Question Number 53324 by 0955083339 last updated on 20/Jan/19 | ||
$$\mathrm{If}\:\mathrm{4}{a}\:+\:\mathrm{5}{b}\:+\:\mathrm{9}{c}=\mathrm{36}\:\mathrm{and}\:\mathrm{7}{a}\:+\:\mathrm{9}{b}\:+\:\mathrm{17}{c}=\mathrm{66}, \\ $$$$\mathrm{then}\:{a}+{b}+{c}=\_\_\_\_\_. \\ $$ | ||
Answered by Kunal12588 last updated on 21/Jan/19 | ||
$${let}\:{a}+{b}+{c}={k} \\ $$$$\mathrm{4}{a}+\mathrm{4}{b}+\mathrm{4}{c}+{b}+\mathrm{5}{c}=\mathrm{36} \\ $$$$\Rightarrow\mathrm{4}{k}+\left({b}+\mathrm{5}{c}\right)=\mathrm{36}\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{7}{a}+\mathrm{7}{b}+\mathrm{7}{c}+\mathrm{2}{b}+\mathrm{10}{c}=\mathrm{66} \\ $$$$\Rightarrow\mathrm{7}{k}+\mathrm{2}\left({b}+\mathrm{5}{c}\right)=\mathrm{66}\:\:\:\:\:\:\:\left(\mathrm{2}\right)\:\:\: \\ $$$${let}\:{b}+\mathrm{5}{c}={h} \\ $$$$\mathrm{4}{k}+{h}=\mathrm{36}\:\:\:\:\left[{multiply}\:{by}\:\mathrm{2}\right] \\ $$$$\mathrm{8}{k}+\mathrm{2}{h}=\mathrm{72} \\ $$$$\mathrm{7}{k}+\mathrm{2}{h}=\mathrm{66} \\ $$$$\Rightarrow{k}={a}+{b}+{c}=\mathrm{72}−\mathrm{66}=\mathrm{6} \\ $$ | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19 | ||
$${excellent}... \\ $$ | ||