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Question Number 517 by Yugi last updated on 25/Jan/15
A person is said to be n years old ( where n is a non−negative integer) if   the person has lived at least n years and has not lived n+1 years. At some point  Tom is 4 years old and John is three times as old as Mary. At another time,  Mary is twice as old as Tom and John is five times as old as Tom. At a third   time, John is twice as old as Mary and Tom is t years old. What is the largest  possible value of t?
$${A}\:{person}\:{is}\:{said}\:{to}\:{be}\:{n}\:{years}\:{old}\:\left(\:{where}\:{n}\:{is}\:{a}\:{non}−{negative}\:{integer}\right)\:{if}\: \\ $$$${the}\:{person}\:{has}\:{lived}\:{at}\:{least}\:{n}\:{years}\:{and}\:{has}\:{not}\:{lived}\:{n}+\mathrm{1}\:{years}.\:{At}\:{some}\:{point} \\ $$$${Tom}\:{is}\:\mathrm{4}\:{years}\:{old}\:{and}\:{John}\:{is}\:{three}\:{times}\:{as}\:{old}\:{as}\:{Mary}.\:{At}\:{another}\:{time}, \\ $$$${Mary}\:{is}\:{twice}\:{as}\:{old}\:{as}\:{Tom}\:{and}\:{John}\:{is}\:{five}\:{times}\:{as}\:{old}\:{as}\:{Tom}.\:{At}\:{a}\:{third}\: \\ $$$${time},\:{John}\:{is}\:{twice}\:{as}\:{old}\:{as}\:{Mary}\:{and}\:{Tom}\:{is}\:{t}\:{years}\:{old}.\:{What}\:{is}\:{the}\:{largest} \\ $$$${possible}\:{value}\:{of}\:{t}? \\ $$
Commented by prakash jain last updated on 23/Jan/15
T, J, M are ages of Tom, John and Mary  p_1   T_1 =4  J_1 =3M_1   p_2  Let us say k years from p_1   T_2 =4+k  M_2 =2T_2 =8+2k  J_2 =5T_2 =20+5k  M_1 =8+2k−k=8+k  J_1 =20+5k−k=20+4k  M_1 =8+2k−k=8+k  J_1 =3M_1 ⇒20+4k=24+3k⇒k=4  M_1 =12, J_1 =36, M_2 =16, J_2 =40, T_2 =8  p_3  Let us say l years from p_1   T_3 =t=4+l  J_3 =2M_3   J_3 =J_1 +l=36+l  M_3 =M_1 +l=12+l  36+l=2(12+l)  36+l=24+2l⇒l=12  So t=4+12=16
$${T},\:{J},\:{M}\:{are}\:{ages}\:{of}\:{Tom},\:{John}\:{and}\:{Mary} \\ $$$${p}_{\mathrm{1}} \\ $$$${T}_{\mathrm{1}} =\mathrm{4} \\ $$$${J}_{\mathrm{1}} =\mathrm{3}{M}_{\mathrm{1}} \\ $$$${p}_{\mathrm{2}} \:{Let}\:{us}\:{say}\:{k}\:{years}\:{from}\:{p}_{\mathrm{1}} \\ $$$${T}_{\mathrm{2}} =\mathrm{4}+{k} \\ $$$${M}_{\mathrm{2}} =\mathrm{2}{T}_{\mathrm{2}} =\mathrm{8}+\mathrm{2}{k} \\ $$$${J}_{\mathrm{2}} =\mathrm{5}{T}_{\mathrm{2}} =\mathrm{20}+\mathrm{5}{k} \\ $$$${M}_{\mathrm{1}} =\mathrm{8}+\mathrm{2}{k}−{k}=\mathrm{8}+{k} \\ $$$${J}_{\mathrm{1}} =\mathrm{20}+\mathrm{5}{k}−{k}=\mathrm{20}+\mathrm{4}{k} \\ $$$${M}_{\mathrm{1}} =\mathrm{8}+\mathrm{2}{k}−{k}=\mathrm{8}+{k} \\ $$$${J}_{\mathrm{1}} =\mathrm{3}{M}_{\mathrm{1}} \Rightarrow\mathrm{20}+\mathrm{4}{k}=\mathrm{24}+\mathrm{3}{k}\Rightarrow{k}=\mathrm{4} \\ $$$${M}_{\mathrm{1}} =\mathrm{12},\:{J}_{\mathrm{1}} =\mathrm{36},\:{M}_{\mathrm{2}} =\mathrm{16},\:{J}_{\mathrm{2}} =\mathrm{40},\:{T}_{\mathrm{2}} =\mathrm{8} \\ $$$${p}_{\mathrm{3}} \:{Let}\:{us}\:{say}\:{l}\:{years}\:{from}\:{p}_{\mathrm{1}} \\ $$$${T}_{\mathrm{3}} ={t}=\mathrm{4}+{l} \\ $$$${J}_{\mathrm{3}} =\mathrm{2}{M}_{\mathrm{3}} \\ $$$${J}_{\mathrm{3}} ={J}_{\mathrm{1}} +{l}=\mathrm{36}+{l} \\ $$$${M}_{\mathrm{3}} ={M}_{\mathrm{1}} +{l}=\mathrm{12}+{l} \\ $$$$\mathrm{36}+{l}=\mathrm{2}\left(\mathrm{12}+{l}\right) \\ $$$$\mathrm{36}+{l}=\mathrm{24}+\mathrm{2}{l}\Rightarrow{l}=\mathrm{12} \\ $$$${So}\:{t}=\mathrm{4}+\mathrm{12}=\mathrm{16} \\ $$
Answered by prakash jain last updated on 23/Jan/15
See comment only one solution t=16
$${See}\:{comment}\:{only}\:{one}\:{solution}\:{t}=\mathrm{16} \\ $$
Commented by prakash jain last updated on 23/Jan/15
T_3 =t=16  M_3 =24  J_3 =48
$${T}_{\mathrm{3}} ={t}=\mathrm{16} \\ $$$${M}_{\mathrm{3}} =\mathrm{24} \\ $$$${J}_{\mathrm{3}} =\mathrm{48} \\ $$

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