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13 Mar 2026, 12:52
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
55% (01:43) correct
45%
(02:02)
wrong
based on 38
sessions
History
Date
Time
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Not Attempted Yet
X of the instructors at Turnbull College have a master’s degree, and the number of instructors with a PhD is 4 greater than the number with a master’s degree.
If n of these instructors have both a master’s degree and a PhD, how many instructors at Turnbull have either a PhD or a master’s degree but not both?
A. 2x + 4 − 2n
B. 2x + 4 − n
C. x + 4 − n
D. 5x − 2n
E. 5x - n
If n of these instructors have both a master’s degree and a PhD, how many instructors at Turnbull have either a PhD or a master’s degree but not both?
A. 2x + 4 − 2n
B. 2x + 4 − n
C. x + 4 − n
D. 5x − 2n
E. 5x - n
13 Mar 2026, 13:18
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Great question — this is a classic Overlapping Sets problem, and the key is to work with "exactly one" rather than "at least one."
Key concept being tested: The Two-Set Formula — specifically isolating elements belonging to exactly one set.
Step 1 — Set up what you know.
Let M = instructors with a master's degree = x
Let P = instructors with a PhD = x + 4 (given as "4 greater than the number with a master's")
Let B = instructors with BOTH = n
Step 2 — Find "master's only" and "PhD only."
Master's only (no PhD) = M − B = x − n
PhD only (no master's) = P − B = (x + 4) − n
Step 3 — Answer the question.
The question asks: how many have either a PhD or a master's but NOT both?
That's exactly (master's only) + (PhD only):
= (x − n) + (x + 4 − n)
= 2x + 4 − 2n
Answer: A
Common trap students fall into: Many jump to using the full union formula |M ∪ P| = M + P − B = x + (x+4) − n = 2x + 4 − n, which gives answer B. That formula counts people with BOTH degrees once — but the question specifically asks for those with one but NOT both. You need to subtract the "both" group entirely from the count, not just avoid double-counting it.
Quick check with numbers: Say x = 10 (10 masters), so PhD = 14, and n = 3 (3 have both).
Masters only = 7, PhD only = 11, total with exactly one = 18.
Plug into A: 2(10) + 4 − 2(3) = 20 + 4 − 6 = 18. ✓
Plug into B: 2(10) + 4 − 3 = 21. ✗ (That 21 includes the 3 "both" people.)
Takeaway: Whenever a question asks for "exactly one" of two categories, calculate each "only" group separately — don't rely on the union formula, which always gives "at least one."
Key concept being tested: The Two-Set Formula — specifically isolating elements belonging to exactly one set.
Step 1 — Set up what you know.
Let M = instructors with a master's degree = x
Let P = instructors with a PhD = x + 4 (given as "4 greater than the number with a master's")
Let B = instructors with BOTH = n
Step 2 — Find "master's only" and "PhD only."
Master's only (no PhD) = M − B = x − n
PhD only (no master's) = P − B = (x + 4) − n
Step 3 — Answer the question.
The question asks: how many have either a PhD or a master's but NOT both?
That's exactly (master's only) + (PhD only):
= (x − n) + (x + 4 − n)
= 2x + 4 − 2n
Answer: A
Common trap students fall into: Many jump to using the full union formula |M ∪ P| = M + P − B = x + (x+4) − n = 2x + 4 − n, which gives answer B. That formula counts people with BOTH degrees once — but the question specifically asks for those with one but NOT both. You need to subtract the "both" group entirely from the count, not just avoid double-counting it.
Quick check with numbers: Say x = 10 (10 masters), so PhD = 14, and n = 3 (3 have both).
Masters only = 7, PhD only = 11, total with exactly one = 18.
Plug into A: 2(10) + 4 − 2(3) = 20 + 4 − 6 = 18. ✓
Plug into B: 2(10) + 4 − 3 = 21. ✗ (That 21 includes the 3 "both" people.)
Takeaway: Whenever a question asks for "exactly one" of two categories, calculate each "only" group separately — don't rely on the union formula, which always gives "at least one."
