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13 Mar 2026, 10:05
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
56% (03:37) correct
44%
(03:58)
wrong
based on 18
sessions
History
Date
Time
Result
Not Attempted Yet
At a certain company, 35 interns are women and 30 interns are economics majors. Exactly 10 percent of the interns are women who are economics majors.
Men must make up at least what fraction of the interns?
A. 2/5
B. 5/12
C. 7/16
D. 1/2
E. 3/5
Men must make up at least what fraction of the interns?
A. 2/5
B. 5/12
C. 7/16
D. 1/2
E. 3/5
13 Mar 2026, 12:02
Kudos
Bookmarks
Great Problem Solving question testing Overlapping Sets — the key is recognizing what "at least" means and setting up the constraint correctly.
Key concept: Overlapping Sets with a minimization condition.
Common trap: Students often jump to thinking the minimum number of interns is exactly 35 + 30 = 65. That's wrong — it ignores the overlap. Once you account for the overlap correctly, the algebra gives a tighter bound.
Step-by-step:
1. Define variables. Let N = total number of interns. We know: Women (W) = 35, Economics majors (E) = 30, Women who are economics majors (W ∩ E) = 10% of N = 0.10N.
2. Apply the inclusion-exclusion principle. W ∪ E = W + E − (W ∩ E) = 35 + 30 − 0.10N = 65 − 0.10N.
3. The union can't exceed the total. Since W ∪ E ≤ N, we get: 65 − 0.10N ≤ N → 65 ≤ 1.10N → N ≥ 65/1.1 = 59.09... So N ≥ 60 (rounding up since N must be a whole number). Minimum N = 60.
4. Find the minimum fraction of men. Men = N − 35 (everyone who isn't a woman). The fraction of men = (N − 35)/N = 1 − 35/N. This fraction is minimized when N is smallest (since 35/N is largest when N is smallest). So plug in N = 60: Men fraction = (60 − 35)/60 = 25/60 = 5/12.
5. Verify: If N = 60, then W ∩ E = 6, W = 35, E = 30. W ∪ E = 35 + 30 − 6 = 59 ≤ 60. ✓ Also W ∩ E = 6 ≤ min(35, 30) = 30. ✓
Answer: B (5/12).
The takeaway: whenever a question asks for the "minimum fraction" of a group, set up your constraint inequality first, find the boundary value of N, and then compute the fraction at that boundary — don't try to guess from the answer choices.
Key concept: Overlapping Sets with a minimization condition.
Common trap: Students often jump to thinking the minimum number of interns is exactly 35 + 30 = 65. That's wrong — it ignores the overlap. Once you account for the overlap correctly, the algebra gives a tighter bound.
Step-by-step:
1. Define variables. Let N = total number of interns. We know: Women (W) = 35, Economics majors (E) = 30, Women who are economics majors (W ∩ E) = 10% of N = 0.10N.
2. Apply the inclusion-exclusion principle. W ∪ E = W + E − (W ∩ E) = 35 + 30 − 0.10N = 65 − 0.10N.
3. The union can't exceed the total. Since W ∪ E ≤ N, we get: 65 − 0.10N ≤ N → 65 ≤ 1.10N → N ≥ 65/1.1 = 59.09... So N ≥ 60 (rounding up since N must be a whole number). Minimum N = 60.
4. Find the minimum fraction of men. Men = N − 35 (everyone who isn't a woman). The fraction of men = (N − 35)/N = 1 − 35/N. This fraction is minimized when N is smallest (since 35/N is largest when N is smallest). So plug in N = 60: Men fraction = (60 − 35)/60 = 25/60 = 5/12.
5. Verify: If N = 60, then W ∩ E = 6, W = 35, E = 30. W ∪ E = 35 + 30 − 6 = 59 ≤ 60. ✓ Also W ∩ E = 6 ≤ min(35, 30) = 30. ✓
Answer: B (5/12).
The takeaway: whenever a question asks for the "minimum fraction" of a group, set up your constraint inequality first, find the boundary value of N, and then compute the fraction at that boundary — don't try to guess from the answer choices.
