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Question Number 78306 by mathocean1 last updated on 15/Jan/20

The sum of age of Hamadou     his wife and theirs son is 100.  n years ago the wife had the   quadruple of his son′s age and   Hamadou was 6 time older than  his son.  Determine theirs ages.    i want that you help me to found  equations.  i found the first : x+y+z=100  please help me for the rest.

$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{age}\:\mathrm{of}\:\mathrm{Hamadou}\:\:\: \\ $$$$\mathrm{his}\:\mathrm{wife}\:\mathrm{and}\:\mathrm{theirs}\:\mathrm{son}\:\mathrm{is}\:\mathrm{100}. \\ $$$$\mathrm{n}\:\mathrm{years}\:\mathrm{ago}\:\mathrm{the}\:\mathrm{wife}\:\mathrm{had}\:\mathrm{the}\: \\ $$$$\mathrm{quadruple}\:\mathrm{of}\:\mathrm{his}\:\mathrm{son}'\mathrm{s}\:\mathrm{age}\:\mathrm{and}\: \\ $$$$\mathrm{Hamadou}\:\mathrm{was}\:\mathrm{6}\:\mathrm{time}\:\mathrm{older}\:\mathrm{than} \\ $$$$\mathrm{his}\:\mathrm{son}. \\ $$$$\mathrm{Determine}\:\mathrm{theirs}\:\mathrm{ages}. \\ $$$$ \\ $$$$\mathrm{i}\:\mathrm{want}\:\mathrm{that}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{found} \\ $$$$\mathrm{equations}. \\ $$$$\mathrm{i}\:\mathrm{found}\:\mathrm{the}\:\mathrm{first}\::\:{x}+{y}+{z}=\mathrm{100} \\ $$$${please}\:{help}\:{me}\:{for}\:{the}\:{rest}. \\ $$

Answered by mfwajoel1 last updated on 15/Jan/20

x age of son;4x age of hamadou wife and 6x age of hamadou  x+4x+6x=100⇔11x=100⇔x=((100)/(11))≈9years  1month  then his mother has 4x=36years=1month  his father has 6x=54years=1month

$${x}\:{age}\:{of}\:{son};\mathrm{4}{x}\:{age}\:{of}\:{hamadou}\:{wife}\:{and}\:\mathrm{6}{x}\:{age}\:{of}\:{hamadou} \\ $$$${x}+\mathrm{4}{x}+\mathrm{6}{x}=\mathrm{100}\Leftrightarrow\mathrm{11}{x}=\mathrm{100}\Leftrightarrow{x}=\frac{\mathrm{100}}{\mathrm{11}}\approx\mathrm{9}{years}\:\:\mathrm{1}{month} \\ $$$${then}\:{his}\:{mother}\:{has}\:\mathrm{4}{x}=\mathrm{36}{years}=\mathrm{1}{month} \\ $$$${his}\:{father}\:{has}\:\mathrm{6}{x}=\mathrm{54}{years}=\mathrm{1}{month} \\ $$$$ \\ $$$$ \\ $$

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