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Difficulty:
95%
(hard)
Question Stats:
19% (02:31) correct
81%
(02:19)
wrong
based on 47
sessions
History
Date
Time
Result
Not Attempted Yet
Last year, among the 200 companies in Growthia,
• 80 companies hired new workers,
• 60 companies laid off workers, and
• 90 companies rehired former workers.
Were there at least 50 companies that engaged in none of these actions?
(1) Exactly 90 companies engaged in at least two of these actions.
(2) Exactly 50 companies engaged in exactly one of these actions.
• 80 companies hired new workers,
• 60 companies laid off workers, and
• 90 companies rehired former workers.
Were there at least 50 companies that engaged in none of these actions?
(1) Exactly 90 companies engaged in at least two of these actions.
(2) Exactly 50 companies engaged in exactly one of these actions.
26 Feb 2026, 13:39
Kudos
Bookmarks
This is a classic three-set Overlapping Sets DS problem, and it's exactly the type where the inclusion-exclusion formula is your best friend. Only 13% of people got this right early on — the trap is trying to work with Venn diagrams intuitively rather than setting up the algebra.
Setting up the framework:
Let A = hired (80), B = laid off (60), C = rehired (90). Total companies = 200.
Inclusion-exclusion formula:
|A∪B∪C| = |A|+|B|+|C| − (pairwise overlaps) + (triple overlap)
= 230 − P + T, where P = |A∩B|+|A∩C|+|B∩C| and T = |A∩B∩C|.
Companies that did none = 200 − |A∪B∪C| = 200 − (230 − P + T) = P − T − 30.
The question asks: Is (P − T − 30) ≥ 50, i.e., is P − T ≥ 80?
Statement 1 — Sufficient
"Exactly 90 engaged in at least two actions." Companies in at least two sets = P − 2T = 90.
So P = 90 + 2T.
Substitute: None = (90 + 2T) − T − 30 = 60 + T.
Since T ≥ 0, None ≥ 60 > 50. The answer to "at least 50?" is definitively YES. Sufficient.
Statement 2 — Sufficient
"Exactly 50 engaged in exactly one action." Companies in exactly one set = |A|+|B|+|C| − 2P + 3T = 50.
→ 230 − 2P + 3T = 50 → 2P − 3T = 180 → P = 90 + 1.5T.
None = P − T − 30 = (90 + 1.5T) − T − 30 = 60 + 0.5T.
Again, T ≥ 0, so None ≥ 60 > 50. Definitively YES. Sufficient.
Common trap: Students often set up the wrong formula for "exactly one" vs "at least two" — these are different expressions and mixing them up kills your work. Memorize both:
- At least two: P − 2T
- Exactly one: (A+B+C) − 2P + 3T
Answer: D
Takeaway: In three-set DS problems, don't draw Venn diagrams hoping to eyeball it — write the inclusion-exclusion formula first, identify what expression equals "none," then check what each statement pins down.
(Kavya | 725 on GMAT Focus Edition)
Setting up the framework:
Let A = hired (80), B = laid off (60), C = rehired (90). Total companies = 200.
Inclusion-exclusion formula:
|A∪B∪C| = |A|+|B|+|C| − (pairwise overlaps) + (triple overlap)
= 230 − P + T, where P = |A∩B|+|A∩C|+|B∩C| and T = |A∩B∩C|.
Companies that did none = 200 − |A∪B∪C| = 200 − (230 − P + T) = P − T − 30.
The question asks: Is (P − T − 30) ≥ 50, i.e., is P − T ≥ 80?
Statement 1 — Sufficient
"Exactly 90 engaged in at least two actions." Companies in at least two sets = P − 2T = 90.
So P = 90 + 2T.
Substitute: None = (90 + 2T) − T − 30 = 60 + T.
Since T ≥ 0, None ≥ 60 > 50. The answer to "at least 50?" is definitively YES. Sufficient.
Statement 2 — Sufficient
"Exactly 50 engaged in exactly one action." Companies in exactly one set = |A|+|B|+|C| − 2P + 3T = 50.
→ 230 − 2P + 3T = 50 → 2P − 3T = 180 → P = 90 + 1.5T.
None = P − T − 30 = (90 + 1.5T) − T − 30 = 60 + 0.5T.
Again, T ≥ 0, so None ≥ 60 > 50. Definitively YES. Sufficient.
Common trap: Students often set up the wrong formula for "exactly one" vs "at least two" — these are different expressions and mixing them up kills your work. Memorize both:
- At least two: P − 2T
- Exactly one: (A+B+C) − 2P + 3T
Answer: D
Takeaway: In three-set DS problems, don't draw Venn diagrams hoping to eyeball it — write the inclusion-exclusion formula first, identify what expression equals "none," then check what each statement pins down.
(Kavya | 725 on GMAT Focus Edition)
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