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A quality inspector accepts a machined part only if its measured length, a whole number of millimeters, differs from the target length of 50 millimeters by no more than 4 millimeters. How many distinct whole-number lengths are acceptable?
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Correct answer
D
Translate the tolerance into an absolute-value inequality, then count integers in the resulting interval. Let L be the measured length. 'L differs from 50 by no more than 4' means the distance from L to 50 is at most 4, so |L − 50| ≤ 4. Removing the absolute value gives both cases at once: −4 ≤ L − 50 and L − 50 ≤ 4. Adding 50 throughout gives 46 ≤ L and L ≤ 54.
The acceptable whole-number lengths run from 46 to 54 inclusive. The count of integers from 46 to 54 is 54 − 46 + 1 = 9, where the '+1' includes both endpoints. So the answer is 9.
Why the wrong choices tempt: (A) 5 keeps only the at-or-above-target side (50 through 54) and drops the negative case, forgetting that lengths below 50 are equally close. (B) 7 treats 'no more than 4' as 'strictly less than 4', dropping the two exact-tolerance endpoints 46 and 54. (C) 8 finds the interval width 54 − 46 = 8 but skips the '+1', the classic fence-post slip that miscounts the values inside an inclusive range. (E) 11 widens the allowed band by one step on each side, counting 45 through 55 as if a 5-millimeter deviation still passed. Only 9 handles both sign cases and counts both endpoints correctly.
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