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Question-194888

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Question Number 194888 by Abdullahrussell last updated on 18/Jul/23
Answered by aleks041103 last updated on 18/Jul/23
The combined ages of Mary and Ann is 44 years. ( M(t_1 )+A(t_1 )=44 ) Mary is twice as old as Ann was ( M(t_1 )=2A(t_2 ) ) when Mary was half as old as Ann will be ( M(t_2 )=(1/2)A(t_3 ) ) when Ann is three times as old as Mary was ( A(t_3 )=3M(t_4 ) ) when Mary was three times as old as Ann. ( M(t_4 )=3A(t_4 ) ) Also we know that M−A=const. let A(t_4 )=x ⇒M(t_4 )=3x ⇒ M−A=2x ⇒A(t_3 )=3M(t_4 )=3.3x=9x ⇒M(t_2 )=(1/2)A(t_3 )=(9/2)x ⇒A(t_2 )=M(t_2 )−2x=(5/2)x ⇒M(t_1 )=2A(t_2 )=5x ⇒A(t_1 )=M(t_1 )−2x=5x−2x=3x ⇒M(t_1 )+A(t_1 )=5x+3x=8x=44 ⇒x=((11)/2) ⇒M(t_1 )=5x=((55)/2)=27.5 ⇒Mary is 27 and a half years old.
$${The}\:{combined}\:{ages}\:{of}\:{Mary}\:{and}\:{Ann}\:{is}\:\mathrm{44}\:{years}. \\ $$$$\left(\:{M}\left({t}_{\mathrm{1}} \right)+{A}\left({t}_{\mathrm{1}} \right)=\mathrm{44}\:\right) \\ $$$${Mary}\:{is}\:{twice}\:{as}\:{old}\:{as}\:{Ann}\:{was} \\ $$$$\left(\:{M}\left({t}_{\mathrm{1}} \right)=\mathrm{2}{A}\left({t}_{\mathrm{2}} \right)\:\right) \\ $$$${when}\:{Mary}\:{was}\:{half}\:{as}\:{old}\:{as}\:{Ann}\:{will}\:{be} \\ $$$$\left(\:{M}\left({t}_{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{2}}{A}\left({t}_{\mathrm{3}} \right)\:\right) \\ $$$${when}\:{Ann}\:{is}\:{three}\:{times}\:{as}\:{old}\:{as}\:{Mary}\:{was} \\ $$$$\left(\:{A}\left({t}_{\mathrm{3}} \right)=\mathrm{3}{M}\left({t}_{\mathrm{4}} \right)\:\right) \\ $$$${when}\:{Mary}\:{was}\:{three}\:{times}\:{as}\:{old}\:{as}\:{Ann}. \\ $$$$\left(\:{M}\left({t}_{\mathrm{4}} \right)=\mathrm{3}{A}\left({t}_{\mathrm{4}} \right)\:\right) \\ $$$$ \\ $$$${Also}\:{we}\:{know}\:{that}\:{M}−{A}={const}. \\ $$$${let}\:{A}\left({t}_{\mathrm{4}} \right)={x} \\ $$$$\Rightarrow{M}\left({t}_{\mathrm{4}} \right)=\mathrm{3}{x}\:\Rightarrow\:{M}−{A}=\mathrm{2}{x} \\ $$$$\Rightarrow{A}\left({t}_{\mathrm{3}} \right)=\mathrm{3}{M}\left({t}_{\mathrm{4}} \right)=\mathrm{3}.\mathrm{3}{x}=\mathrm{9}{x} \\ $$$$\Rightarrow{M}\left({t}_{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{2}}{A}\left({t}_{\mathrm{3}} \right)=\frac{\mathrm{9}}{\mathrm{2}}{x} \\ $$$$\Rightarrow{A}\left({t}_{\mathrm{2}} \right)={M}\left({t}_{\mathrm{2}} \right)−\mathrm{2}{x}=\frac{\mathrm{5}}{\mathrm{2}}{x} \\ $$$$\Rightarrow{M}\left({t}_{\mathrm{1}} \right)=\mathrm{2}{A}\left({t}_{\mathrm{2}} \right)=\mathrm{5}{x} \\ $$$$\Rightarrow{A}\left({t}_{\mathrm{1}} \right)={M}\left({t}_{\mathrm{1}} \right)−\mathrm{2}{x}=\mathrm{5}{x}−\mathrm{2}{x}=\mathrm{3}{x} \\ $$$$\Rightarrow{M}\left({t}_{\mathrm{1}} \right)+{A}\left({t}_{\mathrm{1}} \right)=\mathrm{5}{x}+\mathrm{3}{x}=\mathrm{8}{x}=\mathrm{44} \\ $$$$\Rightarrow{x}=\frac{\mathrm{11}}{\mathrm{2}} \\ $$$$\Rightarrow{M}\left({t}_{\mathrm{1}} \right)=\mathrm{5}{x}=\frac{\mathrm{55}}{\mathrm{2}}=\mathrm{27}.\mathrm{5} \\ $$$$\Rightarrow{Mary}\:{is}\:\mathrm{27}\:{and}\:{a}\:{half}\:{years}\:{old}. \\ $$

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