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Question Number 27568 by MASANJA last updated on 09/Jan/18

carol is 6 times as old as her nephew Hondo.Baby is 22 years yonger than her aunt carol.In four years carol′s age will be twice the sum of Hondo′s and Baby′s age.How old is each perso n now?

$${carol}\:{is}\:\mathrm{6}\:{times}\:{as}\:{old}\:{as}\:{her}\:{nephew} \\ $$$${Hondo}.{Baby}\:{is}\:\mathrm{22}\:{years}\:{yonger}\:{than} \\ $$$${her}\:{aunt}\:{carol}.{In}\:{four}\:{years}\:{carol}'{s} \\ $$$${age}\:{will}\:{be}\:{twice}\:{the}\:{sum}\:{of}\:{Hondo}'{s} \\ $$$${and}\:{Baby}'{s}\:{age}.{How}\:{old}\:{is}\:{each}\:{perso} \\ $$$${n}\:{now}? \\ $$

Answered by Rasheed.Sindhi last updated on 10/Jan/18

((,(Carol),(Hondo),( Baby)),(( Now),( 6x),(x (Say)),(6x−22)),(( +4years),(6x+4),(x+4), { ((6x−22+4)),((=6x−18)) :}) ) C^(+4) =2(H^(+4) +B^(+4) ) { ((C^(+4) :Carol′s age after 4 years.)),((H^(+4) :Hondo′s age after 4 years.)),((B^(+4) :Baby′s age after 4 years.)) :} 6x+4^(C^(+4) ) =2{(x+4^(H^(+4) ) )+(6x−18^(B^(+4) ) )} 6x+4=14x−28 14x−6x=4+28 x=4 (Hondo′s age now) 6x=24 (Carol′s age now) 6x−22=2 (Baby′s age now)

$$\begin{pmatrix}{}&{\mathrm{Carol}}&{\mathrm{Hondo}}&{\:\:\mathrm{Baby}}\\{\:\:\:\:\:\:\mathrm{Now}}&{\:\:\:\mathrm{6x}}&{\mathrm{x}\:\left(\mathrm{Say}\right)}&{\mathrm{6x}−\mathrm{22}}\\{\:+\mathrm{4years}}&{\mathrm{6x}+\mathrm{4}}&{\mathrm{x}+\mathrm{4}}&{\begin{cases}{\mathrm{6x}−\mathrm{22}+\mathrm{4}}\\{=\mathrm{6x}−\mathrm{18}}\end{cases}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{C}^{+\mathrm{4}} =\mathrm{2}\left(\mathrm{H}^{+\mathrm{4}} +\mathrm{B}^{+\mathrm{4}} \right) \\ $$$$\begin{cases}{\mathrm{C}^{+\mathrm{4}} :\mathrm{Carol}'\mathrm{s}\:\mathrm{age}\:\mathrm{after}\:\mathrm{4}\:\mathrm{years}.}\\{\mathrm{H}^{+\mathrm{4}} :\mathrm{Hondo}'\mathrm{s}\:\mathrm{age}\:\mathrm{after}\:\mathrm{4}\:\mathrm{years}.}\\{\mathrm{B}^{+\mathrm{4}} :\mathrm{Baby}'\mathrm{s}\:\mathrm{age}\:\mathrm{after}\:\mathrm{4}\:\mathrm{years}.}\end{cases} \\ $$$$\:\:\:\:\overset{\mathrm{C}^{+\mathrm{4}} } {\mathrm{6x}+\mathrm{4}}=\mathrm{2}\left\{\left(\overset{\mathrm{H}^{+\mathrm{4}} } {\mathrm{x}+\mathrm{4}}\right)+\left(\overset{\mathrm{B}^{+\mathrm{4}} } {\mathrm{6x}−\mathrm{18}}\right)\right\} \\ $$$$\:\:\:\:\:\mathrm{6x}+\mathrm{4}=\mathrm{14x}−\mathrm{28} \\ $$$$\:\:\:\:\:\:\mathrm{14x}−\mathrm{6x}=\mathrm{4}+\mathrm{28} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}=\mathrm{4}\:\left(\mathrm{Hondo}'\mathrm{s}\:\mathrm{age}\:\mathrm{now}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{6x}=\mathrm{24}\:\left(\mathrm{Carol}'\mathrm{s}\:\mathrm{age}\:\mathrm{now}\right) \\ $$$$\:\mathrm{6x}−\mathrm{22}=\mathrm{2}\:\:\:\left(\mathrm{Baby}'\mathrm{s}\:\mathrm{age}\:\mathrm{now}\right) \\ $$

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