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Question Number 170064 by mathlove last updated on 15/May/22
(((a−2))/2)=(((b−3))/3)=(((c−4))/4) and a∙b∙c=192 faind a+b+c=?
$$\frac{\left({a}−\mathrm{2}\right)}{\mathrm{2}}=\frac{\left({b}−\mathrm{3}\right)}{\mathrm{3}}=\frac{\left({c}−\mathrm{4}\right)}{\mathrm{4}} \\ $$$${and}\:\:\:\:{a}\centerdot{b}\centerdot{c}=\mathrm{192} \\ $$$${faind}\:\:\:{a}+{b}+{c}=? \\ $$
Answered by Rasheed.Sindhi last updated on 15/May/22
(((a−2))/2)=(((b−3))/3)=(((c−4))/4) and a∙b∙c=192 find a+b+c=? ((a−2)/2)+1=((b−3)/3)+1=((c−4)/4)+1 (a/2)=(b/3)=(c/4)=((a+b+c)/(2+3+4))=k(say) (a/2)=(b/3)=(c/4)=((a+b+c)/9)=k(say) a=2k, b=3k, c=4k, a+b+c=9k a∙b∙c=2k.3k.4k=192 24k^3 =192 k^3 =8⇒k=2 a+b+c=9k=9(2)=18
$$\frac{\left({a}−\mathrm{2}\right)}{\mathrm{2}}=\frac{\left({b}−\mathrm{3}\right)}{\mathrm{3}}=\frac{\left({c}−\mathrm{4}\right)}{\mathrm{4}} \\ $$$${and}\:\:\:\:{a}\centerdot{b}\centerdot{c}=\mathrm{192} \\ $$$${find}\:\:\:{a}+{b}+{c}=? \\ $$$$\frac{{a}−\mathrm{2}}{\mathrm{2}}+\mathrm{1}=\frac{{b}−\mathrm{3}}{\mathrm{3}}+\mathrm{1}=\frac{{c}−\mathrm{4}}{\mathrm{4}}+\mathrm{1} \\ $$$$\frac{{a}}{\mathrm{2}}=\frac{{b}}{\mathrm{3}}=\frac{{c}}{\mathrm{4}}=\frac{{a}+{b}+{c}}{\mathrm{2}+\mathrm{3}+\mathrm{4}}={k}\left({say}\right) \\ $$$$\frac{{a}}{\mathrm{2}}=\frac{{b}}{\mathrm{3}}=\frac{{c}}{\mathrm{4}}=\frac{{a}+{b}+{c}}{\mathrm{9}}={k}\left({say}\right) \\ $$$${a}=\mathrm{2}{k},\:{b}=\mathrm{3}{k},\:{c}=\mathrm{4}{k},\:{a}+{b}+{c}=\mathrm{9}{k} \\ $$$${a}\centerdot{b}\centerdot{c}=\mathrm{2}{k}.\mathrm{3}{k}.\mathrm{4}{k}=\mathrm{192} \\ $$$$\mathrm{24}{k}^{\mathrm{3}} =\mathrm{192} \\ $$$${k}^{\mathrm{3}} =\mathrm{8}\Rightarrow{k}=\mathrm{2} \\ $$$${a}+{b}+{c}=\mathrm{9}{k}=\mathrm{9}\left(\mathrm{2}\right)=\mathrm{18} \\ $$$$ \\ $$
Answered by som(math1967) last updated on 15/May/22
let ((a−2)/2)=((b−3)/3)=((c−4)/4)=k ⇒a=2(k+1) ,b=3(k+1),c=4(k+1) 2×3×4×(k+1)^3 =abc ⇒(k+1)^3 =((192)/(24))=8 ⇒k+1=2⇒k=1 a+b+c=2(k+1)+3(k+1)+4(k+1) =4+6+8=18 ans
$${let}\:\frac{{a}−\mathrm{2}}{\mathrm{2}}=\frac{{b}−\mathrm{3}}{\mathrm{3}}=\frac{{c}−\mathrm{4}}{\mathrm{4}}=\boldsymbol{{k}} \\ $$$$\Rightarrow{a}=\mathrm{2}\left({k}+\mathrm{1}\right)\:,{b}=\mathrm{3}\left({k}+\mathrm{1}\right),{c}=\mathrm{4}\left({k}+\mathrm{1}\right) \\ $$$$\mathrm{2}×\mathrm{3}×\mathrm{4}×\left({k}+\mathrm{1}\right)^{\mathrm{3}} ={abc} \\ $$$$\Rightarrow\left({k}+\mathrm{1}\right)^{\mathrm{3}} =\frac{\mathrm{192}}{\mathrm{24}}=\mathrm{8} \\ $$$$\Rightarrow{k}+\mathrm{1}=\mathrm{2}\Rightarrow{k}=\mathrm{1} \\ $$$$\:{a}+{b}+{c}=\mathrm{2}\left({k}+\mathrm{1}\right)+\mathrm{3}\left({k}+\mathrm{1}\right)+\mathrm{4}\left({k}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}+\mathrm{6}+\mathrm{8}=\mathrm{18}\:{ans} \\ $$

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