Miscount a range by forgetting to include both endpoints (fencepost error).
An error-pattern drill: practice spotting and avoiding this specific SHSAT trap.
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How many whole numbers are there from 5 to 12? Most students subtract: \(12 - 5 = 7\). The right answer is 8 (5, 6, 7, 8, 9, 10, 11, 12). Subtraction counts the gaps between the numbers, not the numbers themselves.
This is the fencepost trap. A fence with 4 gaps needs 5 posts. The same thing happens with a bus that leaves every 15 minutes, or with the integers between the two ends of an inequality.
Check habits:
Count multiples of 3 from 1 to 84, then take away the multiples from 1 to 44: \(\lfloor 84 \div 3 \rfloor - \lfloor 44 \div 3 \rfloor = 28 - 14 = 14\). Leaving out the endpoint 84 gives \(13\). Leaving out both endpoints gives \(12\). Counting the multiples from 1 to 84 instead of from 45 gives \(28\).
The first time is at 11:45 a.m., so the 8th time comes after \(7\) more gaps, not \(8\): \(7 \times 75 = 525\) minutes, which is 8 hours 45 minutes later, at 8:30 p.m. (Counting \(8\) gaps instead of \(7\) gives 9:45 p.m. Not switching a.m. and p.m. when passing noon gives 8:30 a.m. Reading 525 minutes as 5 h 25 min (as if an hour had 100 minutes) gives 5:10 p.m.)
First inequality: \(4x - 8 < 3x + 5\), so \(x < 13\). Second: multiply both sides by \(3\): \(3 - 3x \le x - 21\), so \(24 \le 4x\) and \(x \ge 6\). Watch the endpoints: \(13\) is excluded and \(6\) is included, so \(x = 6, 7, \dots, 12\). Their sum is \(63\). (Including \(13\) gives a sum of \(76\). Leaving out \(6\) gives a sum of \(57\). In the second inequality, moving the \(- 21\) across without changing its sign gives \(x \ge -9/2\) and a sum of \(68\).)