申論題

120kg 113kg 122kg

(A+B)-(B+C)+(C+D) = 124 - 128 + 124 = 120

(A+D) = 120

(B+C)- (C+D)+(D+E) = 128 - 124 + 119 = 123

(B+E) = 123

(C+D)-(D+E)+(E+F) = 124 - 119 +117 = 122

(C+F) = 122

[(A+B)+(C+D)]-(B+C)=(A+D)

=124+124-128

=120

[(B+C)+(D+E)]-(C+D)=(B+E)

=128+119-124

=123

[(C+D)+(E+F)]-(D+E)=(C+F)

=123+117-119

=122

答：(A+D)=120、(B+E)=123、(C+F)=122

A+D=120 B+E=124 C+F=121

(A+B)+(B+C)+(D+E)+(E+F)=A+2B+C+D+2E+F=124+128+119+117=488; B+E=488-365=123; (F+A)+(A+B)+(C+D)+(D+E)=F+2A+B+C+2D+E=118+124+124+119=485; A+D=485-365=120; (B+C)+(C+D)+(E+F)+(F+A)=128+124+117+118=487; C+F=487-365=122;

(A+B)+(B+C)+(D+E)+(E+F)=A+2B+C+D+2E+F=124+128+119+117=488; B+E=488-365=123kg

將6個男孩體重總和(A+B+C+D+E+F=365) 減去 (B+C) 和 (E+F) 得到 (A+D) 故 (A+D)為365-128-117=120 將6個男孩體重總和(A+B+C+D+E+F=365) 減去 (C+D) 和 (F+A) 得到 (B+E) 故 (B+E) 為365-124-118=123 將6個男孩體重總和(A+B+C+D+E+F=365) 減去 (A+B) 和 (D+E) 得到 (C+F) 故 (C+F) 為365-124-119=122