1. **State the problem:** We have five essences F, G, H, J, K with rules: - If F, then H with $H = 2F$. - If G, then J with $J = G$. - H and J cannot be together. - J and K cannot be together. - If K, then $K > $ sum of all other essences. Check formulas given these rules for questions 14-18. 14. Evaluate acceptability: (A) F=1, K=1: F requires H (absent), reject. (B) G=2, F=2: G requires J (absent), reject. (C) H=3, F=3: F requires H with $H=2F=6$, but here $H=3 eq 6$, reject. (D) J=4, G=4: G requires J equal G, true; J and K not both present; H absent; acceptable. (E) K=5, G=5: G requires J (absent), reject. **Answer: (D)** 15. Adding more H to formulas: (A) F=1, H=1, K=5: F requires $H=2F=2$, currently $H=1$, add $H$ to 2 corrects H; check K rule: $K=5 > 1+2=3$, OK. (B) F=2, H=2, K=2: F requires $H=4$ but have 2; adding more H possible. Add H to 4. K=2 not > others(2+4=6), violates K rule. No fix. (C) G=1, H=1, K=1: G requires J absent, invalid. Adding H won't fix. Reject. (D) G=2, H=1, K=4: G requires J present (absent), no. (E) H=2, J=1, K=3: H and J cannot coexist, reject. **Answer: (A)** 16. For perfume 2H and 1K (unacceptable due to K not > H (1<2)): Try adding: (A) F=1: Then need H=2F=2, already have H=2, no problem. But K=1 not > others (H=2, F=1 total 3), fails. (B) G=1: G requires J=1, J not present, fail. (C) 2H more: H can't combine with J which absent; K=1 not > others (now H=4 + 1K=1) fails. (D) J=1: J with K forbidden. (E) 2K more: K=3, now total others H=2; K=3 > 2 holds. **Answer: (E)** 17. Pairs acceptable except which? (A) F and G: Possible if rules met. (B) F and H: Needed together by rule. (C) F and K: Possible if K rule is met. (D) G and J: Must be together by rule. (E) K and H: H and J can't coexist, no direct rule against K and H. But if K and H coexist, K must be greater than sum others including H. All pairs possible except for H and J; none of the pairs is H and J, so check pairs carefully. Check G and J: always together okay. Check (A) F and G: F requires H, G requires J; no H or J here, so can't be acceptable. Hence (A) is not an acceptable pair alone. **Answer: (A)** 18. Make acceptable by removing essence: (A) F=1, G=1, H=1, K=4: F->H needed $H=2F=2$, have 1, remove H or add H. Removing H: no H, violates F rule. Removing F: then okay? Removing G: G requires J missing. Removing K: no. Removing G or F can help. At least removing one essence can fix. (B) F=1, H=2, J=1, K=4: H and J cannot coexist; removing J or H fixes. (C) F=1, G=1, J=1, K=1: J and K can't coexist; remove one fixes. (D) F=2, H=2, J=1, K=2: H and J can't coexist; remove J or H fixes. (E) G=2, H=1, J=2, K=3: J and K can't coexist; remove one fixes; note G and J together needed. **Answer: (All except possibly (A) because F requires H 2x F, and given H=1 less than 2, removing any can't fix.)** Final: (A) cannot be fixed. **Answer: (A)**