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Question-182928

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Question Number 182928 by yaslm last updated on 16/Dec/22
Answered by manolex last updated on 16/Dec/22
A+B=1040−C ((A+B)/B)=((11)/6) ((B+C)/B)=(4/3) ((1040−C)/B)=((11)/6) B+C=((4B)/3) ((1040−(B−C))/B)=(5/6) 1040−((4B)/3)=((5B)/6) 1040=((13B)/6) B=480 A=((5B)/6)=400 C=(B/3)=160 C.S=(A,B,C)=(400,480,160)
$${A}+{B}=\mathrm{1040}−{C} \\ $$$$\frac{{A}+{B}}{{B}}=\frac{\mathrm{11}}{\mathrm{6}}\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{B}+{C}}{{B}}=\frac{\mathrm{4}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\frac{\mathrm{1040}−{C}}{{B}}=\frac{\mathrm{11}}{\mathrm{6}}\:\:\:\:\:\:\:\:\:\:{B}+{C}=\frac{\mathrm{4}{B}}{\mathrm{3}} \\ $$$$\frac{\mathrm{1040}−\left({B}−{C}\right)}{{B}}=\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\mathrm{1040}−\frac{\mathrm{4}{B}}{\mathrm{3}}=\frac{\mathrm{5}{B}}{\mathrm{6}} \\ $$$$\mathrm{1040}=\frac{\mathrm{13}{B}}{\mathrm{6}} \\ $$$${B}=\mathrm{480} \\ $$$${A}=\frac{\mathrm{5}{B}}{\mathrm{6}}=\mathrm{400} \\ $$$${C}=\frac{{B}}{\mathrm{3}}=\mathrm{160} \\ $$$${C}.{S}=\left({A},{B},{C}\right)=\left(\mathrm{400},\mathrm{480},\mathrm{160}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by HeferH last updated on 17/Dec/22
(A/B) = ((5k)/(6k)) ⇒ (C/B) = ((2k)/(6k)) 5k + 6k + 2k = 1040 k = 80 A = 400; B = 480; C = 160
$$\frac{{A}}{{B}}\:=\:\frac{\mathrm{5}{k}}{\mathrm{6}{k}}\:\Rightarrow\:\:\frac{{C}}{{B}}\:=\:\frac{\mathrm{2}{k}}{\mathrm{6}{k}} \\ $$$$\:\mathrm{5}{k}\:+\:\mathrm{6}{k}\:+\:\mathrm{2}{k}\:=\:\mathrm{1040} \\ $$$$\:{k}\:=\:\mathrm{80} \\ $$$$\:{A}\:=\:\mathrm{400};\:{B}\:=\:\mathrm{480};\:{C}\:=\:\mathrm{160} \\ $$
Answered by Rasheed.Sindhi last updated on 17/Dec/22
determinant ((A,B,C),(5,6, ),( ,3,1),(5,6,2)) A:B:C=5:6:2 A+B+C=5k+6k+2k=13k=1040 k=((1040)/(13))=80 A=5×80=400 B=6×80=480 C=2×80=160_(−) 1040
$$\begin{array}{|c|c|c|c|}{{A}}&\hline{{B}}&\hline{{C}}\\{\mathrm{5}}&\hline{\mathrm{6}}&\hline{\:}\\{\:}&\hline{\mathrm{3}}&\hline{\mathrm{1}}\\{\mathrm{5}}&\hline{\mathrm{6}}&\hline{\mathrm{2}}\\\hline\end{array} \\ $$$${A}:{B}:{C}=\mathrm{5}:\mathrm{6}:\mathrm{2} \\ $$$${A}+{B}+{C}=\mathrm{5}{k}+\mathrm{6}{k}+\mathrm{2}{k}=\mathrm{13}{k}=\mathrm{1040} \\ $$$${k}=\frac{\mathrm{1040}}{\mathrm{13}}=\mathrm{80} \\ $$$${A}=\mathrm{5}×\mathrm{80}=\mathrm{400} \\ $$$${B}=\mathrm{6}×\mathrm{80}=\mathrm{480} \\ $$$${C}=\mathrm{2}×\mathrm{80}=\underset{−} {\mathrm{160}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1040} \\ $$
Answered by Rasheed.Sindhi last updated on 17/Dec/22
(A/B)=(5/6)⇒(A/5)=(B/6) (C/B)=(1/3)⇒(C/1)=(B/3)⇒(C/2)=(B/6) (A/5)=(B/6)=(C/2)=((A+B+C)/(5+6+2))=((1040)/(13))=80 A=5×80=400 B=6×80=480 C=2×80=160_(−) 1040
$$\frac{{A}}{{B}}=\frac{\mathrm{5}}{\mathrm{6}}\Rightarrow\frac{{A}}{\mathrm{5}}=\frac{{B}}{\mathrm{6}} \\ $$$$\frac{{C}}{{B}}=\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow\frac{{C}}{\mathrm{1}}=\frac{{B}}{\mathrm{3}}\Rightarrow\frac{{C}}{\mathrm{2}}=\frac{{B}}{\mathrm{6}} \\ $$$$\frac{{A}}{\mathrm{5}}=\frac{{B}}{\mathrm{6}}=\frac{{C}}{\mathrm{2}}=\frac{{A}+{B}+{C}}{\mathrm{5}+\mathrm{6}+\mathrm{2}}=\frac{\mathrm{1040}}{\mathrm{13}}=\mathrm{80} \\ $$$${A}=\mathrm{5}×\mathrm{80}=\mathrm{400} \\ $$$${B}=\mathrm{6}×\mathrm{80}=\mathrm{480} \\ $$$${C}=\mathrm{2}×\mathrm{80}=\underset{−} {\mathrm{160}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1040} \\ $$
Answered by Rasheed.Sindhi last updated on 17/Dec/22
(A/B)=(5/6)_((i)) , (C/B)=(1/3)_((ii)) ,A+B+C=1040 (i)⇒((A+B)/B)=((5+6)/6) [Componendo ] ⇒((A+B)/B)=((11)/6).....(iii) (iii)+(ii): ((A+B+C)/B)=((11)/6)+(1/3) ((1040)/B)=((13)/6)⇒B=((1040×6)/(13))=480 (i)⇒(A/(480))=(5/6)⇒A=((5×480)/6)=400 (ii)⇒(C/(480))=(1/3)⇒C=((480)/3)=160
$$\underset{\left({i}\right)} {\underbrace{\frac{{A}}{{B}}=\frac{\mathrm{5}}{\mathrm{6}}}}\:,\:\underset{\left({ii}\right)} {\underbrace{\frac{{C}}{{B}}=\frac{\mathrm{1}}{\mathrm{3}}}}\:,{A}+{B}+{C}=\mathrm{1040}\: \\ $$$$\left({i}\right)\Rightarrow\frac{{A}+{B}}{{B}}=\frac{\mathrm{5}+\mathrm{6}}{\mathrm{6}}\:\:\left[{Componendo}\:\right] \\ $$$$\:\:\:\:\:\:\:\Rightarrow\frac{{A}+{B}}{{B}}=\frac{\mathrm{11}}{\mathrm{6}}…..\left({iii}\right) \\ $$$$\left({iii}\right)+\left({ii}\right):\:\frac{{A}+{B}+{C}}{{B}}=\frac{\mathrm{11}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1040}}{{B}}=\frac{\mathrm{13}}{\mathrm{6}}\Rightarrow{B}=\frac{\mathrm{1040}×\mathrm{6}}{\mathrm{13}}=\mathrm{480} \\ $$$$\left({i}\right)\Rightarrow\frac{{A}}{\mathrm{480}}=\frac{\mathrm{5}}{\mathrm{6}}\Rightarrow{A}=\frac{\mathrm{5}×\mathrm{480}}{\mathrm{6}}=\mathrm{400} \\ $$$$\left({ii}\right)\Rightarrow\frac{{C}}{\mathrm{480}}=\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow{C}=\frac{\mathrm{480}}{\mathrm{3}}=\mathrm{160} \\ $$
Answered by Rasheed.Sindhi last updated on 17/Dec/22
(A/B)=(5/6)_((i)) , (C/B)=(1/3)_((ii)) ,A+B+C=1040 (i)+(ii):((A+C)/B)=(7/6) ((A+C+B)/B)=((7+6)/6) [Componendo] ((1040)/B)=((13)/6)⇒B=((1040×6)/(13))=480 (i)⇒(A/(480))=(5/6)⇒A=((480×5)/6)=400 (ii)⇒(C/(480))=(1/3)⇒C=((480×1)/3)=160
$$ \\ $$$$\underset{\left({i}\right)} {\underbrace{\frac{{A}}{{B}}=\frac{\mathrm{5}}{\mathrm{6}}}}\:,\:\underset{\left({ii}\right)} {\underbrace{\frac{{C}}{{B}}=\frac{\mathrm{1}}{\mathrm{3}}}}\:,{A}+{B}+{C}=\mathrm{1040}\: \\ $$$$\left({i}\right)+\left({ii}\right):\frac{{A}+{C}}{{B}}=\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$\:\:\:\:\:\frac{{A}+{C}+{B}}{{B}}=\frac{\mathrm{7}+\mathrm{6}}{\mathrm{6}}\:\:\left[{Componendo}\right] \\ $$$$\:\:\:\frac{\mathrm{1040}}{{B}}=\frac{\mathrm{13}}{\mathrm{6}}\Rightarrow{B}=\frac{\mathrm{1040}×\mathrm{6}}{\mathrm{13}}=\mathrm{480} \\ $$$$\left({i}\right)\Rightarrow\frac{{A}}{\mathrm{480}}=\frac{\mathrm{5}}{\mathrm{6}}\Rightarrow{A}=\frac{\mathrm{480}×\mathrm{5}}{\mathrm{6}}=\mathrm{400} \\ $$$$\left({ii}\right)\Rightarrow\frac{{C}}{\mathrm{480}}=\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow{C}=\frac{\mathrm{480}×\mathrm{1}}{\mathrm{3}}=\mathrm{160} \\ $$

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