Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 1.20
Evaluate the following calculation, using an appropriate number of significant figures and/or decimal places.
0.344 km + 5,500 cm = ?
(A) 0.3495 km.
(B) 0.350 km.
(C) 0.399 km.
(D) 0.40 km.
Correct answer (highlight to unhide): (C)
5,500 cm has two significant figures (as denoted in green, and the digits in yellow are not significant), as the trailing zeroes are ambiguous without an explicit decimal point. Converting this distance measurement to km:
Note that the quantities in these conversion factors (which are definitions of metric powers of ten) are assumed to have an infinite amount of significant figures, and thus do not affect the number of significant figures (two) converting 5,500 cm to 0.055 km.
For addition and subtraction operations, the term with the least number of decimal places determines the number of significant decimal places in the result.
Then the addition/subtraction rule limits the answer to the least number of decimal places (both 0.344 km and 0.055 km have three decimal places, as denoted in gray), such that:
0.344 km + 0.055 km = 0.399 km.
Sections 70854, 70855
Exam code: quiz01unI7
(A) : 1 student
(B) : 7 students
(C) : 40 students
(D) : 9 students
Success level: 70%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.38