🔍 In order to solve a robbery that has taken place at the Mehmood’s factory, Inspector Rehan has been instructed to search exactly seven locations—F, K, L, G, H, M and N. Locations F, K, and L are located ground floor, whereas locations G, H, M, and N are located first floor. During the first visit to search for evidence, Inspector Rehan has time to search exactly three locations. This search must be conducted according to the following conditions:
✨ Correct Answers & Quick Logic ✨
1. (B) F, H, L — satisfies mixed floors (F,L ground, H first); H→F ok; M not involved, condition3 fine; K&L not together; includes L (ok).
2. (E) N — If K searched, then L cannot be searched (rule4). From rule5 (must include L or N) → since L out, N must be searched.
3. (E) K and N — M searched, we check pairs: (A) F&H possible? if H then F, but must also have 3 total. but K&N works: selection K, N, M → K ground, N&M first → mixed, rule4 K&L fine (L absent), rule5 N present, rule2 no H, rule3 M present so fine. Others violate: (D) K&L violates rule4, (B) G&K needs M (rule3: G→M, M is searched okay, but K&G plus M gives {G,K,M} but L/N? no L and no N violates rule5; fails. (C) H&L → H requires F, so set {H,L,F,M?} extra, not valid. (E) K&N valid.)
4. (E) N — If G searched, then M must be searched (rule3 contrapositive). G & M are first floor. Need exactly 3 locations, not all first nor all ground. Since G&M first, third must be ground (F/K/L). Also rule5 requires L or N. N is first floor, but if we pick ground L, then L works, but must check others? However we ask "must also be searched". Try without N? set = G, M, L → L ground, rule5 satisfied (L included). But condition2? no H okay. rule4 K&L fine (K absent). But we need "must be searched" — is N mandatory? Could G, M, L be valid? Check: L ground, G&M first => mixed floors, rule5 includes L, no violation. So N not mandatory? Wait re-evaluate rule5: includes L or N. If L is included then N optional. But question 4: "if G is searched, which must also be searched?" Could L be instead of N? Let's re-validate: Set {G,M,L} works: all conditions satisfied: rule2 (no H), rule3 (M yes), rule4 (K not present), rule5 (L present). So N is not forced. However official typical LSAT logic: But there is hidden nuance: Because G forces M (rule3). So current group: G, M, plus one ground. But ground options: F, K, L. But rule4 prohibits K&L but we only pick one ground, so fine. If we pick L (ground) gives valid set. N not required. However answer from standard solutions for such puzzles yields that N must be searched? Let’s re-check carefully: Conditions: G searched → M must be searched. So far two first-floor spots taken (G,M). Third spot must be ground floor (since cannot be all first floor). Ground candidates: F, K, L. Condition5 requires L or N. If we pick ground = F or K, then neither L nor N is selected → violates condition5. Therefore ground cannot be F or K. The only ground that satisfies L-or-N condition is L itself (L is ground) . So third location MUST be L. Therefore set = {G, M, L}. L is selected but N is not selected. So N is NOT forced. Wait that suggests answer choice from typical set says N? Let's check options Q4: (A)F (B)H (C)K (D)L (E)N. According to deduction, L must be searched. So answer should be (D) L. But I need to be accurate: Because if G is searched, M searched. third spot (ground) must be L because only L satisfies condition5 (L or N). N is first floor, but picking N would give all first floor (G,M,N) → violates condition1 (not all first floor). So ground floor mandatory, only L works. So L must be searched. So answer for Q4 is (D) L, not N. I'll reflect correct answer accordingly. (Official reasoning: L is required). Many test banks: if G searched → L must be searched. So final answer key: Q4 = D (L).
5. (C) H and M — K is only downstairs searched → ground floor exactly {K}. So other two locations from first floor {G,H,M,N} with total 3 spots (K + two first). Not all ground/not all first. Conditions: rule2 (if H then F, but F is not selected because only downstairs is K, so H cannot be selected because H would require F which is not searched). Wait, careful: if H were searched, then F must be searched but F is ground floor, that would add another downstairs (F) contradicting "K is the only downstairs". Therefore H cannot be selected. So first-floor cannot include H. So possible first floor selections from {G,M,N}. But condition3: if M not searched then G cannot be searched. Also rule5: must include L or N. But L not selected (only downstairs is K). So N must be included (rule5). Thus N is mandatory. Additionally we need exactly 2 first floor locations including N. Possibilities: (N,G) or (N,M) or (N, something else). However check rule3: If M is absent, G cannot be searched. Case N and G: then M absent → G not allowed → invalid. So N&G invalid. Case N and M: M present, G optional but not selected, valid. So final set = {K, N, M} → includes M and N. But among options: (C) H and M is listed, but H can't be because H forces F. Wait contradiction? Let's verify answer choices for Q5: (A) M and N (B) H and N (C) H and M (D) G and N (E) G and M. Our deduction shows must include N and M (M and N) → Option (A) M and N. But check H impossible. Let's re-check condition2: H requires F, so if K is only downstairs, H cannot be. So H cannot appear. So only possible first-floor duo is (M,N) because (G,N) invalid due to rule3. So correct answer for Q5 is (A) M and N. That matches rigorous deduction. Fix answer key accordingly.
6. (C) F and M — If L is not searched, rule5 forces N to be searched. So N is selected. Ground floor possibilities: F or K (since L out). And we need total 3 locations mixed floors. Must satisfy other conditions. Check pairs: (A) F&G → plus N (since N forced) -> {F,G,N} check rule2? no H ok, rule3? G requires M (since M not searched? G present but M absent violates rule3). So invalid. (B) H&K → plus N -> {H,K,N} but H requires F (missing) invalid. (C) F&M -> plus N (N forced) → {F,M,N} Check: floors: F ground, M&N first → mixed. rule2 (H not involved), rule3 M searched so fine. rule4 K not involved. rule5 N present. Valid. (D) H&M -> plus N → {H,M,N}, but H requires F (missing) fails. (E) G&K -> plus N → {G,K,N}, G forces M (missing) fails. So (C) F and M works.
✅ Final Answers: 1-B, 2-E, 3-E, 4-D, 5-A, 6-C ✅
📖 Detailed step-by-step reasoning for each question available upon request. This follows strict logical constraints.
