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23 Feb 2026, 16:10
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A
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Difficulty:
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Question Stats:
43% (03:03) correct
57%
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Of the students in a certain anatomy class, 55% of the female and 35% of the male passed an exam. Did more than half of the students in the class pass the exam?
(1) More than 85% of the students who passed the exam were female.
(2) There were 60 more female students than male students.
(1) More than 85% of the students who passed the exam were female.
(2) There were 60 more female students than male students.
13 Mar 2026, 06:04
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Let F be the number of female students and M the number of males.
55% of F and 35% of M passed: total pass rate = (0.55F + 0.35M)/(F + M).
Statement (1) Analysis
More than 85% of passers are female: 0.55F > 0.85(0.55F + 0.35M).
Simplifies to 0.55F > 0.4675F + 0.2975M, or 0.0825F > 0.2975M, so F/M > 3.606.
Pass rate = [0.55F + 0.35M]/(F + M) > 0.55 - 0.2(F/M)/(1 + F/M); since F/M > 3.606, rate > 51.3% (more than half).
(1) alone is sufficient.
Statement (2) Analysis
F = M + 60.
Pass rate = (0.55(M + 60) + 0.35M)/(2M + 60) = (0.9M + 33)/(2M + 60), which equals 50% when M=66 but varies (e.g., 53.8% at M=20, 47.6% at M=200).
(2) alone is insufficient.
Ans =A
55% of F and 35% of M passed: total pass rate = (0.55F + 0.35M)/(F + M).
Statement (1) Analysis
More than 85% of passers are female: 0.55F > 0.85(0.55F + 0.35M).
Simplifies to 0.55F > 0.4675F + 0.2975M, or 0.0825F > 0.2975M, so F/M > 3.606.
Pass rate = [0.55F + 0.35M]/(F + M) > 0.55 - 0.2(F/M)/(1 + F/M); since F/M > 3.606, rate > 51.3% (more than half).
(1) alone is sufficient.
Statement (2) Analysis
F = M + 60.
Pass rate = (0.55(M + 60) + 0.35M)/(2M + 60) = (0.9M + 33)/(2M + 60), which equals 50% when M=66 but varies (e.g., 53.8% at M=20, 47.6% at M=200).
(2) alone is insufficient.
Ans =A
13 Mar 2026, 12:25
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Great question — this is a classic Data Sufficiency problem testing Weighted Average reasoning, and it trips up a lot of people who jump straight into algebra without reframing the question first.
Key concept: When two groups have fixed pass rates (55% female, 35% male), whether the combined rate exceeds 50% depends entirely on the ratio of females to males. The common trap here is testing numbers for Statement (2) without realizing the "60 more" constraint doesn't fix that ratio.
Let F = number of female students, M = number of male students.
Combined pass rate = (0.55F + 0.35M) / (F + M)
First, simplify the question: Is (0.55F + 0.35M)/(F + M) > 0.5?
Step 1: Cross-multiply and rearrange.
0.55F + 0.35M > 0.5F + 0.5M
→ 0.05F > 0.15M
→ F > 3M
So the DS question reduces to: Is F > 3M?
Step 2: Test Statement (1) — "More than 85% of students who passed were female."
Number who passed = 0.55F + 0.35M
Condition: 0.55F > 0.85 × (0.55F + 0.35M)
→ 0.55F > 0.4675F + 0.2975M
→ 0.0825F > 0.2975M
→ F/M > 0.2975/0.0825 ≈ 3.606
Since F/M > 3.606 > 3, we know F > 3M is always true.
Statement (1) alone is SUFFICIENT. ✓
Step 3: Test Statement (2) — "There were 60 more female students than male students."
F = M + 60. We need F > 3M, i.e., M + 60 > 3M → M < 30.
But M is not fixed:
• If M = 20: F = 80, F/M = 4 > 3 → pass rate > 50% → YES
• If M = 200: F = 260, F/M = 1.3 < 3 → pass rate < 50% → NO
Statement (2) alone is NOT SUFFICIENT. ✗
Answer: A
Takeaway: Whenever you see two-group percentage problems in Data Sufficiency, immediately reframe the question as "what does the ratio of the groups need to be?" — then test each statement against that ratio condition directly.
Key concept: When two groups have fixed pass rates (55% female, 35% male), whether the combined rate exceeds 50% depends entirely on the ratio of females to males. The common trap here is testing numbers for Statement (2) without realizing the "60 more" constraint doesn't fix that ratio.
Let F = number of female students, M = number of male students.
Combined pass rate = (0.55F + 0.35M) / (F + M)
First, simplify the question: Is (0.55F + 0.35M)/(F + M) > 0.5?
Step 1: Cross-multiply and rearrange.
0.55F + 0.35M > 0.5F + 0.5M
→ 0.05F > 0.15M
→ F > 3M
So the DS question reduces to: Is F > 3M?
Step 2: Test Statement (1) — "More than 85% of students who passed were female."
Number who passed = 0.55F + 0.35M
Condition: 0.55F > 0.85 × (0.55F + 0.35M)
→ 0.55F > 0.4675F + 0.2975M
→ 0.0825F > 0.2975M
→ F/M > 0.2975/0.0825 ≈ 3.606
Since F/M > 3.606 > 3, we know F > 3M is always true.
Statement (1) alone is SUFFICIENT. ✓
Step 3: Test Statement (2) — "There were 60 more female students than male students."
F = M + 60. We need F > 3M, i.e., M + 60 > 3M → M < 30.
But M is not fixed:
• If M = 20: F = 80, F/M = 4 > 3 → pass rate > 50% → YES
• If M = 200: F = 260, F/M = 1.3 < 3 → pass rate < 50% → NO
Statement (2) alone is NOT SUFFICIENT. ✗
Answer: A
Takeaway: Whenever you see two-group percentage problems in Data Sufficiency, immediately reframe the question as "what does the ratio of the groups need to be?" — then test each statement against that ratio condition directly.
