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Question-136505




Question Number 136505 by rexford last updated on 22/Mar/21
Answered by Dwaipayan Shikari last updated on 22/Mar/21
Assuming soda was x litre in quantity  Your consumption=(x/2)+(x/2^3 )+(x/2^5 )+...=x((1/2)+(1/8)+(1/(32))+...)  =((x/2)/(1−(1/4)))=((2x)/3)  Percentage of Consumption=((2x)/(3x))×100=66.666666666..%
$${Assuming}\:{soda}\:{was}\:{x}\:{litre}\:{in}\:{quantity} \\ $$$${Your}\:{consumption}=\frac{{x}}{\mathrm{2}}+\frac{{x}}{\mathrm{2}^{\mathrm{3}} }+\frac{{x}}{\mathrm{2}^{\mathrm{5}} }+…={x}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{32}}+…\right) \\ $$$$=\frac{\frac{{x}}{\mathrm{2}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\mathrm{2}{x}}{\mathrm{3}} \\ $$$${Percentage}\:{of}\:{Consumption}=\frac{\mathrm{2}{x}}{\mathrm{3}{x}}×\mathrm{100}=\mathrm{66}.\mathrm{666666666}..\% \\ $$
Commented by rexford last updated on 22/Mar/21
thanks very muchfor your..i do understand it now
$${thanks}\:{very}\:{muchfor}\:{your}..{i}\:{do}\:{understand}\:{it}\:{now} \\ $$

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