, and disable ads.

Full question bankExam simulationsFlashcards
From $9.99Unlock all

Prepare for the General AandP Test with comprehensive study materials. Utilize flashcards and multiple-choice questions, each with explanations. Get ready for success in your exam journey!

Multiple Choice

For an outside dimension specified as 1.0625 +.0025/-.0003 inches, what is the total tolerance width?

Explanation:
When an outside dimension has asymmetric plus and minus tolerances, the total tolerance width is the range from the largest allowed size to the smallest allowed size. For 1.0625 +0.0025/-.0003, the maximum size is 1.0625 + 0.0025 = 1.0650 inches, and the minimum size is 1.0625 - 0.0003 = 1.0622 inches. The width of the acceptable range is 1.0650 − 1.0622 = 0.0028 inches. You can also sum the tolerance amounts: 0.0025 + 0.0003 = 0.0028. So the total tolerance width is 0.0028 inches.

When an outside dimension has asymmetric plus and minus tolerances, the total tolerance width is the range from the largest allowed size to the smallest allowed size. For 1.0625 +0.0025/-.0003, the maximum size is 1.0625 + 0.0025 = 1.0650 inches, and the minimum size is 1.0625 - 0.0003 = 1.0622 inches. The width of the acceptable range is 1.0650 − 1.0622 = 0.0028 inches. You can also sum the tolerance amounts: 0.0025 + 0.0003 = 0.0028. So the total tolerance width is 0.0028 inches.

More Multiple Choice Questions
Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy